This book is dedicated to Engineering Students

everywhere.Especially mine.

Preface to the Electronic Edition

In the Spring of 2000, Brooks/Cole (the most recent publisher of record) declared EnergyConversion out of print and returned the copyright to the author. Because of theunavailability of the films used in the printing of the book, the author decided to develop anelectronic edition for the world-wide-web by revising a word processor draft to conform tothe published text, converting it to Adobe Acrobat pdf form, with minor modifications andcorrections where needed.

This electronic textbook is offered in anticipation of the day when students will carry allof their texts and reference materials to class in a single electronic reader no larger than abook of a few hundred pages. Such readers are available now and notebook computersusing the Acrobat reader could be used as such. It is in a spirit of conservation that theauthor hopes that students and other users will download chapters from the internet as theyare needed and store them on their personal computers rather than printing them.

With the cooperation and assistance of Dr. Dale Schoenefeld and Ms. Janet Cairns ofthe University of Tulsa, it was decided to provide the ebook from the university server. Theauthor very much appreciates their participation and that of the University in this effort. Dr.Andrew Dykes again contributed to the updated Chapter 10; and my scholar-wife, Ruth,proof-read the entire book and helped me to avoid a multitude of problems with the text. Iam, of course very grateful for their contributions to this non-profit venture.

As I used the printed text as an instructor, before retiring, and as I worked on theelectronic edition, I found a deeper appreciation of the quantity and quality of the effortsexpended by the production editor of the printed version, Tad W. Bornhoft, and hisassociates. I came to admire and appreciate their work more and more. It was a job well-done. Thank you all very much.

I am grateful to the instructors and students who have used Energy Conversion in thepast. It is for them and future scholars that I am attempting to make this work more readilyavailable. I hope that they find it useful and that it makes a pleasant and meaningfulcontribution to engineering education.

Kenneth C. WestonDecember 2000

Preface to the First Edition

A solid grounding in heat and power has long been a characteristic expected of mechanicalengineers. This text deals with energy conversion topics that should be well understood byall mechanical engineers. It is intended for use in an introductory three-semester-hourcourse in energy conversion, to follow first courses in thermodynamics and fluid mechanicsand, where possible, heat transfer. No attempt is made to treat electrical motors, generators,and other conventional electric power equipment dealt with in electrical engineering powercourses. Rather, it focuses, in the first six chapters, on the three predominant thermal powersystems: the steam power plant, the gas turbine, and the reciprocating engine. These areconsidered the mainstream energy converters in which all mechanical engineers should bewell grounded. The remaining five chapters provide a variety of choices for an instructor toselect to round out a three-hour junior or senior level course. While the latter chaptersdepend on fundamentals appearing in the first six chapters as much as possible, they areorganized and written to be independent of each other, so that they may be used in almostany order, or may be completely ignored.

It is the author�s experience that many students have difficulty with the concepts ofthermodynamics and fail to grasp its importance and power. A course in energy conversion,which applies the concepts of thermodynamics and introduces and analyzes the primeenergy conversion devices, can be an eye-opener to those students who need concreteapplications and are stimulated by them. At the same time, a good course in energyconversion can motivate further study of thermodynamics and the other engineeringsciences.

A number of textbooks are available for the study of energy conversion. Most of thesebooks are best suited for advanced and graduate students because of an abundance of detailthat may distract the undergraduate�fresh from first encounters with thermodynamics andfluid mechanics�from the fundamentals, from the major thrusts of system behavior andoperation, and from engineering analysis. For this reason, the present text is an attempt atuser-friendliness, which trades excessive technical detail for a fuller and easier developmentof mainline topics. Thus this text limits the amount of peripheral information presented andfocuses on topics of current importance and those likely to play a significant role in thecareers of the readers. It seeks to act as a convenient bridge between the engineeringsciences and advanced courses and, at the same time, provide a useful terminal course forstudents pursuing other interests.

The approach taken is to provide a brief review of fundamentals from thermodynamicsand fluid mechanics and to immediately focus on an in-depth study of a major energyconversion system�the steam power plant�together with more advanced fundamentalsneeded for its understanding and analysis. The premise is that a thorough understanding of amajor energy conversion system establishes a point of reference by which the student may

better appreciate and understand other systems. This is the major thrust of the first fourchapters. The first chapter is a brief review and introduction to the text. Instructors maywish to assign the reading of Chapter 1 and a few problems for review at the first classmeeting, and then answer questions and proceed to the important material of Chapter 2 inthe second session.

Chapter 2 starts with the Rankine cycle and examines the important refinements of thecycle, one at a time, culminating in the study of a typical cycle of a large modern steampower plant. Sections of the chapter may be assigned at a pace consistent with the students�prior exposure to the Rankine cycle. The author usually assigns one or more problems ateach class and spends one class each on the basic cycle, reheat, efficiencies and pressurelosses, regeneration and feedwater heaters, combined mass conservation and First Lawanalysis with heaters, and two classes on the study of the power plant flowsheet. At thispoint the student should have the capability to interpret and analyze the flowsheet of anysteam power plant presented to him or her. This may be an appropriate time for a firstexamination to assure that the attention of the student has focused on thermodynamic cycleand system analysis fundamentals before considering the basics of Rankine cycleimplementation.

With a good grasp of the cycles of steam power plants in hand, it is appropriate for thestudent to study some of the fundamentals of fuels and combustion in order to proceedfurther in understanding steam plant design and operation and to prepare him or her forlater chapters. Thus the student should be easily motivated to the study of fuels andcombustion in Chapter 3. To the extent that this and the preceding chapter are review forthe student, the pace of coverage may be adjusted by the instructor in reading and problemassignments and class discussions.

Chapter 4 completes the study of conventional steam power generation by followingand analyzing the major flows in a steam plant: water, fuel, and gases, placing emphasis onthe hardware and systems involved in these flows. A brief introduction to the fundamentalsof engineering economy is included in this chapter, so that financial aspects of power plantoperation may be considered. The latter material is not intended to replace a course inengineering economy. Instead, it provides basic information that may appear too late insome curricula for use in a junior-level energy conversion course. Chapter 4 concludes witha back-of-the-envelope type analysis of plant characteristics, which provides the studentwith an overview of the magnitudes of parameters associated with large steam plantoperation and with an opportunity to reflect on the roles of thermodynamics, economics,and analysis in the plant design.

The succeeding chapters provide opportunities to show the universality of many of thefundamentals and methods presented in the first four. The fifth chapter treats the importantand exciting topics of gas turbines and jet propulsion. The characteristics of stationary gasturbines are studied and contrasted. The jet engine is studied as a form of gas turbine withspecial design and performance characteristics. Care is taken to provide the student with aconceptual understanding of the basic machine before delving into detailed analysis andalternate configurations. Advanced topics such as combined cycles, cogeneration, steaminjection, polytropic efficiencies, and turbofan engines are deferred to Chapter 9, for thoseinstructors who prefer not to treat these topics in an introductory energy conversion course.

As in other chapters, the instructor is largely in control of the level of the presentationby the choice and the pace of the assignments that he or she makes. One or several textsections may be assigned per class meeting, along with problems of appropriate number anddifficulty for the students. At the end of Chapter 5, the instructor may wish to assign thedevelopment of a computer program or a spreadsheet for gas turbine analysis, or the use ofan existing program or spreadsheet for optimization of a configuration for a givenapplication, or for a preliminary design study of a gas turbine or steam plant. Althoughcombined gas turbine and steam cycles are not considered until Chapter 9, some instructorsmay wish to deal with them at this point, and perhaps assign a design problem.

Chapter 6 considers the phenomena and characteristics of the reciprocating engine andthe engineering parameters describing its performance. The chapter develops and comparesmodels of both Otto and Diesel cycles and explores some of the fundamental problems andaspects of their implementation in working engines. Chapter 7 complements Chapter 6 byproviding a brief look at the Wankel rotary engine. While not a major player in energyconversion at this time, the rotary engine offers a unique opportunity for students to analyzeand understand an entirely different, successful, and intriguing implementation of the Ottocycle. For those instructors wishing to substitute other materials, Chapter 7 may be omittedwithout concern for student comprehension of the remainder of the book.

While refrigeration and air conditioning are not power generation technologies, theyclearly involve energy conversion in important ways and represent an important mechanicalengineering discipline. Chapter 8 briefly considers the fundamentals and hardware of thisfield briefly, giving a framework within which HVAC engineering may be understood andupon which engineering analyses may be built. The important topic of the analysis of moistair is introduced by treating moist air as a binary mixture of ideal gases and carried to thepoint where the student is introduced to HVAC system design using the psychrometricchart.

Chapter 9 considers some of the important technological problems of the day and aselected collection of advanced energy conversion techniques that mechanical engineers useto deal with them. The treatments of most of these technologies are largely independent ofeach other, so that instructors may elect to use either the entire chapter or only thoseadvanced topics that they consider most appropriate for their course. A notable exception isthat the presentation on turbofans considers using either isentropic or polytropicefficiencies. It is recommended to cover the preceding section on polytropic efficiencybefore proceeding to turbofans. It is hoped that the study of this chapter will motivatestudents to investigate the selected topics in greater detail, aided by the bibliography.Instructors may wish at this point to assign more in-depth study using the library on one ofthese topics or on other advanced systems.

Certainly, nuclear power is controversial. While the development of new nuclear fissionplants in the United States is currently dormant, the subject remains important and is likelyto become more important to mechanical engineers in the future. Some texts devote a greatdeal of space to nuclear power while others do not choose to treat it at all. The decisionhere was to take the middle ground, to provide a survey that allows mechanical engineers toplace the subject in perspective with respect to other major energy conversion technologies.Chapter 10 thus focuses on relevant aspects of nuclear fission power without going intonuclear reactor physics and design in detail. The chapter may be studied at any time after

the first four chapters on steam power, or may be omitted completely if the instructordesires.

The final chapter offers vistas on a number of promising technologies, some of whichavoid the heat engine approach that dominates today�s modern power and propulsionsystems. An introduction to electrical energy storage-battery technology provides a lead-into fuel-cell energy conversion, now under intensive development as a flexible solution to adiversity of electrical power generation problems. Magnetohydrodynamics (MHD) isconsidered as a means of producing electricity directly from a hot fluid without the use of aturbine. The inspiration of the sun as a massive and eternal energy source continues to driveinterest in solar energy, spurred by the success of small-scale photovoltaic conversion inwatches and calculators, and of remote power applications. The chapter closes with a briefconsideration of the use of hydrogen as a secondary energy source and the transition to asteady-state energy system.

A course in energy conversion may serve as a terminal course in the thermal-fluids-energy stem for those students pursuing other disciplinary interests. These students emergewith an appreciation of the purposes and methods of thermodynamics, of engineeringhardware, and of important segments of industry. For others, the course may serve as amotivating lead-in to electives such as advanced thermodynamics, advanced fluids and heattransfer, refrigeration and air conditioning, solar energy, turbomachinery, gas turbines, andpropulsion. At the University of Tulsa the course is offered to sixth semester students,allowing elective studies in the seventh and eighth semesters.

The goal of this book is to provide a lucid learning tool for good students who may feelinsecure in their understanding of the engineering sciences and their uses. An effort is madeto tie theory, concepts, and techniques to earlier courses, which may not have beenmastered. A special effort is made also to help the student become familiar with andappreciate the important hardware associated with energy conversion, through thepresentation of numerous photographs and diagrams. The author hopes that this study ofenergy conversion will not only prepare the student for work in this field but will alsoprovide motivation for further study of the engineering sciences because their usefulness indesign and analysis is better appreciated.

Numerous examples are provided in the text in connection with the more importanttopics. Photographs and schematic diagrams of modern equipment are included to aid thestudent in hardware visualization. Students should be encouraged to study these carefully asthe useful learning tools that they are. As a further guide and aid to the student, terms whichshould be part of his or her vocabulary are italicized. Many acronyms and abbreviations areused in energy conversion as in other fields of endeavor. There is no point in delaying theinevitable. They are used sparingly but where appropriate in the text and are identified intheir first occurrence and in the list of symbols.

We are in the midst of a protracted transition from one system of engineering units, theEnglish system, to an international system, the Systeme International d�Unites, usuallycalled SI. It is essential that this and other texts provide experience in using both; therefore,an effort has been made to provide discussions and problems expressed in both the SI andEnglish systems.

The importance of units, not only as a tool of communication, but as an analytic toolwarrants repeated emphasis. Using units for dimensional analysis may often be the

difference between obtaining a correct problem solution and falling into quantitative error.To emphasize the importance of units, equations in the text appear with the units of thedependent variable in both English and SI forms in brackets, as for example in the units ofspecific energy: [Btu/lbm | kJ/kg]. When an expression is dimensionless that fact is indicatedby [dl] following the equation.

The personal computer has become a valuable tool for practicing engineers and shouldbe used by engineering students at every appropriate opportunity. The author has foundspreadsheets to be particularly useful in a variety of activities including energy conversion.Spreadsheets are provided for the example problems presented in spreadsheet format in thetext. This allows students to examine these calculations in as much detail as they wish, toconveniently repeat the calculations for alternate inputs, and to substitute alternatecalculation details. Thus instructors may conveniently assign �what if� studies in thesecases. A spreadsheet of data extracted from the JANAF tables is included to allow easyflame temperature and heat transfer calculations in connection with Chapter 3. While thespreadsheets were developed using Quattro Pro, the examples are provided in the WK3format, so that they may be used with any spreadsheet that is compatible with this popularLotus 1-2-3 format.

Those unfamiliar with spreadsheets will find that a few hours with a modernspreadsheet and a good tutorial will enable its use in powerful ways. One of thecharacteristics of spreadsheets is that the new user can usually start solving problems at thefirst sitting. After using the tutorial, the new user should pick a modest problem of currentinterest to solve, or try a few problems that were previously completed by hand. It is wisefor new users to select a spreadsheet that is widely used in their university or company or bytheir friends, so that it will be easy to exchange information with and learn from others.

During the years of preparation of this book, many friends, colleagues, reviewers, andstudents have made constructive suggestions and offered ideas and encouragement. Theyare too numerous to thank individually here. It is, nevertheless, appropriate to express myappreciation to them and to acknowledge their invaluable contributions. It is particularlyfitting, however, to recognize the special contributions of friend and colleague, Dr. AndrewA. Dykes, for both his insightful critique of the first draft and his substantive and continuingcontributions to Chapter 10. Some of his expertise is written into the pages of that chapter;any deficiencies there are mine. I am greatly indebted also to industry for providing many ofthe photographs and illustrations that the reader will find in these pages. Sources areidentified with the figures. Finally, I must acknowledge the patience and support of myfamily and the administration, faculty, and staff of the University of Tulsa, who helped easethe pain and maximize the joys associated with the preparation of this work.

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C H A P T E R O N E

FUNDAMENTALS OF ENERGY CONVERSION

1.1 Introduction

Energy conversion engineering (or heat-power engineering, as it was called prior to theSecond World War), has been one of the central themes in the development of theengineering profession. It is concerned with the transformation of energy from sourcessuch as fossil and nuclear fuels and the sun into conveniently used forms such aselectrical energy, rotational and propulsive energy, and heating and cooling.

A multitude of choices and challenges face the modern energy conversionengineer. A few years ago major segments of the energy conversion industry weresettled into a pattern of slow innovation. Most automobile manufacturers were satisfiedto manufacture engines that had evolved from those produced twenty years earlier,some of which boasted 400 horsepower and consumed a gallon of leaded gasoleneevery eight or nine miles. Many electric power utilities were content with state-of-the-art, reliable, fossil-fuel-consuming steam power plants, except for a fewforward-looking, and in several cases unfortunate, exceptions that risked the nuclearalternative.

Then came the oil embargo of the 1970s, high fuel prices, and threatenedshortages. Also, the public and legislatures began to recognize that air pollutionproduced by factories, power plants, and automobiles and other forms of environmentalpollution were harmful. International competitors, producing quality automobiles withsmaller, lower-pollution engines, exceptional gas mileage, and lower prices shook theautomobile industry. The limitations of the Earth's resources and environment startedto come into clearer focus. These and other influences have been helping to create amore favorable climate for consideration, if not total acceptance, of energy conversionalternatives and new concepts.

There are opposing factors, however. Among them are limited research anddevelopment funding due to budgetary constraints, emphasis on short-term rather thanlong-term goals because of entrepreneurial insistence on rapid payback on investment,and managerial obsession with the bottom line. But more open attitudes have becomeestablished. New as well as previously shelved ideas are now being considered orreconsidered, tested, and sometimes implemented. A few examples are combined steam and gas turbine cycles, rotary combustion engines, solar and windmill powerfarms, stationary and vehicular gas turbine power plants, cogeneration, photovoltaicsolar power, refuse-derived fuel, stratified charge engines, turbocharged engines,fluidized-bed combustors, and coal-gasification power plants. We are living in a

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rapidly changing world that requires continuing adaptation of old technologies and thedevelopment of new ones. Energy conversion engineering is a more stimulating,complex, and viable field today because of this altered climate.

A Look Backward

How did we get where we are today? A good answer requires a study of the history ofscience and engineering worthy of many volumes. Table 1.1 identifies a few pivotalideas and inventions, some of them landmarks to energy conversion engineers, and thenames of the thinkers and movers associated with them. Of course, the table cannotpresent the entire history of energy conversion engineering. Omitted are thecontributions of Newton and Euler, the first rocket engine, the V-8 engine, the ramjetand fanjet. The reader could easily come up with many other glaring omissions andextend the table indefinitely.

While the names of one or two persons are associated with each landmarkachievement, most of these landmarks were the products of teams of unheraldedindividuals whose talents were crucial to success. Moreover, the successes did notoccur in a vacuum, but benefited and followed from the advances and failures of others. Unknown or renowned, each engineer and his or her associates can make a contributionto the progress of mankind.

Table 1.1 can only hint at how the persons, ideas, and events listed there relied ontheir predecessors and on a host of less well-known scientific and technologicaladvances. A brief bibliography of historical sources is given at the end of the chapter.These works chronicle the efforts of famous and unsung heroes, and a few villains, ofenergy conversion and their struggles with ideas and limiting tools and resources toproduce machines for man and industry.

The historical progress of industry and technology was slow until the fundamentalsof thermodynamics and electromagnetism were established in the ninteenth century.The blossoming of energy technology and its central role in the industrial revolution iswell known to all students of history. It is also abundantly clear that the development ofnuclear power in the second half of the twentieth century grew from theoretical andexperimental scientific advances of the first half century. After a little reflection onTable 1.1, there should be no further need to justify a fundamental scientific andmathematical approach to energy conversion engineering.

TABLE 1-1 Some Significant Events in the History of Energy Conversion___________________________________________________________________________Giovanni Branca Impulse steam turbine proposal 1629

Thomas Newcomen Atmospheric engine using steam (first widely used 1700Heat engine)

James Watt Separate steam condenser idea; 1765and first Boulton and Watt condensing steam engine 1775

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Table 1.1 (continued)___________________________________________________________________________ John Barber Gas turbine ideas and patent 1791

Benjamin Thompson Observed conversion of mechanical energy 1798(Count Rumford) to heat while boring cannon

Robert Fulton First commercial steamboat 1807

Robert Stirling Stirling engine 1816

N. L. Sadi Carnot Principles for an ideal heat engine 1824(foundations of thermodynamics)

Michael Faraday First electric current generator 1831

Robert Mayer Equivalence of heat and work 1842

James Joule Basic ideas of the First Law of Thermodynamics; 1847and measured the mechanical equivalent of heat 1849

Rudolph Clausius Second Law of Thermodynamics 1850

William Thompson Alternate form of the Second Law of 1851 (Lord Kelvin) Thermodynamics

Etienne Lenoir Internal combustion engine without without mechanical compression 1860

A. Beau de Rochas Four-stroke cycle internal combustion engine concept 1862

James C. Maxwell Mathematical principles of electromagnetics 1865 Niklaus Otto Four-stroke cycle internal combustion engine 1876

Charles Parsons Multistage, axial-flow reaction steam turbine 1884

Thomas Edison Pearl Street steam-engine-driven electrical power plant 1884

C.G.P. de Laval Impulse steam turbine with convergent-divergent nozzle 1889

Rudolph Diesel Compression ignition engine 1892

___ First hydroelectric power at Niagara Falls 1895

Albert Einstein Mass-energy equivalence 1905

Ernst Schrodinger Quantum wave mechanics 1926

Frank Whittle Turbojet engine patent application; 1930 and first jet engine static test 1937

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Table 1.1 (concluded)___________________________________________________________________________Otto Hahn Discovery of nuclear fission 1938

Hans von Ohain First turbojet engine flight 1939

J. Ackeret, C. Keller Closed-cycle gas turbine electric power generation 1939

Enrico Fermi Nuclear fission demonstration at the University of Chicago 1942

Felix Wankel Rotary internal combustion engine 1954

Production of electricity via nuclear fission by a utility 1957at Shippingport, Pennsylvania .

NASA Rocket-powered landing of man on the moon 1969

Electricité de France Superphénix 1200-MW fast breeder reactor � first 1986grid power

________________________________________________________________

Since energy conversion engineering is deeply rooted in thermodynamics, fluidmechanics, and heat transfer, this chapter briefly reviews those aspects of thesedisciplines that are necessary for understanding, analysis, and design in the field ofenergy conversion.

1.2 Fundamentals of Thermodynamics

The subject of thermodynamics stems from the notions of temperature, heat, and work.Although, the following discussion makes occasional reference to molecules andparticles, useful in clarifying and motivating concepts in thermodynamics,thermodynamics as a science deals with matter as continuous rather than as discrete orgranular. System, Surroundings, and Universe

We define a pure substance as a homogeneous collection of matter. Consider a fixedmass of a pure substance bounded by a closed, impenetrable, flexible surface. Such amass, called a system, is depicted schematically in Figure 1.1(a). For example, thesystem could be a collection of molecules of water, air, refrigerant, or combustion gasconfined in a closed container such as the boundary formed by a cylinder and a fittedpiston, Figure 1.1 (b). A system should always be defined carefully, to ensure that thesame particles are in the system at all times. All other matter which can interact with thesystem is called the surroundings. The combination of the system and the surroundingsis termed the universe, used here not in a cosmological sense, but to include only the

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system and all matter which could interact with the system. Thermodynamics andenergy conversion are concerned with changes in the system and in its interactions withthe surroundings.

State

The mass contained within a system can exist in a variety of conditions call states.Qualitatively, the concept of state is familiar. For example, the system state of agasmight be described qualitatively by saying that the system is at a high temperatureand a low pressure. Values of temperature and pressure are characteristics that identifya particular condition of the system. Thus a unique condition of the system is called astate.

Thermodynamic Equilibrium

A system is said to be in thermodynamic equilibrium if, over a long period of time, nochange in the character or state of the system is observed.

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Thermodynamic Properties

It is a fundamental assumption of thermodynamics that a state of thermodynamicequilibrium of a given system may be described by a few observable characteristicscalled thermodynamic properties, such as pressure, temperature, and volume.Obviously, this approach excludes the possibility of description of the condition of themolecules of the system, a concern that is left to the fields of statistical and quantummechanics and kinetic theory. Nevertheless, it is frequently useful to think ofthermodynamic phenomena in molecular terms.

The Temperature Property. Temperature is a measure of the vigor of the molecularactivity of a system. How can it be observed? A thermometer measures a systemproperty called temperature when it is in intimate and prolonged contact(thermodynamic equilibrium) with the system. A mercury-in-glass thermometer, forinstance, functions by thermal expansion or contraction of mercury within a glass bulb.The bulb must be in intimate thermal contact with the observed system so that thetemperatures of the bulb and the system are the same. As a result of the equilibrium,elongation or contraction of a narrow column of mercury connected to the bulbindicates the temperature change of the system with which it is in contact.

The Pressure Property. Another way to observe changes in the state of a liquid orgaseous system is to connect a manometer to the system and observe the level of thefree surface of the manometer fluid . The manometer free surface rises or drops as theforce per unit area or pressure acting on the manometer-system interface changes.

Defining a State

It has been empirically observed that an equilibrium state of a system containing asingle phase of a pure substance is defined by two thermodynamic properties. Thus, ifwe observe the temperature and pressure of such a system, we can identify when thesystem is in a particular thermodynamic state.

Extensive and Intensive properties

Properties that are dependent on mass are known as extensive properties. For theseproperties that indicate quantity, a given property is the sum of the the correspondingproperties of the subsystems comprising the system. Examples are internal energy andvolume. Thus, adding the internal energies and volumes of subsystems yields theinternal energy and the volume of the system, respectively.

In contrast, properties that may vary from point to point and that do not changewith the mass of the system are called intensive properties. Temperature and pressureare well-known examples. For instance, thermometers at different locations in a systemmay indicate differing temperatures. But if a system is in equilibrium, the temperatures

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of all its subsystems must be identical and equal to the temperature of the system. Thus,a system has a single, unique temperature only when it is at equilibrium.

Work

From basic mechanics, work, W, is defined as the energy provided by an entity thatexerts a force, F, in moving one or more particles through a distance, x. Thus workmust be done by an external agent to decrease the volume, V, of a system of molecules.In the familiar piston-cylinder arrangement shown in Figure 1.1(b), an infinitesimalvolume change of the system due to the motion of the piston is related to thedifferential work through the force-distance product:

dW = Fdx = pAdx = pdV [ft-lbf | n-m] (1.1a)

or

dw = pdv [Btu/lbm | kJ/kg] (1.1b)

where p is the system pressure, and A is the piston cross-sectional area.Note that in Equation (1.1b), the lower case letters w and v denote work and

volume on a unit mass basis. All extensive properties, i.e., those properties of state thatare proportional to mass, are denoted by lowercase characters when on a unit massbasis. These are called specific properties. Thus, if V represents volume, then v denotesspecific volume. Although work is not a property of state, it is dealt with in the sameway.

Also note that the English units of energy in Equation (1.1a) are given inmechanical units. Alternately, the British Thermal unit [Btu] may be used, as inEquation (1.1b). The two sets of units are related by the famous conversion factorknown as the mechanical equivalent of heat, 778 ft-lbf/Btu. The student should payclose attention to the consistency of units in all calculations. Conversion factors arefrequently required and are not explicitly included in many equations. For theconvenience of the reader, Appendix A lists physical constants and conversion factors.

When work decreases the volume of a system, the molecules of the system movecloser together. The moving molecules then collide more frequently with each otherand with the walls of their container. As a result, the average forces (and hence pressures) on the system boundaries increase. Thus the state of the system may bechanged by work done on the system.

Heat

Given a system immersed in a container of hot fluid, by virtue of a difference intemperature between the system and the surrounding fluid, energy passes from the fluidto the system. We say that heat, Q [Btu | kJ], is transferred to the system. The system is

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observed to increase in temperature or to change phase or both. Thus heat transfer toor from the system, like work, can also change the state of the matter within thesystem.

When the system and the surrounding fluid are at the same temperature, no heatis transferred. In this case the system and surroundings are said to be in thermalequilibrium. The term adiabatic is used to designate a system in which no heat crossesthe system boundaries. A system is often approximated as an adiabatic system if it iswell insulated.

Heat and Work Are Not Properties

Mechanics teaches that work can change the kinetic energy of mass and can change theelevation or potential energy of mass in a gravitational field. Thus work performed byan outside agent on the system boundary can change the energy associated with theparticles that make up the system. Likewise, heat is energy crossing the boundary of asystem, increasing or decreasing the energy of the molecules within. Thus heat andwork are not properties of state but forms of energy that are transported across systemboundaries to or from the environment. They are sometimes referred to as energy intransit. Energy conversion engineering is vitally concerned with devices that use andcreate energy in transit.

Internal Energy and The First Law of Thermodynamics

A property of a system that reflects the energy of the molecules of the system is calledthe internal energy, U. The Law of Conservation of Energy states that energy can beneither created nor destroyed. Thus the internal energy of a system can change onlywhen energy crosses a boundary of the system, i.e., when heat and/or work interactwith the system. This is expressed in an equation known as the First Law ofThermodynamics. In differential form the First Law is:

du = dq � dw [Btu/lbm | kJ/kg] (1.2)

Here, u is the internal energy per unit mass, a property of state, and q and w are,respectively, heat and work per unit mass. The differentials indicate infinitesimalchanges in quantity of each energy form. Here, we adopt the common sign conventionof thermodynamics that both the heat entering the system and work done by the systemare positive. This convention will be maintained throughout the text. Thus Equation(1.2) shows that heat into the system (positive) and work done on the system (negative)both increase the system�s internal energy.

Cyclic Process

A special and important form of the First Law of Thermodynamics is obtained by

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integration of Equation (1.2) for a cyclic process. If a system, after undergoingarbitrary change due to heat and work, returns to its initial state, it is said to haveparticipated in a cyclic process. The key points are: (1) the integral of any stateproperty differential is the difference of its limits, and (2) the final state is the same asthe initial state (hence there is no change in internal energy of the system)

�du = uf � ui = 0

where the special integral sign indicates integration over a single cycle and subscripts iand f designate, respectively, initial and final states. As a consequence, the integrationof Equation (1.2) for a cycle yields:

�dq = �dw [Btu/lbm | kJ/kg] (1.3)

This states that the integral of all transfers of heat into the system, taking into accountthe sign convention, is the integral of all work done by the system. The latter is the network of the system. The integrals in Equation (1.3) may be replaced by summations fora cyclic process that involves a finite number of heat and work terms. Because manyheat engines operate in cyclic processes, it is sometimes convenient to evaluate the network of a cycle using Equation (1.3) with heat additions and losses rather than usingwork directly.

Arbitrary Process of a System

Another important form of the First Law of Thermodynamics is the integral ofEquation (1.2) for an arbitrary process involving a system:

q = uf � ui + w [Btu/lbm | kJ/kg] (1.4)

where q and w are, respectively, the net heat transferred and net work for the process,and uf and ui are the final and initial values of the internal energy. Equation (1.4), likeEquation (1.2), shows that a system that is rigid (w = 0) and adiabatic (q = 0) has anunchanging internal energy. It also shows, like Equation (1.3), that for a cyclic processthe heat transferred must equal the work done.

Reversibility and Irreversibility

If a system undergoes a process in which temperature and pressure gradients are alwayssmall, the process may be thought of as a sequence of near-equilibrium states. If each ofthe states can be restored in reverse sequence, the process is said to be internallyreversible. If the environmental changes accompanying the process can also be reversedin sequence, the process is called externally reversible. Thus, a reversible process is

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one that is both internally and externally reversible. The reversible process becomesboth a standard by which we measure the success of real processes in avoiding lossesand a tool that we can use to derive thermodynamic relations that approximate reality.

All real processes fail to satisfy the requirements for reversibility and are thereforeirreversible. Irreversibility occurs due to temperature, pressure, composition, andvelocity gradients caused by heat transfer, solid and fluid friction, chemical reaction,and high rates of work applied to the system. An engineer�s job frequently entailsefforts to reduce irreversibility in machines and processes.

Entropy and Enthalpy

Entropy and enthalpy are thermodynamic properties that, like internal energy, usuallyappear in the form of differences between initial and final values. The entropy change ofa system, �s [Btu/lbm-R | kJ/kg-K], is defined as the integral of the ratio of the systemdifferential heat transfer to the absolute temperature for a reversible thermodynamicpath, that is, a path consisting of a sequence of well-defined thermodynamic states. Indifferential form this is equivalent to:

ds = dqrev /T [Btu/lbm-R | kJ/kg-K] (1.5)

where the subscript rev denotes that the heat transfer must be evaluated along areversible path made up of a sequence of neighboring thermodynamic states. It isimplied that, for such a path, the system may be returned to its condition before theprocess took place by traversing the states in the reverse order.

An important example of the use of Equation (1.5) considers a thermodynamic cyclecomposed of reversible processes. The cyclic integral, Equation (1.3), may then be usedto show that the net work of the cycle is:

wn = �dq = �Tds [Btu/lbm | kJ/kg]

This shows that the area enclosed by a plot of a reversible cyclic process on atemperature-entropy diagram is the net work of the cycle.

The enthalpy, h, is a property of state defined in terms of other properties:

h = u + pv [Btu/lbm | kJ/kg] (1.6)

where h, u and v are, respectively, the system specific enthalpy, specific internalenergy, and specific volume, and p is the pressure.

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Two other important forms of the First Law make use of these properties.Substitution of Equations (1.1) and (1.5) in Equation (1.2) yields, for a reversibleprocess Tds = du + pdv [Btu/lbm | kJ/kg] (1.7)

and differentiation of Equation (1.6), combined with elimination of du in Equation(1.7), gives

Tds = dh - vdp [Btu/lbm | kJ/kg] (1.8)

Equations (1.7) and (1.8) may be regarded as relating changes in entropy for reversibleprocesses to changes in internal energy and volume in the former and to changes inenthalpy and pressure in the latter. The fact that all quantities in these equations areproperties of state implies that entropy must also be a thermodynamic property.

Because entropy is a state property, the entropy change between two equilibriumstates of a system is the same for all processes connecting them, reversible orirreversible. Figure 1.2 depicts several such processes 1-a-b-c-2, 1-d-2, and a sequenceof nonequilibrium states not describable in thermodynamic terms indicated by thedashed line (an irreversible path). To use Equation (1.5) directly or as in Equations(1.7) and (1.8), a reversible path must be employed. Because of the path independenceof state property changes, any reversible path will do. Thus the entropy change, s2 � s1,

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may be evaluated by application of Equations (1.5), (1.7), or (1.8) to either of thereversible paths shown in Figure 1.2 or to any other reversible path connecting states 1and 2.

The Second Law of Thermodynamics

While Equation (1.5) may be used to determine the entropy change of a system, theSecond Law of Thermodynamics, is concerned with the entropy change of the universe,i.e., of both the system and the surroundings. Because entropy is an extensive property,the entropy of a system is the sum of the entropy of its parts. Applying this to theuniverse, the entropy of the universe is the sum of the entropy of the system and itssurroundings. The Second Law may be stated as "The entropy change of the universe isnon-negative":

�Suniv � 0 [Btu/R | kJ/K] (1.9)

Note that the entropy change of a system may be negative (entropy decrease) if theentropy change of its environment is positive (entropy increase) and sufficiently largethat inequality (1.9) is satisfied.

As an example: if the system is cooled, heat is transferred from the system. Theheat flow is therefore negative, according to sign convention. Then, according toEquation (1.5), the system entropy change will also be negative; that is, the systementropy will decrease. The associated heat flow, however, is into the environment,hence positive with respect to the environment (considered as a system). Then Equation(1.5) requires that the environmental entropy change must be positive. The Second Lawimplies that, for the combined process to be possible, the environmental entropy changemust exceed the magnitude of the system entropy change.

The First Law of Thermodynamics deals with how the transfer of heat influencesthe system internal energy but says nothing about the nature of the heat transfer, i.e.,whether the heat is transferred from hotter or colder surroundings. Experience tells usthat the environment must be hotter to transfer heat to a cooler object, but the FirstLaw is indifferent to the condition of the heat source. However, calculation of theentropy change for heat transfer from a cold body to a hot body yields a negativeuniverse entropy change, violates the Second Law, and is therefore impossible. Thusthe Second Law provides a way to distinguish between real and impossible processes.This is demonstrated in the following example:

EXAMPLE 1.1

(a) Calculate the entropy change of an infinite sink at 27°C temperature due to heattransfer into the sink of 1000 kJ.

(b) Calculate the entropy change of an infinite source at 127°C losing the same amountof heat.

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(c) What is the entropy change of the universe if the aforementioned source supplies1000 kJ to the sink with no other exchanges?

(d) What are the entropy changes if the direction of heat flow is reversed and thesource becomes the sink?

Solution (a) Because the sink temperature is constant, Equation (1.5) shows that the entropychange of the sink is the heat transferred reversibly divided by the absolute temperatureof the sink. This reversible process may be visualized as one in which heat is transferredfrom a source which is infinitesimally hotter than the system:

�Ssink = 1000/(273 + 27) = + 3.333 kJ/K.

(b) Treating the source in the same way:

�Ssource = � 1000/(273 + 127) = � 2.5 kJ/K.

(c) Because the entropy change of the universe is the sum of the entropy changes ofsource and sink, the two acting together to transfer 1000kJ irreversibly give:

�Suniverse = 3.333 � 2.5 = +0.833 kJ/K > 0

which satisfies the Second Law inequality (1.9).

(d) A similar approach with the direction of heat flow reversed, taking care to observethe sign convention, gives

�Ssink = (� 1000 )/(273 + 27) = � 3.333 kJ/K

�Ssource = (1000)/(273 + 127) = + 2.555 kJ

�Suniv = � 3.333 + 2.5 = � 0.833 kJ/K.

Thus we see that heat flow from a low to a high temperature reduces the entropy of theuniverse, violates the Second Law, and therefore is not possible.____________________________________________________________________Parts a, b, and c of Example 1.1 show that the entropy change of the universe dependson the temperature difference driving the heat transfer process:

�Suniv = Q(1/Tsink � 1/ Tsource) = Q( Tsource � Tsink) / Tsource Tsink

Note that if the temperature difference is zero, the universe entropy change is also zeroand the heat transfer is reversible. For finite positive temperature differences, �Suniv

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exceeds zero and the process is ireversible. As the temperature difference increases,�Suniv increases. This exemplifies the fact that the entropy change of the universeproduced by a process is a measure of the irreversibility of the process.

For an isolated system, there is no change in the entropy of the surroundings.Hence the system entropy change is the entropy change of the universe and thereforemust be non-negative. In other words, the entropy of an isolated system can onlyincrease or at best stay constant.

1.3 Control Volumes and Steady Flows

In many engineering problems it is preferrable to deal with a flow of fluid particles asthey pass through a given region of space rather than following the flow of a fixedcollection of particles. Thus, putting aside the system concept (fixed collection) for themoment, consider a volume with well-defined spatial boundaries as shown in Figure 1-3. This is called a control volume. Mass at state 1 enters at a rate m1 and leaves atstate 2 with mass flow m2. If one mass flow rate exceeds the other, mass eitheraccumulates in the volume or is depleted. The important special case of steady flow, inwhich no accumulation or depletion of mass occurs in the control volume, is consideredhere. In steady flow, the conservation of mass requires equal mass flows in and out, i.e.,m1 = m2, [lbm /s, | kg /s].

If Q-dot is the rate of heat flow into the control volume and W-dot is the rate atwhich shaft work is delivered from the control volume to the surroundings,conservation of energy requires that the excess of inflowing heat over outgoing workequal the net excess of the energy (enthalpy) flowing out of the ports, i.e.,

[Btu/s | kJ/s] (1.10)

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where summations apply to inflows i and outflows o, and where other types of energyterms, such as kinetic and potential energy flows, are assumed negligible. For clarity,the figure shows only one port in and one port out. Kinetic and potential energy termsmay be added analogous to the enthalpy termsat each port, if needed.

Equation (1.10) may be the most important and frequently used equation in thisbook. Mastery of its use is therefore essential. It is known as the steady flow form ofthe First Law of Thermodynamics. It may be thought of as a bookkeeping relation forkeeping track of energy crossing the boundaries of the control volume.

The Second Law of Thermodynamics applied to steady flow through an adiabaticcontrol volume requires that m2s2 � m1s1, or by virtue of mass conservation:

s2 � s1 [Btu/lbm-R | kJ/kg-K] (1.11)

That is, because entropy cannot accumulate within the control volume in a steady flow,the exit entropy must equal or exceed the inlet entropy. In steady flows, heat transfercan increase or decrease the entropy of the flow, depending on the direction of heattransfer, as long as the entropy change of the surroundings is such that the net effect isto increase the entropy of the universe.

We will often be concerned with adiabatic flows. In the presence of fluid frictionand other irreversibilities, the exit entropy of an adiabatic flow exceeds its inlet entropy.Adiabatic flows that have no irreversibilities also have no entropy change and thereforeare called isentropic flows.

1.4 Properties of Vapors: Mollier and T-s Diagrams

When heated, liquids are transformed into vapors. The much different physicalcharacter of liquids and vapors makes engines in which phase change takes placepossible. The Newcomen atmospheric engine, for instance condensed steam to liquidwater in a piston-cylinder enclosure to create a partial vacuum. The excess ofatmospheric pressure over the low pressure of the condensed steam, acting on theopposite face of the piston, provided the actuating force that drove the first successfulengines in the early eighteenth century. In the latter half of the eighteenth century,engines in which work was done by steam pressure on the piston rather than by theatmosphere, replaced Newcomen-type engines. Steam under pressure in reciprocatingengines was a driving force for the industrial revolution for about two centuries. By themiddle of the twentieth century, steam turbines and diesel engines had largely replacedthe steam engine in electric power generation, marine propulsion, and railroadlocomotives. .

Figure 1.4 shows typical saturation curves for a pure substance plotted intemperature and entropy coordinates. A line of constant pressure (an isobar) is shownin which the subcooled liquid at state 1 is heated, producing increases in entropy,temperature, and enthalpy, until the liquid is saturated at state 2. Isobars in the

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subcooled region of the diagram lie very close to the saturated liquid curve. Theseparation of the two is exaggerated for clarity.

Once the substance has reached state 2, further transfer of heat fails to increase thesystem temperature but is reflected in increased enthalpy and entropy in a vaporizationor boiling process. During this process the substance is converted from a saturatedliquid at state 2 to a mixture of liquid and vapor, and finally to a saturated vapor atstate 3. The enthalpy difference between the saturation values, h3 � h2, is called theenthalpy of vaporization or heat of vaporization.

Continued addition of heat to the system, starting at state 3, superheats the steamto state 4, again increasing temperature, enthalpy, and entropy.

Several observations about the isobaric process may be made here. Equation (1.5)and Figure 1.4 show that the effect of adding heat is to always increase system entropyand that of cooling to always decrease it. A similar conclusion can be drawn fromEquation (1.10) regarding heat additions acting to increase enthalpy flow through acontrol volume in the absence of shaft work.

A measure of the proximity of a superheated state (state 4 in the figure) to thesaturated vapor line is the degree of superheat. This is the difference between thetemperature T4 and the saturated vapor temperature T3, at the same pressure. Thus thedegree of superheat of superheated state 4 is T 4 - T 3.

In the phase change from state 2 to state 3, the temperature and pressure give noindication of the relative quantities of liquid and vapor in the system. The quality x isdefined as the ratio of the mass of vapor to the mass of the mixture of liquid and vaporat any point between the saturation curves at a given pressure. By virtue of thisdefinition, the quality varies from 0 for a saturated liquid to 1 for a saturated vapor.

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Because extensive properties are proportional to mass, they vary directly with thevapor quality in the mixed region. The entropy, for example, varies from the entropy ofthe saturated liquid sl at state 2 to the saturated vapor entropy sv at state 3 inaccordance with the following quality equation:

s = sl + x(sv � sl) [Btu/lbm-R | kJ/kg-K] (1.12)

where s is the entropy per unit mass. Other extensive properties such as enthalpy andvolume vary with quality in the same way.

A variable closely related to the quality is moisture fraction (both quality andmoisture fraction can be expressed as percentages). Moisture fraction, M, is defined asthe ratio of the mass of liquid to the total mass of liquid and vapor. It can be easilyshown that the sum of the quality and the moisture fraction of a mixture is one.

A Mollier chart, a diagram with enthalpy as ordinate and entropy as abscissa, ismuch like the temperature-entropy diagram. A Mollier diagram for steam is included inAppendix B. An isobar on a Mollier chart, unlike that on a T-s diagram, has acontinuous slope. It shows both enthalpy and entropy increasing monotonically withheat addition. Such a diagram is frequently used in energy conversion and other areasbecause of the importance of enthalpy in applying the steady-flow First Law.

1.5 Ideal Gas Basics

Under normal ambient conditions, the average distance between molecules in gases islarge, resulting in negligible influences of intermolecular forces. In this case, molecularbehavior and, therefore, system thermodynamics are governed primarily by moleculartranslational and rotational kinetic energy. Kinetic theory or statistical thermodynamicsmay be used to derive the ideal gas or perfect gas law:

pv = RT [ft-lbf /lbm | kJ/kg] (1.13) where p [lbf /ft2 | kN/m2], v [ft3/lbm | m3/kg] and T [°R | °K] are pressure, specificvolume, and temperature respectively and R [ft-lbf /lbm-°R | kJ/kg-°K] is the ideal gasconstant. The gas constant R for a specific gas is the universal gas constant R dividedby the molecular weight of the gas.

Thus, the gas constant for air is (1545 ft-lbf /lb-mole-°R) / (29 lbm/lb-mole) = 53.3 ft-lbf /lbm-°R in the English system and (8.31 kJ/kg-mole-°K) / (29 kg/kg-mole)= 0.287 kJ/kg-°K in SI units.

The specific heats or heat capacities at constant volume and at constant pressure,respectively, are:

cv = (�u / �T)v [Btu/lbm-°R | kJ/kg-°K] (1.14)

and

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cp = (�h /�T)p [Btu/lbm-°R | kJ/kg-°K] (1.15)

As thermodynamic properties, the heat capacities are, in general, functions of twoother thermodynamic properties. For solids and liquids, pressure change has littleinfluence on volume and internal energy, so that to a very good approximation: cv = cp.

A gas is said to be thermally perfect if it obeys Equation (1.13) and its internalenergy, enthalpy, and heat capacities are functions of temperature only. Then

du = cv(T) dT [Btu/lbm | kJ/kg] (1.16)and

dh = cp(T) dT [Btu/lbm | kJ/kg] (1.17)

A gas is said to be calorically perfect if in addition to being thermally perfect italso has constant heat capacities. This is reasonably accurate at low and moderatepressures and at temperatures high enough that intermolecular forces are negligible butlow enough that molecular vibrations are not excited and dissociation does not occur.For air, vibrational modes are not significantly excited below about 600K, anddissociation of oxygen does not occur until the temperature is above about 1500K.Nitrogen does not dissociate until still higher temperatures. Excitation of molecularvibrations causes specific heat to increase with temperature increase. Dissociationcreates further increases in heat capacities, causing them to become functions of pressure.

It can be shown (see Exercise 1.4) that for a thermally perfect gas the heatcapacities are related by the following equation:

cp = cv + R [Btu/lbm-R | kJ/kg-K] (1.18)

This relation does not apply for a dissociating gas, because the molecular weight of thegas changes as molecular bonds are broken. Note the importance of assuring that R andthe heat capacities are in consistent units in this equation.

Another important gas property is the ratio of heat capacities defined by k = cp /cv.It is constant for gases at room temperatures but decreases as vibrational modesbecome excited. The importance of k will be seen in the following example.

EXAMPLE 1.2

(a) Derive an expression for the entropy change of a system in terms of pressure andtemperature for a calorically perfect gas. (b) Derive a relation between p and T for an isentropic process in a calorically perfectgas.

Solution (a) For a reversible process, Equation (1.8) gives Tds = dh - vdp. Dividing by T andapplying the perfect gas law gives ds = cp dT/T - Rdp/p.

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Then integration between states 1 and 2 yields

s2 - s1 = cp ln(T2 /T1) - R ln( p2 /p1)

(b) For an isentropic process, s2 = s1. Then the above equation gives

T2 /T1 = (p2/p1)(R/cp)

But R/cp = (cp - cv )/cp = (k - 1)/k. Hence T2 /T1 = (p2 /p1)(k - 1)/k. ____________________________________________________________________

This and other important relations for an isentropic process in a calorically perfect gasare summarized as follows

T2 /T1 = (p2 /p1)(k - 1)/k [dl] (1.19)

T2 /T1 = (v2 /v1)(k - 1) [dl] (1.20)

p2 /p1 = (v1 /v2)k [dl] (1.21)

These relations show that the ratio of heat capacities governs the variation ofthermodynamic properties in an isentropic process. For this reason the ratio of heatcapacities is sometimes called the isentropic exponent.

1.6 Fundamentals of Fluid Flow

Almost all energy conversion devices involve the flow of some form of fluid. Air, liquidwater, steam, and combustion gases are commonly found in some of these devices.Here we review a few of the frequently used elementary principles of fluid flow.

The volume flow rate, Q [ft3/s | m3/s] at which a fluid flows across a surface is theproduct of the area, A [ft2 | m2], of the surface and the component of velocity normal tothe area, V [ft/s | m/s]. The corresponding mass flow rate is the ratio of the volume rateand the specific volume, v [ft3/lbm | m3/kg]:

m = AV/v = Q/v [lbm /s | kg /s] (1.22)

Alternatively the flow rate can be expressed in terms of the reciprocal of the specificvolume, the density, � [lbm /ft3 | kg /m3]:

m = AV� = Q� [lbm /s | kg /s] (1.23)

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The first important principle of fluid mechanics is the conservation of mass, aprinciple that we have already used in Section 1.3. For a steady flow, the net inflow to acontrol volume must equal the net outflow. Any imbalance between the inflow andoutflow implies an accumulation or a reduction of mass within the control volume, i.e.,an unsteady flow. Given a control volume with n ports, the conservation of massprovides an equation that may be used to solve for the nth port flow rate, given theother n-1 flow rates. These flows may be (1) given, (2) calculated from data at theports using Equation (1.22) or (1.23), (3) obtained by solving n-1 other equations, or(4) a combination of the preceding three.

For isentropic flow of an incompressible (constant density, � ) fluid, the Bernoulliequation applies:

p1 /� + V12/2 = p2 /� + V2

2/2 [ft-lbf/lbm | kJ/kg] (1.24)

This is an invariant form, i.e. an equation with the same terms on both sides, p/� +V2/2. The subscripts identify the locations in the flow where the invariants areevaluated. The first term of the invariant is sometimes called the pressure head, and thesecond the velocity head. The equation applies only in regions where there are noirreversibilities such as viscous losses or heat transfer.

The invariant sum of the two terms on either side of Equation (1.24) may be calledthe total head or stagnation head. It is the head that would be observed at a pointwhere the velocity approaches zero. The pressure associated with the total head istherefore called the total pressure or stagnation pressure, po = p + �V2/2. Each pointin the flow may be thought of as having its own stagnation pressure resulting from animaginary isentropic deceleration.

In the event of significant irreversibilities, there is a loss in total head and theBernoulli equation may be generalized to:

p1 /� + V12/2 = p2/� + V2

2/2 + loss [ft-lbf/lbm | kJ/kg] (1.25a) or

po1 /� = po2 /� + loss [ft-lbf/lbm | kJ/kg] (1.25b)

Stagnation pressure or head losses in ducts, such as due to flow turning or suddenarea change, are tabulated in reference books as fractions of the upstream velocity headfor a variety of geometries. Another example is the famous Darcy-Weisbach equationwhich gives the head loss resulting from fluid friction in a pipe of constant cross-section.

1.7 Compressible Flow

While many engineering analyses may reasonably employ incompressible flowprinciples, there are cases where the compressibility of gases and vapors must beconsidered. These are situations where the magnitude of the kinetic energy of the flow

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is comparable to its enthalpy such as in supersonic nozzles and diffusers, in turbines andcompressors, and in supersonic flight. In these cases the steady-flow First Law must begeneralized to include kinetic energy per unit mass terms. For two ports:

[Btu /s | kJ/s] (1.26a)

Care should be taken to assure consistency of units, because enthalpy is usually statedin thermal units [Btu /lbm | kJ/kg] and velocity in mechanical units [ft /s | m /s].

Another invariant of significance appears in Equation (1.26a). The form

ho = h + V2/2 [Btu/lbm | kJ/kg] (1.27)

is seen to be invariant in applications where heat transfer and shaft work areinsignificant. The invariant, ho, is usually given the name stagnation enthalpy because itis the enthalpy at a point in the flow (real or imagined) where velocity approaches zero.In terms of stagnation enthalpy, Equation (1.26a) may be rewritten as

[Btu/s | kJ/s] (1.26b)

where conservation of mass with steady flow through two ports has been assumed.Writing dho = cp dTo with cp constant, we get

ho2 - ho1 = cp(To2 - To1)

Combining this with Equation (1.27), we are led to define another invariant, thestagnation temperature for a calorically perefect gas:

To = T + V2/2cp [ R | K] (1.28)

The stagnation temperature may be regarded as the temperture at a real or imaginarypoint where the gas velocity has been brought to zero adiabatically. For this specialcase of a constant heat capacity, Equation (1.26b) may be written as

[Btu /s | kJ/s] (1.26c)

In both incompressible and compressible flows, the mass flow rates at all stationsin a streamtube are the same. Because the specific volume and density are constant inincompressible flow, Equation (1.22) shows that the volume flow rates are the same atall stations also. However for compressible flow, Equation (1.23) shows that densitychange along a streamtube implies volume flow rate variation. Thus, while it isfrequently convenient to think and talk in terms of volume flow rate when dealing withincompressible flows, mass flow rate is more meaningful in compressible flows and ingeneral.

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A measure of the compressiblity of a flow is often indicated by a Mach number, definedas the dimensionless ratio of a flow velocity to the local speed of sound in the fluid. Forideal gases the speed of sound is given by

a = (kp/�)½ = (kRT)½ [ft /s | m /s] (1.29)

Compressible flows are frequently classified according to their Mach number:

M = 0 Incompressible0 < M < 1 SubsonicM = 1 SonicM > 1 Supersonic

Studies of compressible flows show that supersonic flows have a significantly differentphysical character than subsonic and incompressible flows. For example, the velocityfields in subsonic flows are continuous, whereas discontinuities known as shock wavesare common in supersonic flows. Thus the student should not be surprised to find thatdifferent relations hold in supersonic flows than in subsonic flows.

1.8 Energy Clasification

Energy exists in a variety of forms. All human activities involve conversion of energyfrom one form to another. Indeed, life itself depends on energy conversion processes.The human body, through complex processes, transforms the chemical energy stored infood into external motion and work produced by muscles as well as electrical impulsesthat control and activate internal functions.

It is instructive to examine some of the processes for transformation betweentechnically important forms of energy. Table 1.2 shows a matrix of energy forms andthe names of some associated energy converters.

Table 1.2 Energy Transformation MatrixFrom: To: Thermal Energy Mechanical Energy Electrical Energy

Chemical Energy Furnace Diesel engine Fuel cell

Thermal Energy Heat exchanger Steam turbine Thermocouple

Mechanical Energy Refrigerator, heatpump

Gearbox Electrical generator

Nuclear Energy Fission reactor Nuclear steam turbine Nuclear power plant

The table is far from complete, and other energy forms and energy converterscould readily be added. However, it does include the major energy converters ofinterest to mechanical engineers. It is a goal of this book to present important aspectsof the design, analysis, performance, and operation of most of these devices.

One of the major criteria guiding the design of energy conversion systems is

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efficiency. Each of the conversions in Table 1.2 is executed by a device that operateswith one or more relevant efficiencies. The following section explores some of thevariety of definitions of efficiency used in design and performance studies of energyconversion devices.

1.9 Efficiencies

Efficiency is a measure of the quality of an operation or of a characteristic of a device.Several types of efficiencies are widely used. It is important to clearly distinguishamong them. Note that the terms work and power are equally applicable here.

The efficiency of a machine that transmits mechanical power is measured by itsmechanical efficiency, the fraction of the power supplied to the transmission devicethat is delivered to another machine attached to its output, Figure 1.5(a). Thus a gear-box for converting rotational motion from a power source to a device driven at anotherspeed dissipates some mechanical energy by fluid and/or dry friction, with a consequentloss in power transmitted to the second machine. The efficiency of the gearbox is theratio of its power output to the power input, a value less than one. For example, aturboprop engine with a gearbox efficiency of 0.95 will transmit only 95% of its poweroutput to its propeller.

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Another type of efficiency that measures internal losses of power is used toindicate the quality of performance of turbomachines such as pumps, compressors andturbines. These devices convert flow energy to work (power), or vice versa. Here theefficiency compares the output with a theoretical ideal in a ratio [Figure 1.5(b)]. Theresulting efficiency ranges from 0 to 1 as a measure of how closely the processapproaches a relevant isentropic process. A turbine with an efficiency of 0.9, will, forexample, deliver 90% of the power of a perfect (isentropic) turbine operating under thesame conditions. This efficiency, sometimes referred to as isentropic efficiency or turbine efficiency, will be considered in more detail in the next chapter.

Another form of isentropic efficiency, sometimes called compressor efficiency (or pump efficiency), is defined for compressors (or pumps). It is the ratio of the isentropicwork to drive the compressor (or pump) to the actual work required. Because theactual work required exceeds the isentropic work, this efficiency is also less than to 1.

A third type of efficiency compares the magnitude of a useful effect to the cost ofproducing the effect, measured in comparable units. An example of this type ofefficiency compares the net work output, wn, of a heat engine to the heat supplied, qa, to operate the engine. This is called the to (�th = wn /qa) [Figure 1.5(c)]. For example,the flow of natural gas to an electrical power plant provides a chemical energy flowrate or heat flow rate to the plant that leads to useful electric power output. It is knownfrom basic thermodynamics that this efficiency is limited by the Carnot efficiency, aswill be discussed in the next section.

Another example of this type of efficiency as applied to refrigerators and heatpumps [Figure 1.5(d)] is called the coefficient of performance, COP. In this case theuseful effect is the rate of cooling or heating, and the cost to produce the effect is thepower supplied to the device. The term "coefficient of performance" is used instead ofefficiency for this measure of quality because the useful effect usually exceeds the cost in comparable units of measure. Hence, unlike other efficiencies, the COP can exceedunity. As seen in Figure 1.5(d), there are two definitions for COP, one for a refrigeratorand another for a heat pump. It may be shown using the First Law of Thermodynamics,that a simple relationship exists between the two definitions: COPhp = COPrefr + 1.

1.10 The Carnot Engine

On beginning the study of the energy conversion ideas and devices that will serve us inthe twenty-first century, it is appropriate to review the theoretical cycle that stands asthe ideal for a heat engine. The ideas put forth by Sadi Carnot in 1824 in his�Reflections on the Motive Power of Heat� (see Historical Bibliography) expressed thecontent of the Second Law of Thermodynamics relevant to heat engines, which, inmodern form (attributed to Kelvin and Planck) is: �It is impossible for a device whichoperates in a cycle to receive heat from a single source and convert the heat completelyto work.� Carnot�s great work also described the cycle that today bears his name andprovides the theoretical limit for efficiency of heat engine cycles that operate betweentwo given temperature levels: the Carnot cycle.

The Carnot cycle consists of two reversible, isothermal processes separated by two

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reversible adiabatic or isentropic processes, as shown in Figure 1.6. All of the heat transferred to the working fluid is supplied isothermally at the high temperature TH = T3, and all heat rejected is transferred from the working medium at the lowtemperature TL = T1. No heat transfer takes place, of course, in the isentropicprocesses. It is evident from Equation (1.5) and the T-s diagram that the heat added isT3(s3 - s2 ), the heat rejected is T1 (s1 - s4 ), and, by the cyclic integral relation, the network is T3(s3 - s2 ) + T1 (s1 - s4 ). The thermal efficiency of the Carnot cycle, like that ofother cycles, is given by wn / qa and can be expressed in terms of the high and low cycletemperatures as :

26

because both isothermal processes operate between the same entropy limits.The Carnot efficiency equation shows that efficiency rises as TL drops and as

TH increases. The message is clear: a heat engine should operate between the widestpossible temperature limits. Thus the efficiency of a heat engine will be limited by themaximum attainable energy-source temperature and the lowest available heat-sinktemperature.

Students are sometimes troubled by the idea of isothermal heat transfer processesbecause they associate heat transfer with temperature rise. A moments reflection,however, on the existence of latent heats�e.g., the teakettle steaming on the stove atconstant temperature-makes it clear that one should not always associate heat transferwith temperature change.

It is important here, as we start to consider energy conversion devices, to recall thefamous Carnot Theorem, the proof of which is given the most thermodynamics texts. It states that it is impossible for any engine operating in a cycle between two reservoirsat different temperatures to have an efficiency that exceeds the Carnot efficiencycorresponding to those temperatures. It can also be shown that all reversible enginesoperating between two given reservoirs have the same efficiency and that all irreversibleengines must have lower efficiencies. Thus the Carnot efficiency sets an upper limit onthe performance of heat engines and therefore serves as a criterion by which otherengines may be judged.

1.11 Additional Second-Law Considerations

The qualitative relationship between the irreversibility of a process and the entropyincrease of the universe associated with it was considered in section 1.2. Let us nowconsider a quantitative approach to irreversibility and apply it to a model of a powerplant.

Reversible Work

Instead of comparing the work output of the power plant with the energy supplied fromfuel to run the plant, it is instructive to compare it with the maximum work achievableby a reversible heat engine operating between the appropriate temperature limits, thereversible work. It has been established that any reversible engine would have the sameefficiency as a Carnot engine. The Carnot engine provides a device for determining thereversible work associated with a given source temperature, TH, and a lower sinktemperature,TL.

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Irreversibility

The irreversibility, I, of a process is defined as the difference between the reversiblework and the actual work of a process:

I = Wrev - Wact [Btu | kJ]

It is seen that the irreversibility of a process vanishes when the actual work is the same

as that produced by an appropriate Carnot engine. Moreover, the irreversibility of anon-work-producing engine is equal to the reversible work. It is clear that I � 0,because no real engine can produce more work than a Carnot engine operating betweenthe same limiting temperatures.

Second-Law Efficiency

We can also define a �second-law efficiency,� �II, as the ratio of the actual work of aprocess to the reversible work:

[dl]

This is an efficiency that is limited to 100%, as opposed to the thermal efficiency of aheat engine, sometimes referred to as a �first-law efficiency,� which may not exceedthat of the appropriate Carnot engine. Note that an engine which has no irreversibilityis a reversible engine and has a second-law efficiency of 100%.

A Power Plant Model

Let us consider a model of a power plant in which a fuel is burned at a hightemperature, TH, in order to transfer heat to a working fluid at an intermediatetemperature, TINT. The working fluid, in turn, is used in an engine to produce work andreject heat to a sink at the low temperature, TL. Figure 1.7 presents a diagram of themodel that shows explicitly the combustion temperature drop from the sourcetemperature, TH, to the intermediate temperature, TINT, the actual work-producingengine, and a Carnot engine used to determine the reversible work for the situation. The Carnot engine has an efficiency of �C = 1 - TL /TH and develops work in thefollowing amount:

Wrev = WC = QIN - QC = �CQIN [Btu|kJ]

Suppose that the engine we are considering is a Carnot engine that operates from theintermediate source at TINT. Its efficiency and work output are, respectively, �I = 1 - TL /TINT and Wact = QIN � QI = �I QIN . The irreversibility, I, of the power plantis then

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I = Wrev - Wact = (�C - �I )QIN = [1 - TL /TH - ( 1 - TL /TINT)]QIN

= TLQIN ( TH - TINT ) /TINTTH [Btu|kJ]

and the second-law efiiciency is:

�II = Wact /Wrev = (1 - TL/TINT ) / (1 - TL/TH) [dl]

Note that when TINT = TH, the irreveribility vanishes and the second-law efficiencybecomes 100%. Also, when TINT = TL, the irreversibility is equal to the reversible workof the Carnot engine and the second- law efficiency is zero. The latter conditionindicates that a pure heat transfer process or any process that produces no useful workcauses a loss in the ability to do work in the amount of Wrev . Thus the reversible workassociated with the extremes of a given process is a measure of how much capability todo work can be lost, and the irreversibility is a measure of how much of that work-producing potential is actually lost. The following example illustrates these ideas.

EXAMPLE 1.3

Through combustion of a fossil fuel at 3500°R, an engine receives energy at a rate of3000 Btu/s to heat steam to 1500°R. There is no energy loss in the combustionprocess. The steam, in turn, produces 1000 Btu/s of work and rejects the remainingenergy to the surroundings at 500°R.

(a) What is the thermal efficiency of the plant?(b) What are the reversible work and the Carnot efficiency corresponding to the source

29

and sink temperatures?(c) What is the irreversibility?(d) What is the second-law efficiency?(e) What would the irreversibility and the second-law efficiency be if the working fluidwere processed by a Carnot engine rather than by the real engine?

Solution(a) The thermal efficiency, or first-law efficiency, of the plant is �I = Wact/QIN =

1000/3000 = 0.333 or 33.3%.(b) The relevant Carnot efficiency is 1 - 500/3500 = 0.857, or 85.7%. The engine�sreversible work is then �CQIN = 0.857(3000) = 2571 Btu/s.

(c) The plant irreversibility is 2571 - 1000 = 1571 Btu/s.

(d) The second-law efficiency is then �I /�C = 0.333/0.857 = 0.389, or 38.9%orWact /Wrev = 1000/2571 = 0.389, or 38.9%.

(e) The Carnot efficiency corresponding to the maximum temperature of the workingfluid is 1 - 500/1500 = 0.667 or 66.7%. The second-law efficiency for this system isthen 0.667/0.857 = 0.778, compared with 0.389 for the actual engine. The actual workproduced by the irreversibly heated Carnot engine is 0.667(3000) = 2001 Btu/s, and itsirreversibility is then I = 2571 -2001 = 570 Btu/s.___________________________________________________________________

In summary, the thermal efficiency, or first-law efficiency, of an engine is ameasure of how well the engine converts the energy in its fuel to useful work. It saysnothing about energy loss, because energy is conserved and cannot be lost: it can onlybe transformed. The second-law efficiency, on the other hand, recognizes that some ofthe energy of a fuel is not available for conversion to work in a heat engine andtherefore assesses the ability of the engine to convert only the available work into usefulwork. This is a reason why some regard the second-law efficiency as more significantthan the more commonly used first-law efficiency.

Bibliography and References

1. Van Wylen, Gordon J., and Sonntag, Richard E., Fundamentals of ClassicalThermodynamics, 3rd ed. New York: Wiley, 1986.

2. Balmer, Robert, Thermodynamics. Minneapolis: West, 1990. 3. Cengel, Yunus A., and Boles, Michael A., Thermodynamics. New York: McGraw-Hill, 1989.

30

4. Faires, Virgil Moring, Thermodynamics, 5th ed. New York: Macmillan, 1970.

5. Silver, Howard F., and Nydahl, John E., Engineering Thermodynamics.Minneapolis: West, 1977.

6. Bathie, William W., Fundamentals of Gas Turbines. New York: Wiley, 1984

7. Wilson, David Gordon, The Design of High Efficiency Turbomachinery and GasTurbines. Boston: MIT Press, 1984.

8. Anderson, John D., Modern Compressible Flow. New York: McGraw-Hill, 1982. 9. Anderson, John D., Introduction to Flight. New York: McGraw-Hill, 1978.

10. Chapman, Alan J. and Walker, William F., Introductory Gas Dynamics. New York:Holt, Rinehart, and Winston, 1971.

Historical Bibliography

1. Barnard, William N., Ellenwood, Frank E., and Hirshfeld, Clarence F., Heat-PowerEngineering. Wiley, 1926.

2. Bent, Henry, The Second Law.NewYork: Oxford University Press, 1965

3. Cummins, C. Lyle, Jr., Internal Fire, rev. ed. Warrendale, Penna.: Society ofAutomotive Engineers, 1989.

4. Tann, Jennifer, The Selected Papers of Boulton and Watt. Boston: MIT Press, 1981.

5. Carnot, Sadi, Reflexions Sur la Puissance Motrice de Feu. Paris: Bachelor,1824.

6. Potter, J. H., �The Gas Turbine Cycle.� ASME Paper presented at the Gas TurbineForum Dinner, ASME Annual Meeting, New York, Nov. 27, 1972.

7. Grosser, Morton, Diesel: The Man and the Machine. New York: Atheneum, 1978.

8. Nitske, W. Robert, and Wilson, Charles Morrow, Rudolph Diesel: Pioneer of theAge of Power. Norman, Okla: University of Oklahoma Press, 1965.

9. Rolt, L.T. C., and Allen, J. S., The Steam Engine of Thomas Newcomen. New York:Moorland Publishing Co., 1977.

10. Briggs, Asa, The Power of Steam. Chicago: University of Chicago Press, 1982.

31

11. Thurston, Robert H., A History of the Growth of the Steam Engine. 1878; rpr.Ithaca, N.Y.: Cornell University Press, 1939.

EXERCISES

1.1 Determine the entropy of steam at 1000 psia and a quality of 50%.

1.2 Show that the moisture fraction for a liquid water-steam mixture, defined as theratio of liquid mass to mixture mass, can be written as 1 - x, where x is steam quality.

1.3 Write expressions for the specific entropy, specific enthalpy, and specific volume asfunctions of the moisture fraction. Determine the values of these properties for steam at500°F and a moisture fraction of 0.4.

1.4 Show that for a thermally perfect gas, cp - cv = R.

1.5 During a cyclic process, 75kJ of heat flow into a system and 25kJ are rejected fromthe system later in the cycle. What is the net work of the cycle.

1.6 Seventy-five kJ of heat flow into a rigid system and 25 kJ are rejected later. Whatare the magnitude and sign of the change in internal energy? What does the signindicate?

1.7 The mass contained between an insulated piston and an insulated cylinderdecreases in internal energy by 50 Btu. How much work is involved, and what is thesign of the work term? What does the sign indicate?

1.8 Derive Equation (1.8) from Equation (1.7).

1.9 Use Equation (1.8) to derive an expression for the finite enthalpy change of anincompressible fluid in an isentropic process. If the process is the pressurization ofsaturated water initially at 250 psia, what is the enthalpy rise, in Btu /lbm and in ft-lbf / lbm, if the final pressure is 4000 psia? What is the enthalpy rise if the initialpressure is 100 kPa and the final pressure is 950 kPa?

1.10 Sixty kg /s of brine flows into a device with an enthalpy of 200 kJ/kg. Brine flowsout of the other port at a flow rate of 20 kg /s. What is the net inflow? Is the system insteady flow? Explain.

1.11 Use the steam tables in Appendices B and C to compare the heats of vaporizationat 0.01, 10, and 1000 psia. Compare the saturated liquid specific volumes at thesepressures. What do you conclude about the influence of pressure on these properties?

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1.12 Using the heat capacity equation for nitrogen:

cp = 9.47 - 3.47 x 103/T + 1.16 x 106/T2

where cp is in Btu / lb-mole and T is in degrees Rankine (From Gordon J. Van Wylenand Richard E. Sonntag, Fundamentals of Classical Thermodynamics, 3rd ed. NewYork: Wiley, 1986). Compare the enthalpy change per mole of nitrogen between540°R and 2000°R for nitrogen as a thermally perfect gas and as a calorically perfectgas.

1.13 Use Equation (1.7) to derive Equation (1.20) for a calorically perfect gas.

1.14 Use Equation (1.7) to derive a relation for the entropy change as a function oftemperature ratio for a constant-volume process in a calorically perfect gas.

1.15 Use Equation (1.8) to derive a relation for the entropy change as a function oftemperature for an isobaric process in a calorically perfect gas.

1.16 A convergent nozzle is a flow passage in which area decreases in the streamwisedirection (the direction of the flow). It is used for accelerating the flow from a lowvelocity to a higher velocity. Use the generalized form of the steady-flow First Lawgiven in Equation (1.26a) to derive an equation for the exit velocity for an adiabaticnozzle.

1.17 Derive an equation for the pressure drop for a loss-free incompressible flow in avarying-area duct as a function of area ratio.

1.18 Two units of work are required to transfer 10 units of heat from a refrigerator tothe environment. What is the COP of the refrigerator? Suppose that the same amountof heat transfer instead is by a heat pump into a house. What is the heat pump COP?

1.19 A power plant delivers 100 units of work at 30% thermal efficiency. How manyheat units are supplied to operate the plant? How many units of heat are rejected to thesurroundings?

1.20 A steam turbine has an efficiency of 90% and a theoretical isentropic power of'100 kW. What is the actual power output?

1.21 Thomas Newcomen used the fact that the specific volume of saturated liquid ismuch smaller than the specific volume of saturated steam at the same pressure in hisfamous "atmospheric engine." Calculate the work done on the piston by the atmosphereif steam is condensed at an average pressure of 6 psia by cooling in a tightly fittedpiston-cylinder enclosure if the piston area is 1 ft2 and the piston stroke is 1 ft. If theprocess takes place 10 times a minute, what is the power delivered? Discuss what can

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be done to increase the power of the engine. Describe the characteristics of aNewcomen engine that would theoretically deliver 20 horsepower.

1.22 Expand Table 1.2 to include solar and geothermal sources.

1.23 Twenty pounds of compressed air is stored in a tank at 200 psia and 80�F. Thetank is heated to bring the temperature to 155�F. What is the final tank pressure, andhow much heat was added?

1.24 Ten kilograms of compressed air is stored in a tank at 250 kPa and 50�C. Thetank is heated to bring the air temperature to 200 °C. What is the final tank pressure,and how much heat was added?

1.25 An 85-ft3 tank contains air at 30 psia and 100� F. What mass of air must be addedto bring the pressure to 50 psia and the temperature to 150�F?

1.26 A 20-L tank contains air at 2 bar and 300K. What mass of air must be added tobring the pressure to 2 bar and the temperature to 375K?

1.27 Air enters a wind tunnel nozzle at 160�F, 10 atm, and a velocity of 50 ft/s. Theentrance area is 5ft2. If the heat loss per unit mass is 10 Btu/lbm and the exit pressureand velocity are, respectively, 1.5 atm and 675 ft/s, what are the exit temperature andarea?

1.28 Air enters a wind tunnel nozzle at 90°C, 250 kPa, and a velocity of 40 m/s. Theentrance area is 3 m2. If the heat loss per unit mass is 7 kJ/kg and the exit pressure andvelocity are, respectively, 105 kPa and 250 m/s, what are the exit temperature andarea?

1.29 Air enters a wind tunnel nozzle at 160�F, 10 atm, and a velocity of 50 ft/s. Theentrance area is 5ft2. If the heat loss per unit mass is 8 Btu/lbm and the exit pressure andtemperature are, respectively, 1.25 atm and 120�F, what are the exit velocity and area?

1.30 Air enters a wind tunnel nozzle at 90�C, 250 kPa, and a velocity of 40 m/s. The entrance area is 3 m2. If the heat loss per unit mass is 5 kJ/kg and the exit pressure andtemperature are, respectively, 120 kPa and 43°C, what are the exit velocity and area?

1.31 Sketch a Mollier diagram showing the character of three isotherms and threeisobars for a calorically perfect gas. Label each curve with a value in SI units to showthe directions of increasing temperature and pressure. Explain how the diagram woulddiffer if the gas were not calorically perfect.

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C H A P T E R T W O

Fundamentals of Steam Power

2.1 Introduction

Much of the electricity used in the United States is produced in steam power plants.Despite efforts to develop alternative energy converters, electricity from steam willcontinue, for many years, to provide the power that energizes the United States andworld economies. We therefore begin the study of energy conversion systems with thisimportant element of industrial society.

Steam cycles used in electrical power plants and in the production of shaft powerin industry are based on the familiar Rankine cycle, studied briefly in most courses inthermodynamics. In this chapter we review the basic Rankine cycle and examinemodifications of the cycle that make modern power plants efficient and reliable.

2.2 A Simple Rankine-Cycle Power Plant

The most prominent physical feature of a modern steam power plant (other than itssmokestack) is the steam generator, or boiler, as seen in Figure 2.1. There thecombustion, in air, of a fossil fuel such as oil, natural gas, or coal produces hotcombustion gases that transfer heat to water passing through tubes in the steamgenerator. The heat transfer to the incoming water (feedwater) first increases itstemperature until it becomes a saturated liquid, then evaporates it to form saturated vapor, and usually then further raises its temperature to create superheated steam.

Steam power plants such as that shown in Figure 2.1, operate on sophisticatedvariants of the Rankine cycle. These are considered later. First, let�s examine thesimple Rankine cycle shown in Figure 2.2, from which the cycles of large steam powerplants are derived.

In the simple Rankine cycle, steam flows to a turbine, where part of its energy isconverted to mechanical energy that is transmitted by rotating shaft to drive anelectrical generator. The reduced-energy steam flowing out of the turbine condenses toliuid water in the condenser. A feedwater pump returns the condensed liquid(condensate) to the steam generator. The heat rejected from the steam entering thecondenser is transferred to a separate cooling water loop that in turn delivers therejected energy to a neighboring lake or river or to the atmosphere.

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As a result of the conversion of much of its thermal energy into mechanical energy,or work, steam leaves the turbine at a pressure and temperature well below the turbineentrance (throttle) values. At this point the steam could be released into theatmosphere. But since water resources are seldom adequate to allow the luxury of one-time use, and because water purification of a continuous supply of fresh feedwater iscostly, steam power plants normally utilize the same pure water over and over again. We usually say that the working fluid (water) in the plant operates in a cycle orundergoes of cyclic process, as indicated in Figure 2.2. In order to return the steam tothe high-pressure of the steam generator to continue the cycle, the low- pressure steamleaving the turbine at state 2 is first condensed to a liquid at state 3 and thenpressurized in a pump to state 4. The high pressure liquid water is then ready for itsnext pass through the steam generator to state 1 and around the Rankine cycle again.

The steam generator and condenser both may be thought of as types of heatexchangers, the former with hot combustion gases flowing on the outside of water-

37

filled tubes, and the latter with external cooling water passing through tubes on whichthe low- pressure turbine exhaust steam condenses. In a well-designed heat exchanger,both fluids pass through with little pressure loss. Therefore, as an ideal, it is commonto think of steam generators and condensers as operating with their fluids atunchanging pressures.

It is useful to think of the Rankine cycle as operating between two fixed pressurelevels, the pressure in the steam generator and pressure in the condenser. A pumpprovides the pressure increase, and a turbine provides the controlled pressure dropbetween these levels.

Looking at the overall Rankine cycle as a system (Figure 2.2), we see that work isdelivered to the surroundings (the electrical generator and distribution system) by theturbine and extracted from the surroundings by a pump (driven by an electric motor ora small steam turbine). Similarly, heat is received from the surroundings (combustiongas) in the steam generator and rejected to cooling water in the condenser.

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At the start of the twentieth century reciprocating steam engines extracted thermalenergy from steam and converted linear reciprocating motion to rotary motion, toprovide shaft power for industry. Today, highly efficient steam turbines, such as shownin Figure 2.3, convert thermal energy of steam directly to rotary motion. Eliminatingthe intermediate step of conversion of thermal energy into the linear motion of a pistonwas an important factor in the success of the steam turbine in electric powergeneration. The resulting high rotational speed, reliability, and power output of theturbine and the development of electrical distribution systems allowed the centralizationof power production in a few large plants capable of serving many industrial andresidential customers over a wide geographic area.

The final link in the conversion of chemical energy to thermal energy to mechanicalenergy to electricity is the electrical generator. The rotating shaft of the electricalgenerator usually is directly coupled to the turbine drive shaft. Electrical windingsattached to the rotating shaft of the generator cut the lines of force of the statorwindings, inducing a flow of alternating electrical current in accordance with Faraday'sLaw. In the United States, electrical generators turn at a multiple of the generationfrequency of 60 cycles per second, usually 1800 or 3600 rpm. Elsewhere, where 50cycles per second is the standard frequency, the speed of 3000 rpm is common.Through transformers at the power plant, the voltage is increased to several hundredthousand volts for transmission to distant distribution centers. At the distributioncenters as well as neighborhood electrical transformers, the electrical potential isreduced, ultimately to the 110- and 220-volt levels used in homes and industry.

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Since at present there is no economical way to store the large quantities ofelectricity produced by a power plant, the generating system must adapt, from momentto moment, to the varying demands for electricity from its customers. It is thereforeimportant that a power company have both sufficient generation capacity to reliablysatisfy the maximum demand and generation equipment capable of adapting to varyingload.

2.3 Rankine-Cycle Analysis

In analyses of heat engine cycles it is usually assumed that the components of theengine are joined by conduits that allow transport of the working fluid from the exit ofone component to the entrance of the next, with no intervening state change. It will beseen later that this simplification can be removed when necessary.

It is also assumed that all flows of mass and energy are steady, so that the steadystate conservation equations are applicable. This is appropriate to most situationsbecause power plants usually operate at steady conditions for significant lengths oftime. Thus, transients at startup and shutdown are special cases that will not beconsidered here.

Consider again the Rankine cycle shown in Figure 2.2. Control of the flow can beexercised by a throttle valve placed at the entrance to the turbine (state 1). Partial valveclosure would reduce both the steam flow to the turbine and the resulting poweroutput. We usually refer to the temperature and pressure at the entrance to the turbineas throttle conditions. In the ideal Rankine cycle shown, steam expands adiabaticallyand reversibly, or isentropically, through the turbine to a lower temperature andpressure at the condenser entrance. Applying the steady-flow form of the First Law ofThermodynamics [Equation (1.10)] for an isentropic turbine we obtain:

q = 0 = h2 � h1 + wt [Btu/lbm | kJ/kg]

where we neglect the usually small kinetic and potential energy differences between theinlet and outlet. This equation shows that the turbine work per unit mass passingthrough the turbine is simply the difference between the entrance enthalpy and thelower exit enthalpy:

wt = h1 � h2 [Btu/lbm | kJ/kg] (2.1)

The power delivered by the turbine to an external load, such as an electrical generator,is given by the following:

Turbine Power = mswt = ms(h1 � h2) [Btu/hr | kW]

where ms [lbm /hr | kg/s] is the mass flow of steam though the power plant.

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Applying the steady-flow First Law of Thermodynamics to the steam generator,we see that shaft work is zero and thus that the steam generator heat transfer is

qa = h1 � h4 [Btu/lbm | kJ/kg] (2.2)

The condenser usually is a large shell-and-tube heat exchanger positioned below oradjacent to the turbine in order to directly receive the large flow rate of low-pressureturbine exit steam and convert it to liquid water. External cooling water is pumpedthrough thousands of tubes in the condenser to transport the heat of condensation ofthe steam away from the plant. On leaving the condenser, the condensed liquid (calledcondensate) is at a low temperature and pressure compared with throttle conditions.Continued removal of low-specific-volume liquid formed by condensation of the high-specific-volume steam may be thought of as creating and maintaining the low pressurein the condenser. The phase change in turn depends on the transfer of heat released tothe external cooling water. Thus the rejection of heat to the surroundings by thecooling water is essential to maintaining the low pressure in the condenser. Applyingthe steady-flow First Law of Thermodynamics to the condensing steam yields:

qc = h3 � h2 [Btu/lbm | kJ/kg] (2.3)

The condenser heat transfer qc is negative because h2 > h3. Thus, consistent with signconvention, qc represents an outflow of heat from the condensing steam. This heat isabsorbed by the cooling water passing through the condenser tubes. The condenser-cooling-water temperature rise and mass-flow rate mc are related to the rejected heatby:

ms|qc| = mc cwater(Tout - Tin) [Btu/hr | kW]

where cwater is the heat capacity of the cooling water [Btu/lbm-R | kJ/kg-K]. Thecondenser cooling water may be drawn from a river or a lake at the temperature Tin andreturned downstream at Tout, or it may be circulated through cooling towers where heatis rejected from the cooling water to the atmosphere.

We can express the condenser heat transfer in terms of an overall heat transfercoefficient, U, the mean cooling water temperature, Tm = (Tout + Tin)/2, and thecondensing temperature T3:

ms|qc| = UA(T3 - Tm) [Btu/hr | kJ/s]

It is seen for given heat rejection rate, the condenser size represented by the tubesurface area A depends inversely on (a) the temperature difference between thecondensing steam and the cooling water, and (b) the overall heat-transfer coefficient.

For a fixed average temperature difference between the two fluids on oppositesides of the condenser tube walls, the temperature of the available cooling watercontrols the condensing temperature and hence the pressure of the condensing steam.

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Therefore, the colder the cooling water, the lower the minimum temperature andpressure of the cycle and the higher the thermal efficiency of the cycle.

A pump is a device that moves a liquid from a region of low pressure to one ofhigh pressure. In the Rankine cycle the condenser condensate is raised to the pressureof the steam generator by boiler feed pumps, BFP. The high-pressure liquid waterentering the steam generator is called feedwater. From the steady-flow First Law ofThermodynamics, the work and power required to drive the pump are:

wp = h3 � h4 [Btu/lbm | kJ/kg] (2.4)

and

Pump Power = mswp = ms(h3 � h4) [Btu/hr | kW]

where the negative values resulting from the fact that h4 > h3 are in accordance with thethermodynamic sign convention, which indicates that work and power must be suppliedto operate the pump.

The net power delivered by the Rankine cycle is the difference between the turbinepower and the magnitude of the pump power. One of the significant advantages of theRankine cycle is that the pump power is usually quite small compared with the turbinepower. This is indicated by the work ratio, wt / wp, which is large compared with onefor Rankine cycle. As a result, the pumping power is sometimes neglected inapproximating the Rankine cycle net power output.

It is normally assumed that the liquid at a pump entrance is saturated liquid. Thisis usually the case for power-plant feedwater pumps, because on the one handsubcooling would increase the heat edition required in the steam generator, and on theother the introduction of steam into the pump would cause poor performance anddestructive, unsteady operation. The properties of the pump inlet or condenser exit(state 3 in Figure 2.2) therefore may be obtained directly from the saturated-liquidcurve at the (usually) known condenser pressure.

The properties for an isentropic pump discharge at state 4 could be obtained froma subcooled-water property table at the known inlet entropy and the throttle pressure. However, such tables are not widely available and usually are not needed. Theenthalpy of a subcooled state is commonly approximated by the enthalpy of thesaturated-liquid evaluated at the temperature of the subcooled liquid. This is usuallyquite accurate because the enthalpy of a liquid is almost independent of pressure. Anaccurate method for estimating the pump enthalpy rise and the pump work is given later(in Example 2.3).

A measure of the effectiveness of an energy conversion device is its thermalefficiency, which is defined as the ratio of the cycle net work to the heat supplied fromexternal sources. Thus, by using Equations (2.1), (2.2), and (2.4) we can express theideal Rankine-cycle thermal efficiency in terms of cycle enthalpies as:

42

�th = (h1 � h2 + h3 � h4)/(h1 � h4) [dl] (2.5)

In accordance with the Second Law of Thermodynamics, the Rankine cycleefficiency must be less than the efficiency of a Carnot engine operating between thesame temperature extremes. As with the Carnot-cycle efficiency, Rankine-cycleefficiency improves when the average heat-addition temperature increases and the heat-rejection temperature decreases. Thus cycle efficiency may be improved by increasingturbine inlet temperature and decreasing the condenser pressure (and thus thecondenser temperature).

Another measure of efficiency commonly employed by power plant engineers is theheat rate, that is, the ratio of the rate of heat addition in conventional heat units to thenet power output in conventional power units. Because the rate of heat addition isproportional to the fuel consumption rate, the heat rate is a measure of fuel utilizationrate per unit of power output. In the United States, the rate of heat addition is usuallystated in Btu/hr, and electrical power output in kilowatts, resulting in heat rates being expressed in Btu/kW-hr. The reader should verify that the heat rate in English units isgiven by the conversion factor, 3413 Btu/kW-hr, divided by the cycle thermal efficiencyas a decimal fraction, and that its value has a magnitude of the order of 10,000Btu/kW-hr. In the SI system of units, the heat rate is usually expressed in kJ/kW-hr, isgiven by 3600 divided by the cycle efficiency as a decimal fraction, and is of the sameorder of magnitude as in the English system. It is evident that a low value of heat raterepresents high thermal efficiency and is therefore desirable.

EXAMPLE 2.1

An ideal Rankine cycle (see Figure 2.2) has a throttle state of 2000 psia/1000°F andcondenser pressure of 1 psia. Determine the temperatures, pressures, entropies, andenthalpies at the inlets of all components, and compare the thermal efficiency of thecycle with the relevant Carnot efficiency. Neglect pump work. What is the quality ofthe steam at the turbine exit?

SolutionThe states at the inlets and exits of the components, following the notation of

Figure 2.2, are listed in the following table. The enthalpy and entropy of state 1 may beobtained directly from tables or charts for superheated steam (such as those inAppendices B and C) at the throttle conditions. A Mollier chart is usually moreconvenient than tables in dealing with turbine inlet and exit conditions.

For an ideal isentropic turbine, the entropy is the same at state 2 as at state 1. Thusstate 2 may be obtained from the throttle entropy (s2 = s1 = 1.5603 Btu/lbm-R) and thecondenser pressure (1 psia). In general, this state may be in either the superheated-steam region or the mixed-steam-and-liquid region of the Mollier and T-s diagrams. Inthe present case it is well into the mixed region, with a temperature of 101.74°F and anenthalpy of 871 Btu/lbm.

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The enthalpy, h3 = 69.73 Btu/lbm, and other properties at the pump inlet are obtainedfrom saturated-liquid tables, at the condenser pressure. The steady-flow First Law ofThermodynamics, in the form of Equation (2.4), indicates that neglecting isentropicpump work is equivalent to neglecting the pump enthalpy rise. Thus in this caseEquation (2.4) implies that h3 and h4 shown in Figure (2.2) are almost equal. Thus wetake h4 = h3 as a convenient approximation.

State Temperature (°F)

Pressure (psia)

Entropy (Btu/lbm-°R)

Enthalpy (Btu/lbm)

1 1000.0 2000 1.5603 1474.1

2 101.74 1 1.5603 871.0

3 101.74 1 0.1326 69.73

4 101.74 2000 0.1326 69.73

The turbine work is

h1 � h2 = 1474.1 � 871 = 603.1 Btu/lbm.

The heat added in the steam generator is

h1 � h4 = 1474.1 � 69.73 = 1404.37 Btu/lbm.

The thermal efficiency is the net work per heat added = 603.1/1404.37 = 0.4294(42.94%). This corresponds to a heat rate of 3413/0.4294 = 7946 Btu/kW-hr. Asexpected, the efficiency is significantly below the value of the Carnot efficiency of 1 � (460 + 101.74)/(460 + 1000) = 0.6152 (61.52%), based on a sourcetemperature of T1 and a sink temperature of T3.

The quality of the steam at the turbine exit is

(s2 � sl)/(sv � sl) = (1.5603 � 0.1326)/(1.9781 � 0.1326) = 0.7736

Here v and l indicate saturated vapor and liquid states, respectively, at pressure p2. Note that the quality could also have been obtained from the Mollier chart for steam as1 - M, where M is the steam moisture fraction at entropy s2 and pressure p2.__________________________________________________________________

Example 2-2If the throttle mass-flow is 2,000,000 lbm/hr and the cooling water enters the condenserat 60°F, what is the power plant output in Example 2.1? Estimate the cooling-watermass-flow rate.

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Solution: The power output is the product of the throttle mass-flow rate and the powerplant net work. Thus

Power = (2 × 106)(603.1) = 1.206 × 109 Btu/hr

or

Power = 1.206 × 109 / 3413 = 353,413 kW.

The condenser heat-transfer rate is

msqc = ms ( h3 � h2 ) = 2,000,000 × (69.73 � 871) = � 1.603×109 Btu/hr

The condensing temperature, T3 = 101.74 °F, is the upper bound on the coolingwater exit temperature. Assuming that the cooling water enters at 60°F and leaves at95°F, the cooling-water flow rate is given by

mc = ms|qc| / [ cwater(Tout � Tin)] = 1.603×109 /[(1)(95 - 60)] = 45.68×106 lbm/hr

A higher mass-flow rate of cooling water would allow a smaller condenser cooling-water temperature rise and reduce the required condenser-heat-transfer area at theexpense of increased pumping power.____________________________________________________________________

2.4 Deviations from the Ideal � Component Efficiencies

In a power plant analysis it is sometimes necessary to account for non-ideal effects suchas fluid friction, turbulence, and flow separation in components otherwise assumed tobe reversible. Decisions regarding the necessity of accounting for these effects arelargely a matter of experience built on familiarity with the magnitudes of the effects,engineering practices, and the uses of the calculated results.

Turbine

In the case of an adiabatic turbine with flow irreversibilities, the steady-flow First Lawof Thermodynamics gives the same symbolic result as for the isentropic turbine inEquation (2.1), i.e.,

wt = h1 � h2 [Btu/lb | kJ/kg] (2.6)

except that here h2 represents the actual exit enthalpy and wt is the actual work of anadiabatic turbine where real effects such as flow separation, turbulence, irreversibleinternal heat transfers, and fluid friction exist.

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An efficiency for a real turbine, known as the isentropic efficiency, is defined asthe ratio of the actual shaft work to the shaft work for an isentropic expansion betweenthe same inlet state and exit pressure level. Based on the notation of Figure 2.4, we seethat the turbine efficiency is:

�turb = (h1 � h2 )/(h1 � h2s ) [dl] (2.7)

where h2s, the isentropic turbine-exit enthalpy, is the enthalpy evaluated at the turbineinlet entropy and the exit pressure. For the special case of an isentropic turbine, h2 = h2s and the efficiency becomes 1. Note how state 2 and the turbine work changein Figure 2.4 as the efficiency increases toward 1. The diagram shows that thedifference between the isentropic and actual work, h2 � h2s, represents work lost due toirreversibility. Turbine isentropic efficiencies in the low 90% range are currentlyachievable in well-designed machines.

Normally in solving problems involving turbines, the turbine efficiency is knownfrom manufacturers� tests, and the inlet state and the exhaust pressure are specified.State 1 and p2 determine the isentropic discharge state 2s using the steam tables. Theactual turbine-exit enthalpy can then be calculated from Equation (2.7). Knowing bothp2 and h2, we can then fully identify state 2 and account for real turbine behavior in anycycle analysis.

Pump

Work must be supplied to a pump to move liquid from a low pressure to a high pressure. Some of the work supplied is lost due to irreversibilities. Ideally the remainingeffective work to raise the pressure is necessarily less than that supplied. In order for

46

the efficiency of a pump to be less than or equal to 1, it is defined in inverse fashion toturbine efficiency. That is, pump efficiency is the ratio of the isentropic work to theactual work input when operating between two given pressures. Applying Equation(2.4) and the notation of Figure (2.5), the isentropic pump work, wps = h3 � h4s, and thepump isentropic efficiency is

�pump = wps /wp = (h4s � h3)/(h4 � h3) [dl] (2.8)

Note the progression of exit states that would occur in Figure 2.5 as pump efficiencyincreases for a fixed inlet state and exit pressure. It is seen that the pump lost work,given by h4 � h4s decreases and that the actual discharge state approaches the isentropicdischarge state.

States 4 and 4s are usually subcooled liquid states. As a first approximation theirenthalpies may be taken to be the saturated liquid enthalpy at T3. More accurateapproximations for these enthalpies may be obtained by applying the First Law for aclosed system undergoing a reversible process, Equation (1.8): Tds = dh - vdp. For anisentropic process it follows that dh = vdp. Because a liquid is almost incompressible,its specific volume, v, is almost independent of pressure. Thus, using the notation ofFigure 2.5, integration with constant specific volume yields

h4s = h3 + v3 ( p4 � p3 ) [Btu/lbm | kJ/kg]

where a knowledge of state 3 and p4 determines h4s.

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Using Equation (2.8), and without consulting tables for subcooled water, we canthen calculate the pump work from

wp = v3(p3 � p4)/�p [ft-lbf/lbm | kN-m/kg] (2.9)

Note that the appropriate conversion factors must be applied for dimensionalconsistency in Equation (2.9).

EXAMPLE 2.3

Calculate the actual work and the isentropic and actual discharge enthalpies for an 80%efficient pump with an 80°F saturated-liquid inlet and an exit pressure of 3000 psia.

SolutionFrom the saturated-liquid tables, for 80°F, the pump inlet conditions are 0.5068 psia,48.037 Btu/lbm, and 0.016072 ft3/lbm.

Using Equation (2.9), we find that the pump work is

wp = [0.016072(0.5068 � 3000)(144)]/0.8 = � 8677 ft-lbf / lbmor

wp = � 8677/778 = � 11.15 Btu/lbm.

Note the importance of checking units here.The actual discharge enthalpy is

h4 = h3 � wp = 48.037 � (�11.15) = 59.19 Btu/lbm.

and the isentropic discharge enthalpy is

h4s = h3 � �p wp = 48.037 � (0.8)(� 11.15) = 56.96 Btu/lbm. ____________________________________________________________________

EXAMPLE 2.4

What is the turbine work, the net work, the work ratio, and the cycle thermal efficiencyfor the conditions of Example 2.1 if the turbine efficiency is 90% and the pumpefficiency is 85%? What is the turbine exit quality?

SolutionBy the definition of isentropic efficiency, the turbine work is 90% of the isentropicturbine work = (0.9)(603.1) = 542.8 Btu/lbm.By using Equation (2.9), the isentropic pump work is

[(0.01614)(1 � 2000)(144)] / 778 = � 5.97 Btu/lbm.

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The actual pump work is then � 5.97/.85 = � 7.03 Btu/lbm and the work ratio is542.8/| � 7.03| = 77.2

The cycle net work is wt + wp = 542.8 � 7.03 = 535.8 Btu/lbm..

Applying the steady-flow First Law of Thermodynamics to the pump, we get theenthalpy entering the steam generator to be

h4 = h3 � wp = 69.73 � (� 7.03) = 76.76 Btu/lbm.

The steam-generator heat addition is then reduced to 1474.1 � 76.76 = 1397.3Btu/lbm. and the cycle efficiency is 535.8/1397.3 = 0.383. Study of these examplesshows that the sizable reduction in cycle efficiency from that in Example 2.1 is largelydue to the turbine inefficiency, not to the neglect of pump work.

From Equation (2.6), the true turbine exit enthalpy is the difference between thethrottle enthalpy and actual turbine work = 1474.1 - 542.8 = 931.3 Btu/lbm.

The quality is then x = (h2 � hl)/(hv � hl) = (931.3 � 69.73)/(1105.8 � 69.73) =0.832.

Thus the turbine inefficiency increases the turbine exhaust quality over theisentropic turbine value of 0.774.____________________________________________________________________

2.5 Reheat and Reheat Cycles

A common modification of the Rankine cycle in large power plants involvesinterrupting the steam expansion in the turbine to add more heat to the steam beforecompleting the turbine expansion, a process known as reheat. As shown in Figure 2.6,steam from the high-pressure (HP) turbine is returned to the reheat section of the steamgenerator through the "cold reheat" line. There the steam passes through heated tubeswhich restore it to a temperature comparable to the throttle temperature of the highpressure turbine. The reenergized steam then is routed through the "hot reheat" line toa low-pressure turbine for completion of the expansion to the condenser pressure.

Examination of the T-s diagram shows that reheat increases the area enclosed bythe cycle and thus increases the net work of the cycle by virtue of the cyclic integral,Equation (1.3). This is significant, because for a given design power output higher network implies lower steam flow rate. This, in turn, implies that smaller plant componentsmay be used, which tends to reduce the initial plant cost and to compensate for addedcosts due to the increased complexity of the cycle.

Observe from Figure 2.6 that the use of reheat also tends to increase the averagetemperature at which heat is added. If the low-pressure turbine exhaust state issuperheated, the use of reheat may also increase the average temperature at which heatis rejected. The thermal efficiency may therefore increase or decrease, depending onspecific cycle conditions. Thus the major benefits of reheat are increased net work,

49

drying of the turbine exhaust (discussed further later), and the possibility of improvedcycle efficiency.

Note that the net work of the reheat cycle is the algebraic sum of the work of thetwo turbines and the pump work. Note also that the total heat addition is the sum of theheat added in the feedwater and reheat passes through the steam generator. Thus the

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thermal efficiency of the reheat cycle is:

(h1 � h2) + (h3 � h4) + (h5 � h6) �th = ----------------------------------- [dl] (2.10)

(h1 � h6) + (h3 � h2)

Relations such as this illustrate the wisdom of learning to analyze cycles usingdefinitions and applying fundamentals to components rather than memorizing equationsfor special cases such as Equation (2.5) for the efficiency of the simple Rankine cycle.

Note that the inclusion of reheat introduces a third pressure level to the Rankinecycle. Determination of a suitable reheat pressure level is a significant design problemthat entails a number of considerations. The cycle efficiency, the net work, and otherparameters will vary with reheat pressure level for given throttle and condenserconditions. One of these may be numerically optimized by varying reheat pressure levelwhile holding all other design conditions constant.

Reheat offers the ability to limit or eliminate moisture at the turbine exit. Thepresence of more than about 10% moisture in the turbine exhaust can cause erosion ofblades near the turbine exit and reduce energy conversion efficiency. Study of Figure2.6 shows that reheat shifts the turbine expansion process away from the two-phaseregion and toward the superheat region of the T-s diagram, thus drying the turbineexhaust.

EXAMPLE 2.5

Reanalyze the cycle of Example 2.1 (2000 psia/1000°F/1 psia) with reheat at 200 psiaincluded. Determine the quality or degree of superheat at the exits of both turbines.Assume that reheat is to the HP turbine throttle temperature.

SolutionReferring to Figure 2.6, we see that the properties of significant states are the

following:

State Temperature (°F)

Pressure (psia)

Entropy (Btu/lbm-°R)

Enthalpy (Btu/lbm)

1 1000.0 2000 1.5603 1474.1

2 400.0 200 1.5603 1210.0

3 1000.0 200 1.84 1527.0

4 101.74 1 1.84 1028.0

5 101.74 1 0.1326 69.73

6 101.74 2000 0.1326 69.73

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Properties here are obtained from the steam tables and the Mollier chart as follows:

1. The enthalpy and entropy at state 1 are read from the superheated-steam tablesat the given throttle temperature and pressure.

2. State 2 is evaluated from the Mollier diagram at the given reheat pressure andthe same entropy as in state 1 for the isentropic turbine expansion.

3. Reheat at constant pressure p3 = p2 to the assumed throttle temperature T3 = T1gives s3 and h3. Normally, T3 is assumed equal to T1 unless otherwise specified.

4. The second turbine flow is also specified as isentropic with expansion at s4 = s3to the known condenser pressure p4.

5. The condenser exit (pump entrance) state is assumed to be a saturated liquid atthe known condenser pressure.

6. Pump work is neglected here. The steady-flow First Law then implies thath6 = h5, which in turn implies the T6 = T5.

The turbine work is the sum of the work of both turbines:(1474.1 - 1210) + (1527 - 1028) = 763.1 Btu/lbm.

The heat added in the steam generator feedwater and reheat passes is(1474.1 - 69.73) + (1527 - 1210) = 1721.4 Btu/lbm.

The thermal efficiency then is 763.1/1721.4 = 0.443, or 44.3%. Both the net work and the cycle efficiency are higher than in the simple Rankine

cycle case of Example 2.1. From the Mollier chart in Appendix B it is readily seen thatstate 2 is superheated, with 400 - 381.8 = 18.2 Fahrenheit degrees of superheat; andstate 4 is wet steam, with 7.4% moisture, or 0.926 (92.6%) quality. Thus the firstturbine has no moisture and the second is substantially drier than 0.774 quality value inExample 2.1. ____________________________________________________________________

Reheat is an important feature of all large, modern fossil-fueled steam power plants. We now consider another key feature of these plants, but temporarily omit reheat, forthe purpose of clarity.

2.6 Regeneration and Feedwater Heaters

The significant efficiency advantage of the Carnot cycle over the Rankine cycle is dueto the fact that in the Carnot cycle all external heat addition is at a single high

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temperature and all external heat rejection at a single low temperature. Examination ofFigures 2.2 and 2.6 shows that heat addition in the steam generator takes place over awide range of water temperature in both the simple and reheat Rankine cycles.Presumably, the Rankine-cycle thermal efficiency could be improved by increasing theaverage water temperature at which heat is received. This could be accomplished by aninternal transfer of heat from higher-temperature steam to low-temperature feedwater.An internal transfer of heat that reduces or eliminates low-temperature additions ofexternal heat to the working fluid is known as regeneration.

Open Feedwater Heaters

Regeneration is accomplished in all large-scale, modern power plants through the useof feedwater heaters. A feedwater heater (FWH) is a heat exchanger in which the latentheat (and sometimes superheat) of small amounts of steam is used to increase thetemperature of liquid water (feedwater) flowing to the steam generator. This providesthe internal transfer of heat mentioned above.

An open feedwater heater is a FWH in which a small amount of steam mixesdirectly with the feedwater to raise its temperature. Steam drawn from a turbine forfeedwater heating or other purposes is called extraction steam. Feedwater heaters inwhich extraction steam heats feedwater without fluid contact will be discussed later.

Consider the regenerative Rankine-cycle presented in Figure 2.7. The steamleaving the high-pressure (HP) turbine is split with a small part of the mass flowextracted to an open FWH and the major part of the flow passing to a low pressure(LP) turbine. The T-s diagram shows that steam entering the FWH at state 2 is at ahigher temperature than the subcooled feedwater leaving the pump at state 5. When thetwo fluids mix in the FWH, the superheat and the heat of vaporization of the extractionsteam are transferred to the feedwater, which emerges with the condensed extractionsteam at a higher temperature, T6.. It is assumed that all streams entering and leavingthe FWH are the same pressure so that the mixing process occurs at constant pressure.

The T-s and flow diagrams show that heat from combustion gases in the steamgenerator need only raise the water temperature from T7 to T1 rather than from T5 whenextraction steam is used to heat the feedwater. The average temperature for externalheat addition must therefore increase. Despite the reduced flow rate through the low-pressure turbine, we will see by example that the thermal efficiency of the steam cycle isimproved by the transfer of energy from the turbine extraction flow to the feedwater.

The analysis of cycles with feedwater heaters involves branching of steam flows.In Figure 2.7, for example, conservation of mass must be satisfied at the flow junctiondownstream of the high-pressure-turbine exit. Thus, assuming a mass flow of 1 at theHP turbine throttle and a steam mass-flow fraction, m1, through the feedwater heater,the low-pressure-turbine mass-fraction must be 1 - m1. Note that the latter flow passesthrough the condenser and pump and is reunited with the extraction flow, m1, in theFWH at state 6, where the exit-flow-rate fraction is again unity.

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It will be seen later that it is common for more than one FWH to be used in asingle power plant. When more than one FWH is present, mass flows m1, m2...mn aredefined for each of the n FWHs. Conservation of mass is used to relate these flows to

54

condenser flow rate and the reference throttle flow rate. This is accomplished by takinga mass flow of 1 at the high-pressure-turbine throttle as a reference, as in the case of asingle FWH discussed above. After solving for each of the thermodynamic states andFWH mass fractions, actual mass flow rates are obtained as the products of the known(or assumed) throttle flow rate and FWH mass-flow fractions.

The function of feedwater heaters is to use the energy of extraction steam toreduce the addition of low-temperature external heat by raising the temperature of thefeedwater before it arrives at the steam generator. Feedwater heaters are thereforeinsulated to avoid heat loss to the surroundings. Because the resulting heat loss isnegligible compared with the energy throughflow, feedwater heaters are usually treatedas adiabatic devices.

In order to avoid irreversibility associated with unrestrained expansion, constantpressure mixing of the streams entering the FWH is necessary. Returning to Figure 2.7, this implies that the pressures of the feedwater at state 5 and at the FWH exit state 6are chosen to be the same as that of the extraction steam at state 2.

Note that, as with reheat, the inclusion of a FWH also introduces an additionalpressure level into the Rankine cycle as seen in the T-s diagram. In the figure, theextraction pressure level, p2, is another parameter under the control of the designer.The extraction mass flow rate, m1, is in turn controlled by the designer�s choice of p2.The mass-flow rate is determined by the physical requirement that the feedwaterentering the FWH at state 5 increase in temperature to T6 through absorption of theheat released by the condensing extraction steam. This is accomplished by applying thesteady-flow First Law of Thermodynamics, using appropriate mass fractions, to theinsulated open FWH:

q = 0 = (1)h6 � m1h2 � (1 � m1 )h5 + 0 [Btu/lbm | kJ/kg]

Every term in this equation has dimensions of energy per unit throttle mass, thusreferring all energy terms to the mass-flow rate at the throttle of the high-pressureturbine. For example, the second term on the right is of the form:

FWH Extraction mass Enthalpy at state 2 Enthalpy at state 2-------------------------- × -------------------------- = ---------------------- Throttle mass FWH Extraction mass Throttle mass

Similarly, the structure of the third term on the right has the significance of

Pump mass Enthalpy at state 5 Enthalpy at state 5---------------- × --------------------- = ----------------------Throttle mass Pump mass Throttle mass

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Solving for the extraction mass fraction, we obtain

m1 = (h6 � h5) / (h2 � h5) [dl] (2.11)

For low extraction pressures, the numerator is usually small relative to thedenominator, indicating a small extraction flow. The T-s diagram of Figure 2.7 showsthat increasing the extraction pressure level increases both h6 and h2. Thus, because thesmall numerator increases faster than the large denominator, we may reason, fromEquation (2.11), that the extraction mass-flow fraction must increase as the extractionpressure level increases. This conforms to the physical notion that suggests the needfor more and hotter steam to increase the feedwater temperature rise. While suchintuitions are valuable, care should be exercised in accepting them without proof.

The total turbine work per unit throttle mass flow rate is the sum of the work ofthe turbines referenced to the throttle mass-flow rate. Remembering that 1 - m1 is theratio of the low-pressure turbine mass flow to the throttle mass flow, we obtain:

wt = (h1 � h2 ) + (1 � m1 )(h2 � h3 ) [Btu/lbm | kJ/kg] (2.12) The reader should examine the structure of each term of Equation (2.12) in the light ofthe previous discussion. Note that it is not important to remember these specificequations, but it is important to understand, and be able to apply, the reasoning bywhich they are obtained.

For a given throttle mass flow rate, mthr [lbm/s | kg/s], the total turbine poweroutput is given by mthrwt [Btu/s | kW].

We see in Figure 2.7 that the heat addition in the steam generator is reduced, dueto extraction at pressure p6 = p2, by about h7 � h5 to

qa = h1 � h7 [Btu/lbm | kJ/kg] (2.13)

At the same time, the net work also decreases, but more slowly, so that the net effect isthat the cycle efficiency increases with increased extraction.

EXAMPLE 2.6

Solve Example 2.1 (2000 psia /1000°F/1 psia) operating with an open feedwater heaterat 200 psia.

SolutionReferring to Figure 2.7, we find that the properties of significant states are:

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State Temperature (°F)

Pressure (psia)

Entropy (Btu/lbm-°R)

Enthalpy (Btu/lbm)

1 1000.0 2000 1.5603 1474.1

2 400.0 200 1.5603 1210.0

3 101.74 1 1.5603 871.0

4 101.74 1 0.1326 69.73

5 101.74 200 0.1326 69.73

6 381.8 200 0.5438 355.5

7 381.8 2000 0.5438 355.5

States 1 through 4 are obtained in the same way as in earlier examples. Constantpressure mixing requires that p5 = p6 = p2, the extraction pressure level. State 6, apump entrance state, is assumed to be a saturated-liquid state as usual. Subcooled-liquid states are approximated, as before, consistent with the neglect of pump work.

The extraction mass fraction obtained by applying the steady-flow First Law ofThermodynamics to the FWH, Equation (2.11), is

m1 = (355.5 � 69.73)/(1210 � 69.73) = 0.251.

The net work (neglecting pump work) by Equation (2.12), is then

wn = (1474.1 � 1210) + (1 � 0.251)(1210 � 871) = 518.1 Btu/lbm

This may be compared with the simple-cycle net work of 603.1 Btu/lbm.The heat added in the steam generator by Equation (2.13) is

qa = h1 � h7 = 1474.1 � 355.5 = 1118.6 Btu/lbm.

The resulting cycle efficiency is �th = 518.1/1118.6 = 0.463, or 46.3%, asignificantly higher value than the 42.94% for the corresponding simple Rankine cycle.

Note, however, that the LP-turbine exhaust quality is the same as for the simpleRankine cycle, an unacceptable 77.4%. This suggests that a combination of reheat andregeneration through feedwater heating may be desirable. We will investigate thispossibility later after looking at closed feedwater heaters. _____________________________________________________________________

Closed Feedwater Heaters

We have seen that feedwater heating in open feedwater heaters occurs by mixing ofextraction steam and feedwater. Feedwater heating also is accomplished in shell-and-

57

tube-type heat exchangers, where extraction steam does not mix with the feedwater.Normally, feedwater passes through banks of tubes whereas steam condenses on theoutside of the tube surfaces in these heaters. Such heat exchangers are called closedfeedwater heaters.

Pumped Condensate. Closed feedwater heaters normally are employed in twoconfigurations in power plants. In the configuration shown in figure 2.8, condensate ispumped from the condenser through the FWH and the steam generator directly to theturbine along the path 4-5-8-9-1. Ideally, p5 = p1 assuming no pressure drop in theFWH and steam generator.

Note that if m1 mass units of steam are extracted from the turbine for use in theFWH, only 1 - m1 units of feedwater pass throught the condenser, pump, and the tubesof the FWH. The condensed extraction steam (condensate) emerging from the FWH atstate 6 is pumped separately from p6 = p2 to throttle pressure p7 = p1, where it becomespart of the steam generator feedwater. The pumped condensate at state 7 thus mixeswith the heated feedwater at state 8 to form the total feedwater flow at state 9. Constant pressure mixing ( p7 = p8 = p9) is required at this junction to avoid lossesassociated with uncontrolled flow expansion.

The enthalpy of the feedwater entering the steam generator can be determined byapplying the steady-flow First Law of Thermodynamics to the junction of the feedwaterand FWH streams:

h9 = (1 � m1 )h8 + m1h7 [Btu/lbm | kJ/kg]

As in the open FWH analysis, the extraction mass fraction depends on the choiceof intermediate pressure p2 and is obtained by applying the steady-flow First Law ofThermodynamics to the feedwater heater.

Throttled Condensate. The second closed FWH configuration is shown in Figure 2.9 where the FWH condensate drops in pressure from p6 = p2 through a trap into thecondenser at pressure p7 = p3 = p4. The trap allows liquid only to pass from the FWH atstate 6 in a throttling process to state 7. As usual, it is assumed that the throttlingprocess is adiabatic. The T-s diagram shows that the saturated liquid at state 6 flashesinto a mixture of liquid and vapor in the condenser with no change in enthalpy, h7 = h6.

For this configuration, the closed FWH condensate mass-flow rate is equal to theextraction mass-flow rate. As a result, conservation of mass applied to the condenser shows that the mass-flow rate leaving the condenser and passing through the pump andFWH tubes is the same as the throttle mass-flow rate. The throttled-condensate, closedfeedwater heater is the preferred configuration in power plants, because it isunnecessary for each FWH to have a condensate pump.

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59

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EXAMPLE 2.7

Rework Example 2.1 (2000 psia/1000°F/1 psia) with reheat and a closed feedwaterheater with extraction from the cold reheat line and FWH condensate throttled to the condenser. Both reheat and extraction are at 200 psia. Assume that the feedwater leaving the FWH is at the temperature of the condensing extraction stream. Drawappropriate T-s and flow diagrams.

SolutionReferring to the notation of Figure 2.10, verify that the significant the

thermodynamic state properties are:

State Temperature (°F)

Pressure (psia)

Entropy (Btu/lbm-°R)

Enthalpy (Btu/lbm)

1 1000.0 2000 1.5603 1474.1

2 400.0 200 1.5603 1210.0

3 1000.0 200 1.84 1527.0

4 101.74 1 1.84 1028.0

5 101.74 1 0.1326 69.73

6 101.74 2000 0.1326 69.73

7 381.8 200 0.5438 355.5

8 101.74 1 __ 355.5

9 381.8 2000 __ 355.5

Applying the steady-flow First Law of Thermodynamics to the FWH, we obtain:

0 = h9 + m1h7 � m1h2 � h6 + 0

which, solved for m1, yields:

m1 = ( h9 � h6 )/( h2 � h7 ) = (355.5 � 69.73)/(1210 � 355.5) = 0.3344

The total net work per unit of mass flow at the throttle of the HP turbine is the sum ofthe specific work of each of the turbines adjusted for the HP turbine throttle mass flow:

wn = h1 � h2 + (1 � m1)( h3 � h4 )

= 1474.1 � 1210 + (1 � 0.3344)(1527 � 1028) = 596.2 Btu/lbm

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62

As in the earlier examples in this series, pump work has been neglected.The heat addition per unit HP-turbine-throttle mass is the sum of the heat addition

in the main pass and reheat pass through the steam generator, the latter as adjusted forthe reduced mass flow. Thus the steady-flow First Law of Thermodynamics yields

qa = h1 � h9 + (1 � m1 )( h3 � h2 )

= 1474.1 � 355.5 + (1 � 0.3344)(1527 � 1210) = 1329.6 Btu/lbm

The thermal efficiency of the cycle is wn / qa = 596.2 / 1329.6 = 0.448, or 44.8%.The Mollier chart shows that the discharge of the first turbine (state 2) has 20

degrees of superheat and the second turbine (state 4) 7.4% moisture, or a quality of0.926.___________________________________________________________________

In the above calculation it was assumed that the feedwater temperature leaving theFWH had risen to the temperature of the condensing extraction steam. Since the FWHis a heat exchanger of finite area, the feedwater temperature T9 usually differs from thecondensing temperature of the extraction steam T7. If the surface area of the FWH issmall, the feedwater will emerge at a temperature well below the extraction-steamcondensing temperature. If the area were increased, the feedwater temperature wouldapproach the condensing temperature. This aspect of FWH design is reflected in theparameter known as the terminal temperature difference, TTD, defined as

TTD = Tsat - Tfw [R | K]

where Tfw is the temperature of the feedwater leaving the tubes and Tsat is thecondensing temperature of the extraction steam in the closed FWH. In Figure 2.10, forinstance, Tfw = T9 and Tsat = T7. Thus, if the TTD and the extraction pressure areknown, the true FWH exit temperature may be determined. An application of the TTDwill be considered in a later example.

Table 2.1 summarizes, for comparison, the results of the calculations for theseveral plant configurations that we have considered. The reader is cautioned that sincethese calculations have not accounted for turbine inefficiency, the thermal efficienciesare unusually high. While the efficiency differences with respect to the simple cycle mayseem insignificant, they are of great economic importance. It must be realized thathundreds of millions of dollars may be spent on fuel each year in a power plant and thatcapital costs are equally impressive. As a result, the choice of cycle and designcharacteristics are of great significance. Some further improvement in net work andefficiency could be shown by selecting extraction and reheat pressure levels tomaximize these parameters.

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Table 2.1 Comparison of Rankine Cycle Modifications

Net Work (Btu/lbm)

Efficiency %

Heat Rate (Btu/kW-hr)

Turbine Exit Quality

Simple cycle 603.1 42.9 7956 0.774

Reheat cycle 763.1 44.3 7704 0.926

One open FWH 518.1 46.3 7371 0.774

One closed FWH and reheat

596.2 44.8 7618 0.926

Multistage Extraction

It has been shown that increases in cycle efficiency may be accomplished in a steampower plant through regeneration via the feedwater heater. Large steam power plantstypically employ large numbers of feedwater heaters for this purpose. Multistageextraction refers to the use of multiple extractions to supply steam to these feedwaterheaters. Earlier discussions of examples involved extractions taken only from the flowsbetween turbines. However, the number of extractions is not limited by the number ofturbines. In fact, large turbines are designed with several extraction points throughwhich steam may be withdrawn for feedwater heating and other purposes.

Assigning Extraction-Pressure Levels. Given n feedwater heaters, it is necessary toassign values to the n associated extraction pressures. For preliminary design purposes,the extraction-pressure levels assigned may be those that give equal feedwatertemperature rises through each heater and through the steam generator to the boilingpoint. Thus, for n heaters the appropriate temperature rise is given by

�Topt = ( Tsl � Tcond )/( n + 1) [R | K] (2.14)

where Tsl is the temperature the saturated liquid at the throttle pressure and Tcond is thetemperature the feedwater leaving the condenser. The corresponding steam condensingtemperature in the ith heater is then

Ti = Tcond + ( i )�Topt

= Tcond + i ( Tsl � Tcond )/( n + 1) [R | K] (2.15)

where i = 1, 2..., n. Steam tables may then be used to evaluate the correspondingextraction-pressure levels. It is, of course, possible and sometimes necessary to assignextraction-pressure levels in other ways.

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EXAMPLE 2.8

Evaluate the recommended extraction-pressure levels for single heater for the 1000° F/2000 psia throttle and one psia condenser that have been used throughout thischapter.

SolutionThe feedwater temperature rise to establish an appropriate extraction-pressure

level for a single heater for a plant such as that shown in Figures 2.7 through 2.9 is (Tsl � T4 )/2 = (635.8 � 101.74)/2 = 267.05°F where Tsl was evaluated at p1 = 2000psia. This would make T6 = 101.74 + 267.05 = 368.79°F and the correspondingextraction pressure level p6 = p2 = 171 psia, using the saturated-steam tables.____________________________________________________________________

At this point we have the tools necessary to evaluate the performance and penaltiesassociated with a given configuration. The following example examines the gains thatfollow from the use a single feedwater heater and the sensitivity of the thermalefficiency to the assigned feedwater temperature rise.

EXAMPLE 2.9

Consider a single open feedwater heater operating in a Rankine cycle with a 2000 psiasaturated-vapor throttle and a 1 psia condenser. Evaluate the thermal efficiency as afunction of feedwater temperature rise. Compare the temperature rise that maximizesthe thermal efficiency with the results of Equation (2.14).

Solution

Utilizing the notation of Figure 2.7 and taking the throttle state as a saturated vapor,we get the results that are summarized in spreadsheet format in Table 2.2. (This table isa direct reproduction of a Quattro Pro spreadsheet used in the analysis. Care should betaken if this spreadsheet is used for "what if" studies, because it is dependent on manualentry of thermodynamic properties. To explore other cases, appropriate properties mustbe obtained from steam tables or charts and inserted in the spreadsheet. Despite thisdrawback, the spreadsheet provides a convenient means of organizing, performing, anddisplaying calculations.) Details of the methodology are given in the right-most column.It is seen that the net work drops, as expected, as more extraction steam is used to heatthe feedwater. Figure 2.11 shows the percentage increase in thermal efficiency as afunction of the feedwater temperature rise for this case. Over a 9% increase in thermalefficiency is achieved with feedwater temperature rises between 200�F and 300°F.Thus the prediction of Topt = 267°F using Equation (2.14) in Example 2.8 is clearly inthis range._____________________________________________________________________

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Example 2.9 shows that improved thermal efficiency is achieved over a broad range offeedwater temperature rise and therefore extraction pressure. This gives the designerfreedom to assign extraction-pressure levels so as to make use of existing designs forfeedwater heaters and turbines without severely compromising the efficiency of theplant design.

Calculation Methodology. Once the extraction- and reheat-pressure levels areestablished for a cycle with multistage extraction, and once throttle and condenserconditions, turbomachine efficiencies, and FWH terminal temperature differences areknown, significant state properties should be determined. Symbols for extractionmass-fraction variables should be assigned for each heater and related to otherunknown flows using mass conservation assuming unit mass flow at the high-pressure-turbine throttle. The steady-flow First Law of Thermodynamics should thenbe applied to each of the FWHs, starting with the highest extraction pressure andprogressing to the lowest-pressure FWH. Analyzing the heaters in this order allowseach equation to be solved immediately for a mass fraction rather than solving all ofthe equations simultaneously. Important performance parameters such as thermalefficiency, net work, and work ratio may then be evaluated taking care to accountproperly for component mass flows. The following example illustrates thismethodology.

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EXAMPLE 2.10

Consider a power plant with 1000�F/2000-psia throttle, reheat at 200 psia back to1000�F, and 1-psia condenser pressure. The plant has two closed feedwater heaters,both with terminal temperature differences of 8�F. The high-pressure (HP) heatercondensate is throttled into the low-pressure (LP) heater, which in turn drains into thecondenser. Turbomachine efficiencies are 0.88, 0.9, and 0.8 for the HP turbine, the LPturbine, and the boiler feed pump, respectively. Draw relevant T-s and flow diagramsand evaluate FWH mass fractions, thermal efficiency, net work, and work ratio.

SolutionThe notation used to study this plant is shown in Figure 2.12. The pertinent

thermodynamic properties and part of the analysis are presented in the spreadsheetgiven in Table 2.3. The earlier-stated caution (Example 2.9) about using spreadsheetsthat incorporate external data applies here as well, because changing parameters mayrequire changes in steam-table lookup values.

To start the analysis we first determine the extraction-pressure levels. The idealFWH temperature rise is given by

( Tsl � T7 )/3 = ( 635.8 � 101.74)/3 = 178.02°F

where the saturation temperature is evaluated at the HP-turbine throttle pressure of2000 psia. The corresponding extraction condensing temperatures and extraction-pressure levels are

101.74 + 178.02 = 280�F � p9 = p5 = 49 psia

and

101.74 + (2)(178.02) = 457.8°F � p12 = p2 = 456 psia

where the extraction pressures have been evaluated using the saturated-steam tables.After the entropy and enthalpy at state 1 are evaluated, the enthalpy h3s at the HP-

turbine isentropic discharge state 3s is determined from s1 and p3. The HP-turbine efficiency then yields h3 and the steam tables give s3. The entropy and enthalpy at theHP-turbine extraction state 2 may be approximated by drawing a straight line on thesteam Mollier diagram connecting states 1 and 3 and finding the intersection with theHP-extraction pressure P2. This technique may be used for any number of extractionpoints in a turbine.

Once the hot reheat properties at state 4 are determined from the steam tables, theLP-turbine exit and extraction states at 6 and 5 may be obtained by the same methodused for the HP turbine.

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The determination of the FWH condensate temperatures and pressures at states 9and 12 have already been discussed. The temperatures of the heated feedwater leavingthe FWHs may be determined from the terminal temperature differences:

T11 = T9 - TTD = 281 - 8 = 273°F

T14 = T12 - TTD = 457.5 - 8 = 449.5°F

Recalling that the enthalpy of a subcooled liquid is almost independent of pressure, wenote that the enthalpies h11 and h14 may be found in the saturated-liquid tables at T11 andT14, respectively.

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The pump discharge state 8 is a subcooled-liquid state, which may beapproximated in the same way as in Examples 2.3 and 2.4. Thus

h8s = h7 + ( p8 � p7 )v7

= 69.7 + (2000 � 1)(144)(0.016136)/778 = 75.7 Btu/lbm

and

h8 = h7 + ( h8s � h7 )/�p = 69.7 + (75.7 � 69.7)/0.8 = 77.2 Btu/lbm

The pump work is then

wp = h7 � h8 = 69.7 � 77.2 = � 7.5 Btu/lbm

The extraction mass-flow fractions designated m1 and m2 relate other flows to theunit mass flow at the high-pressure-turbine throttle. For example, the condensate flowrate from the LP heater at state 10 is given by m1 + m2.

The steady-flow First Law of Thermodynamics may now be applied to the heaters.For the HP FWH:

0 = m1h12 + (1)h14 – m1h2 � (1)hll

may be rewritten as

m1 = ( h14 � hll )/( h2 � h12 ) [dl]

This and the T-s diagram show that the HP extraction-flow enthalpy drop fromstate 2 to state 12 provides the heat to raise the enthalpy in the feedwater from state 11to state 14. Also, for the LP FWH:

0 = (1)h ll + (m2 + m1 )h9 – (1)h

8 – m2h5 � mlhl3

becomes

m2 = [ m1( h9 - h13 ) + h11 � h8 ]/( h5 � h9 ) [dl]

This and the T-s diagram show that the discharge from the HP FWH at state 13 aidsthe mass flow m2 in heating the LP FWH flow from state 8 to state 11. The values ofm1 and m2 are evaluated at the bottom of spreadsheet in Table 2.3.

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With all states and flows known, we may now determine some plant performanceparameters. The turbine work referenced to the throttle mass-flow rate is easilyobtained by summing the flow contributions through each section of the turbines:

wt = h1 � h2 + (1 � ml )( h2 � h3) + (1 � ml )( h4 � h5) + (1 � ml � m2 )( h5 � h6) [Btu/lbm | kJ /kg]

The net work is then wt + wp, and the heat added in the steam generator is the sum ofheat additions in the feedwater pass and the reheat pass:

qa = h1 - h14 + (1 - ml )( h4 - h3) [Btu/lbm | kJ/kg]

These parameters and the work ratio are evaluated in Table 2.3.____________________________________________________________________

Example 2.10 shows that a good thermal efficiency and net work output arepossible with the use of two feedwater heaters despite taking into account realisticturbomachine inefficiencies. The high work ratio clearly demonstrates the low-compression work requirements of Rankine cycles.

2.7 A Study of a Modern Steam Power Plant

Modern steam power plants incorporate both reheat and feedwater heating. Aflowsheet for the Public Service Company of Oklahoma (PSO) Riverside Station Unit#1, south of Tulsa, is shown in Figure 2.13. This natural-gas-burning plant was sizedfor two nominal 500-megawatt units. Several other plants in the PSO system havesimilar unit flowsheets, including a coal-burning plant. Note the flowsheet coding W, H,F, and A for flow rate in lbm/hr, enthalpy in Btu/lbm, temperature in °F, and pressure inpsia, respectively.

The steam generator, not shown on the flowsheet, interacts through the feedwaterand steam lines on the right-hand side of the diagram. The high pressure turbine throttleis at 1000°F and 3349 psia and has a mass-flow rate of 2,922,139 lbm/hr. This type ofunit is called supercritical, because the pressure in the main steam line to the HP-turbine throttle exceeds the 3208.2-psia critical pressure of steam. Note that a largefraction of the HP-turbine mass-flow rate enters the cold reheat line at 630 psia and isreheated to the intermediate-pressure (IP) turbine throttle conditions of 1000°F and567 psia.

Most of the steam flow through the IP turbine passes through the crossover at 186psia to the double-flow low-pressure (DFLP) turbine. The term double-flow refers tothe fact that the incoming flow enters at the middle, splits, and flows axially in opposite

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directions through the turbine. This causes the large axial force components on theblades and shaft to oppose each other so that the resultant axial thrust is small and doesnot necessitate heavy thrust bearings. The combined HP and IP turbines are similarlyconfigured.

The plant is equipped with six closed FWHs and one open FWH (the deaerator).Note that the condensate of each of the closed feedwater heaters is throttled to the nextlowest pressure FWH or, in the case of the lowest-pressure heater, to the condenser.The extraction steam for the four lowest-pressure FWHs flows from the DFLP turbine.Extraction steam for the highest pressure FWH is provided by the HP turbine, and theIP turbine supplies heater HTR1-6 and the open feedwater heater identified as thedeaerator. The deaerator is specially designed to remove non-condensable gases fromthe system, in addition to performing its feedwater heating duties.

The feedwater starts at the "hot well" of the condenser on the left of the diagram,enters the condensate pump at 101.1°F and 2"Hg abs., and starts its passage throughthe FWHs. Note that the feedwater increases in temperature from 102.1° to 180°,227.2°, 282.7°, and 314.4° in passing through the 4 lowest pressure FWHs. Thefeedwater from the deaerator is pumped to 405 psia by the booster pump andsubsequently to 3933 psia by the boiler feed pump (BFP). The BFP exit pressureexceeds the HP-turbine throttle pressure of 3349 psia in order to overcome flow lossesin the high pressure heater, the boiler feed line, the steam generator main steam pass,and the main steam line, all of which operate at supercritical pressure.

The boiler feed pump turbine (BFPT) shown in the upper left of the diagramsupplies the shaft power to drive the BFP at the lower right. The BFPT receives steamfrom an extraction line of the DFLP turbine and exhausts directly to the condenser.

The reader should study Figure 2.13 thoroughly in the light of the precedingdiscussions of reheat and feedwater heating. It is particularly useful to consider the flowrates with respect to mass and energy conservation. Mastery of this flow sheet willmake it possible to quickly understand flowsheets of other major power plants.

Example 2.11

Verify that the steam generator feedwater flow rate satisfies the conservation of massinto all the feedwater heaters shown for the Riverside Unit #1 in Figure 2.13. You mayneglect all flows of less than 2000 lbm/hr.

SolutionThe shell side of the low pressure heater, labeled HTR1-1, receives condensate

from heaters 2, 3 and 4 as well as steam entering from the LP turbine. The totalcondensate from the low-pressure heaters into the condenser are:

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Source Flow rate, lbm/hr _______________________________________________Condensate from HTR1-4 75,005

Extraction steam into HTR1-3 125,412

Extraction steam into HTR1-2 102,897 ----------

Total condensate into HTR1-1 303,314

Extraction steam into HTR1-1 157,111---------

Total condensate leaving HTR1-1 460,425

The feedwater flow rate through the four low-pressure heaters (the condensercondensate pump flow rate) is the sum of the flows into the condenser:

460,425 + 162,701 + 1,812,971 = 2,436,097 lbm/hr.

An easier approach to evaluating this flow rate is by imagining a control volumearound the entire left side of the diagram that cuts it in two parts between the deaeratorand HTR1-4 and through the crossover steam line. Because these are the only pointswhere the control volume is penetrated by large mass flows, the two flows must beequal. Consequently the crossover mass-flow rate of 2,434,357 lbm/hr agrees very wellwith our above calculation of the feedwater flow rate into the deaerator.

Now, observing that the boiler feedwater all flows from the deaerator through thebooster pump, we sum all of the flows into the deaerator:

Feedwater into deaerator 2,434,357

Steam to deaerator 148,321

Steam to HTR1-6 107,661

Steam to HTR1-7 222,876 -----------

Total feedwater into HTR1-7 2,913,215 lbm/hr

This compares well with the tabulated value of 2,922,139 lbm /hr to the steamgenerator. Accounting for the small flows should improve the agreement.______________________________________________________________

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2.8 Deviations from the Ideal - Pressure Losses It is evident from study of Figure 2.13 that there are significant pressure drops in theflows through the steam generator between the HP FWH and the HP-turbine throttleand in the reheat line between the HP and IP turbines. While we have neglected suchlosses in our calculations, final design analysis requires their consideration. A firstattempt at this may be made by applying a fractional pressure drop based onexperience. Two per cent pressure drops through the main steam and feedwater linesand a 3.7% loss through the steam generator would, for instance, account for theindicated 14.8% loss from the boiler feed pump to the HP turbine.

In the final analysis, of course, when realistic values are available for flow rates andproperties, known fluid mechanic relations for pressure drop may be employed toaccount for these losses.

Bibliography and References

1. Anon., Steam, Its Generation and Use. New York: Babcock and Wilcox, 1978.

2. Singer, J. G., (Ed.), Combustion/Fossil Power Systems. Windsor, Conn.:Combustion Engineering, 1981.

3. Wood, Bernard, Applications of Thermodynamics, 2nd ed. Reading, Mass.: Addison-Wesley, 1981.

4. Li, Kam W., and Priddy, A. Paul, Powerplant System Design. New York: Wiley,1985.

5. El-Wakil, M. M., Power Plant Technology. New York: McGraw-Hill, 1984.

6. Skrotzi, B. G. A. and Vopat, W. A., Power Station Engineering and Economy. New York: McGraw-Hill, 1960.

EXERCISES

2.1 An ideal Rankine-cycle steam power plant has 800-psia saturated steam at theturbine throttle and 5-psia condenser pressure. What are the turbine work, pump work,net work, steam generator heat addition, thermal efficiency, maximum cycletemperature, and turbine exit quality? What is the Camot efficiency corresponding tothe temperature extremes for this cycle?

2.2 A Rankine-cycle steam power plant has an 800-psia/900�F throttle and 5-psiacondenser pressure. What are the net work, turbine work, pump work, steam generator

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heat addition, thermal efficiency, and turbine exit quality? What is the Carnot efficiencycorresponding to the temperature extremes for this cycle?

2.3 Solve Exercise 2.2 for the cases of (a) an 85% efficient turbine, (b) an 85%efficient pump, and (c) both together. Tabulate and discuss your results together withthose of Exercise 2.2.

2.4 Solve Exercise 2.2 for the case of (a) 1000�F throttle, (b) 2000-psia throttle, (c) 2-psia condenser, and (d) all three changes simultaneously. Make a table comparingnet work, quality, and thermal efficiency, including the results of Exercise 2.2. Whatconclusions can you draw from these calculations?

2.5 Sketch coordinated, labeled flow and T-s diagrams for the ideal Rankine cycle.Tabulate the temperatures, entropies, pressures, enthalpies, and quality or degree ofsuperheat for each significant state shown on the diagram for a throttle at 1000 psia and1000�F and a condenser at 5 psia. Determine the net work, heat added, thermalefficiency, heat rate, and heat rejected in the condenser. If the power plant output is100 megawatts and the condenser cooling-water temperature rise is 15 Rankinedegrees, what is the steam flow rate and cooling-water flow rate? Neglect pump work.

2.6 Consider a simple Rankine cycle with a 2000-psia/1100�F throttle and 1-psiacondenser. Compare the thermal efficiencies and net work for cycles with a perfectturbine and one having 86% turbine isentropic efficiency. Assume isentropic pumping.

2.7 A boiling-water reactor operates with saturated vapor at 7500 kPa at the throttleof the high-pressure turbine. What is the lowest turbine exit pressure that ensures thatthe turbine exit moisture does not exceed 12% if the turbine is isentropic? What wouldthe lowest pressure be if the turbine isentropic efficiency were 85%?

2.8 Consider a steam plant with a single reheat and a single open feedwater heater thattakes extraction from the cold reheat line. Sketch carefully coordinated and labeled T-sand flow diagrams. If the throttle is at 1000�F and 3000psia, the condenser is at 1 psia,and reheat is to 1000�F at 400 psia, what is the extraction mass fraction, the heat rate,and the thermal efficiency? The turbine efficiency is 89%. Neglect pump work.

2.9 A Rankine-cycle power plant condenses steam at 2 psia and has 1000�F and 500psia at the turbine throttle. Assume an isentropic turbine.(a) Tabulate the temperature, pressure, entropy, and enthalpy of all states. Determinethe quality and moisture fraction for all mixed states.(b) Calculate the heat transferred in the condenser and the steam generator and theturbine work, all per unit mass. What is the thermal efficiency?(c) Calculate the pump work. What is the ratio of turbine to pump work?

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(d) What is the turbine work and thermal efficiency if the turbine efficiency is 85%?Include pump work.

2.10 For throttle conditions of 1000�F and 1000 psia and a condenser pressure of 2psia, compare the net work, thermal efficiency, and turbine discharge quality or degreeof superheat for a simple cycle and two reheat cycles with reheat to 1000�F at 50 and200 psia. Tabulate your results. Sketch a single large, labeled T-s diagram comparingthe cycles. Turbine isentropic efficiencies are 85%.

2.11 Consider a regenerative Rankine cycle with a 1000�F and 500-psia throttle, 2-psiacondenser, and an open feedwater heater operating between two turbines at 50 psia.Turbine efficiencies are 85%. Neglect pump work.(a) Draw labeled, coordinated T-s and flow diagrams.(b) Determine the fraction of the throttle mass flow that passes through the extractionline.(c) Calculate the turbine work per unit mass at the throttle.(d) Calculate the cycle efficiency, and compare it to the simple-cycle efficiency.

2.12 Consider a 1120°F, 2000-psia, 10-psia steam cycle with reheat at 200 psia to1000�F and a closed feedwater heater taking extraction from a line between twoturbines at 100psia. The FWH condensate is throttled to the condenser, and thefeedwater in the FWH is raised to the condensing temperature of the extraction steam.(a) Draw labeled T-s and flow diagrams for this plant.(b) Tabulate the enthalpies for each significant state point.(c) What is the extraction fraction to the FWH?(d) What are the net work and work ratio?(e) What are the thermal efficiency and the heat rate?

2.13 A turbine operates with a 860�F, 900-psia throttle. Calorimetric measurementsindicate that the discharge enthalpy is 1250 Btu/lbm at 100 psia. What is the isentropicefficiency?

2.14 An ideal Rankine cycle has 1000-psia saturated steam at the turbine throttle. Thecondenser pressure is 10psia. What are the turbine work, steam generator heat addition,maximum cycle temperature, turbine exit quality, and Carnot efficiency correspondingto the temperature extremes of the cycle? Neglect pump work.

2.15 Assume that the extraction mass-flow rate to FWH #7 in Figure 2.13 is notknown. Calculate the FWH extraction mass fraction (relative to the HP-turbine throttleflow) and the extraction mass-flow rate. Compare the extraction-steam energy loss ratewith the feedwater energy gain rate.

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2.16 Compare the inflow and outflow of steam of the DFLP turbine in Figure 2.13,and calculate the percentage difference. Calculate the power output of the DFLPturbine in Btu/hr and in kW.

2.17 Calculate the power delivered by the PSO Riverside Unit #1 boiler feed pumpturbine, BFPT. Based on the feedwater enthalpy rise across the BFP, determine itspower requirements, in kilowatts. What fraction of the plant gross output is used by theBFPT?

2.18 Without performing a detailed analysis of the FWHs, determine the PSORiverside Unit #1 feedwater flow rate from heater number 4 to the deaerator. Explainyour methodology.

2.19 Total and compare the inflows and outflows of mass and energy to the PSORiverside Unit #1 deaerator.

2.20 Rework Example 2.4 neglecting pump work. Repeat your calculations for an 80%efficient pump. Compare and comment on the significance of accounting for pumpwork and turbomachine efficiency.

2.21 For a 1080�F, 2000-psia, 5-psia Rankine cycle with 85% turbine efficiency and60% pump efficiency:(a) Compare the actual net work and the isentropic turbine work and the isentropic network.(b) Calculate the actual heat transfer and work for each component, and evaluate thecyclic integrals of Q and W.(c) Compare the real cycle efficiency with that for the ideal Rankine cycle.

2.22 For a 1080�F, 2000-psia, 5-psia Rankine cycle with 85% turbine efficiency and60% pump efficiency, evaluate the effect of a single reheat to 1080�F at 500 psia on:(a) Heat addition in the steam generator.(b) Work of each turbine, total turbine work, and net work. Compare the net workwith the cyclic integral of the external transfers of heat.(c) Cycle efficiency and heat rate.(d) Quality or degree of superheat at the exit of the turbines.Draw labeled flow and T-s diagrams.

2.23 Consider a 1080�F, 2000-psia, 5-psia Rankine cycle with 85% turbine efficiencyand 60% pump efficiency. Compare the simple cycle with the same cycle operating witha single reheat to 1080�F at 1000 psia with respect to:(a) Heat addition in the steam generator.(b) Work of each turbine, total turbine work and net work, condenser heat rejection,and cyclic integral of heat added.

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(c) Cycle efficiency.(d) Quality or degree of superheat at the exit of the turbines.Draw labeled flow and T-s diagrams.

2.24 For a 1080�F, 2000-psia, 5-psia Rankine cycle with 85% turbine efficiencies and60% pump efficiencies and using a single open feedwater heater operating at 500 psia:(a) Draw labeled and coordinated flow and T-s diagrams.(b) Evaluate the feedwater heater mass fraction.(c) Evaluate heat addition in the steam generator, work of each turbine, total turbinework, and net work, all per pound of steam at the HP-turbine throttle.(d) Evaluate condenser heat transfer per unit mass at the HP-turbine throttle.(e) Evaluate cycle efficiency and heat rate. Compare with simple-cycle efficiency.(f) Evaluate the cyclic integral of the differential heat addition, and compare it withthe net work.

2.25 Consider a 1080�F, 2000-psia, 5-psia Rankine reheat-regenerative cycle withperfect turbomachinery and a closed feedwater heater taking extraction from the coldreheat line at 500 psia. FWH condensate is pumped into the feedwater line downstreamof the feedwater heater. Assume that the enthalpy of the feedwater entering the steamgenerator is that of the saturated liquid leaving the FWH.(a) Draw coordinated and labeled flow and T-s diagrams.(b) Determine the extraction mass fraction, the net work, and the total heat addition.(c) Determine the thermal efficiency and heat rate.(d) Determine the superheat or quality at the turbine exhausts:

2.26 Taking the reheat-pressure level as a variable, plot net work, thermal efficiency,and turbine exhaust superheat and/or moisture against reheat pressure for theconditions of Example 2.5. Select a suitable design value based on your analysis.

2.27 Solve Example 2.6 for 1200�F throttle substituting a closed FWH for the openheater. Consider two cases in which the FWH condensate is (a) throttled to thecondenser, and (b) pumped to throttle pressure.

2.28 Solve Example 2.6 using the method for assigning extraction-pressure levelsgiven in the subsection of Section 2.6 on multistage extraction systems.

2.29 Solve Example 2.7 using the method for assigning extraction-pressure levelsgiven in the section on multistage extraction systems, and determine by trial and errorthe reheat-pressure level that maximizes the thermal efficiency.

2.30 Solve Example 2.7 with the extraction condensate from the closed FWH pumpedahead to the feedwater-pressure level.

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2.31 Solve Example 2.6 for 900�F throttle temperature with the open FWH replacedby a closed FWH where the feedwater is (a) throttled to the condenser, and (b) pumpedinto the feedwater line downstream of the FWH.

2.32 Compare the work and exhaust quality of 90% efficient turbines with 2500-psiathrottle pressure and 1000�F and 1200�F throttle temperatures exiting to a 2-psiacondenser.

2.33 Draw a large T-s diagram showing the states associated with the important flowsof the PSO Riverside Unit #1 (Figure 2.13).

2.34 A Rankine-cycle steam power plant has 5-MPa saturated steam at the turbinethrottle and 25-kPa condenser pressure. What are the net work, steam generator heataddition, thermal efficiency, heat rate, maximum cycle temperature, and turbine exitquality? What is the Carnot efficiency corresponding to the temperature extremes forthis cycle?

2.35 A Rankine-cycle steam power plant has a 5-MPa/450�C throttle and 10-kPacondenser pressure. What are the net work, steam generator heat addition, thermalefficiency, heat rate, and turbine exit quality? What is the Carnot efficiencycorresponding to the temperature extremes for this cycle?

2.36 Solve Exercise 2.35 for the cases of (a) an 85% efficient turbine, (b) an 85%efficient pump, and (c) both together. What conclusions may be inferred from yourresults?

2.37 Solve Exercise 2.35 for the case of (a) a 550�C throttle, (b) a 15-MPa throttle, (c) a 5-kPa condenser, and (d) all three changes simultaneously. What conclusions canyou draw from these calculations?

2.38 Sketch coordinated, labeled flow and T-s diagrams for the following Rankinecycle. Tabulate the temperatures, entropies, pressures, enthalpies, and quality or degreeof superheat for each significant state shown on the diagram for a throttle at 10 MPaand 550�C and condenser at 5 kPa. Determine the net work, heat added, thermalefficiency, and heat rejected in the condenser. If the power plant output is 100megawatts and the condenser cooling water temperature rise is 15�Rankine, what is thesteam flow rate and cooling-water flow rate? Neglect pump work.

2.39 Consider a Rankine cycle with a 20MPa/600�C throttle and 3-kPa condenser.Compare the thermal efficiencies and net work for cycles with a perfect turbine and onehaving 86% turbine isentropic efficiency.

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2.40 Consider a steam plant, with a single reheat and a single open feedwater heater,that takes extraction from the cold reheat line. Sketch carefully coordinated and labeledT-s and flow diagrams. If the throttle is at 550�C and 15 MPa, the condenser is at 5kPa, and reheat is to 3 MPa and 550° C, what are the extraction mass fraction, workratio, and thermal efficiency? The pump and turbine efficiencies are 82% and 89%,respectively.

2.41 A Rankine-cycle power plant condenses steam at 10 kPa and has 550�C and 5MPa at the turbine throttle. Assume an isentropic turbine.(a) Tabulate the temperature, pressure, entropy, and enthalpy of all states. Determinethe quality and moisture fraction for all mixed states.(b) Calculate the heat transferred in the condenser and steam generator and the turbinework, all per unit mass. What is the thermal efficiency?(c) Calculate the pump work. What is the ratio of turbine to pump work?(d) What is the turbine work and thermal efficiency if the turbine efficiency is 85%?Include pump work.

2.42 For throttle conditions of 550°C and 5 MPa and a condenser pressure of 10 kPa,compare the net work, thermal efficiency, and turbine discharge quality or degree ofsuperheat for a simple cycle and two reheat cycles. Consider reheat to 500�C at (a)4MPa and (b) 1 MPa. Tabulate and compare your results. Sketch a large, labeled T-sdiagram for a reheat cycle. Turbine efficiencies are 85%.

2.43 Consider a regenerative Rankine cycle with a 600�C and 4-MPa throttle, a 5-kPacondenser, and an open feedwater heater at 500 kPa. Turbine efficiencies are 85%.Neglect pump work.(a) Draw labeled, coordinated T-s and flow diagrams.(b) Determine the fraction of the throttle mass flow that passes through the extractionline.(c) Calculate the turbine work per unit mass at the throttle.(d) Calculate the cycle efficiency, and compare it with the simple-cycle efficiency.(e) Calculate the heat rate.

2.44 Consider a 600�C, 15-MPa steam cycle with reheat at 2 MPa to 600�C andextraction to a closed feedwater heater at 600 kPa. The FWH condensate is throttled tothe condenser at 5 kPa, and the feedwater in the FWH is raised to the condensingtemperature of the extraction steam. Neglecting pump work:(a) Draw labeled T-s and flow diagrams for this plant.(b) Tabulate the enthalpies for each significant state point.(c) What is the extraction fraction to the FWH?(d) What is the net work?(e) What is the thermal efficiency?

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(f) What is the heat rate?

2.45 A turbine operates with a 600°C, 7-MPa throttle. Calorimetric measurementsindicate that the discharge enthalpy is 3050 kJ/kg at 0.8 MPa. What is the turbineisentropic efficiency?

2.46 A pressurized water-reactor nuclear power plant steam generator has separateturbine and reactor water loops. The steam generator receives high-pressure hot waterfrom the reactor vessel to heat the turbine feedwater. Steam is generated from thefeedwater in the turbine loop. The water pressure in the reactor is 15 MPa, and thewater temperature in and out of the reactor is 289°C and 325°C, respectively. The planthas one turbine with a single extraction to a closed FWH with condensate throttled tothe condenser. Throttle conditions are 300�C and 8 MPa. The extraction and condenserpressures are 100 kPa and 5 kPa, respectively. The reactor-coolant flow rate is 14,000kg/s. Assume no heat losses in heat exchangers and isentropic turbomachines. Neglectpump work.(a) What is the rate of heat transfer from the reactor in MWt?(b) Draw coordinated flow and T-s diagrams that show both loops.(c) Determine the extraction mass fraction of the throttle flow rate.(d) Determine the cycle net work, heat rate, and thermal efficiency.(e) Calculate the steam flow rate.(f) Assuming the electrical generator has 97% efficiency, calculate the power output,in MWe (electric).

2.47 Perform an optimization of the extraction pressure of a Rankine cycle with a2000-psia saturated-vapor throttle, a 1-psia condenser with a single closed feedwaterheater, as in Example 2.9. Compare the optimum extraction temperature given byEquation (2.14) with your results.

2.48 Prepare an optimization study of thermal efficiency with a table and plot of network and thermal efficiency as a function of reheat pressure level for Example 2.5.Discuss the selection of reheat pressure for this case. How does the reheat pressureused in Example 2.5 compare with your results?

2.49 Solve Exercise 2.25 for reheat and extraction at 200 psia. Compare the extractionmass fraction, net work, thermal efficiency, heat rate, and turbine exit conditions withthose of Exercise 2.25.

2.50 Rework Exercise 2.25, accounting for 90% turbine efficiencies and a 10�Fterminal temperature difference.

2.51 A 1000�F/2000-psia-throttle high-pressure turbine discharges into a cold reheatline at 200 psia. Reheat is to 1000�F. The low-pressure turbine discharges into the

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condenser at 0.5 inches of mercury absolute. Both turbines are 90% efficient. Designthe cycle for the use of three feedwater heaters. Draw coordinated T-s and flowdiagrams. State and discuss your decisions on the handling of the feedwater heaterdesign.

2.52 A steam turbine receives steam at 1050�F and 3000 psia and condenses at 5 psia.Two feedwater heaters are supplied by extraction from the turbine at pressures of 1000psia and 200 psia. The low-pressure heater is an open FWH, and the other is closedwith its condensate throttled to the open heater. Assuming isentropic flow in the turbineand negligible pump work:(a) Sketch accurately labeled and coordinated T-s and flow diagrams for the system,and create a table of temperature, pressure, and enthalpy values for each state.(b) What are the extraction flows to each feedwater heater if the throttle mass flowrate is 250,000 pounds per hour?(c) How much power, in kW, is produced by the turbine?(d) Compare the thermal efficiency of the system with the efficiency if valves of bothextraction lines are closed.(e) What is the heat rate of the system with both feedwater heaters operative?

2.53 Apply the steady-flow First Law of Thermodynamics to a single control volumeenclosing the two turbines in Example 2.7. Show that the same equation is obtained forthe turbine work as when the work of individual turbines is summed.

2.54 Apply the steady-flow First Law of Thermodynamics to a single control volumeenclosing the two turbines in Example 2.10. Show that the same equation is obtainedfor the turbine work as when the work of individual turbines is summed.

2.55 Resolve Example 2.7 with 4% pressure drops in the main steam pass and reheatpass through the steam generator. Make a table comparing your results with those ofthe example to show the influence of the losses on plant performance. Calculate anddisplay the percentage differences for each parameter. Assume turbine throttleconditions are unchanged.

2.56 Draw labeled and coordinated T-s and flow diagrams for a steam power plantwith 1000°F / 3000-psia / 2" Hg absolute conditions, assuming isentropicturbomachinery. The plant has reheat at 500 psia to ll00�F. The plant has the followingfeedwater heaters:

1. A closed FWH with extraction at 1000 psia and pumped condensate.2. A closed FWH at 400 psia with condensate throttled into the next-lowest-pressure FWH.3. An open FWH at 20 psia.

Define mass fraction variables. Show mass-flow variable expressions, with arrowsindicating mass fractions along the various process paths on the T-s and flow diagrams.

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Write equations for conservation of energy for the FWHs that allow you to solve easilyfor the mass fractions in terms of known state enthalpies and other mass fractions.Indicate a solution method for the mass fractions that involves simple substitution only.

2.57 A pressurized-water nuclear-reactor steam generator has separate turbine andreactor loops. The steam generator linking the two loops cools high-pressure hot waterfrom the reactor vessel and transfers the heat to the turbine feedwater producing steam.The water pressure in the reactor is 2250 psia, and the water temperatures in and out ofthe reactor are 559°F and 623°F, respectively. The plant has one turbine with a singleextraction to an open FWH. Throttle conditions are 555�F and 1100 psia. Theextraction and condenser pressures are 100 psia and 1 psia, respectively. The reactor-coolant flow rate is 147,000,000 lbm/hr. Assume no heat losses in heat exchangers andisentropic turbomachines.(a) What is the rate of heat transfer from the reactor, in Btu/hr and in MWt?(b) Draw coordinated flow and T-s diagrams that show both loops with states in theirproper relations with respect to each other.(c) Determine the extraction mass fraction relative to the throttle flow rate.(d) Determine the cycle net work.(e) What are the cycle thermal efficiency and heat rate?(f) Calculate the turbine-steam flow rate.(g) Assuming the electrical generator has 100% efficiency, calculate the turbine power,in Btu/hr and in MWe.

2.58 Determine the efficiencies of the boiler feed pump and boiler feed pump turbine ofthe PSO Riverside Station Unit 4/1 (Figure 2.13).

2.59 A Rankine cycle with a single open feedwater heater has a 1040�F and 550-psiathrottle. Extraction from the exit of the first turbine (assumed isentropic) is at 40�F ofsuperheat. The second turbine has an efficiency of 85% and expands into the condenserat 5 psia.(a) Draw matched, labeled T-s and flow diagrams.(b) Accurately calculate and tabulate the enthalpies of all significant states. Neglectpump work.(c) What is the feedwater-heater mass fraction relative to the mass flow at the firstthrottle?(d) What is the quality or degree of superheat at the condenser inlet?(e) What are the net work, thermal efficiency, and heat rate?(f) Estimate the feedwater-heater condensate pump work and its percentage of turbinework.

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C H A P T E R 3

FUELS AND COMBUSTION

3.1 Introduction to Combustion

Combustion Basics

The last chapter set forth the basics of the Rankine cycle and the principles of operationof steam cycles of modern steam power plants. An important aspect of powergeneration involves the supply of heat to the working fluid, which in the case of steampower usually means turning liquid water into superheated steam. This heat comes froman energy source. With the exception of nuclear and solar power and a few other exoticsources, most power plants are driven by a chemical reaction called combustion, whichusually involves sources that are compounds of hydrogen and carbon. Processindustries, businesses, homes, and transportation systems have vast heat requirementsthat are also satisfied by combustion reactions. The subject matter of this chaptertherefore has wide applicability to a variety of heating processes.

Combustion is the conversion of a substance called a fuel into chemical compoundsknown as products of combustion by combination with an oxidizer. The combustionprocess is an exothermic chemical reaction, i.e., a reaction that releases energy as itoccurs. Thus combustion may be represented symbolically by:

Fuel + Oxidizer � Products of combustion + Energy

Here the fuel and the oxidizer are reactants, i.e., the substances present before thereaction takes place. This relation indicates that the reactants produce combustionproducts and energy. Either the chemical energy released is transferred to thesurroundings as it is produced, or it remains in the combustion products in the form ofelevated internal energy (temperature), or some combination thereof.

Fuels are evaluated, in part, based on the amount of energy or heat that theyrelease per unit mass or per mole during combustion of the fuel. Such a quantity isknown as the fuel's heat of reaction or heating value.

Heats of reaction may be measured in a calorimeter, a device in which chemicalenergy release is determined by transferring the released heat to a surrounding fluid. The amount of heat transferred to the fluid in returning the products of combustion totheir initial temperature yields the heat of reaction.

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In combustion processes the oxidizer is usually air but could be pure oxygen, anoxygen mixture, or a substance involving some other oxidizing element such asfluorine. Here we will limit our attention to combustion of a fuel with air or pureoxygen.

Chemical fuels exist in gaseous, liquid, or solid form. Natural gas, gasoline, andcoal, perhaps the most widely used examples of these three forms, are each a complexmixture of reacting and inert compounds. We will consider each more closely later inthe chapter. First let's review some important fundamentals of mixtures of gases, suchas those involved in combustion reactions. Mass and Mole Fractions

The amount of a substance present in a sample may be indicated by its mass or by thenumber of moles of the substance. A mole is defined as the mass of a substance equal toits molecular mass or molecular weight. A few molecular weights commonly used incombustion analysis are tabulated below. For most combustion calculations, it issufficiently accurate to use integer molecular weights. The error incurred may easily beevaluated for a given reaction and should usually not be of concern. Thus a gram-moleof water is 18 grams, a kg-mole of nitrogen is 28 kg, and a pound-mole of sulfur is 32lbm. _____________________________________________________________________

Molecule Molecular Weight -------------------------------------------

C 12N2 28O2 32S 32H2 2

_____________________________________________________________________

The composition of a mixture may be given as a list of the fractions of each of thesubstances present. Thus we define the mass fraction, of a component i, mfi, as theratio of the mass of the component, mi, to the mass of the mixture, m:

mfi = mi /m

It is evident that the sum of the mass fractions of all the components must be 1. Thus

mf1 + mf2 + ... = 1

Analogous to the mass fraction, we define the mole fraction of component i, xi, asthe ratio of the number of moles of i, ni, to the total number of moles in the mixture, n:

xi = ni /n

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The total number of moles, n, is the sum of the number of moles of all the componentsof the mixture:

n = n1 + n2 + ...

It follows that the sum of all the mole fractions of the mixture must also equal 1.

x1 + x2 + ... = 1 The mass of component i in a mixture is the product of the number of moles of i and itsmolecular weight, Mi. The mass of the mixture is therefore the sum, m = n1M1 + n2M2 +..., over all components of the mixture. Substituting xin for ni, the total mass becomes

m = (x1M1 + x2M2 + ...)n

But the average molecular weight of the mixture is the ratio of the total mass to thetotal number of moles. Thus the average molecular weight is

M = m /n = x1M1 + x2M2 + ...

EXAMPLE 3.1

Express the mass fraction of component 1 of a mixture in terms of: (a) the number ofmoles of the three components of the mixture, n1, n2, and n3, and (b) the mole fractionsof the three components. (c) If the mole fractions of carbon dioxide and nitrogen in athree component gas containing water vapor are 0.07 and 0.38, respectively, what arethe mass fractions of the three components?

Solution (a) Because the mass of i can be written as mi = niMi , the mass fraction of componenti can be written as:

mfi = niMi /(n1M1 + n2M2 + ..) [dl]

For the first of the three components, i = 1, this becomes:

mf1 = n1M1/(n1M1 + n2M2 + n3M3)

Similarly, for i = 2 and i = 3:

mf2 = n2M2/(n1M1 + n2M2 + n3M3)

mf3 = n3M3/(n1M1 + n2M2 + n3M3)

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(b) Substituting n1 = x1 n, n2 = x2 n, etc. in the earlier equations and simplifying, weobtain for the mass fractions:

mf1 = x1M1/(x1M1 + x2M2 + x3M3)

mf2 = x2M2/(x1M1 + x2M2 + x3M3)

mf3 = x3M3 /(x1M1 + x2M2 + x3M3)

(c) Identifying the subscripts 1, 2, and 3 with carbon dioxide, nitrogen, and watervapor, respectively, we have x1 = 0.07, x2 = 0.38 and x3 = 1 � 0.07 � 0.038 = 0.55.Then:

mf1 = (0.07)(44)/[(0.07)(44) + (0.38)(28) + (0.55)(18)]

= (0.07)(44)/(23.62) = 0.1304

mf2 = (0.38)(28)/(23.62) = 0.4505

mf3 = (0.55)(18)/(23.62) = 0.4191

As a check we sum the mass fractions: 0.1304 + 0.4505 + 0.4191 = 1.0000.________________________________________________________________

For a mixture of gases at a given temperature and pressure, the ideal gas lawshows that pVi = ni�T holds for any component, and pV = n�T for the mixture as awhole. Forming the ratio of the two equations we observe that the mole fractions havethe same values as the volume fraction:

xi = Vi /V = ni /n [dl]

Similarly, for a given volume of a mixture of gases at a given temperature, piV = ni�Tfor each component and pV = n�T for the mixture. The ratio of the two equationsshows that the partial pressure of any component i is the product of the mole fractionof i and the pressure of the mixture:

pi = pni /n = pxi

EXAMPLE 3.2

What is the partial pressure of water vapor in Example 3.1 if the mixture pressure istwo atmospheres?

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SolutionThe mole fraction of water vapor in the mixture of Example 3.1 is 0.55. The partial

pressure of the water vapor is therefore (0.55)(2) = 1.1 atm. _____________________________________________________________________

Characterizing Air for Combustion Calculations

Air is a mixture of about 21% oxygen, 78% nitrogen, and 1% other constituents byvolume. For combustion calculations it is usually satisfactory to represent air as a 21% oxygen, 79% nitrogen mixture, by volume. Thus for every 21 moles of oxygen thatreact when air oxidizes a fuel, there are also 79 moles of nitrogen involved. Therefore,79/21 = 3.76 moles of nitrogen are present for every mole of oxygen in the air.

At room temperature both oxygen and nitrogen exist as diatomic molecules, O2and N2, respectively. It is usually assumed that the nitrogen in the air is nonreacting atcombustion temperatures; that is, there are as many moles of pure nitrogen in theproducts as there were in the reactants. At very high temperatures small amounts ofnitrogen react with oxygen to form oxides of nitrogen, usually termed NOx. These smallquantities are important in pollution analysis because of the major role of even smalltraces of NOx in the formation of smog. However, since these NOx levels areinsignificant in energy analysis applications, nitrogen is treated as inert here.

The molecular weight of a compound or mixture is the mass of 1 mole of thesubstance. The average molecular weight, M, of a mixture, as seen earlier, is the linearcombination of the products of the mole fractions of the components and theirrespective molecular weights. Thus the molecular weight for air, Mair, is given by thesum of the products of the molecular weights of oxygen and nitrogen and theirrespective mole fractions in air. Expressed in words:

Mair = Mass of air/Mole of air = (Moles of N2 /Mole of air)(Mass of N2 /Mole of N2)

+ (Moles of O2/Mole of air)(Mass of O2 /Mole of O2) or

Mair = 0.79 Mnitrogen + 0.21 Moxygen

= 0.79(28) + 0.21(32) = 28.84

The mass fractions of oxygen and nitrogen in air are then

mfoxygen = (0.21)(32)/28.84 = 0.233, or 23.3%

and

mfnitrogen = (0.79)(28)/28.84 = 0.767, or 76.7%

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3.2 Combustion Chemistry of a Simple Fuel

Methane, CH4, is a common fuel that is a major constituent of most natural gases.Consider the complete combustion of methane in pure oxygen. The chemical reactionequation for the complete combustion of methane in oxygen may be written as:

CH4 + 2O2 � CO2 + 2H2O (3.1)

Because atoms are neither created nor destroyed, Equation (3.1) states that methane(consisting of one atom of carbon and four atoms of hydrogen) reacts with four atomsof oxygen to yield carbon dioxide and water products with the same number of atomsof each element as in the reactants. This is the basic principle involved in balancing allchemical reaction equations.

Carbon dioxide is the product formed by complete combustion of carbon throughthe reaction C + O2 � CO2. Carbon dioxide has only one carbon atom per molecule.Since in Equation (3.1) there is only one carbon atom on the left side of the equation,there can be only one carbon atom and therefore one CO2 molecule on the right.Similarly, water is the product of the complete combustion of hydrogen. It has twoatoms of hydrogen per molecule. Because there are four hydrogen atoms in thereactants of Equation (3.1), there must be four in the products, implying that twomolecules of water formed. These observations require four atoms of oxygen on theright, which implies the presence of two molecules (four atoms) of oxygen on the left.

The coefficients in chemical equations such as Equation (3.1) may be interpreted asthe number of moles of the substance required for the reaction to occur as written.Thus another way of interpreting Equation (3.1) is that one mole of methane reactswith two moles of oxygen to form one mole of carbon dioxide and two moles of water.While not evident in this case, it is not necessary that there be the same number ofmoles of products as reactants. It will be seen in numerous other cases that a differentnumber of moles of products is produced from a given number of moles of reactants.

Thus although the numbers of atoms of each element must be conserved during areaction, the total number of moles need not. Because the number of atoms of eachelement cannot change, it follows that the mass of each element and the total mass mustbe conserved during the reaction. Thus, using the atomic weights (masses) of eachelement, the sums of the masses of the reactants and products in Equation (3.1) areboth 80:

CH4 + 2O2 � CO2 + 2H2O[12 + 4(1)] + 4(16) � [12 + 2(16)] + 2[2(1) + 16] = 80

Other observations may be made with respect to Equation (3.1). There are 2 moles ofwater in the 3 moles of combustion products, and therefore a mole fraction of water inthe combustion products of xwater = 2/3 = 0.667. Similarly, xCarbon dioxide = 1/3 = 0.333moles of CO2 in the products.

There are 44 mass units of CO2 in the 80 mass units of products for a mass

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fraction of CO2 in the products,

mfcarbon dioxide = 44/80 = 0.55

Likewise, the mass fraction of water in the products is 2(18)/80 = 0.45.We also observe that there are 12 mass units of carbon in the products and

therefore a carbon mass fraction of 12/80 = 0.15. Note that because the mass of anyelement and the total mass are conserved in a chemical reaction, the mass fraction ofany element is also conserved in the reaction. Thus the mass fraction of carbon in thereactants is 0.15, as in the products.

Combustion in Air

Let us now consider the complete combustion of methane in air. The same combustionproducts are expected as with combustion in oxygen; the only additional reactantpresent is nitrogen, and it is considered inert. Moreover, because we know that in airevery mole of oxygen is accompanied by 3.76 moles of nitrogen, the reaction equationcan be written as

CH4 + 2O2 + 2(3.76)N2 � CO2 + 2H2O + 2(3.76)N2 (3.2)

It is seen that the reaction equation for combustion in air may be obtained from thecombustion equation for the reaction in oxygen by adding the appropriate number ofmoles of nitrogen to both sides of the equation.

Note that both Equations (3.1) and (3.2) describe reactions of one mole ofmethane fuel. Because the same amount of fuel is present in both cases, both reactionsrelease the same amount of energy. We can therefore compare combustion reactions inair and in oxygen. It will be seen that the presence of nitrogen acts to dilute thereaction, both chemically and thermally. With air as oxidizer, there are 2 moles of watervapor per 10.52 moles of combustion products, compared with 2 moles of water per 3moles of products for combustion in oxygen. Similarly, with air, there is a mass fractionof CO2 of 0.1514 and a carbon mass fraction of 0.0413 in the combustion products,compared with 0.55 and 0.15, respectively, for combustion in oxygen.

The diluting energetic effect of nitrogen when combustion is in air may be reasonedas follows: The same amount of energy is released in both reactions, because the sameamount of fuel is completely consumed. However, the nonreacting nitrogen moleculesin the air have heat capacity. This added heat capacity of the additional nitrogenmolecules absorbs much of the energy released, resulting in a lower internal energy perunit mass of products and hence a lower temperature of the products. Thus the energyreleased by the reaction is shared by a greater mass of combustion products when thecombustion is in air.

Often, products of combustion are released to the atmosphere through a chimney,stack, or flue. These are therefore sometimes referred to as flue gases. The flue gascomposition may be stated in terms of wet flue gas (wfg) or dry flue gas (dfg), because

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under some circumstances the water vapor in the gas condenses and then escapes as aliquid rather than remaining as a gaseous component of the flue gas. When liquid wateris present in combustion products, the combustion product gaseous mass fractions maybe taken with respect to the mass of flue gas products, with the product water presentor omitted. Thus, for Equation (3.2), the mass of dry combustion products is 254.56.Hence the mass fraction of carbon dioxide is 44/254.56 = 0.1728 with respect to dryflue gas, and 44/290.56 = 0.1514 with respect to wet flue gas.

In combustion discussions reference is frequently made to higher and lower heatingvalues. The term higher heating value, HHV, refers to a heating value measurement inwhich the product water vapor is allowed to condense. As a consequence, the heat ofvaporization of the water is released and becomes part of the heating value. The lowerheating value, LHV, corresponds to a heating value in which the water remains avapor and does not yield its heat of vaporization. Thus the energy difference betweenthe two values is due to the heat of vaporization of water, and

HHV = LHV + (mwater /mfuel)hfg [Btu/lbm | kJ/kg]

where mwater is the mass of liquid water in the combustion products, and hfg is the latentheat of vaporization of water.

Air-Fuel Ratio

It is important to know how much oxygen or air must be supplied for completecombustion of a given quantity of fuel. This information is required in sizing fans andducts that supply oxidizer to combustion chambers or burners and for numerous otherdesign purposes. The mass air-fuel ratio, A/F, or oxygen-fuel ratio, O/F, for completecombustion may be determined by calculating the masses of oxidizer and fuel from theappropriate reaction equation. Let�s return to Equation (3.2):

CH4 + 2O2 + 2(3.76)N2 � CO2 + 2H2O + 2(3.76)N2 (3.2)

The A/F for methane is [(2)(32) + (2)(3.76)(28)]/(12 + 4) = 17.16 and the O/F is 2(32)/(12 + 4) = 4. Thus 4 kg of O2 or 17.16 kg of air must be supplied for eachkilogram of methane completely consumed.

Of course it is possible, within limits, to supply an arbitrary amount of air to aburner to burn the fuel. The terms stoichiometric or theoretical are applied to thesituation just described, in which just enough oxidizer is supplied to completely convertthe fuel to CO2 and H2O. Thus the stoichiometric O/F and A/F ratios for methane are4.0 and 17.16, respectively. If less than the theoretical amount of air is supplied, theproducts will contain unburned fuel. Regardless of the magnitude of A/F, whenunburned fuel remains in the products (including carbon, carbon monoxide, orhydrogen), combustion is said to be incomplete. Because air is virtually free and fuel isexpensive, it is usually important to burn all of the fuel by using more air than thetheoretical air-fuel ratio indicates is needed. Thus most burners operate with excess air.

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The actual air-fuel ratio used in a combustor is frequently stated as a percentage ofthe theoretical air-fuel ratio

% theoretical air = 100(A/F)actual /(A/F)theor (3.3)

Thus, for methane, 120% of theoretical air implies an actual mass air-fuel ratio of(120/100)(17.16) = 20.59.

Excess air is defined as the difference between the actual and the theoretical airsupplied. Accordingly, the percentage of excess air is

% excess air = 100[(A/F)actual � (A/F)theor ]/(A/F)theor (3.4)

Thus, for methane, 120% of theoretical air implies

% excess air = (100)(20.59 � 17.16)/17.16 = 20%.

Note also that combining Equations (3.4) and (3.3) yields the following general result:

% excess air = % theoretical air � 100% (3.5)

Again, the excess air percentage is 120% � 100% = 20%. Table 3.1 shows examples ofranges of excess air used with certain fuels and combustion systems.

The air/fuel parameters just discussed emphasize the amount of air supplied to burna given amount of fuel relative to the theoretical requirement. An alternate approachconsiders a given amount of air and indicates the mass of fuel supplied , the fuel-airratio, F/A, which is the inverse of the air-fuel ratio. A measure of how much fuel isactually supplied, called the equivalence ratio, is the ratio of the actual fuel-air ratio tothe theoretical fuel-air ratio:

� = (F/A)actual / (F/A)theor = (A/F)theor / (A/F)actual

= 100/( % theoretical air)

Thus 100% theoretical air corresponds to an equivalence ratio of 1, and 20% excess airto � = 100/120 = 0.833. When the equivalence ratio is less than 1, the mixture is calledlean; when greater than 1, it is called rich.

This section has dealt with the application of combustion chemistry orstoichiometry applied to methane gas. Other fuels for which a reaction equation such asEquation (3.1) or (3.2) is available may be treated in a similar way. Before consideringmore complex combustion problems, it is appropriate to investigate the nature anddescription of the various types of fossil fuels.

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3.3 Fossil Fuel Characteristics

Most chemical fuels are found in nature in the form of crude oil, natural gas, and coal.These fuels are called fossil fuels because they are believed to have been formed by thedecay of vegetable and animal matter over many thousands of years under conditions ofhigh pressure and temperature and with a deficiency or absence of oxygen. Other fuelssuch as gasoline, syngas (synthetic gas), and coke may be derived from fossil fuels bysome form of industrial or chemical processing. These derived fuels are also calledfossil fuels.

Coal

Coal is an abundant solid fuel found in many locations around the world in a variety offorms. The American Society for Testing Materials, ASTM, has established a rankingsystem (ref. 3) that classifies coals as anthracite (I), bituminous (II), subbituminous(III), and lignite (IV), according to their physical characteristics. Table 3.2 lists

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seventeen of the many United States coals according to this class ranking.Coal is formed over long periods of time, in a progression shown from left to right

in Figure 3.1. The bars on the ordinate show the division of the combustibles betweenfixed carbon and volatile matter in the fuels. �Fixed carbon� and �volatile matter�indicate roughly how much of the fuel burns as a solid and as a thermally generated gas,respectively. It is seen that the volatile matter and oxygen contained in the fuelsdecrease with increasing age.

Peat is a moist fuel, at the geologically young end of the scale, that has a relatively low heating value. It is not considered a coal but, nevertheless, follows the patterns of characteristics shown in the figure. Peat is regarded as an early stage or precursor ofcoal. At the other extreme, anthracite is a geologically old, very hard, shiny coal with high carbon content and high heating value. Bituminous is much more abundant thananthracite, has a slightly lower carbon content, but also has a high heating value.Subbituminous coal, lignite, and peat have successively poorer heating values andhigher volatile matter than bituminous.

Coal is a highly inhomogeneous material, of widely varying composition, found inseams (layers) of varying thickness at varying depths below the earth's surface. Thewide geographic distribution of coal in the United States is shown in Figure 3.2.

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According to reference 1, the average seam in the United States is about 5.5 ft. thick.The largest known seam is 425 ft. thick and is found in Manchuria.

Coal Analyses

It is often difficult to obtain representative samples of coal because of compositionvariations from location to location even within a given seam. As a result there arelimits on the accuracy and adequacy of coal analyses in assessing coal behavior in agiven application. Before discussing the nature of these analyses, it is important toestablish the basis on which they are conducted.

Coal contains varying amounts of loosely held moisture and noncombustiblematerials or mineral matter (ash), which are of little or no use. The basis of an analysishelps to specify the conditions under which the coal is tested. The coal sample may befreshly taken from the mine, the as-mined basis. It may have resided in a coal pile formonths, and be analyzed just before burning, the as-fired basis. It may be examinedimmediately after transport from the mine, the as-received basis. Exposure to rain ordry periods, weathering, and separation and loss of noncombustible mineral matterthrough abrasion and the shifting of loads during transport and storage may cause thesame load of coal to have changing mineral matter and moisture content over time. It istherefore important to specify the basis for any test that is conducted. Publishedtabulations of coal properties are frequently presented on a dry, ash-free, or dry andash-free basis, that is, in the absence of water and/or noncombustible mineral matter.

Coal ranking and analysis of combustion processes rely on two types of analysis ofcoal composition: the proximate analysis and the ultimate analysis. The proximateanalysis starts with a representative sample of coal. The sample is first weighed, then raised to a temperature high enough to drive off water, and then reweighed. The weight

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loss divided by the initial weight gives the coal moisture content, M. The remainingmaterial is then heated at a much higher temperature, in the absence of oxygen, for atime long enough to drive off gases. The resulting weight-loss fraction gives thevolatile matter content, VM, of the coal. The remainder of the sample is then burned inair until only noncombustibles remain. The weight loss gives the fixed carbon, FC, andthe remaining material is identified as non-combustible mineral matter or ash, A.

The proximate analysis may be reported as percentages (or fractions) of the fourquantities moisture, ash, volatile matter, and fixed carbon, as in Table 3.2, or without ash and moisture and with the FC and VM normalized to 100%. Sulfur, as a fraction ofthe coal mass, is sometimes reported with the proximate analysis. The proximateanalysis, while providing very limited information, can be performed with limitedlaboratory resources.

A more sophisticated and useful analysis is the ultimate analysis, a chemicalanalysis that provides the elemental mass fractions of carbon, hydrogen, nitrogen,oxygen, and sulfur, usually on a dry, ash-free basis. The ash content of the coal andheating value are sometimes provided also.

Data from a dry, ash-free analysis can be converted to another basis by using thebasis adjustment factor, 1 - A - M, as follows. The mass of coal is the mass of ultimateor proximate analysis components plus the masses of water (moisture) and ash:

m = mcomp + mash + mmoist [lbm | kg]

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Dividing through by the total mass m and rearranging, we get the following as the ratioof the mass of components to the total mass:

mcomp / m = 1 � A � M [dl]

where A is the ash fraction and M is the moisture fraction of the total coal mass. Acomponent of a coal analysis may be converted from the dry, ash-free basis to someother basis by forming the product of the component fraction and the basis adjustmentfactor. Thus an equation for the wet and ashy volatile matter fraction in the proximateanalysis may be determined from the dry, ash-free proximate analysis by using

VMas-fired = (Mass of combustibles/Total mass)VMdry,ashfree

= ( 1 - A - M ) VMdry,ash-free (3.6)

where A and M are, respectively, the ash and moisture fractions for the as-fired coal.Here the as-fired (wet, ashy) mass fraction of volatile matter is the product of the dry,ash-free mass fraction and the basis adjustment factor. Fixed carbon, heating values,and components of the ultimate analysis may be dealt with in a similar way.

Table 3.3 gives proximate and ultimate analyses for a number of United Statescoals on a dry basis. Another extensive tabulation of the characteristics of Americanand world coals is given in Appendix E.

EXAMPLE 3.3

If the as-fired moisture fraction for Schuylkill, Pa. anthracite culm characterized inTable 3.3 is 4.5%, determine the as-fired proximate and ultimate analysis and heatingvalue of the coal. (Culm is the fine coal refuse remaining from anthracite mining.)

SolutionThe FC, VM, and ash contents are given in Table 3.3. Because ash is already present inthe analysis, the appropriate adjustment factor is 1 - A - M = 1 � 0.0 � 0.045 = 0.955.Using Equation (3.6) and the data from Table 3.3, we get

VMas-fired = (0.955)(8.3) = 7.927FCas-fired = (0.955)(32.6) = 31.133 Aas-fired = (0.955)(59.1) = 56.411Mas-fired = 4.500

Check Sum = 99.971Heating valueas-fired = (0.955)(4918) = 4697 Btu/lbm

Similarly, the as-fired ultimate analysis is 32% C, 1.15% H2, 4.87% O2, 0.57% N2,0.48% S, 56.44% ash, and 4.5% moisture, with a checksum of 100.01._____________________________________________________________________

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As a solid fuel, coal may be burned in a number of ways. Starting with the smallestof installations, coal may be burned in a furnace, in chunk form on a stationary ormoving grate. Air is usually supplied from below with combustion gases passingupward and ash falling through a stationary grate or dropping off the end of a movinggrate into an ash pit. A wide variety of solid fuels can be burned in this way.

Though all furnaces were onced fired manually, today many are fired by or with theassistance of mechanical devices called stokers. Figure 3.3 shows a spreader stoker,which scatters coal in a uniform pattern in the furnace, the finer particles burning insuspension in the rising combustion air stream while the heavier particles drop to thegrate as they burn. The particles that reach the grate burn rapidly in a thin layer, and theremaining ash drops off the end into the ash pit. This type of combustion system hasbeen in use for over fifty years for hot water heating and steam generation.

In large installations, coal is crushed to a particular size, and sometimes pulverizedto powder immediately before firing, to provide greater surface exposure to the

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oxidizer and to ensure rapid removal of combustion gases. Because of the widevariation in the characteristics of coals, specialized types of combustion systemstailored to a specific coal or range of coal characteristics are used.

Natural Gas

Natural gas is a mixture of hydrocarbons and nitrogen, with other gases appearing insmall quantities. Table 3.4 shows the composition of samples of natural gases found inseveral regions of the United States. For these samples, it is seen that the gases contain83-94% methane (CH4), 0-16% ethane (C2H6), 0.5-8.4% nitrogen and small quantitiesof other components, by volume. The ultimate analysis shows that the gases containabout 65-75% carbon, 20-24% hydrogen, 0.75-13% nitrogen, and small amounts ofoxygen and sulfur in some cases. The higher heating values are in the neighborhood of1000 Btu/ft3 on a volume basis and 22,000 Btu/lbm on a mass basis. In regions where itis abundant, natural gas is frequently the fuel of choice because of its low sulfur and ashcontent and ease of use.

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EXAMPLE 3.4

Determine the molecular weight and stoichiometric mole and mass air-fuel ratios for theOklahoma gas mole composition given in Table 3.4.

SolutionEquation (3.2),

CH4 + 2O2 + 2(3.76)N2 � CO2 + 2H2O + 2(3.76)N2 (3.2)

shows that there are 2 + 2(3.76) = 9.52 moles of air required for complete combustionof each mole of methane. Similarly for ethane, the stoichiometric reaction equation is:

C2H6 + 3.5O2 + (3.5)(3.76)N2 � 2CO2 + 3H2O + 13.16N2

where 2 carbon and 6 hydrogen atoms in ethane require 2 CO2 molecules and 3 H2Omolecules, respectively, in the products. There are then 7 oxygen atoms in theproducts, which implies 3.5 oxygen molecules in the reactants. This in turn dictates thepresence of (3.5)(3.76) = 13.16 nitrogen molecules in both the reactants and products.The reaction equation then indicates that 3.5(1 + 3.76) = 16.66 moles of air arerequired for complete combustion of one mole of ethane.

In Table 3.5, the molecular weight of the gas mixture, 18.169, is found in thefourth column by summing the products of the mole fractions of the fuel componentsand the component molecular weights. This is analogous to the earlier determination ofthe average air molecular weight from the nitrogen and oxygen mixture mole fractions.

The products of the mole fractions of fuel components and the moles of airrequired per mole of fuel component (as determined earlier and tabulated in the fifthcolumn of Table 3.5) then yield the moles of air required for each combustible per moleof fuel (in the sixth column). Summing these, the number of moles of air required permole of fuel yields the stoichiometric mole air-fuel ratio, 9.114.

The stoichiometric mass A/F is then given by the mole A/F times the ratio of airmolecular weight to fuel molecular weight: (9.114)(28.9)/18.169 = 14.5.

Table 3.5 Calculations for Example 3.4

i Mi xi xiMi Moles air permole i

Moles air per mole fuel

Methane 16 0.841 13.456 9.52 (0.841)(9.52) = 7.998

Ethane 30 0.067 2.010 16.66 (0.067)(16.66) = 1.116

CO2 44 0.008 0.351 0.0

Nitrogen 28 0.084 2.352 0.0

Totals 1.000 18.169 moles air /mole fuel = 9.114

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Liquid Fuels

Liquid fuels are primarily derived from crude oil through cracking and fractionaldistillation. Cracking is a process by which long-chain hydrocarbons are broken up intosmaller molecules. Fractional distillation separates high-boiling-point hydrocarbonsfrom those with lower boiling points. Liquid fuels satisfy a wide range of combustionrequirements and are particularly attractive for transportation applications because oftheir compactness and fluidity. Table 3.6 gives representative analyses of some of theseliquid fuels. Compositions of liquid and solid fuels, unlike gaseous fuels, are usuallystated as mass fractions.

3.4 Combustion Reactions and Analysis

Mechanism of Combustion

Details of the mechanics of combustion depend to a great extent on the fuel and thenature of the combustion system. They are sometimes not well understood and arelargely beyond the scope of this book. There are, however, certain fundamentals that are useful in dealing with combustion systems.

The chemical reaction equations presented here do not portray the actualmechanism of combustion; they merely indicate the initial and final chemicalcompositions of a reaction. In most cases the reactions involve a sequence of steps,leading from the reactants to the products, the nature of which depends on thetemperature, pressure, and other conditions of combustion. Fuel molecules, forinstance, may undergo thermal cracking, producing more numerous and smaller fuelmolecules and perhaps breaking the molecules down completely into carbon andhydrogen atoms before oxidation is completed.

In the case of solid fuels, combustion may be governed by the rate at whichoxidizer diffuses from the surrounding gases to the surface and by the release ofcombustible gases near the surface. Combustion of solids may be enhanced byincreasing the fuel surface area exposed to the oxidizer by reducing fuel particle size.

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The following simple model illustrates the effect.

Example 3.5 is, of course, an idealized example. In reality, the reacting surface area ofsolid fuels is usually much larger than the spherical surface area implied by their size.

We have seen that, for combustion to occur, molecules of oxidizer must affiliatewith fuel molecules, an action enhanced by the three T�s of combustion: turbulence,time, and temperature. Chemical reactions take place more rapidly at high temperaturesbut nevertheless require finite time for completion. It is therefore important that burnersbe long enough to retain the fuel-air mixture for a sufficiently long time so thatcombustion is completed before the mixture leaves. Turbulence, or mixing, enhancesthe opportunities for contact of oxidizer and fuel molecules and removal of products ofcombustion.

A flame propagates at a given speed through a flammable mixture. It willpropagate upstream in a flow of a combustible mixture if its flame speed exceeds theflow velocity. If a fixed flame front is to exist at a fixed location in a duct flow in whichthe velocity of the combustion gas stream exceeds the propagation speed, some form offlame stabilization is required. Otherwise the flame front is swept downstream andflameout occurs. Stabilization may be achieved by using fixed flameholders (partial

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flow obstructions that create local regions of separated flow in their bases where theflame speed is greater than the local flow velocity) or by directing a portion of the flowupstream to provide a low-speed region where stable combustion may occur.

Each combination of oxidizer and fuel has been seen to have a particularstoichiometric oxidizer-fuel ratio for which the fuel is completely burned with aminimum of oxidizer. It has also been pointed out that it is usually desirable to operateburners at greater than the theoretical air-fuel ratio to assure complete combustion ofthe fuel and that this is sometimes referred to as a lean mixture. Occasionally it may bedesirable to have incomplete combustion, perhaps to produce a stream of products inwhich carbon monoxide exists or to assure that all the oxidizer in the mixture isconsumed. In that case a burner is operated at less than the stoichiometric air-fuel ratiowith what is called a rich mixture.

There are limits to the range of air-fuel ratios for which combustion will occurcalled limits of flammability. Here the density of the mixture is important. The limits offlammability around the stoichiometric A/F are reduced at low densities. If combustionis to occur reliably in mixtures at low densities, it is necessary to closely control theair-fuel ratio.

Combustion Analysis of Solid Fuels

In the determination of the air-fuel ratio and flue gas composition for the combustion ofsolid fuels, it is important to account for the ash and moisture in the fuel in the as-firedcondition. In the following analyses, all of the elements of the reactants in the fuel andoxidizer are assumed to be present in the flue gas products except for the ash, which isassumed to fall as a solid or flow as molten slag to the furnace bottom. Nitrogen andoxygen are present in many solid fuels and should be accounted for in predicting theflue gas composition. While both carbon monoxide and oxygen may be present incombustion products at the same time because of imperfect mixing of combustibles andoxygen in some instances, we will assume for prediction of the flue gas compositionthat perfect mixing occurs such that no carbon monoxide is present when excess air issupplied.

EXAMPLE 3.6

A coal with a dry, ash-free composition of 0.87 C, 0.09 H2, 0.02 S, and 0.02 O2 isburned with 25% excess air. The as-fired ash and moisture contents are 6% and 4%, respectively.

(a) What are the stoichiometric and actual air-fuel ratios?(b) What is the flue gas composition?

Solution(a) Before performing combustion calculations, it is necessary to convert coalcomposition data to an as-fired basis. The ratio of as-fired to dry, ash-free

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mfj = (kg j / kg fg) = (kg j / kg coal) / (kg fg / kg coal)

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(Note: The reference conditions of the 1985 JANAF Thermochemical tables used differslightly from those of preceding editions.) Heats of formation are usually determinedbased on statistical thermodynamics and spectroscopic measurements.

By definition, heats of formation are zero for all elements in the standard state. Hence, from the steady-flow First Law of Thermodynamics, the heat transferred in aformation reaction of a compound created from elements in the standard state is theheat of formation for the compound, as in the hydrogen-to-water example just mentioned.

Heat Transfer in a Chemically Reacting Flow

Consider now the combustion problem in which fuel and oxidizer flow into a controlvolume and combustion products flow out. The steady-flow First Law of Thermo-dynamics applied to the control volume may be written as

Q = Hp � Hr + Ws [Btu | kJ] (3.7)

where Q is heat flow into the control volume, Ws is the shaft work delivered by thecontrol volume, and the enthalpies, H, include chemical as well as thermal energy. Thesubscripts r and p refer to the reactants entering and products leaving the controlvolume, respectively. The enthalpy Hp is the sum of the enthalpies of all productstreams leaving the control volume. A similar statement applies to Hr for the enteringreactant streams.

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The individual enthalpies may each be written as the product of the number ofmoles of the component in the reaction equation and its respective enthalpy per mole ofthe component. For example, for k products:

Hp = n1h1 + n2h2 +...+ nkhk [Btu | kJ] (3.8)

where the n�s are the stoichiometric coefficients of the chemical equation for thecombustion reaction, and the enthalpies are on a per-mole bases.

EXAMPLE 3.9

Write an equation for the enthalpy of the products of complete combustion of octane inoxygen.

Solution The balanced equation for the complete combustion of octane is

C8H18 + 12.5O2 � 8CO2 + 9H2O

The mole coefficients, 8 and 9, of the products are stoichiometric coefficients that yield

Hp = 8h(CO2) + 9h(H2O) [Btu | kJ]

per mole of octane consumed. ___________________________________________________________________

The enthalpy of any component of the reactants or products may be written as thesum of (1) its enthalpy of formation at the standard temperature, To, and standardpressure, and (2) its enthalpy difference between the actual state and the standard stateof the components. Thus, for each component:

h(T) = hf (To) + [h(T) � h(To)] [Btu /mole | kJ /mole] (3.9)

where it is assumed that the sensible gas enthalpy difference is independent of pressure.Sensible enthalpies (those that depend on temperature but do not involve reactions orphase change) relative to the standard reference state are given in Appendix D.

Thus, returning to the formation reaction for the combustion of hydrogen inoxygen at the standard state to produce water, as discussed in the preceding section,we see that the steady-flow First Law of Thermodynamics becomes

Q = (1)hf, H2O � (1)hf, H2 � (0.5)hf, O2 = �103,996 � 0 � 0 = � 103,996 Btu/lb-mole

with water in the vapor phase as the product of combustion of one mole of H2. Herethe sensible enthalpy differences are zero, because both the products and the reactantsare at the standard state. Note that because the stoichiometric coefficients of both

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hydrogen and water are the same in this reaction, the resulting heat transfer may beinterpreted as either per mole of water formed or per mole of hydrogen consumed.

If, instead, liquid water is the combustion product, the heat transfer is given by

Q = � 122,885 � 0 � 0 = � 122,885 Btu/lb-mole

of H2O. The difference between the two cases, 18,919 Btu/lb-mole H2O, is equivalentto 18,935/18 = 1,051.9 Btu/lbm of water, the enthalpy of vaporization of water. Thisresult compares with the enthalpy or latent heat of vaporization of water at 77°F,1050.1 Btu/lbm, given in the steam tables.

With either liquid or gaseous water as the product, the heat transfer term for thecontrol volume is negative, indicating, in accordance with the usual sign convention,that the heat flows from the control volume to the surroundings. The two calculationsabove illustrate the fact that the heat transfer in a formation reaction at the standardstate is the heat of formation of the compound created.

EXAMPLE 3.10

What is the enthalpy of water vapor at 1800°R and 1 bar? What is the heat transfer inthe formation reaction of water from hydrogen and oxygen if the products are at1800°R and the reactants are (a) at the standard state, and (b) at 900°R?

SolutionThe heat of formation of water vapor at the standard state of 298.15K. (536.7°R)

and one bar is � 103,966 Btu/lb-mole. The enthalpy of water vapor at 1800°R is thesum of the heat of formation at the standard state and the sensible enthalpy differenceof water vapor between 536.7°R and 1800°R. Thus:

Hp = � 103,966 + 11,185 = � 92,781 Btu/lb-mole of water.

(a) In this case, the reactants, oxygen and hydrogen, have zero enthalpies because theyare in the standard state and, as elements, their heats of formation are zero. Thus theheat transferred is � 92,781 Btu/lb-mole, or 5154.5 Btu/lbm of water. (b) For reactants at 900°R , Appendix D gives hH2(900) � hH2(536.7) = 2530 Btu/moleof H2 and hO2(900) � hO2(536.7) = 2617 Btu/mole of O2. The enthalpy of the reactantsis then Hr = (1.0)(2530) + (0.5)(2617) = 3838.5 Btu/lb-mole H2O. The heat transferredis then:

Q = Hp � Hr = � 92,781 � 3,838.5 = � 96,619.5 Btu/lb-mole of water, or Q = � 96,619.5 / 18 = � 5,367.8 Btu/lbm of water.

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Thus more heat must be transferred from the control volume to form water vapor at1800°R if the reactants are at 900°R than if they are in the 536.7°R standard state.

Combustion Flame Temperature

In many combustion problems, the reactants enter the combustor near room tempera-ture and products emerge at an elevated temperature. The temperature of the productsflowing from the control volume depends on the energy released in the combustionreaction and heat gain or loss through the boundary of the control volume. Theresulting combustion product temperature is sometimes called the flame temperature.

EXAMPLE 3.11

Methane and stoichiometric air enter a combustor at the standard state. Using a spread-sheet, calculate the heat transfer as a function of the exit (flame) temperature of theproducts of complete combustion. Assume the water in the products remains a vapor.

SolutionThe reaction equation for complete combustion of methane in air is:

CH4 + 2O2 + (2)3.76N2 � CO2 + 2H2O + 7.52N2 The enthalpy of the products at temperature T and of the reactants at the standard stateis

Hp = (1)hf, CO2 + (1)[hCO2 (T) � hCO2 (537)] + (2)hf, H2O

+ (2)[hH2O (T) � hH2O (537)] + (7.52)[hN2 (T) � hN2 (537)]

Hr = hf, CH4 = � 32,189.6 Btu/lb-mole

of methane, where the heats of formation of elemental nitrogen and oxygen are zeroand the heat of formation of water is for the vapor phase. Writing the enthalpydifferences as �h�s and applying the steady-flow First Law of Thermodynamics, we get

Q = Hp � Hr = � 169,184 + �hCO2(T) + (2)(� 103,966)+ (2) �hH2O(T) + 7.52 �hN2(T) � (� 32,189.6)

= � 344,926.4 + �hCO2(T) + (2)�hH2O(T) + 7.52 �hN2 Btu/lb-mole

of methane. This function is tabulated in the spreadsheet in Table 3.10 and plotted inFigure 3.4 using values of enthalpies at the temperature T from the the JANAF tables. Negative values of Q indicate that heat must be rejected from the control volume tomaintain product effluent temperature below about 4200°R. Beyond 4200°R, the CO2,

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N2, and H2O outflow carries more energy than is released in the control volume bychemical reaction; hence, heat must flow into the control volume to achieve theresulting high exit temperatures. Thus the final flame temperature clearly depends onthe chemical composition of the flow and on the consequent control volume heattransfer. ____________________________________________________________________

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Heat of Combustion and Heating Value

The heat of combustion, or enthalpy of combustion, of a fuel is defined as the energytransferred during a steady-flow process in which the fuel is completely burned andwhere the products are returned to the temperature and pressure of the reactants. It willbe seen that the enthalpy of combustion evaluated at the standard state may be deter-mined from the heats of formation. The heat of combustion of hydrogen has, in fact,been determined in a preceding section that examined the formation reaction for water.

The negative of the enthalpy of combustion of a fuel burned in air is usuallyreferred to as the heating value of the fuel. When water in the combustion products iscondensed, the heat of vaporization of the water adds to the chemical energy released,and the resulting heating value is called the higher heating value, HHV. Recall alsothat the heating value obtained when the product water stays a vapor is called the lowerheating value, LHV. The difference between HHV and LHV has been illustrated in theprevious section for the formation reaction of water resulting from the combustion of hydrogen.

For methane, note also that the heat of combustion, and thus the magnitude of thelower heating value, appears in the value of Q in the top row of table 3.10, since therethe combustion products are at the reference temperature.

EXAMPLE 3.12

Illinois no. 6 raw coal has the following dry mass composition: 61.6% C, 4.2% H2,9.7% O2, 1.3% N2, 4.6% S, and 18.5% ash. Using heats of formation, determine thehigher and lower heating values, in kJ / kg, of the as-fired coal with 10% moisture, andcompare them with the heating value in Table 3.3.

SolutionTo adjust the composition for 10% moisture, the factor 1 � A � M becomes 1 � 0

� 0.1 = 0.9. The resulting moist coal composition is given in the following table.It was seen earlier that the heat of reaction of hydrogen in its standard state, and

thus its heat of combustion, is the heat of formation of its product of combustion. Thereaction equation

H2 + 0.5O2 � H2O

shows that one mole of hydrogen produces one mole of water. Thus, from Table 3.9,the heat of formation of steam, � 241,826 kJ per kg-mole of water formed or per kg-mole of hydrogen burned, is also the heat of combustion of hydrogen in the standardstate. Thus the hydrogen contributes 241,826/2 kJ per kg of hydrogen in the coal. Thetotal energy released by the hydrogen in the coal is then 241,826/2 times the massfraction of hydrogen in the coal, as shown in the following table. Similar argumentsmay be made for hydrogen with product water in the liquid phase and the carbon andsulfur components of the coal.

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Element i Dry mfi Wet mfi Heat of Combustion kJ/kg coal

H2:

For LHV 0.042 0.0378 (241,826)(0.0378)/2 = 4,570.5 (v)

For HHV (285,830)(0.0378)/2 5402.2 (l)

C 0.616 0.5544 (393,522)(0.5544)/12 = 18,180.7

O2 0.097 0.0873

N2 0.013 0.0117

S 0.046 0.0414 (296,842)(0.0414)/32 = 384.0

Ash 0.185 0.1665

H2O _____ 0.1000

0.999 0.9991 23,135.2 (v)

23,966.9 (l)

Thus the lower and higher heating values of the coal are 23,135.2 and 23,966.9 kJ/kg,respectively. Table 3.3 lists a heating value of dry Illinois no. 6 raw coal of 11,345Btu/lbm. The corresponding heating value for the wet coal is (0.9)(11,345) = 10,210.5Btu/lbm. This corresponds to (10,210.5)/(0.43) = 23,745 kJ/kg.____________________________________________________________________

Adiabatic Flame Temperature

The results of Example 3.11, tabulated in Figure 3.10, show that for a given air-fuelmixture there is a unique product temperature for which the control volume isadiabatic. This temperature is known as the adiabatic flame temperature. It can bedetermined as in Example 3.11, or it may be calculated from the steady-flow First Lawof Thermodynamics by setting Q = 0. The resulting First Law equation for the adiabaticflame temperature, designated T*, becomes:

Hp (T*) = Hr (T r) [Btu | kJ] (3.10)

where the reactants are at the temperature Tr . The enthalpy terms depend on theindividual enthalpies of the components as functions of temperature. Thus a trial-and-error solution is required using data on heat of formation from Table 3.9 and theenthalpy tables in Appendix D. Given a known Tr , the adiabatic flame temperature mayalso be obtained as the intercept (Q = 0) on a graph of Q versus temperature, T, such asFigure 3.4.Adiabatic Flame Temperature for Solid Fuels

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As a final example, we will determine the combustion products, heat of combustion,and adiabatic flame temperature for a solid fuel specified by its ultimate analysis. Thesolution is presented in a spreadsheet in which enthalpies are tabulated as a function oftemperature for the relevant chemical components as given in the JANAF tables.

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The number of moles of excess O2 in the flue gas per pound of coal may also beobtained from the excess air-fuel ratio:

[(A/F)actual � (A/F)theor ](0.233)/32 = (8.288 � 6.375)(0.233)/32 = 0.0139

The number of moles of flue gas nitrogen is also given by

[(A/F)actual ](1 � 0.233)/28 + d = 8.288(0.767)/28 + 0.000357 = 0.227

With the actual balanced reaction equation known, the mole coefficients may then beused to write an equation for the enthalpy of the products per unit mass of coal. Because all the reactants are assumed to be elements at the reference temperature, theenthalpy of the reactants is zero. By setting Q = Hp = 0, we can solve this First Lawequation for the adiabatic flame temperature by trial and error. However, with aspreadsheet, it is convenient to calculate the heat transfer at each temperature-enthalpydata point and determine the adiabatic flame temperature by inspection and, ifadditional precision is required, by explicit interpolation. The heat transfer equation is shown on the spreadsheet, and the values of the flue gas mole coefficients and heats offormation are shown above the appropriate JANAF enthalpies to which they relate.

It is seen that the adiabatic flame temperature for combustion with 30% excess airis about 2110K, or 3800°R. Note also that the heat of combustion of the coal, 8506.7 Btu/lbm of coal, may be read from the spreadsheet at the JANAF table referencetemperature. This is possible because the heat of reaction is independent of the amountof excess air employed in the reaction._____________________________________________________________________

The spreadsheet for Example 3.13 is easily modified to compute and plot thecomposition of the flue gas as a function of the percentage of excess air. Figure 3.5shows that the mole fractions of excess oxygen and of nitrogen increase while thefractions of other products decrease. Excess oxygen measured from a flue gas sampleis commonly used as a measure of excess air in adjusting the air-fuel ratio ofcombustion systems.

3.6 Molecular Vibration, Dissociation, and Ionization

The temperature of a gas is a measure of the random translational kinetic energy ofmolecules. The simple kinetic theory of the heat capacity of a gas predicts heatcapacities that are independent of temperature and determined by the number ofdegrees of freedom of the molecules. The kinetic theory is usually regarded asapplicable at low pressures and moderate and high temperatures, conditions at whichcollisions between molecules are rare.

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For monatomic gases, kinetic theory predicts an internal energy per atom of

u = 3kT/2

and an enthalpy of

h = u + pv = 3kT/2 + kT = 5kT/2

Here k, the Boltzmann constant, is the ideal-gas constant per molecule, which can becalculated from the universal gas constant, �, and Avogadro�s number of moleculesper mole. Thus

k = �/No = 8.32/(6.025 × 1023) = 1.38 × 10 -23 J/K

From their definitions, Equations (1.14) and (1.15), this leads to cv = 3k/2 and cp = 5k/2per atom and to a heat capacity ratio of k = cp /cv = 5/3 = 1.667.

The concept of equipartition of energy assumes that the energy of a particle is equally divided among its various degrees of freedom. Each mode of energy storage ofa molecule is assigned an energy kT/2. The theory then represents the internal energyas nkT/2, the enthalpy as (n + 2)kT/2, and the heat capacity ratio as k = (n + 2)/n,where n is the number of modes of energy storage, or degrees of freedom, of themolecule. For atoms with three translational and no rotational degrees of freedom, n = 3 (an atom is presumed to be a point and to have no rotational kinetic energy

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because it has zero radius); and these relations reduce to the findings described in thepreceding paragraph.The simple kinetic theory suggests that a diatomic molecule has five degrees offreedom: three degrees of translational freedom and two modes having significantrotational kinetic energy. For this case we obtain u = 5kT/2, h = 7kT/2, and k = 7/5 = 1.4, in agreement with experiment for oxygen and nitrogen at moderatetemperatures and densities.

At higher temperatures, diatomic molecules start to vibrate, adding additionaldegrees of freedom that reduce the heat capacity ratio k below 1.4. As the temperatureincreases and the collisions between molecules become more vigorous, molecules notonly vibrate but they start to be torn apart, each forming separate atoms, the processknown as dissociation. The energy required to break the bonds between atoms inmolecules is called the dissociation energy. At low temperatures, few molecules havesufficient kinetic energy to provide the energy needed to cause dissociation by collision. At higher temperatures, when more molecules have energies exceeding the dissociationenergy, the chemical equilibrium shifts to a composition in which there are more atomsand fewer molecules in the gas. This trend continues as temperature increases.

At still higher temperatures, when particle kinetic energies exceed the ionizationenergy of the gas, outer electrons are separated from atoms, forming positively chargedions, in a process known as the ionization. Particles lose kinetic energy when causingdissociation and ionization. This lost energy is then not reflected in the temperature. Thus, when vibration, dissociation, and ionization occur, the internal energy andenthalpy increase more rapidly than the temperature. Then the simple linear relationsjust given for u and T are no longer correct. Stated another way, the temperature of agas rises less rapidly with heat addition when it is at temperature levels wheresignificant vibration, dissociation, or ionization take place. The phenomenon may bethought of as analogous to phase change, in which enthalpy increases with heat additionwhile temperature does not.

As a result of these phenomena, high flame temperatures determined with ideal gasenthalpies may be overestimated. At temperatures exceeding 2000K (3600°R), flametemperature calculations based on the JANAF gas enthalpies may start to becomeinaccurate. At these temperature levels, dissociation (and at still higher temperatures,ionization) starts to influence the composition of the gases and hence theirthermodynamic properties. The progression of dissociation and ionization withtemperature is shown for air at sea level density in Figure 3.6. It is seen that, at thisdensity, little dissociation of nitrogen occurs below about 6000 K. but that oxygenstarts to dissociate significantly above 3000 K.. At lower densities, the onset ofdissociation occurs at progressively lower temperatures. In Figure 3.6, the number ofparticles per initial atom of air may be obtained by multiplying the ordinates by 1.993.

Usually, dissociation does not seriously influence combustion calculations when theoxidizer is air, but the high combustion temperatures resulting from use of pure oxygenmay be significantly influenced by dissociation. The reader is referred to advancedthermodynamics, physical chemistry, and advanced engineering texts for methods ofpredicting the effects of dissociation and ionization.

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Bibliography and References

1. Anon, Steam, Its Generation and Use, 39th ed. New York,: Babcock and Wilcox,1978. 2. Singer, Joseph G. (Ed.), Combustion / Fossil Power Systems. Windsor, Conn.:Combustion Engineering Inc., 1981.

3. Anon., Classification of Coals by Rank, Standard D-388. Philadelphia: AmericanSociety for Testing Materials, Section 5, Volume 05.05, 1983.

4. Chase, M. W. Jr., et.al., JANAF Thermochemical Tables, 3rd ed., J. Phys. Chem.Ref. Data 14, Supplement No. 1, 1985.

5. Van Wylen, Gordon J., and Sonntag, Richard E., Fundamentals of ClassicalThermodynamics. New York: Wiley, 1986.

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6. El-Wakil, M. M., Powerplant Technology. New York: McGraw-Hill, 1984.

7. Campbell, Ashley S., Thermodynamic Analysis of Combustion Engines. New York: Wiley, 1979.

8. Culp, Archie W., Principles of Energy Conversion. New York: McGraw-Hill, 1979.

9. Wood, Bernard D., Applications of Thermodynamics. Reading, Mass.: Addison-Wesley, 1982.

10. Lefebvre, Arthur H., Gas Turbine Combustion, New York: McGraw-Hill, 1983.

11. Baumeister, Theodore, and Marks, Lionel S. (Eds.), Standard Handbook forMechanical Engineers, 7th ed. New York: McGraw-Hill, 1967.

12. Sorenson, Harry A., Energy Conversion Systems. New York: Wiley, 1983.

13. Hill, Phillip G., Power Generation. Cambridge, Mass.: MIT Press, 1977.

14. Anon., Coal Quality Information Book, CS-5421, Interim Report. Palo Alto:Electric Power Research Institute, December 1987.

15. Moeckel, W. E., and Weston, Kenneth C., �Composition and ThermodynamicProperties of Air in Chemical Equilibrium,� NACA TN 4265, April 1958.

EXERCISES

3.1 Determine the mass fractions of a mixture of six grams of carbon, three grams ofsulfur, and one gram of sodium chloride.

3.2 Determine the mole fractions of a gas consisting of a mole of oxygen, eight molesof nitrogen, a mole of CO, and two moles of CO2. Determine also the mass fractions. What is the average molecular weight of the gas?

3.3 Write the balanced reaction equation for the complete combustion of sulfur inoxygen. What are the mass and mole fractions of oxygen in the reactants?

3.4 Write the balanced reaction equation for the complete combustion of carbon inoxygen. What are the mass and mole fractions of oxygen in the reactants?

3.5 Write the balanced reaction equation for the complete combustion of carbon in air.What are the mass and mole fractions of fuel, air, oxygen, and nitrogen in the reactants?

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3.6 Write the balanced reaction equation for the complete combustion of ethane, C2H6,in air. What are the mass and mole fractions of fuel, air, oxygen, and nitrogen in thereactants? What are the mass and mole fractions of carbon dioxide and water vapor inthe combustion products?

3.7 Write the balanced reaction equation for the complete combustion of propane,C3H8,, in air. What are the mass and mole fractions of fuel, air, oxygen, and nitrogen inthe reactants? What are the mass and mole fractions of carbon dioxide and water vaporin the combustion products. What are the mass and mole air-fuel ratios?

3.8 Write the balanced reaction equation for the complete combustion of C8H18 in air. What are the mass and mole fractions of fuel, air, oxygen, and nitrogen in the reactants? What are the mass and mole fractions of carbon dioxide and water vapor in thecombustion products? What are the mass and mole air-fuel ratios?

3.9 Gasoline, sometimes represented as C8H18, is burned in 25% excess air mass. Whatare the mass and mole stoichiometric and actual air-fuel ratios? Determine the massand mole fractions of the combustion products.

3.10 Determine the lower and higher heating values of methane using the JANAF tableof heats the formation.

3.11 Determine the as-fired stoichiometric and actual air-fuel ratios for Greene,Pennsylvania raw coal (Table 3.3) with 5% moisture and the mass and mole flue gascompositions for combustion with 20% excess air.

3.12 Compare the stoichiometric and actual air-fuel ratios and the mole flue gascomposition for combustion with 20% excess air for the following raw and clean(process #1) coals (Table 3.3): (a) Freestone, Texas, big brown lignite; (b) Indiana,Pennsylvania, Freeport (upper); (c) British Columbia, Hat Creek (A zone); (d) Perry,Illinois no. 6; (e) Muhlenberg, Kentucky no. 9; (f) Nicholas, West Virginia, Kittanning;(g) Belmont, Ohio, Pittsburgh; (h) Big Horn, Montana, Robinson; (i) Greene,Pennsylvania, Sewickley; (j) Kanawha, West Virginia, Stockton-Lewiston; (k) Belmont,Ohio, Waynesburg.

3.13 If Union, Kentucky no. 11 raw coal has 10% moisture, as mined, determine the as-mined proximate and ultimate analyses for this coal.

3.14 If Big Horn, Montana, Robinson raw coal has 15% moisture, as-mined, what areits as-mined proximate and ultimate analyses?

3.15 Determine the ultimate analyses of the raw coals listed in Exercise 3.12 (a-e)assuming 10% as-mined moisture.

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3.16 Write the balanced chemical equation for the combustion of methane instoichiometric air. Use the table of heats of formation to determine the heats ofreaction of methane (in Btu/lbm and Btu/lb-mole), with products and reactants all at thestandard state and product water as liquid. What are the values if the product water isvapor? Would the heat of reaction be different if the combustion were in pure oxygen? Compare your results with tabulated heating values for methane.

3.17 Solve Exercise 3.16 in SI units: kJ/gm-mole and kJ/kg.

3.18 Calculate the heating values, in Btu/lbm and Btu/lb-mole, for the completecombustion of hydrogen, with product water in liquid and in vapor phases. Comparewith tabulated heating values. Repeat the calculations in SI units.

3.19 Determine the heat transferred when ethane is burned (a) in stoichiometric air, and(b) in 100% excess air. In both cases the reactants are in the standard state andproducts at 1000K. Use a heat of formation of �36,420 Btu/lb-mole.

3.20 What is the adiabatic flame temperature for the combustion of ethane in air,ignoring dissociation? Use a heat of formation of �36,420 Btu/lb-mole of ethane.

3.21 Compare the adiabatic flame temperatures for the stoichiometric combustion ofhydrogen in air and in pure oxygen, ignoring dissociation.

3.22 Determine the adiabatic flame temperature for the stoichiometric combustion inair of Illinois no. 6 coal after the clean #2 process. Determine also its heat ofcombustion, and compare with the tabulated value.

3.23* Develop a spreadsheet that determines the air-fuel ratio for coal characterized bya dry ultimate analysis, such as given in Table 3.3. Apply it to several of the coals inthe table, as assigned by your instructor.

3.24* Develop a spreadsheet that determines the air-fuel ratio for a coal characterizedby a dry ultimate analysis, such as given in Table 3.3, and a given moisture content. Apply it to a coal in the table and several different moisture contents, as assigned byyour instructor.

3.25* Develop a spreadsheet that determines the mass and mole, wet and dry flue gascompositions for a coal characterized by a dry ultimate analysis, such as given in Table3.3, and a given percentage of excess air. Apply it to several coals in the table for 20%and 40% excess air.

______________________*Exercise numbers with an asterisk involve computer usage.

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3.26* Develop a spreadsheet that determines the average molecular weight, and themass and mole, wet and dry flue gas compositions for a coal characterized by a dryultimate analysis, such as given in Table 3.3, a given percentage of excess air, and agiven moisture content. Apply it to a coal in the table for 10% moisture and 10% and20% excess air.

3.27* Develop a spreadsheet that determines the theoretical air-fuel ratio for a gascharacterized by any combination of the components of Table 3.4. Apply it to thegases in the table, as assigned by your instructor.

3.28* Develop a spreadsheet that determines the average molecular weight, and themass and mole, wet and dry flue gas compositions for a gas characterized by anycombination of the components of Table 3.4 and a given percentage of excess air. Apply it to several the gases in the table for 20% excess air.

3.29 What are the theoretical and actual air-fuel ratios and the wet and dry mole fluegas compositions for 20% excess air for the Oklahoma natural gas in Table 3.4?

3.30 What are the theoretical and actual air-fuel ratios and the wet and dry mole fluegas compositions for 15% excess air for the Ohio natural gas in Table 3.4?

3.31 An adiabatic gas turbine combustor burns methane at 77°F with air at 400°F. Thecombustion products emerge from the combustion chamber at 3200°F. What is the air-fuel ratio? What is the equivalent external heat transferred per lbm of air to produce thistemperature rise, assuming a mean heat capacity of 0.24 Btu/lbm-R?

3.32 An adiabatic gas turbine combustor burns methane at 25°C with air at 250°C. Thecombustion products emerge from the combustion chamber at 2000K. What is the air-fuel ratio? What is the equivalent external heat transferred per kilogram of air toproduce this temperature rise, assuming a mean heat capacity of 1.005 kJ/kg-K?

3.33 An adiabatic combustor burns methane with 400% excess air. Both air andmethane are initially at 298K. What is the exit temperature?

3.34 An adiabatic combustor burns methane with 500% theoretical air. Both air andmethane are initially at 77°F. What is the flame temperature?

3.35 An adiabatic combustor burns methane in 100% excess oxygen. Both fuel andoxidizer enter at the JANAF tables reference temperature. What is the flametemperature?

3.36 Calculate and tabulate the higher and lower heating values of methane, in kJ/kg-mole and in kJ/kg.

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3.37 Determine the theoretical air-fuel ratio and the air-fuel ratio for 20% excess airfor Muhlenberg, Kentucky, raw coal with 10% moisture.

3.38 Determine the air-fuel ratio and wet and dry flue gas mole and mass fractions fordry Muhlenberg, Kentucky no. 9 raw coal with 20% excess air.

3.39 Solve Exercise 3.38 for the coal having 10% moisture in the as-fired condition.

3.40 Solve Exercise 3.38 for 40% excess air.

3.41* Set up a spreadsheet to solve Exercises 3.38�3.40 where it is necessary tochange only one parameter for each of the latter cases.

3.42 Using the compound composition data of Table 3.4, calculate the ultimate(elemental) analysis, and compare with the tabular results for the following naturalgases: (a) Pennsylvania, (b) Southern California, (c) Ohio, (d) Louisiana, (e) Oklahoma.

3.43* Develop an interactive computer program to solicit and receive ultimate analysisdata for an arbitrary coal, arbitrary as-fired ash and moisture fractions, an excess airpercentage, and output the appropriate air-fuel ratio and mass and mole, wet and dryflue gas compositions.

3.44* Develop an interactive computer program to solicit and receive ultimate analysisdata for an arbitrary coal, arbitrary as-fired ash and moisture fractions, and Orsat CO,CO2, and O2 data. The program should determine the actual operating air-fuel ratio.

3.45* Apply the spreadsheet of Example 3.13 to determine the adiabatic flametemperature, heat of combustion, and wet flue gas composition for stoichiometriccombustion of the liquid fuels in Table 3.6.

3.46* Develop a well-organized spreadsheet in which the user may enter a coalultimate analysis for C, H, O, N, and S; dry flue gas mole compositions for CO, CO2,and 02; and as-fired moisture and ash mass fractions to determine theoretical and actualair-fuel ratios and percentage of excess air.

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C H A P T E R F O U R

ASPECTS of STEAM POWER PLANT DESIGN

4.1 Introduction

After studying the fundamental thermodynamic cycles of steam power plants and con-sidering the characteristics and thermochemistry of fuels, it is appropriate to considerthe design of the systems and flow processes that are operative in steam plants andother large-scale power production facilities. This chapter will focus first on theprocessing of several fundamental streams that play a major role in power plantoperation. Up to this point, a great deal of attention has been focused on the water pathfrom the point of view of the thermodynamics of the steam cycle. Additional aspects ofthe water path related to plant design are considered here.

Another fundamental flow in the power plant, the gas stream, includes the intakeof combustion air, the introduction of fuel to the air stream, the combustion process,combustion gas cooling in the furnace heat exchange sections, and processing anddelivery of the gas stream to the atmosphere through a chimney or stack. A thirdimportant stream involves the transportation and preparation of fuel up to the point thatit becomes part of the combustion gas.

A major non-physical aspect of power production is the economics of power plantdesign and operation. This is considered in conjunction with some preliminary designanalyses of a prototype plant. Environmental considerations also play an important partin planning and design. The chapter concludes with back-of-the-envelope typecalculations that define the magnitudes of the flows in a large plant and identify majordesign aspects of steam power plants.

4.2 The Water Path

The Liquid-Water-to-Steam Path Several pumps are employed in the feedwater path of a steam power plant to push the working fluid through its cycle by progressively elevating the pressure of the waterfrom the condenser to above the turbine throttle pressure. These pumps are usually driven by electric motors powered by electricity generated in the plant or by steamturbines powered by steam extracted from the main power cycle.

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The power requirement of a pump is proportional to the liquid mass-flow rate and thepump work, as given by Equation 2.9, and inversely proportional to the pumpefficiency:

Power = mvsat �p/�pump [ft-lbf /s | kW]

The pumps are required to overcome frictional pressure losses in water-flow andsteam-flow passages, to provide for the pressure differences across turbines, and toelevate the liquid to its highest point in the steam generator. The pump powerrequirements are typically a small percentage of the gross power output of the plant.

Thus condensate leaving the condenser passes through one or more pumps andfeedwater heaters on its way to the steam generator. A typical shell-and-tube closedfeedwater heater is shown in Figure 4.1. Normally, feedwater passes through the tubeswhile extracton steam enters at the top and condenses as it flows over the tubes to thebottom exit.

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After passing through the chain of feedwater heaters and pumps, the feedwaterenters the steam generator through the economizer. An economizer is a combustion-gas-to-feedwater tubular heat exchanger that shares the gas path in a steam generator,as seen in Figure 4.2. The economizer heats the feedwater by transferring to it some ofthe remaining energy from the cooled exhaust gas before the gas passes to the airheater, the pollution control equipment, and the stack.

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Steam Generators

Figures 4.3 and 4.4 show two-drum steam generators in which the designvaporization pressure is below the critical pressure of water. Hot, subcooled liquidfeedwater passes from the economizer and through the boiler tube walls to the drum loop located near the top of the steam generator. Liquid water circulates by freeconvection through the many boiler tubes between the drums until it is vaporized by thehot gas stream flowing over the tubes. A fixed liquid level is maintained in the uppersteam drum, where the steam separates from the liquid and passes to the superheater.Solids settle in the bottom of the so-called �mud drum� below it.

A steam drum mounted on a railroad flat car en route to a construction site isshown in Figure 4.5. The many stub tubes around the bottom and on top are to be

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connected to steam-generating loops and to steam-superheating pipes, respectively, asseen at the top of Figure 4.4. The large pipes on the bottom and the ends are forconnection to downcomers, which supply recirculated liquid water to various heatingcircuits in the steam generator.

Steam produced in the steam drum, at a boiling temperature corresponding to thevapor pressure in the drum, passes to superheater tube or plate heat-exchanger banks.The superheater tube banks are located in the gas path upstream of the drum loop, asseen in Figures 4.3 and 4.4, taking advantage of the highest gas temperatures there tosuperheat the steam to throttle temperature. The hottest gases are used to heat the

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hottest water-tube banks, to minimize the irreversibility associated with the heattransfer through the large temperature differences between the combustion gas and thesteam or liquid water. The dry steam from the superheater then passes from the steamgenerator through the main steam line to the HP turbine. The progression of tubebanks, with decreasing water temperatures exposed to successively cooler gas temp-eratures from the secondary superheater to the economizer to the air heater are alsoseen in the universal-pressure (supercritical pressure) steam generator in Figure 4.2.

As the design throttle steam pressure increases toward the critical pressure ofwater, the density difference between liquid water and vapor decreases, finallyvanishing at the critical point (3208.2 psia, 705.47°F.). As a consequence, in steam-generator boiling loops, natural convection water circulation�which is driven by thedensity difference between liquid and steam�becomes impractical at pressures aboveabout 2500 psia. Thus modern high-throttle-pressure power plants use circulatingpumps to provide forced to circulation to augment or replace natural circulation ofwater in the steam generator.

In single-drum steam generators, water flows downward from the steam drumthrough large pipes called downcomers located outside the furnace wall, then throughcirculating pumps to headers at the bottom of the steam generator. From the headers,water flows upward in vertical tubes forming the inside of the furnace walls. The wateris heated by the furnace gases as it rises, and eventually boils and forms a two-phaseflow that returns to the steam drum. There, vapor separates and passes to thesuperheater.

Steam generators may utilize natural convection flow through downcomers and

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vapor-laden upward flow through the tube walls alone or may combine naturalconvection with the use of booster pumps to provide adequate circulation for a widerrange of operating loads. It is important to recognize that at the same time steam isbeing generated in the boiler, the tube walls are being cooled by the water. Adequatewater circulation must be ensured to provide waterside heat transfer rates high enoughto maintain tube wall temperatures below their limiting design values and thereby toavoid tube failure.

A once-through supercritical steam generator, operates at a throttle pressureabove the critical pressure of water as in the Riverside station discussed in Chapter 2.There are no drums and no water recirculation in a once-through steam generator.Water from the economizer passes to the bottom of the furnace, where it starts itsupward flow through the furnace tube walls. Steam formed in the tubes flows upwardto be collected in headers and mixed to provide a unifrom feed to the superheater.

The feedwater passes directly from the liquid to the vapor phase as it is heated at apressure above the saturation pressure. It may be compared to a flow of water pumpedthrough a highly heated tube with a downstream valve. The state of the steam emergingfrom the tube depends on the valve setting, the heat addition rate, and the feedwaterflow rate. In the same way, the steam conditions at the turbine throttle may be adjustedby changing the turbine throttle valve setting, the fuel firing rate, and the feedwaterflow rate. If the flow rate is decreased by closing the throttle valve, it is necessary todecrease the fuel firing rate to maintain the same thermodynamic conditions at thethrottle. On the other hand, if the rate of heat transfer is increased without changing theflow rate, the steam discharge temperature will increase. Other adjustments, such asincreasing condenser cooling-water flow rate, may then be appropriate to avoid anincrease in condenser temperature and pressure. Similarly, an increase in fuel flow ratemust be accompanied by an increase in air flow rate to maintain a constant air-fuelratio.

In cycles with reheat, the reduced-pressure steam from the HP turbines passesthrough the cold reheat line to the reheater section in the steam generator, where thesteam temperature is returned to approximately the original throttle temperature. Thesteam then returns to the next turbine through the hot reheat steam line, as Figure 2.13indicates.

After leaving the LP turbine, low-pressure steam then passes over thewater-cooled tubes in the condenser and returns to the feedwater heating system assaturated liquid condensate. A condensate pump then raises the pressure of the liquidand transports it to the first low-pressure feedwater heater, where it begins another tripthrough the cycle.

In order to avoid corrosion, scaling and the deposits of solids along the water pathcan result in losses of efficiency and unscheduled shutdowns, water of extreme purity isrequired in the steam cycle. Chemical and filtration processes are employed to ensurethat high water quality is maintained, to avoid deterioration or clogging of water pathcomponents. An example of the potential deposits when proper water treatment isneglected is seen in Figure 4.6. The deaerator, an open feedwater heater mentioned in

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Chapter 2, provides for the removal of noncondensable gases, particularly oxygen, from the working fluid. The deaerator allows noncondensable gases to escape to theatmosphere through a vent condenser, while accompanying steam is retained bycondensing it on cool surfaces and returning it to the feedwater heater stream bygravity flow.

The turbine room at the Bull Run coal-burning power plant of the TennesseeValley Authority (TVA) is shown in Figure 2.3. Electrical generators are seen in theleft and right foreground. Behind them, high-pressure turbines on the left are seenjoined to low-pressure turbines on the right by two large, vee-shaped crossover steamlines. The side-by-side condensers are seen on either side of the low pressure turbines,a departure from the usual practice of locating the condenser below the low-pressureturbines. Figure 4.7 shows the turbogenerator room at TVA�s Brown�s Ferry nuclearpower plant with a turbine in the foreground.

The Condenser Cooling-Water Loop

The cooling loop, in which water passes through tubes in the condenser removing heatfrom the condensing steam, is an important water path in large steam plants. Thiscooling water, clearly separate from the working fluid, is usually discharged into anearby body of water (a river, or a natural or man-made lake) or into the atmosphere.

Figure 4.8 shows a typical wood-framed, induced-draft cooling tower used todissipate heat from the condenser cooling water into the atmosphere. The tower isusually located a few hundred yards from a plant. Typically, the cooling water enteringthe tower is exposed to a flow of air created by upward-blowing fans at the bases of thefunnels at the top of the towers. A fraction of the condenser cooling water, whichpasses over extensive aerating surfaces in the tower, evaporates and exits to the

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atmosphere, cooling the rest of the water. The remaining chilled water is then returnedto the condenser by a cooling-water circulating pump. A continuing supply of liquidmakeup water is required for these towers to compensate for vapor loss to theatmosphere.

In areas where large structures associated with power plants are acceptable, largenatural-draft cooling towers may be used. Figure 4.9 shows two large natural-drafthyperbolic cooling towers serving a large power plant. The height of these towers,which may reach over 500 feet, creates an upward draft due to the difference in densitybetween the warm air in the tower and the cooler ambient air. Heat from the condensercooling water warms the air, inducing an upward air flow through the heat transfersurfaces at the base. These towers offer long-term fan-power savings over mechanicaldraft towers. Under some conditions, these power savings may offset high constructioncosts of hyperbolic towers.

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4.3 The Fuel Path for a Coal-Burning Plant

The supply and handling of fuel for a modern coal-burning power plant is a complexand expensive undertaking. In contrast to the relatively simple steady flow of fluid fuelsin power plants that consume natural gas and fuel oil, solid-fuel-burning plants offermajor and continuing challenges to engineers. The discussion here focuses on theseoperations and their challenges.

Getting the Coal to the Plant

The source of coal for a plant may be a surface mine or a deep underground mine.Power plants are sometimes located adjacent to mines, where conveyors may providethe only transportation required. This significantly reduces coal transportation costswhich otherwise can be higher than the cost of the coal alone. Such plants are called mine-mouth plants. A mine-mouth plant may be an attractive option if its selectiondoes not result in significant transmission costs to bring the electrical power to distantload centers where the utilities� customers are located.

Today the power plant and coal mines are likely to be a considerable distanceapart, perhaps a thousand miles or more. The most widely used modern transportationlink between mine and plant is the unit train, a railroad train of about a hundred carsdedicated to transporting a bulk product such as coal. Although slurry pipelines (aslurry is a fluid mixture of solid lumps and liquid, usually water, which can be pumpedthrough pipes by continuous motion) sometimes offer attractive technical solutions tocoal transport problems, economic and political forces frequently dictate against theiruse. Dedicated truck transport is an occasional short-haul solution, and barges aresometimes used for water transport. Here, we will focus on unit trains.

Several unit trains may operate continuously to supply a single plant. Trainscarrying low-sulfur coal from Wyoming, Montana, and the Dakotas supply coal to plants as distant as Michigan, Illinois, and Oklahoma. This strange situation, in whichutilities located in states with large quantities of coal purchase coal from distant states,was a response to pollution control requirements. It was preferred to purchase low-sulfur coal from a distant state rather than pay a high price for sulfur removalequipment (perhaps 10% of the cost of the plant construction), some of which has areputation for unreliability. In response to such choices, the Oklahoma legislaturepassed a law requiring utilities to burn at least 10% Oklahoma coal in theircoal-burning steam generators. Such mixing of small quantities of high-sulfur coal withlow-sulfur coal is an expedient to protect local businesses and to spread out resourceutilization geographically.

New plants, however, no longer have the option to choose low-sulfur coalor sulfur removal equipment. �Best available control technology,� BACT, has becomethe rule. The Environmental Protection Agency now requires new plants to havescrubbers (sulfur removal equipment) even if the plants use low-sulfur coal, and theyare required to employ the currently most effective pollution control technology.

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Coal Unloading and Storage

On arrival at the plant, the unit train passes through an unloading station. Some coalcars have doors on the bottom that open and dump their load to a conveyor below.Others have couplings between cars that allow the rotation of individual cars abouttheir coupling-to-coupling axis, by a dumping machine, without detachment from thetrain, as seen in Figure 4.10. The figure shows a breaker in the dumping facility thatreduces large coal chunks to a smaller, more uniform size for transport on a beltconveyor. The under-track conveyor at the unloading station then carries the newlyarrived coal up and out to a bunker or to a stacker-reclaimer in the coal yard as seen in Figure 4.11. The stacker-reclaimer either feeds the coal through a crusher to the plantor adds it to the live storage pile.

A permanent coal storage pile sufficient to supply the plant for several months isusually maintained. While the first-in-first-out approach common in handling perishablegoods seems logical, a first-in-last-out storage system is usually used. A primaryreason for this approach is the hazard and expense of coal pile fires, which can occurdue to spontaneous combustion. Once a stable storage pile is achieved by packing andother treatment to restrict air acccess, it is usually not disturbed unless coal must bewithdrawn by the stacker-reclaimer to satisfy unusual demands caused by labor strikes,extreme weather, rail accidents, or the like.

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A conveyor transports coal from the reclaimer to a crusher house, where hammermills, ball crushers, or roller crushers break up large chunks to a more manageablesize. Another conveyor may then carry the crushed coal to one of several bunkers orsilos for temporary storage prior to firing. Some of these features may be seen in thephotograph of the PSO Northeastern Station in Figure 2.1.

The rate of feeding coal from the silos is controlled to maintain the desired steamgenerator energy-release rate. In a pulverized-coal plant, the coal is fed from the silosto pulverizers, where it is further reduced in size to a powdery form. Warm air drawnthrough an air preheater in the steam generator by the primary air fan flows throughthe pulverizer, where it picks up the fine coal particles and transports them pneumatic-ally through piping to the steam generator burners. Several arrangements of silos,feeders, pulverizers, and pneumatic transport systems are seen in Figures 4.2 to 4.4. 4.4 The Gas Path

Fans

While natural or free convection may be used to provide combustion air to small boilersand heaters, modern power plants employ large fans or blowers to circulate air to theburners and to assist flue gas in escaping from the furnace. These fans are calledforced-draft fans and induced-draft fans, respectively. A common arrangement ofthese fans is shown in Figure 4.4. Atmospheric air drawn into the steam generator byone or more forced-draft fans is heated as it passes through the cold gas side of an airheater on its way to the furnace. At the same time, combustion gases that have passedthrough the furnace heat transfer sections are cooled as they passed through the hotside of the air heater on their way to induced-draft fans and thence to the stack. In thecase of pulverized-coal-burning plants, primary air fans, as seen in Figures 4.3 and 4.4,supply enough pre-heated air to pulverizers to transport coal pneumatically to theburners. Primary air usually pre-heated to 300-600°F to dry the coal as it passesthrough the pulverizer.

With a forced-draft fan alone, the furnace pressure is above atmospheric pressure,causing large outward forces on the furnace walls and a tendency for leakage ofcombustion gas from the furnace. On the other hand, the use of an induced-draft fanalone would cause the furnace pressure to be below atmospheric pressure, producinglarge inward forces on the walls and possible air leakage into the furnace. The forceson the walls, which can be significant, can be minimized by keeping the furnacepressure near atmospheric by using balanced draft, that is, the use of both forced- andinduced-draft fans, which produce a gas path pressure distribution such as shown bythe heavy line in Figure 4.12. In this design the forced-draft fan raises the pressure to15 in. of water gauge entering the steam generator, and the induced-draft fan depressesits inlet pressure about 21 in. of water below atmospheric. As a result, the furnaceinside-wall pressure is less than an inch of water below atmospheric. This substantially reduces both the potential for furnace leakage and the forces on the furnace walls.

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The power requirements for fans may be determined in much the same way thatpump power requirements are determined. Fans are primarily gas-moving devices thatproduce small pressure rises. Pressure and density changes across fans are usuallysmall fractions of the fan inlet values. This justifies the approximation that the fanprocess is incompressible. Fan power requirements then closely follow the pumppower prediction method discussed earlier. Thus, for a forced-draft fan, power may beestimated by using

PowerFD = Qair �pair/�FDfan [ft-lbf/s | kW]

For the induced-draft fan

PowerID = �airQair �pgas(1 + F/A)/�gas�FDfan [ft-lbf/s | kW]

where Qair is the volume flow rate of air entering the forced-draft fan. The secondequation accounts for the additional fuel mass handled by the induced-draft fan and the

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density of the gas leaving the furnace, assuming no leakage or diversion of air from thatleaving the forced-draft fan.

The drawing of a centrifugal forced-draft fan is shown in Figure 4.13; a photograph of a rotor and open housing is presented in Figure 4.14. In a centrifugal fan, air is spun by the rotor blades, producing tangential motion and pressure rise andleaving behind a vacuum for air to flow in along the axis of the fan. It the fan entranceis open to the atmosphere, its exhaust is pressurized; if its exhaust is atmospheric, itsentrance pressure is below atmospheric. Fans typically do not produce large pressurerises but do produce large flows of gases.

A diagram of an axial-flow fan designed for induced-draft use is shown in Figure4.15. Figure 4.16 presents a photo of the same type of fan. Induced-draft fansmust be able to withstand high-temperature service and erosion due to airborneparticulates. Large electrostatic precipitators located upstream of the induced-draftfans remove most of the flyash by inducing a static charge on the flowing particlesand collecting them on plates of opposite charge. The plates are periodically rappedmechanically to free ash deposits that drop to the bottom of the precipitator and arecollected and removed. Figure 2.1 shows electrostatic precipitators, to the right of thesteam generator. The large structure behind the fans, air heater, and ducting and belowthe stack in Figure 4.4 is an electrostatic precipitator.

Air Preheaters

The air leaving the forced-draft fan usually flows through an air preheater to a windboxaround the furnace and then to the burners. A Ljungstrom rotary air preheater, used

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in many large plants, is shown in Figures 4.17 and 4.18. The rotary air heater is aslowly rotating wheel with many axial-flow passages, having large surface area andheat capacity, through which air and flue gas pass in counterflow parallel to the wheelaxis. When the wheel surfaces heated by the flue gas rotate to the air side, they arecooled by the air from the forced-draft fan. As result, the air temperature rises severalhundred degrees before passing to the furnace windbox.

Although the Ljungstrom rotary air heater is widely used in utility and industrialpower plants, heat-pipe air and process heaters are now being considered and appliedfor use in power plants and industry. A heat pipe, shown in Figure 4.19, is a sealedtube in which energy is transported from one end to the other by a thermally drivenvapor. The heat-pipe working fluid absorbs heat and vaporizes at the lower, hot end. After rising to the higher, cold end, the vapor condenses, releasing its heat ofvaporization, which is carried away by conduction and convection through external finsto the combustion air. The liquid then returns to the hot end by gravity and/or bycapillary action through wicking, to complete the cycle. The wicking may be spiralgrooves around the inside of the tube that ensure that the entire inside surface is wettedfor maximum heat transfer. The wicking in the cold section is particularly important,because it provides increased surface area that increases inside-gas heat transfer rates. The outside the tube is usually finned to provide adequate external heat transfer rates,both from the flue gas and to the incoming air.

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As a heat transfer device, a well-designed heat pipe has an effective thermalconductance many times that of a copper rod. Note that the energy transfer inside theheat pipe is essentially isothermal, since the liquid and the vapor are in near equilibrium. Although heat pipes will operate in a horizontal orientation, theireffectiveness is augmented by gravity by inclining them about 5°� 10° to assist in liquidreturn to the hot end.

In power plant air heater applications (see Figures 4.19 and 4.20), finned heatpipes supported by a central partition between the incoming air stream and the flue gas

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stream are free to expand outward. This reduces thermal expansion problems andvirtually eliminates the possibility of leakage between the flows. Such heaters oftenhave been installed in process plants and have been retrofitted in power plants originallybuilt without air heaters, because of their ease of installation and compact sizecompared with stationary tubular air heaters.

EXAMPLE 4.1

The heat-pipe air heater in Figure 4.20 has an air flow of 360,800 lbm/hr and a flue gasflow rate of 319,000 lbm/hr. The flue gas enters at 705°F and leaves at 241°F; thecombustion air enters at 84°F. What is the rate of energy recovery from the flue gas,and what is the air temperature entering the windbox? Assume a flue gas heat capacityof 0.265 Btu/lbm-R.

SolutionThe rate of heat recovery from the flue gas is

mcp (Tout � Tin) = 319,000(0.265)(705 � 241) = 39,224,240 Btu/hr

The air temperature is then

84 + 39,224,240/[(0.24)(360,800)] = 537°F_____________________________________________________________________

An analysis of the gas flow through a steam generator must take into account thestreamwise pressure rise through the fans and pressure losses due to friction and lossesthrough flow restrictions and turns. These include losses due to flow through furnacetube and plate heat exchanger banks and other passages, such as in both the air and thegas passes through the air heater.

Power Plant Burners

Burner design depends on the choice of fuel and the steam generator design. Figure4.21 shows a burner designed for forced-draft applications burning natural gas and oil. Oil and steam under pressure are mixed in the central feed rod to atomize the oil to a fine mist coming out of the oil tip. The cone at the oil tip stabilizes the flame in thesurrounding air flow. The gas pilot next to the oil feed rod provides a continuousignition source. A separate duct for the natural gas supply feeds gas to the two typesof gas tip. Separate air registers control the flow of air to the gas and oil tips. Figure4.22 shows an oil burner, with a water-cooled throat, installed in a furnace wall. Registers that control the flow of air from the windbox are also visible.

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In the case of the pulverized-coal plant, primary air flows through the pulverizer and carries the fuel directly to the burners (Figure 4.23). Secondary (and sometimestertiary) air helps to control the temperature of the control nozzle and of the furnacewall, and mixes with the combustion gases to provide for essentially completecombustion of the fuel. Features of a burner designed for pulverized-coal firing inplanar furnace walls are shown in Figure 4.24. Note that the secondary air flowthrough the windbox registers helps to cool the nozzle through which the coal and theprimary air flow.

Another approach to burning pulverized coal uses corner burners in tangentiallyfired steam generators. A plan view of such a furnace is shown in Figure 4.25. Theburners induce a circular motion, in the horizontal plane, on the upward-risingcombustion gases, promoting vigorous mixing, which hastens completion ofcombustion in the furnace. For control purposes, the corner burners can be tilted in thevertical direction to adjust the furnace heat transfer distribution.

A cyclone furnace type of burner installation, used for burning slagging coals(those that form liquid ash, or slag, at moderate temperatures) in steam generators, is shown in Figure 4.2. The cyclone furnace is a cylindrical furnace with very large

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volumetric heat release rates that lies adjacent to and opens onto the main furnace. Details of a cyclone furnace are shown in Figures 4.26 and 4.27.

The coal supplied to cyclone furnaces, which is crushed but not pulverized, is fedto the cyclone by a mechanical feeder. The coal and primary air entering the cyclone

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move tangentially to the inside of the furnace cylinder. There the momentum of thecoal carried by the swirl flows forces the coal pieces toward the cylindrical burner wall. The very high temperature and vigorous mixing produce a high rate of burning. As aresult, combustion is virtually complete by the time the combustion gas flow enters the

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main furnace. The cooled walls stimulate formation of a protective slag layer on thecylinder walls. Because the main furnace is only required for steam generation and tocool the combustion gases, and not to provide time for completion of combustion, thecyclone furnace steam generator can be significantly smaller than the pulverized-coalsteam generator. A steady flow of slag drains from the cyclone furnace into a slag tankat the bottom of the main furnace.

In both cyclone and pulverized-coal steam generators the combustion gases flowupward from the burners, transferring heat to the tube walls by radiation andconvection. The cooling gases then flow through superheater, reheat, and boiler tube orplate sections. The combustion gas temperature drops as it passes through these steamgenerator sections in essentially a counterflow arrangement with the water flow. Thecombustion gases undergo their final cooling as they pass through the economizer andthen the air preheater. From there they pass through an electrostatic precipitator forremoval of airborne particles and through scrubbers for control of oxides of nitrogenand sulfur (NOx and SOx) and through an induced-draft fan before entering the stack.

The serious degradation of the environment caused by oxides of sulfur andnitrogen in the flue gas of power plants and from other sources has led to widespreadchemical processing of flue gases. Figure 4.28 shows a schematic diagram of the gasflow path for removal of NOx and SOx after particulate removal in the precipitator andpassage through an induced-draft fan. In this scheme, gas-to-gas heat exchangers(GGH) provide the proper temperatures for the flue gas desulfurization (FGD) unit andthe DENOX catalyst unit. This additional equipment increases the pressure dropthrough the system, sometimes necessitating an additional fan.

With high smokestacks, the stack effect also influences the gas path and must betaken into account.The stack effect is the upward movement of exhaust gas producedby the density difference between the hot gases inside the stack and the surroundingcooler atmospheric air. Because of hot gas buoyancy, a smaller pressure gradient alongthe stack length is required to expel the combustion gases from the stack. This effect isopposed by the usual viscous friction pressure losses. The diameter and height of thestack control the relative influence of frictional forces in opposing the stack effect.Other considerations, such as cost and a possible need to disperse the stack gas above aparticular height, also have a significant influence on these dimensions.

It is obvious that the air heater of the steam generator should extract as muchenergy from the combustion gas as possible to maximize its regenerative heat transfereffect. This implies cooling the gas to a low tempeature. However, practical limits existon the minimum combustion gas temperature, to avoid the condensation of water vaporin the presence of sulfur and nitrogen compounds in the gas and to meet the temp-erature requirements of the pollution control equipment. Condensation of water vaporin the presence of gaseous oxides of sulfur and nitrogen leads to the formation of acidsthat erode the materials on which the liquid condenses. The temperature at which thevapor condenses is called the acid dew point. Typical acid dew points for coal range toabout 320°F. As a result, stack gas design temperatures may exceed that value,depending on the coal and flue gas treatment.

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4.5 Introduction to Engineering Economics The success of any engineering undertaking depends on adequate financial planningto ensure that the proceeds of the activity will exceed the costs. The construction of anew power plant or the upgrading of an old one involves a major financial investmentfor any energy company. Financial planning therefore starts long before ground isbroken, detailed design is begun, and orders are placed for equipment. Cost analysisand fiscal control activities continue throughout the construction project and theoperating life of the plant. This section briefly introduces fundamentals of engineeringeconomics, with a slant toward power plant cost analysis as well as issues ofmaintenance and equipment replacement.

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The cost to construct a power plant, waterworks, dam, bridge, factory, or othermajor engineering work is called its capital cost. It is common to discuss the capitalcost of building a power plant in terms of dollars per kilowatt of plant power output. A plant may cost $1100 per kilowatt of installed power generation capacity, for instance.

In addition to the cost of building the plant, there are many additional expendituresrequired to sustain its operation. These are called operating costs. They may beoccasional, or they may occur regularly and continue throughout the life of the plant.Often these costs are periodic, or are taken to be periodic for convenience of analysis.There are, for instance, annual fuel costs, salary expenses, and administrative andmaintenance costs that are not associated with the initial cost of the plant but are thecontinuing costs of generating and selling power. Operating costs are sometimesrelated to the amount of electrical energy sold. Usually they are expressed in cents perkilowatt-hour of energy distributed to customers.

Thus the expenses associated with power generation and other business endeavorsmay be thought of as two types: (1) initial costs usually associated with the purchase ofland, building site preparation, construction, and the purchase of plant equipment; and(2) recurring operating costs of a periodic or cyclic nature.

It is frequently desirable to express all costs on a common basis. The company andits investors may wish to know what annual sum of money is equivalent to both thecapital and operating costs. The company may, for example, borrow money to financethe capital cost of the plant and then pay the resulting debt over the expected useful lifeof the plant, say, 30 or 40 years. On the other hand, they may wish to know whatpresent sum would be required to ensure the payment of all future expenses of theenterprise.

It is clear that $100 in hand today is not the same as $100 in hand ten years fromnow. One difference is that money can earn interest. One hundred dollars investedtoday at 8% annual compound interest will become $215.89 in ten years. Clearly, animportant aspect of engineering economics is the time value of money.

Compound Interest

If Alice lends Betty $500, who agrees to pay $50 each year for five years for the use ofthe money, together with the original $500, then at the end of the fifth year, Alice willhave earned $250 in simple interest and receive a total of $750 in return. The annualinterest rate is

i = Annual interest / Capital = 50 / 500 = 0.1

or (0.1)(100) = 10% rate of return.If, however, Betty keeps the interest instead of paying it to Alice annually, and

eventually pays 10% on both the retained interest and the capital, the deal involvescompound interest. The total sum to be returned to Alice after 5 years is computed asfollows: At the end of the first year Alice has earned $50 in interest. The interest for the

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next year should be paid on the original sum and on the $50 interest earned in the firstyear, or $550. The interest on this sum for the second year is 0.1 × $550 = $55. Thefollowing table shows the calculation of the annual debt for the five-year loan of $500at 10% interest:

At the Endof:

The Accumulated Debt is: This Sum

First year $500 + 0.1 x $500 = $550.00

Second year $550 + 0.1 x $550 = $605.00

Third year $605 + 0.1 x $605 = $665.50

Fourth year $665.50 + 0.1 x $665.50 = $732.05

Fifth year $732.05 + 0.1 x $732.05 = $805.26

It is evident that the interest earned on the preceding interest accumulation causesthe annual indebtedness to grow at a increasing rate. It can be shown that the futuresum, S, is given by

S = P (1 + i)n

where P is the principal, the initial sum invested; i is the interest rate, and n is thenumber of investment periods, in this case the number of years. Here the factormultiplying the principal,

S / P = (1 + i)n

is called the compound amount factor, CAF. The difference between simple andcompound interest may not be spectacular for short investment periods but it is very impressive for long periods of time such as the operating life of a power plant. For ourexample, the CAF is (1 + 0.1)5 = 1.6105, and S = 500(1.6105) = $805.26. Now,consider the following closely related problem.

EXAMPLE 4.2

What sum is required now, at 8% interest compunded annually, to produce one milliondollars in 25 years?

SolutionThe future sum is

S = P (1 + i)n = 1,000,000 = P (1 + 0.08)25

Solving for P, the present sum is 1,000,000/(1.08)25 = $146,017.90. Thus, compoundinterest brings a return of almost over seven times the original investment here. The

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same present sum invested at 8% simple interest for twenty-five years would produce afuture sum of less than half a million dollars. _____________________________________________________________________

In the example, the inverse of the CAF was used to determine the present worth ofa future sum. The inverse of the CAF is called the present-worth factor, ( PWF):

PWF = P/ S = 1 / (1 + I)n

Thus we see that the time value of money is related to the compound interest thatcan be earned, and that taking compounding into account can be important. Torecklessly adapt an old adage, �A dollar in the hand is worth two (or more) in thefuture (if invested wisely).�

Capital Recovery

Another important aspect of compound interest is the relationship between apresent sum of money and a regular series of uniform payments. Consider a series offive annual payments of R dollars each, when the interest rate is i. What is the present dollar equivalent of these payments? Applying the CAF as in the preceding example,with R as the future sum, the present sum associated with the first payment is R/(1 + i).The present sum associated with the second payment is R/(1 + i)2. Thus the presentworth of the five payments is

P = R [ (1 + i) � 1 + (1 + i) � 2 + (1 + i) � 3 + (1 + i) � 4 + (1 + i) � 5 ]

It may be shown that this expression can be written as

P = R [(1 + i)5 � 1]/[i(1 + i)5].

The factor multiplying the annual sum R is called the series present-worth factor,SPWF, which for n years is:

SPWF = P/ R = [(1 + i)n � 1]/[i(1 + i)n] (4.1)

Solving for R, we obtain an expression for the regular annual payment for n yearsneeded to fund a present expenditure of P dollars at an interest rate i. The resultingfactor is called the capital recovery factor, CRF, which is the reciprocal of the seriespresent worth factor:

CRF = R / P = i(1 + i) n / [(1 + i) n � 1] (4.2)

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EXAMPLE 4.3

What uniform annual payments are required for forty years at 12% interest to retire thedebt associated with the purchase of a $500,000,000 power plant?

SolutionUsing equation (4.2), we get

R = Pi (1 + i)/[(1 + i) n � 1] = 5×108(0.12)(1 + 0.12) 40/(1.1240 � 1) = $60,651,813

This sum may be regarded as part of the annual operating expense of the plant. It mustbe recovered annually by the returns from the sale of power. _____________________________________________________________________

4.6 A Preliminary Design Analysis of a 500-MW Plant

Consider the design of a 500-megawatt steam power plant with a heat rate of10,000 Btu/kW-hr and a water-cooled condenser with a 20°F cooling-watertemperature rise produced by heat transfer from the condensing steam. The plant usescoal with a heating value of 10,000 Btu/lbm. Let us estimate the magnitude of some ofthe parameters that characterize the design of the plant. The reader should verifycarefully each of the following calculations.

A 500-megawatt plant operating at full load produces 500,000 kW and an annualelectrical energy generation of

500,000 � 365 � 24 = 4.38 × 109 kW-hr

With a heat rate of 10,000 Btu/kW-hr, this requires a heat addition rate of

500,000 � 10,000 = 5 × 109 Btu/hr

Coal with an assumed heating value of 10,000 Btu/lbm must therefore be supplied at arate of 5 × 109 / 104 = 500,000 lbm/hr or 500,000 / 2000 = 250 tons/hr. A dedicated coal car carries about 100 tons. Hence the plant requires 250 /100 = 2.5 cars per hourof continuous operation. A coal unit train typically has about 100 cars. Then the plantneeds 2.5 � 24 / 100 = 0.6 unit trains per day, or a unit train roughly every two days.

If coal costs $30 per ton, the annual cost of fuel will be

30 � 250 � 24 � 365 = $65,700,000

The cost of fuel alone per kW-hr, based on 100% annual plant capacity, will be

65,700,000/(500,000 � 365 � 24) = $0.015/kW-hr � 1.5 cents/kW-hr

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The annual plant factor, or annual capacity factor, expressed as a decimal fraction , isthe ratio of the actual annual generation to the annual generation at 100 % capacity.

If the coal has 10% ash, the plant will produce 250 � 0.1 = 25 tons of ash per hour.Under some circumstances the ash may be used in the production of cement or otherpaving materials. If it is not marketable, it is stabilized and stored in nearby ash pondsuntil it can be moved to a permanent disposal site.

Similarly, if 2% of the coal is sulfur and half of it is removed from the combustionproducts, 2.5 tons per hour is produced for disposal. If the sulfur is of sufficient purity,it may be sold as an industrial chemical.

With an air-fuel ratio of 14, an air flow rate of 14 � 500,000 = 7,000,000 lbm/hr isrequired for combustion. This information is important in determining the size of theinduced- and forced-draft fans, that of their driving motors or turbines, and of theplant�s gas path flow passages.

The heat rate of 10,000 Btu/kW-hr corresponds to a thermal efficiency of3413/10,000 = 0.3413 or 34.13%. If we approximate the heat of vaporization of wateras 1000 Btu/lbm, the throttle steam flow rate, with no superheat, would be about

10,000 � 500,000 / 1000 = 5,000,000 lbm/hr

This determines the required capacity of the feedwater pumps and is important in sizingthe passages for the water path. The above thermal efficiency implies that about 65% ofthe energy of the fuel is rejected into the environment, mostly through the condenserand the exiting stack-gas energy. As an upper limit, assume that all of the heat isrejected in the condenser. Thus

(1 � 0.3413)(5 × 109) = 3.29 × 109 Btu/hr

must be rejected to condenser cooling water. With 20° water temperature rise in thecondenser, this rate of cooling requires a cooling-water flow rate to the condenser of

3.29 × 109/(1.0 × 20) = 1.65 × 108 lbm / hr

assuming a water heat capacity of 1.0 Btu/lbm-R. This gives information relevant to thedesign sizing of cooling-water lines, cooling towers, and water pump capacities.

These back-of-the-envelope calculations should not be regarded as precise, butthey are reasonable estimates of the magnitudes of important power plant parameters.Such estimates are useful in establishing a conceptual framework of the relationshipsamong design factors and of the magnitude of the design problem.

EXAMPLE 4.4

Relating to the above rough design of a 500-MW plant, and assuming the capital costinformation of Example 4.3, determine the capital cost per kW of generation capacityand estimate the minimum cost of generation for the plant if it is predicted to have an

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annual plant factor of 80% and maintenance and administrative costs of $0.007 /kW-hr.

SolutionThe unit cost of the power plant is

$500,000,000/(500 �1000) = $1000 per kW-hr

of capacity. The capital cost part of the annual cost of power generation is

(60,651,813 �100)/(365 � 24 � 0.8 �500,000) = 1.73 cents per kW-hr

The cost of coal was determned to be 1.5 cents/kW-hr. The minimum cost ofproducing electricity is then

1.73 + 1.5 + 0.7 = 3.93 cents per kW-hr_____________________________________________________________________

Bibliography and References

1. Singer, Joseph G., (Ed.), Combustion /Fossil Power Systems. Windsor, Conn.:Combustion Engineering, Inc.,, 1981.

2. Anon., Steam, It�s Generation and Use. New York: Babcock and Wilcox, 1978.

3. Hensley, John C., Cooling Tower Fundamentals. Mission, Kan.: Marley CoolingTower Co., 1985.

4. Li, Kam W., and Priddy, A. Paul, Power Plant Systems Design. New York: Wiley,1985.

EXERCISES

4.1 Derive an equation for the sum, S, resulting from P dollars invested at simpleinterest rate i for a period of n years.

4.2 For the power plant design discussed in Section 4.6, estimate the horsepower of amotor required to drive the fans used to overcome a steam generator gas-path pressuredrop of 1 psia. Assume a fan efficiency of 80%. What is the fractional and percentagereduction in power plant output due to the fans?

4.3 Estimate the horsepower required by the feedwater pumps in the Section 4.6design if the HP-turbine throttle pressure is 3200 psia. Assume a pump efficiency of70%. What fractional and percentage reduction of the power plant output does thisrepresent?

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4.4 What are the annual savings in fuel costs in the Section 4.6 plant design if the plantheat rate can be reduced to 8500 Btu/kW-hr?

4.5 If the total capital cost of the Section 4.6 plant design is $600,000,000 and theannual administrative and maintenance costs are one cent per kW-hr, what is the minimum cost of electricity per kW-hr, assuming an annual interest rate of 9% and anexpected plant lifetime of thirty-five years?

4.6 What is the present worth of a sequence of five annual payments of $4500, $6500,$3500, $7000, and $10,000 at an annual interest rate of 8%?

4.7 You have collected the following data on 1.5-MW steam turbines, as alternatives tothe purchase of utility power, for a new process plant to operate at 60% plant factor:

Turbine Number

1 2 3

Heat rate, Btu/kW-hr 12,000 10,600 9,500

Installed cost $124,000 $190,000 $245,000

Estimated annual maintenance cost

$2,000 $1,800 $2,550

Coal (14,000 Btu/lbm) is the fuel to be used, at a cost of $26 per ton. Assuming anannual interest rate of 8%, compare the annual cost of of the turbines for thirty-yearturbine lifetimes. Which turbine would you select? What other factors would youconsider before making a decision?

4.8 Using the data of Exercise 4.7, compare the turbines on the basis of present worthof all costs.

4.9 For the power plant design discussed in Section 4.6, estimate the motor horse-power required to drive a 75% efficient fan that is used to overcome a steam generatorgas-path pressure drop of 50 kPa.

4.10 Estimate the total power required by 65% efficient feedwater pumps operating inparallel in the Section 4.6 design for a throttle pressure of 20 MPa.

4.11 Work out a back-of-the-envelope analysis similar to that of Section 4.6 in SI units.

4.12* Develop an interactive computer program that implements a steam power plantsystem analysis of the type presented in Section 4.6 at one of the following levels, to be

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assigned by your instructor.Level 1: User supplies parameters in response to screen prompts, and one or moreoutput screens display the resulting input and output parameters.Level 2: Same as Level 1, but also provide a capability for the user to change thedesign by varying one input while holding all others constant.Level 3: Allow the user to select a dependent variable from a list of outputs, and a

parameter to be varied and its range from another list. Display a graph of the variationof user-selected outputs over the range of the parameter.

4.13* Construct a spreadsheet that systematizes the computations for a steam powerplant along the lines presented in this chapter. Set up a version of the spreadsheet thatallows easy variation of input parameters. Use the spreadsheet to develop graphs thatshow the influence of plant heat rate on fuel costs and sulfur byproduct production.

4.14 An electric utility, expecting to increase its system capacity by 400 megawatts,must choose between a high-technology combined-cycle plant, at a cost of $1800 perkW of installed capacity, and an oil-burning steam plant, at $1150 per kW. Thecombined-cycle plant has a variable cost of 18 mills per kW-hr, while the oil-burningplant variable cost is 39 mills per kW-hr. For an annual plant factor of 0.6 and fixedcharges of 15% of the capital cost, determine (a) the total annual cost of operation ofeach plant, and (b) the cost of electricity, in cents per kW-hr, for each plant.

4.15 Estimate the mass-flow rate of makeup water required by an evaporative coolingtower satisfying the cooling requirements of the example power plant of Section 4.6.

4.16 An 800-MW steam power plant operates at a heat rate of 8700 Btu/kW-hr. It hasa 16°F rise in condenser cooling-water temperature. Neglecting energy losses, estimatethe condenser cooling-water flow rate and the flow-rate of cooling-tower makeupwater. Estimate the amount of pump power required to circulate the cooling water tothe cooling tower.

4.17 Plot a curve of condenser cooling-water flow rate and makeup-water flow rate asa function of condenser cooling-water temperature rise for Riverside Station Unit #1.

4.18 The average temperature in an 800 ft. high power plant exhaust gas stack is 350°Fand the ambient temperature is 60°F. Neglecting fluid friction and exhaust gas kineticenergy, estimate the pressure inside the base of the stack.

4.19 The average temperature in a 200 meter power plant exhaust gas stack is 150°C,and the ambient temperature is 20°C. Neglecting fluid friction and exhaust gasmomentum, estimate the pressure, in kPa, at the inside of the base of the stack.

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4.20 Estimate the heat transfer rates in the Riverside Station Unit #1 in the air pre-heater, an economizer that heats liquid water to saturation, the boiling surfaces, the re-heater, and the superheater. Estimate the temperature drops in the combustion gasacross each of these, assuming that they are arranged in the same order as just listed.

4.21 After completing Exercise 4.20, estimate the flow area of combustion gas througha crossflow economizer in the Riverside Station Unit #1, and define a suitable design.

4.22 After completing Exercise 4.20, estimate the flow area of combustion gas througha pendant superheater consisting of parallel U-tubes in cross flow, and define a suitabledesign.

4.23 The following is a list of ten air and gas path components of a steam power plant. Number the components so as to put them in order, starting with the air into the plantas number 1 and concluding with the flue gas out as number 10.

_________ superheater __________ boiler tubewall

_________ economizer __________ windbox

_________ induced-draft fan __________ air heater, gas side

_________ burner __________ electrostatic precipitator

_________ forced-draft fan __________ air heater, air side

4.24 Upon hiring on with Hot Stuff Engineering Company after graduation, youpurchase a $30,000 automobile to establish an image as a prosperous engineer. Youpay no money down, but 1% interest per month, compounded monthly, for four years. What are your monthly payments? What will your payments be if you are payingsimple interest?

4.25 A forced-draft fan with an efficiency of 70% supplies 1,000,000 ft3 per minute ofair to a furnace that produces a pressure drop of 0.7 psia. What is the fan powerrequirement, in horsepower and in kilowatts?

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C H A P T E R F I V E

GAS TURBINES AND JET ENGINES

5.1 Introduction

History records over a century and a half of interest in and work on the gas turbine.However, the history of the gas turbine as a viable energy conversion device beganwith Frank Whittle's patent award on the jet engine in 1930 and his static test of a jetengine in 1937. Shortly thereafter, in 1939, Hans von Ohain led a Germandemonstration of jet-engine-powered flight, and the Brown Boveri company intro-duced a 4-MW gas-turbine-driven electrical power system in Neuchatel, Switzerland.The success of the gas turbine in replacing the reciprocating engine as a power plant forhigh-speed aircraft is well known. The development of the gas turbine was less rapid asa stationary power plant in competition with steam for the generation of electricity andwith the spark-ignition and diesel engines in transportation and stationary applications.Nevertheless, applications of gas turbines are now growing at a rapid pace as researchand development produces performance and reliability increases and economic benefits.

5.2 An Ideal Simple-Cycle Gas Turbine

The fundamental thermodynamic cycle on which gas turbine engines are based is calledthe Brayton Cycle or Joule cycle. A temperature-entropy diagram for this ideal cycleand its implementation as a closed-cycle gas turbine is shown in Figure 5.1. The cycleconsists of an isentropic compression of the gas from state 1 to state 2; a constantpressure heat addition to state 3; an isentropic expansion to state 4, in which work isdone; and an isobaric closure of the cycle back to state 1.

As Figure 5.1(a) shows, a compressor is connected to a turbine by a rotating shaft.The shaft transmits the power necessary to drive the compressor and delivers thebalance to a power-utilizing load, such as an electrical generator. The turbine is similarin concept and in many features to the steam turbines discusssed earlier, except that it isdesigned to extract power from a flowing hot gas rather than from water vapor. It isimportant to recognize at the outset that the term "gas turbine" has a dual usage: Itdesignates both the entire engine and the device that drives the compressor and theload. It should be clear from the context which meaning is intended. The equivalentterm �combustion turbine� is also used occasionally, with the same ambiguity.

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Like the simple Rankine-cycle power plant, the gas turbine may be thought of as adevice that operates between two pressure levels, as shown in Figure 5.1(b). Thecompressor raises the pressure and temperature of the incoming gas to the levels of p2and T2. Expansion of the gas through the turbine back to the lower pressure at thispoint would be useless, because all the work produced in the expansion would berequired to drive the compressor.

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Instead, it is necessary to add heat and thus raise the temperature of the gas beforeexpanding it in the turbine. This is achieved in the heater by heat transfer from anexternal source that raises the gas temperature to T3, the turbine inlet temperature.Expansion of the hot gas through the turbine then delivers work in excess of thatneeded to drive the compressor. The turbine work exceeds the compressor requirementbecause the enthalpy differences, and hence the temperature differences, alongisentropes connecting lines of constant pressure increase with increasing entropy (andtemperature), as the figure suggests.

The difference between the turbine work, Wt, and the magnitude of the compressorwork, |Wc|, is the net work of the cycle. The net work delivered at the output shaft maybe used to drive an electric generator, to power a process compressor, turn an airplanepropeller, or to provide mechanical power for some other useful activity.

In the closed-cycle gas turbine, the heater is a furnace in which combustion gasesor a nuclear source transfer heat to the working fluid through thermally conducting tubes. It is sometimes useful to distinguish between internal and external combustionengines by whether combustion occurs in the working fluid or in an area separate fromthe working fluid, but in thermal contact with it. The combustion-heated, closed-cyclegas turbine is an example, like the steam power plant, of an external combustionengine. This is in contrast to internal combustion engines, such as automotive engines,in which combustion takes place in the working fluid confined between a cylinder and apiston, and in open-cycle gas turbines.

5.3 Analysis of the Ideal Cycle

The Air Standard cycle analysis is used here to review analytical techniques and toprovide quantitative insights into the performance of an ideal-cycle engine. AirStandard cycle analysis treats the working fluid as a calorically perfect gas, that is, aperfect gas with constant specific heats evaluated at room temperature. In Air Standardcycle analysis the heat capacities used are those for air.

A gas turbine cycle is usually defined in terms of the compressor inlet pressure andtemperature, p1 and T1, the compressor pressure ratio, r = p2/p1, and the turbine inlettemperature, T3, where the subscripts correspond to states identified in Figure 5.1. Starting with the compressor, its exit pressure is determined as the product of p1 andthe compressor pressure ratio. The compressor exit temperature may then be deter-mined by the familiar relation for an isentropic process in an ideal gas, Equation (1.19):

T2 = T1( p2 /p1)(k�1)/k [R | K] (5.1)

For the two isobaric processes, p2 = p3 and p4 = p1. Thus the turbine pressure ratio,p3/p4, is equal to the compressor pressure ratio, r = p2 /p1. With the turbine inlettemperature T3 known, the turbine discharge temperature can be determined from

T4 = T3/( p2/p1)(k�1)/k [R | K] (5.2)

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and the temperatures and pressures are then known at all the significant states.Next, taking a control volume around the compressor, we determine the shaft

work required by the compressor, wc, assuming negligible heat losses, by applying thesteady-flow energy equation:

0 = h2 � h1 + wc

or

wc = h1 � h2 = cp( T1 � T2) [Btu/lbm | kJ/kg] (5.3)

Similarly, for the turbine, the turbine work produced is

wt = h3 - h4 = cp ( T3 � T4) [Btu/lbm | kJ/kg] (5.4) The net work is then

wn = wt + wc = cp ( T3 � T4 + T1 � T2) [Btu/lbm | kJ/kg] (5.5)

Now taking the control volume about the heater, we find that the heat addition perunit mass is

qa = h3 � h2 = cp ( T3 � T2) [Btu/lbm | kJ/kg] (5.6)

The cycle thermal efficiency is the ratio of the net work to the heat supplied to theheater:

�th = wn /qa [dl] (5.7)

which by substitution of Equations (5.1), (5.2), (5.5), and (5.6) may be simplified to

�th = 1 � ( p2/p1) � (k�1)/k [dl] (5.8)

It is evident from Equation (5.8) that increasing the compressor pressure ratio increasesthermal efficiency.

Another parameter of great importance to the gas turbine is the work ratio, wt /|wc|.This parameter should be as large as possible, because a large amount of the powerdelivered by the turbine is required to drive the compressor, and because the engine network depends on the excess of the turbine work over the compressor work. A littlealgebra will show that the work ratio wt /|wc| can be written as:

wt /|wc| = (T3 /T1) / ( p2 /p1) (k�1)/k [dl] (5.9)

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Note that, for the ideal cycle, the thermal efficiency and the work ratio depend on onlytwo independent parameters, the compressor pressure ratio and the ratio of the turbineand compressor inlet temperatures. It will be seen that these two design parameters areof utmost importance for all gas turbine engines.

Equation (5.9) shows that the work ratio increases in direct proportion to the ratioT3 /T1 and inversely with a power of the pressure ratio. On the other hand, Equation(5.8) shows that thermal efficiency increases with increased pressure ratio. Thus, thedesirability of high turbine inlet temperature and the necessity of a tradeoff involvingpressure ratio is clear. Equation (5.9) also suggests that increases in the ratio T3 /T1allow the compressor pressure ratio to be increased without reducing the work ratio.This is indicative of the historic trend by which advances in materials allow higherturbine inlet temperatures and therefore higher compressor pressure ratios.

It was shown in Chapter 1 that the area of a reversible cycle plotted on a T�sdiagram gives the net work of the cycle. With this in mind, it is interesting to consider afamily of cycles in which the compressor inlet state, a, and turbine inlet temperaturesare fixed, as shown in Figure 5.2. As the compressor pressure ratio pb/pa approaches 1,the cycle area and hence the net work approach 0, as suggested by the shaded cyclelabeled with single primes. At the other extreme, as the compressor pressure ratioapproaches its maximum value, the net work also approaches 0, as in the cycle denotedby double primes. For intermediate pressure ratios, the net work is large and positive,indicating that there is a unique value of compressor pressure ratio that maximizes thenet work. Such information is of great significance in gas turbine design,

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because it indicates the pressure ratio that yields the highest power output for giventurbomachine inlet temperatures and mass flow rate. This is an important approach tothe pressure ratio tradeoff mentioned earlier. It will be considered from an analyticviewpoint for a more realistic gas turbine model in a later section.

Up to this point the discussion has focused on the closed-cycle gas turbine, anexternal combustion or nuclear-heated machine that operates with a circulating fixedmass of working fluid in a true cyclic process. In fact, the same Air Standard cycleanalysis may be applied to the open-cycle gas turbine. The open cycle operates withatmospheric air that is pressurized by the compressor and then flows into a combustionchamber, where it oxidizes a hydrocarbon fuel to produce a hot gas that drives theturbine. The turbine delivers work as in the closed cycle, but the exiting combustiongases pass into the atmosphere, as they must in all combustion processes.

A diagram of the cycle implementation is shown in Figure 5.3. Clearly, the open-cycle gas turbine is an internal combustion engine, like the automotive engine. Notethat the diagram is consistent with Figure 5.1 and all the preceding equations in thischapter. This is true because (1) the atmosphere serves as an almost infinite source andsink that may be thought of as closing the cycle, and (2) the energy released bycombustion has the same effect as the addition of external heat in raising thetemperature of the gas to the turbine inlet temperature. A cutaway of an open-cycleutility gas turbine is presented in Figure 5.4.

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5.4 Realistic Simple-Cycle Gas Turbine Analysis

The preceding analysis of the Air Standard cycle assumes perfect turbomachinery, anunachievable but meaningful ideal, and room-temperature heat capacities. Realisticquantitative performance information can be obtained by taking into accountefficiencies of the compressor and the turbine, significant pressure losses, and morerealistic thermal properties.

Properties for Gas Turbine Analysis

It is pointed out in reference 1 that accurate gas turbine analyses may be performedusing constant heat capacities for both air and combustion gases. This appears to be aspecialization of a method devised by Whittle (ref. 4). The following properties aretherefore adopted for all gas turbine analyses in this book: Air:

cp = 0.24 Btu/lbm-R or 1.004 kJ /kg-K

k = 1.4 implies k /(k � 1) = 3.5

Combustion gas:cp,g = 0.2744 Btu/lbm-R or 1.148 kJ /kg-K

kg = 1.333 implies kg/(kg � 1) = 4.0

The properties labeled as combustion gas above are actually high-temperature-airproperties. Because of the high air-fuel ratio required by gas turbines, the

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thermodynamic properties of gas turbine combustion gases usually differ little fromthose of high-temperature air. Thus the results given below apply equally well toclosed-cycle machines using air as the working fluid and to open-cycle engines.

Analysis of the Open Simple-Cycle Gas Turbine

A simple-cycle gas turbine has one turbine driving one compressor and apower-consuming load. More complex configurations are discussed later. It is assumedthat the compressor inlet state, the compressor pressure ratio, and the turbine inlettemperature are known, as before. The turbine inlet temperature is usually determinedby the limitations of the high-temperature turbine blade material. Special metals orceramics are usually selected for their ability to withstand both high stress at elevatedtemperature and erosion and corrosion caused by undesirable components of the fuel.

As shown in Figure 5.3, air enters the compressor at a state defined by T1 and p1.The compressor exit pressure, p2, is given by

p2 = rp1 [lbf /ft2 | kPa] (5.10)

where r is the compressor pressure ratio. The ideal compressor discharge temperature,T2s is given by the isentropic relation

T2s = T1 r (k�1)/k [R | K] (5.11)

The compressor isentropic efficiency, defined as the ratio of the compressor isentropicwork to the actual compressor work with both starting at the same initial state andending at the same pressure level, may be written as

�c = ( h1 � h2s )/( h1 � h2 ) = ( T1 � T2s )/( T1 � T2 ) [dl] (5.12)

Here the steady-flow energy equation has been applied to obtain expressions for thework for an irreversible adiabatic compressor in the denominator and for an isentropiccompressor in the numerator. Solving Equation (5.12) for T2, we get as the actualcompressor discharge temperature:

T2 = T1 + ( T2s � T1 ) / � c [R | K] (5.13)

Equation (5.3) then gives the work needed by the compressor, wc:

wc = cp ( T1 � T2 ) = cp ( T1 � T2s )/� c [Btu /lbm | kJ/kg] (5.14)

Note that the compressor work is negative, as required by the sign convention thatdefines work as positive if it is produced by the control volume. The compressor powerrequirement is, of course, then given by mawc [Btu/hr | kW], where ma is the

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compressor mass flow rate [lbm / hr | kg / s].After leaving the compressor at an elevated pressure and temperature, the air then

enters the combustion chamber, where it completely oxidizes a liquid or a gaseous fuelinjected under pressure. The combustion process raises the combustion gas temperatureto the turbine inlet temperature T3. One of the goals of combustion chamber design isto minimize the pressure loss from the compressor to the turbine. Ideally, then, p3 = p2,as assumed by the Air Standard analysis. More realistically, a fixed value of thecombustor fractional pressure loss, fpl, (perhaps about 0.05 or 5%) may be used toaccount for burner losses:

fpl = (p2 � p3)/p2 [dl] (5.15)

Then the turbine inlet pressure may be determined from

p3 = (1 � fpl) p2 [lbf /ft2 | kPa] (5.16) Rather than deal with its complexities, we may view the combustion process simply asone in which heat released by exothermic chemical reaction raises the temperature ofcombustion gas (with hot-air properties) to the turbine inlet temperature. The rate ofheat released by the combustion process may then be expressed as:

Qa = ma(1 + f )cp,g(T3 � T2) [Btu/hr | kW] (5.17)

where f is the mass fuel-air ratio. The term ma(1 + f) is seen to be the sum of the air andfuel mass flow rates, which also equals the mass flow rate of combustion gas. For gasturbines it will be seen later that f is usually much less than the stoichiometric fuel-airratio and is often neglected with respect to 1 in preliminary analyses.

The turbine in the open-cycle engine operates between the pressure p3 andatmospheric pressure, p4 = p1, with an inlet temperature of T3. If the turbine wereisentropic, the discharge temperature would be

T4s = T3( p4 /p3 ) (kg� 1) / kg [R | K] (5.18)

From the steady-flow energy equation, the turbine work can be written as

wt = cp,g (T3 � T4 ) = �t cp,g( T3 � T4s) [Btu/lbm | kJ/kg] (5.19)

referenced to unit mass of combustion gas, and where �t is the turbine isentropicefficiency. The turbine power output is then ma(1 + f)wt, where, as seen earlier, ma(1 +f) is the mass flow rate of combustion gas flowing through the turbine. The net workbased on the mass of air processed and the net power output of the gas turbine, Pn, arethen given by

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wn = (1 + f )wt + wc [Btu/lbm air| kJ/kg air] (5.20)

and

Pn = ma [(1 + f )wt + wc ] [Btu/hr | kW] (5.21)

and the thermal efficiency of the engine is

�th = Pn /Qa [dl] (5.22)

EXAMPLE 5.1

A simple-cycle gas turbine has 86% and 89% compressor and turbine efficiencies,respectively, a compressor pressure ratio of 6, a 4% fractional pressure drop in thecombustor, and a turbine inlet temperature of 1400°F. Ambient conditions are 60°F andone atmosphere. Determine the net work, thermal efficiency, and work ratio for theengine. Assume that the fuel-mass flow rate is negligible compared with the air flowrate.

SolutionThe notation for the solution is that of Figure 5.3. The solution details are given in

Table 5.1 in a step-by-step spreadsheet format. Each line presents the parameter name, symbol, and units of measure; its value; and the right-hand side of its specificdetermining equation.

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When an entire cycle is to be analyzed, it is best to start at the compressor with theinlet conditions and proceed to calculate successive data in the clockwise direction onthe T-s diagram. The compressor isentropic and actual discharge temperatures andwork are determined first using Equations (5.11), (5.13), and (5.14). The turbinepressure ratio is determined next, accounting for the combustor pressure loss, usingEquation (5.16). The isentropic relation, Equation (5.18), gives the isentropic turbineexit temperature, and the turbine efficiency and Equation (5.19) yields the true turbineexit temperature and work. Once all the turbomachine inlet and exit temperatures areknown, other cycle parameters are easily determined, such as the combustor heattransfer, net work, thermal efficiency, and work ratio._____________________________________________________________________

An important observation may be made on the basis of this analysis regarding the magnitude of the compressor work with respect to the turbine work. Much of theturbine work is required to drive the compressor. Compare the work ratio of 1.66, forexample, with the much higher values for the steam cycles of Chapter 2 (the Rankine-cycle pumps have the same function there as the compressor here). Example 2.4 for theRankine cycle with a 90% turbine efficiency has a work ratio of 77.2. Thus the gasturbine�s pressurization handicap relative to the Rankine cycle is substantial.

The unimpressive value of the thermal efficiency of the example gas turbine, 25%(not typical of the current state of the art) compares with a Carnot efficiency for thesame cycle temperature extremes of 72% The large amount of compressor work requi-red clearly contributes to this weak performance. Nevertheless, current gas turbines arecompetitive with many other engines on an efficiency basis, and have advantages suchas compactness and quick-start capability relative to Rankine cycle power plants. Oneapproach to the improvement of thermal efficiency of the gas turbine will be addressedlater in Section 5.5. First let�s look at what can be done about gas turbine work.

Maximizing the Net Work of the Cycle

Using Equations (5.14) and (5.19), we can rewrite the cycle net work as

wn = �t cp,g ( T3 � T4s) � cp( T2s � T1 )/�c [Btu/lbm | kJ/kg] (5.23)

where the fuel-air ratio has been neglected with respect to 1. In the following, thecombustor pressure losses and the distinction between hot-gas and air heat capacities will be neglected but the very important turbomachine efficiencies are retained.Nondimensionalizing the net work with the constant cpT1 we get:

wn/cpT1 = �t (T3 /T1)(cp,g/cp)(1 � r � (k�1)/k) + (1 � r (k�1)/k)/�c [dl] (5.24)

By differentiating wn with respect to the compressor pressure ratio r and setting the

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result equal to 0, we obtain an equation for r*, the value of r that maximizes the network with fixed turbomachine efficiencies and with a constant ratio of the temperaturesof the turbomachine inlets, T3 /T1. For constant gas properties throughout, the result is

r* = (�c �t T3/T1)k/2(k�1) [dl] (5.25)

This relation gives a specific value for the compressor pressure ratio that defines anoptimum cycle, in the sense of the discussion of Figure 5.2. There it was establishedqualitatively that a cycle with maximum net work exists for a given value of T3/T1.Equation (5.25) defines the condition for this maximum and generalizes it to includeturbomachine inefficiency.

The pressure ratio r* given by Equation (5.25) increases with increasingturbomachine efficiencies and with T3/T1. This is a clear indicator that increasingturbine inlet temperature favors designs with higher compressor pressure ratios. Thisinformation is important to the gas turbine designer but does not tell the whole designstory. There are other important considerations; for example, (1) compressors andturbines become more expensive with increasing pressure ratios, and (2) the pressureratio that maximizes thermal efficiency is different from that given by Equation (5.25).Figure 5.5 shows the influence of compressor pressure ratio on both efficiency and network and the position of the value given by Equation (5.25). Thus, when all factors aretaken into account, the final design pressure ratio is likely to be in the vicinity of, butnot necessarily identical to, r*.

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5.5 Regenerative Gas Turbines

It was shown in Chapter 2 that the efficiency of the Rankine cycle could be improvedby an internal transfer of heat that reduces the magnitude of external heat addition, afeature known as regeneration. It was also seen in Chapter 2 that this is accomplishedconveniently in a steam power plant by using a heat exchanger known as a feedwaterheater.

Examination of Example 5.1 shows that a similar opportunity exists for the gasturbine cycle. The results show that the combustion process heats the incoming air from924°R to 1860°R and that the gas turbine exhausts to the atmosphere at 1273°R. Thusa maximum temperature potential of 1273 � 924 = 349°F exists for heat transfer. As inthe Rankine cycle, this potential for regeneration can be exploited by incorporation of aheat exchanger. Figure 5.6 shows a gas turbine with a counterflow heat exchanger thatextracts heat from the turbine exhaust gas to preheat the compressor discharge air to Tcahead of the combustor. As a result, the temperature rise in the combustor is reduced toT3 � Tc, a reduction reflected in a direct decrease in fuel consumed.

Note that the compressor and turbine inlet and exit states can be the same as for asimple cycle. In this case the compressor, turbine, and net work as well as the workratio are unchanged by incorporating a heat exchanger.

The effectiveness of the heat exchanger, or regenerator, is a measure of how wellit uses the available temperature potential to raise the temperature of the compressordischarge air. Specifically, it is the actual rate of heat transferred to the air divided bythe maximum possible heat transfer rate that would exist if the heat exchanger hadinfinite heat transfer surface area. The actual heat transfer rate to the air is mcp(Tc � T2),and the maximum possible rate is mcp(T4 � T2). Thus the regenerator effectiveness canbe written as

�reg = ( Tc � T2 )/( T4 � T2 ) [dl] (5.26)

and the combustor inlet temperature can be written as

Tc = T2 + �reg( T4 � T2 ) [R | K] (5.27)

It is seen that the combustor inlet temperature varies from T2 to T4 as theregenerator effectiveness varies from 0 to 1. The regenerator effectiveness increases asits heat transfer area increases. Increased heat transfer area allows the cold fluid toabsorb more heat from the hot fluid and therefore leave the exchanger with a higher Tc.

On the other hand, increased heat transfer area implies increased pressure losseson both air and gas sides of the heat exchanger, which in turn reduces the turbinepressure ratio and therefore the turbine work. Thus, increased regenerator effectivenessimplies a tradeoff, not only with pressure losses but with increased heat exchanger sizeand complexity and, therefore, increased cost.

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The exhaust gas temperature at the exit of the heat exchanger may be determinedby applying the steady-flow energy equation to the regenerator. Assuming that the heatexchanger is adiabatic and that the mass flow of fuel is negligible compared with the airflow, and noting that no shaft work is involved, we may write the steady-flow energyequation for two inlets and two exits as

q = 0 = he + hc � h2 � h4 + w = cp,gTe + cpTc � cpT2 � cp,gT4 + 0

Thus the regenerator combustion-gas-side exit temperature is:

Te = T4 � (cp/cp,g)( Tc � T2 ) [R | K] (5.28)

While the regenerator effectiveness does not appear explicitly in Equation (5.28),the engine exhaust temperature is reduced in proportion to the air temperature rise inthe regenerator, which is in turn proportional to the effectiveness. The dependence ofthe exhaust temperature on �reg may be seen directly by eliminating Tc from Equation(5.28), using Equation (5.27) to obtain

T4 � Te = �reg (cp/cp,g)(T4 � T2) [R | K] (5.29)

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The regenerator exhaust gas temperature reduction, T4 � Te, is seen to be jointlyproportional to the effectiveness and to the maximum temperature potential, T4 � T2.

The regenerator, like other heat exchangers, is designed to have minimal pressurelosses on both air and gas sides. These may be taken into account by the fractionalpressure drop approach discussed in connection with the combustor. EXAMPLE 5.2

Let�s say we are adding a heat exchanger with an effectiveness of 75% to the enginestudied in Example 5.1. Assume that the same frictional pressure loss factor applies toboth the heat exchanger air-side and combustor as a unit, and that gas-side pressureloss in the heat exchanger is negligible. Evaluate the performance of the modifiedengine.

SolutionThe solution in spreadsheet format, expressed in terms of the notation of Figures

5.6 and 5.7, is shown in Table 5.2. Examination of the spreadsheet and of the T-sdiagram in Figure 5.7 shows that the entry and exit states of the turbomachines are notinfluenced by the addition of the heat exchanger, as expected. (There would have beena slight influence if a different pressure loss model had been assumed.)

With the heat exchanger, it is seen that the combustor inlet temperature hasincreased about 262° and the exhaust temperature reduced 229°. The net work and

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work ratio are clearly unchanged. Most importantly, however, the thermal efficiencyhas increased 10 percentage counts over the simple cycle case in Example 5.1. Such again must be traded off against the added volume, weight, and expense of theregenerator. The efficiency gain and the associated penalties may be acceptable instationary power and ground and marine transportation applications, but are seldomfeasible in aerospace applications. Each case, of course, must be judged on its ownmerits. _____________________________________________________________________

Figure 5.8 shows the influence of regenerator effectiveness and turbine inlettemperature on the performance of the gas turbine, all other conditions being the sameas in the example. The values for �reg = 0 correspond to a gas turbine withoutregenerator. The abscissa is arbitrarily truncated at �reg = 0.8 because gas turbine heatexchanger effectivenesses usually do not exceed that value. The impressive influence ofboth design parameters is a strong motivator for research in heat exchangers and high-temperature materials. The use of regeneration in automotive gas turbines is virtuallymandated because good fuel economy is so important.

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Figure 5.9 shows the layout of a regenerative gas turbine serving a pipelinecompressor station. Gas drawn from the pipeline may be used to provide the fuel forremotely located gas-turbine-powered compressor stations. (A later figure, Figure 5.12,shows details of the turbomachinery of this gas turbine.)

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5.6 Two-Shaft Gas Turbines

Problems in the design of turbomachinery for gas turbines and in poor part-load oroff-design performance are sometimes avoided by employing a two-shaft gas turbine, inwhich the compressor is driven by one turbine and the load by a second turbine. Bothshafts may be contained in a single structure, or the turbines may be separatelypackaged. Figure 5.10 shows the flow and T-s diagrams for such a configuration. Theturbine that drives the compressor is called the compressor turbine. The compressor,combustor, compressor-turbine combination is called the gas generator, or gasifier, because its function is to provide hot, high-pressure gas to drive the second turbine, thepower turbine. The compressor-turbine is sometimes also referred to as the gasifierturbine or gas-generator turbine.

The analysis of the two-shaft gas turbine is similar to that of the single shaftmachine, except in the determination of the turbine pressure ratios. The pressure riseproduced by the compressor must be shared between the two turbines. The manner inwhich it is shared is determined by a power, or work, condition. The work conditionexpresses mathematically the fact that the work produced by the gasifier turbine is usedto drive the compressor alone. As a result, the gas generator turbine pressure ratio,p3/p4, is just high enough to satisfy the compressor work requirement.

Thus the compressor power (work) input is the same as the delivered gas-generator turbine power(work) output:

|wc| = �mech (1 + f )wt [Btu /lbm | kJ/kg] (5.30)

where f is the fuel�air ratio and �mech is the mechanical efficiency of transmission ofpower from the turbine to the compressor. The mechanical efficiency is usually close tounity in a well-designed gas turbine. For this reason, it was not included in earlieranalyses.

The gasifier turbine work may be written in terms of the turbine pressure ratio:

wt = �t cp,gT3( 1 � T4s /T3 )

= �t cp,gT3[1 � 1/( p3/p4)(kg-1)/kg] [Btu /lbm | kJ/kg] (5.31)

With the compressor work determined, as before, by the compressor pressure ratio andthe isentropic efficiency, the compressor-turbine pressure ratio, p3/p4, is obtained bycombining Equations (5.30) and (5.31):

p3/p4 = [ 1 � |wc| /�mech�t cp,g (1 + f )T3]�kg/(kg-1)

= ( 1 � wf )�kg/(kg-1) [dl] (5.32)

where wf is the positive, dimensionless work factor, |wc| /�mech�t cp,g (1 + f )T3, used as

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a convenient intermediate variable. The power turbine pressure ratio may then bedetermined from the identity p4/p5 = p4/p1 = ( p4/p3)( p3/p2)( p2/p1). This shows that thepower turbine pressure ratio is the compressor pressure ratio divided by the gasifierturbine pressure ratio when there is no combustion chamber pressure loss ( p3 = p2).With the pressure ratios known, all the significant temperatures and performanceparameters may be determined.

EXAMPLE 5.3

Let�s consider a two-shaft gas turbine with a regenerative air heater. The compressorpressure ratio is 6, and the compressor and gas generator turbine inlet temperatures are520°R and 1860°R, respectively. The compressor, gasifier turbine, and power turbineisentropic efficiencies are 0.86, 0.89, and 0.89, respectively. The regeneratoreffectiveness is 75%, and a 4% pressure loss is shared by the high-pressure air side ofthe regenerator and the combustor. Determine the pressure ratios of the two turbines,and the net work, thermal efficiency, and work ratio of the engine.

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SolutionThe solution in spreadsheet form shown in Table 5.3 follows the notation of Figure

5.10. The solution proceeds as before, until the calculation of the turbine pressureratios. The available pressure ratio shared by the two turbines is p3/p5 = p3/p1 = (p2/p1)(p3/p2) = r ( 1 � fpl) = 5.76. The gasifier turbine pressure ratio is determined bythe work-matching requirement of the compressor and its driving turbine, as expressedin Equation (5.32), using the dimensionless compressor work factor, wf. The resultinggas generator and power turbine pressure ratios are 2.61 and 2.2, respectively.

Comparison shows that the design point performance of the two-shaft gas turbinestudied here is not significantly different from that of the single-shaft machineconsidered an Example 5.2. While the performance of the two machines is found to beessentially the same, the single-shaft machine is sometimes preferred in applications

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with fixed operating conditions where good part-load performance over a range of speeds is not important. On the other hand, the independence of the speeds of the gasgenerator and power turbine in the two-shaft engine allows acceptable performanceover a wider range of operating conditions._____________________________________________________________________

Let us examine further the characteristics of regenerative two-shaft gas turbines,starting with the spreadsheet reproduced in Table 5.3. By copying the value column ofthat spreadsheet to several columns to the right, a family of calculations with identicalmethodologies may be performed.. The spreadsheet /EDIT-FILL command may thenbe used to vary a parameter in a given row by creating a sequence of numbers with aspecified starting value and interval. Such a parametric study of the influence ofcompressor pressure ratio on two-shaft regenerative gas turbine performance is shownin Table 5.4, where the pressure ratio is varied from 2 to 7. The fifth numeric columncontains the values from Table 5.3. The data of Table 5.4 are included in the Example

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5-4.wk3 spreadsheet that accompanies this text.Table 5.4 shows that, for the given turbine inlet temperature, the thermal efficiency

maximum is at a pressure ratio between 4 and 5, while the net work maximum is at apressure ratio of about 7. The work ratio is continually declining because themagnitude of the compressor work requirement grows faster with compressor pressureratio than the turbine work does.

Figure 5.11, plotted using the spreadsheet, compares the performance of theregenerative two-shaft gas turbine with a nonregenerative two-shaft engine (�reg = 0 ). Net work for both machines has the same variation with pressure ratio. But notice thehigh efficiency attained with a low-compressor pressure ratio, a significant advantageattributable to regeneration.

A cutaway view of a two-shaft regenerative gas turbine of the type used in pipelinecompressor stations such as that shown in Figure 5.9 is seen in Figure 5.12. The figureshows that the compressor blade heights decrease in the direction of flow as the gas iscompressed. The exhaust from the last of the sixteen compressor stages is reduced invelocity by a diffusing passage and then exits through the right window-like flange,which connects to a duct (not shown) leading to the regenerator. The heated air fromthe external regenerator reenters the machine combustor casing, where it flows aroundand into the combustor cans, cooling them. The air entering near the combustor fuelnozzles mixes with the fuel and burns locally in a near-stoichiometric mixture. As themixture flows downstream, additional secondary air entering the combustor throughslots in its sides mixes with, and reduces the temperature of, the combustion gas beforeit arrives at the turbine inlet.

A cutaway view of an industrial two-shaft gas turbine with dual regenerators ispresented in Figure 5.13. From the left, the air inlet and radial compressor and axialflow gasifier turbine and power turbine are seen on the axis of the machine, with the combustion chamber above and one of the rotary regenerators at the right. Due to therelatively low pressure ratios required by regenerative cycles, centrifugal compressorsare normally used in regenerative machines because of their simplicity, good efficiency,compactness, and ruggedness.

Performance data for the turbine of Figure 5.13 is graphed in Figure 5.14. The GT404 gas turbine delivers about 360 brake horsepower at 2880-rpm output shaft speed. The torque-speed curve of Figure 5.14 shows an important characteristic of two-shaftgas turbines with respect to off-design point operation. Whereas the compressorpressure ratio and output torque of a single-shaft gas turbine drop as the shaft speeddrops, the compressor speed and pressure ratio in a two-shaft machine is independentof the output speed. Thus, as the output shaft speed changes, the compressor maymaintain its design speed and continue to develop high pressure and mass flow. Thusthe torque at full stall of the output shaft of the GT404 is more than twice the full-loaddesign torque. This high stall torque is superior to that of reciprocating engines and isimportant in starting and accelerating rotating equipment that has high initial turningresistance. This kind of engine may be used in truck, bus, and marine applications aswell as in an industrial setting.

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A unique patented feature of some of the Allison gas turbines, called �powertransfer,� is the ability to link the duel shafts. A hydraulic clutch mechanism betweenthe two turbine shafts acts to equalize their speeds. This tends to improve part-loadfuel economy, and provides engine braking and overspeed protection for the powerturbine. When the clutch mechanism is fully engaged, the shafts rotate together as asingle-shaft machine.

5.7 Intercooling and Reheat

Intercooling

It has been pointed out that the work of compression extracts a high toll on the outputof the gas turbine. The convergence of lines of constant pressure on a T-s diagramindicates that compression at low temperatures reduces compression work. The idealcompression process would occur isothermally at the lowest available temperature. Isothermal compression is difficult to execute in practice. The use of multistagecompression with intercooling is a move in that direction.

Consider replacing the isentropic single-stage compression from p1 to p2 = p2* inFigure 5.15 with two isentropic stages from p1 to pis and pi* to p2s. Separation of thecompression processes with a heat exchanger that cools the air at Tis to a lowertemperature Ti* acts to move the final compression process to the left on the T-s

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diagram and reduces the discharge temperature following compression to T2s. A heatexchanger used to cool compressed gas between stages of compression is called anintercooler.

The work required to compress from p1 to p2s = p2* = p2 in two stages is

wc = cp [(T1 � Tis) + ( T1* � T2s)] [Btu/lbm | kJ/kg]

Note that intercooling increases the net work of the reversible cycle by the areais�i*�2s�2*�is. The reduction in the work due to two-stage intercooled compressionis also given by this area. Thus intercooling may be used to reduce the work ofcompression between two given pressures in any application. However, the favorableeffect on compressor work reduction due to intercooling in the gas turbine applicationmay be offset by the obvious increase in combustor heat addition, cp (T2* � T2s), and byincreased cost of compression system. The next example considers the selection of theoptimum pressure level for intercooling, pi = pis = pi*.

EXAMPLE 5.4

Express the compressor work, for two-stage compression with intercooling back to theoriginal inlet temperature, in terms of compressor efficiencies and pressure ratios. Develop relations for the compressor pressure ratios that minimize the total work ofcompression in terms of the overall pressure ratio.

Solution

Taking Ti* = T1 as directed in the problem statement, and letting r = p2/p1, r1 = pis/p1 andr2 = p2s/pi* = r/r1 as in Figure 5.15, we get for the compression work,

wc = cp [(T1 � Tis)/�c1 + ( T1 � T2s)/�c2 ]

= cp T1[(1 � Tis /T1)/�c1 + ( 1 � T2s / T1)/�c2 ]

= cp T1[(1 � r1(k � 1 )/k )/�c1 + ( 1 � r2

(k � 1 )/k )/�c2 ] [Btu /lbm | kJ/kg]

Eliminating r2, using r = r1r2, yields

wc = cp T1[(1 � r1(k � 1 )/k )/�c1 + ( 1 � (r/r1) (k � 1 )/k )/�c2 ]

Differentiating with respect to r1 for a fixed r and setting the result equal to zero, weobtain

� r1� 1/k /�c1 + (r (k � 1 )/k /�c2) r1

� (2k � 1 )/k = 0

which simplifies to

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r1opt = (�c1/�c2)k/(2k � 1) r1/2

Using this result we find also that

r2opt = (�c2/�c1)k/(2k � 1) r1/2

Examination of these equations shows that, for compressors with equal efficiencies,both compressor stages have the same pressure ratio, which is given by the square rootof the overall pressure ratio. For unequal compressor efficiencies, the compressor withthe higher efficiency should have the higher pressure ratio._____________________________________________________________________

Reheat

Let us now consider an improvement at the high-temperature end of the cycle. Figure5.16 shows the replacement of a single turbine by two turbines in series, each withappropriately lower pressure ratios, and separated by a reheater. The reheater may be acombustion chamber in which the excess oxygen in the combustion gas leaving the firstturbine burns additional fuel, or it may be a heater in which external combustionprovides the heat necessary to raise the temperature of the working fluid to Tm*. Thehigh temperature at the low-pressure turbine inlet has the effect of increasing the areaof the cycle by m�m*�4�4*m and hence of increasing the net work.

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Like intercooling, the increase in net work is made possible by the spreading of theconstant pressure lines on the T-s diagram as entropy increases. Thus the increase inturbine work is

�wt = cp,g [(Tm* � T4) � ( Tm � T4*) ] [Btu/lbm | kJ/kg] (5.33)

Also as with intercooling, the favorable effect in increasing net work is offset by thereduction of cycle efficiency resulting from increased addition of external heat from thereheater:

qrh = cp,g (Tm* � Tm) [Btu/lbm | kJ/kg] (5.34)

As with intercooling, the question arises as to how the intermediate pressure forreheat will be selected. An analysis similar to that of Example 5.4 shows the unsur-prising result that the reheat pressure level should be selected so that both turbines havethe same expansion ratio if they have the same efficiencies and the same inlettemperatures.

Combining Intercooling, Reheat, and Regeneration

Because of their unfavorable effects on thermal efficiency, intercooling and reheat aloneor in combination are unlikely to be found in a gas turbine without another feature thathas already been shown to have a favorable influence on gas turbine fuel economy: a re-generator. The recuperator or regenerator turns disadvantage into advantage in a cycleinvolving intercooling and/or reheat. Consider the cycle of Figure 5.17, whichincorporates all three features.

The increased turbine discharge temperature T4 produced by reheat and thedecreased compressor exit temperature T2 due to intercooling both provide an enlargedtemperature potential for regenerative heat transfer. Thus the heat transfer cp (Tc � T2)is accomplished by an internal transfer of heat from low pressure turbine exhaust gas. This also has the favorable effect of reducing the temperature of the gas discharged tothe atmosphere. The requisite external heat addition for this engine is then

qa = cp [(T3 � Tc) + ( Tm* � Tm)] [Btu/lbm | kJ/kg] (5.35)

Thus the combination of intercooling, reheat, and regeneration has the net effect ofraising the average temperature of heat addition and lowering the average temperatureof heat rejection, as prescribed by Carnot for an efficient heat engine.

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The Ericsson cycle

Increasing the number of intercoolers and reheaters without changing the overall pres-sure ratio may be seen to cause both the overall compression and the overall expansionto approach isothermal processes. The resulting reversible limiting cycle, consisting oftwo isotherms and two isobars, is called the Ericsson cycle. With perfect internal heat

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transfer between isobaric processes, all external heat addition would be at the maximumtemperature of the cycle and all heat rejected at the lowest temperature. Analysis of thelimiting reversible cycle reveals, as one might expect, that its efficiency is that of theCarnot cycle. Plants with multistage compression, reheat, and regeneration can havehigh efficiencies; but complexity and high capital costs have resulted in few plants thatactually incorporate all these features.

5.8 Gas Turbines in Aircraft �Jet engines

Gas turbines are used in aircraft to produce shaft power and hot, high-pressure gas forjet propulsion. Turbine shaft power is used in turboprop aircraft and helicopters todrive propellers and rotors. A modern turboprop engine and an aircraft that uses it areshown in Figures 5.18 through 5.20. While its jet exhaust provides some thrust, thebulk of the propulsive thrust of the turboprop is provided by its propeller. The rear-

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ward acceleration of a large air mass by the propeller is responsible for the good fueleconomy of turboprop aircraft. Thus the turboprop is popular as a power plant forsmall business aircraft. At higher subsonic flight speeds, the conventional propellerloses efficiency and the turbojet becomes superior.

Auxiliary power units, APUs, are compact gas turbines that provide mechanicalpower to generate electricity in transport aircraft while on the ground. The thermo-dynamic fundamentals of these shaft-power devices are the same those of stationary gasturbines, discussed earlier. Their design, however, places a premium on low weight and volume and conformance to other constraints associated with airborne equipment. Thus their configuration and appearance may differ substantially from those of otherstationary gas turbines.

The jet engine consists of a gas turbine that produces hot, high-pressure gas buthas zero net shaft output. It is a gasifier. A nozzle converts the thermal energy of the hot, high-pressure gas produced by the turbine into a high-kinetic-energy exhauststream. The high momentum and high exit pressure of the exhaust stream result in a forward thrust on the engine.

Although the analysis of the jet engine is similar to that of the gas turbine, theconfiguration and design of jet engines differ significantly from those of most stationary gas turbines. The criteria of light weight and small volume, mentionedearlier, apply here as well. To this we can add the necessity of small frontal area tominimize the aerodynamic drag of the engine, the importance of admitting air into the

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engine as efficiently (with as little stagnation pressure loss) as possible, and the efficientconversion of high-temperature turbine exit gas to a high-velocity nozzle exhaust. Theresulting configuration is shown schematically in Figure 5.21.

Up to now we have not been concerned with kinetic energy in the flows in gasturbines, because the flows at the stations of interest are usually designed to have lowvelocities. In the jet engine, however, high kinetic energy is present in the free streamahead of the engine and in the nozzle exit flow. The analysis here will therefore bepresented in terms of stagnation, or total, temperatures and pressures, where kinetic

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energy is taken into account implicitly, as discussed in Section 1.7. The precedinganalyses may be readily adapted to deal with the stagnation properties associated with compressible flow. In the following discussion, engine processes are first described andthen analyzed.

It should be recalled that if there are no losses, as in an isentropic flow, thestagnation pressure of a flow remains constant. All loss mechanisms, such as fluidfriction, turbulence, and flow separation, decrease stagnation pressure. Only by doingwork on the flow (with a compressor, for example) is it possible to increase stagnationpressure.

In Figure 5.21, free-stream ambient air, denoted by subscript a, enters an engineinlet that is carefully designed to efficiently decelerate the air captured by its frontalarea to a speed low enough to enter the compressor, at station 1, with minimalaerodynamic loss. There is stagnation pressure loss in the inlet, but efficientdeceleration of the flow produces static and stagnation pressures at the compressorentrance well above the ambient free-stream static pressure. This conversion of relativekinetic energy of ambient air to increased pressure and temperature in the engine inlet issometimes called ram effect.

The compressor raises the stagnation pressure of the air further to its maximumvalue at station 2, using power delivered by the turbine. Fuel enters the combustionchamber and is burned with much excess air to produce the high turbine inlettemperature at station 3. We adopt here, for simplicity, the familiar idealization that nopressure losses occur in the combustion chamber. The hot gases then expand throughthe turbine and deliver just enough power to drive the compressor (the work conditionagain). The gases leave the turbine exit at station 4, still hot and at a stagnation pres-sure well above the ambient. These gases then expand through a nozzle that convertsthe excess pressure and thermal energy into a high-kinetic-energy jet at station 5. The forward thrust on the engine, according to Newton�s Second Law, is produced by thereaction to the internal forces that accelerate the internal flow rearward to a high jet velocity and the excess of the nozzle exit plane pressure over the upstream ambientpressure.

Inlet Analysis

Given the flight speed, Va , and the free-stream static temperature and pressure, Ta andpa, at a given altitude, the free-stream stagnation temperature and pressure are

Toa = Ta + Va2 / 2cp [R | K] (5.36)

and

poa = pa( Toa/Ta ) k/(k � 1) [lbf /ft2 | kPa] (5.37)

Applying the steady-flow energy equation to the streamtube entering the inlet, we find

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that the stagnation enthalpy hoa = ho1 for adiabatic flow. For subsonic flight andsupersonic flight at Mach numbers near one, the heat capacity of the air is essentiallyconstant. Thus constancy of the stagnation enthalpy implies constancy of the stagnationtemperature. Hence, using Equation (5.36),

To1 = Toa = Ta + Va2 / 2cp [R | K] (5.38)

The effects of friction, turbulence, and other irreversibilities in the inlet flow are

represented by the inlet pressure recovery, PR, defined as

PR = po1 / poa [dl] (5.39)

where an isentropic flow through the inlet has a pressure recovery of 1.0. Lower valuesindicate reduced inlet efficiency and greater losses. For subsonic flow, values on theorder of 0.9 to 0.98 are typical. At supersonic speeds the pressure recovery decreaseswith increasing Mach number.

Compressor Analysis

With the stagnation conditions known at station 1 in Figure 5.21, the compressorpressure ratio, r = po2 /po1, now yields po2; and the isentropic relation, Equation (1.19),gives the isentropic temperaure, To2s:

To2s = To1r(k � 1)/k [R | K] (5.40)

The actual compressor discharge stagnation temperature is then obtained from thedefinition of the compressor efficiency in terms of stagnation temperatures:

�comp = ( To1 � To2s ) / ( To1 � To2 ) [dl] (5.41)

Combustor and Turbine Analysis

The turbine inlet temperature, T03, is usually assigned based on turbine blade materialconsiderations. For preliminary analysis it may be assumed that there are negligiblepressure losses in the combustion chamber, so that p03 = p02. As with the two-shaft gasturbine, the condition that the power absorbed by the compressor equal the power delivered by the turbine determines the turbine exit temperature, T04:

cp(To2 � To1 ) = (1 + f )cp,gTo3 ( 1 � To4 / To3 ) [Btu | kJ] (5.42)

where f is the engine fuel-air ratio, which often may be neglected with respect to 1 (asin our earlier studies) when high precision is not required.

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The turbine efficiency equation then yields the isentropic discharge temperatureT04s, and Equation (1.19) yields the turbine pressure ratio:

To4s = To3 � ( To3 � To4) /�turb [R | K] (5.43)

po3 /po4 = ( To3/To4s)kg/(kg � 1) [dl] (5.44)

Thus the stagnation pressure and temperature at station 4 are known. Note that the turbine pressure ratio is usually significantly lower than the compressor pressure ratio.

Nozzle Analysis

The flow is then accelerated to the jet velocity at station 5 by a convergent nozzle thatcontracts the flow area. A well-designed nozzle operating at its design condition hasonly small stagnation pressure losses. Hence the nozzle here is assumed to be loss-freeand therefore isentropic.

Under most flight conditions the exhaust nozzle is choked; that is, it is passing themaximum flow possible for its upstream conditions. A choked nozzle has the localflow velocity at its minimum area, or throat, equal to the local speed of sound. As aresult, simple relations exist between the upstream stagnation conditions at station 4and the choked conditions at the throat. Thus, for a choked isentropic nozzle,

To4 = To5 = T5 + V5 2 / 2cp,g

= T5 + kgRT5/2cp,g = T5 ( 1 + kgR/2cp,g)

= T5( kg + 1 )/2 [R | K] (5.45)

where kgR/cp,g = kg � 1. With kg = 1.333 for the combustion gas, this determines the exittemperature T5. Combining the isentropic relation with Equation (5.45) then gives thenozzle exit static pressure p5:

p5 /po4 = (T5 /T04)kg/(kg � 1)

= [2/(kg + 1)]kg/(kg � 1) [dl] (5.46)

EXAMPLE 5.5

The stagnation temperature and pressure leaving a turbine and entering a convergentnozzle are 970.2K. and 2.226 bar, respectively. What is the static pressure andtemperature downstream if the nozzle is choked? If the free-stream ambient pressure is0.54 bar, is the nozzle flow choked? Compare the existing nozzle pressure ratio withthe critical pressure ratio.

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SolutionIf the nozzle is choked, then from Equations (5.45) and (5.46),

T5 = 2To4/( kg + 1) = 2(970.2)/2.333 = 831.6 K

and the static pressure at the nozzle throat is

p5 = po4 [2/(kg + 1)]kg/(kg � 1) = 2.226( 2 / 2.333 )4 = 1.202 bar

The fact that p5 > pa = 0.54 indicates that the nozzle is choked.The critical pressure ratio of the nozzle is

po4 /p5 = [(kg + 1)/2]kg/(kg � 1) = (2.333/2)4 = 1.852

and the applied pressure ratio is 2.226/0.54 = 4.12. Thus the applied pressure ratioexceeds the critical pressure ratio.This also indicates that the nozzle is choked._____________________________________________________________________

For the isentropic nozzle, the steady-flow energy equation gives

0 = h5 + V52/2 � ho4

or, with cp,g constant,

V5 = [2cp,g(To4 � T5)]½ [ft/s | m/s] (5.47)

Thus the jet velocity is determined from Equation (5.47), where T5 is obtained fromEquation (5.45).

The thrust of the engine is obtained by applying Newton�s Second Law to acontrol volume, as shown in Figure 5.22. If the mass flow rate through the engine is m,the rates of momentum flow into and out of the control volume are mVa and mV5,respectively. The net force exerted by the exit pressure is (p5 � pa)A5, where A5 is thenozzle exit area. Thus, applying Newton�s Second Law to the control volume, we canrelate the force exerted by the engine on the gases flowing through, F, and the net exitpressure force to the rate of increase of flow momentum produced by the engine:

m(V5 � Va) = F � (p5 � pa)A5 [lbf | kN]

or

F = m(V5 � Va) + ( p5 � pa)A5 [lbf | kN] (5.48)

Here, F is the engine force acting on the gas throughflow. The reaction to this force isthe thrust on the engine acting in the direction of flight. Thus the magnitude of the

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thrust is given by Equation (5.48). It is the sum of all the pressure force componentsacting on the inside the engine in the direction of flight.

The exit area, A5, is related to the mass flow rate by

m = A5�5V5 [lbm /s | kg /s] (5.49)

where the density at station 5 is obtained from the perfect gas law using p5 and T5 fromEquations (5.45) and (5.46). If A5 is known, the mass flow rate through the engine maybe determined from Equation (5.49) and the thrust from Equation (5.48).

Another type of nozzle used in high-performance engines and in rocket nozzles is aconvergent-divergent nozzle, one in which the flow area first contracts and thenincreases. It differs from the convergent nozzle in that it can have supersonic flow atthe exit. For such a fully expanded, convergent-divergent nozzle operating at its designcondition, p5 = pa, and the engine thrust from Equation (5.48) reduces to m(V5 � Va).

Jet Engine Performance

It is seen that engine thrust is proportional to the mass flow rate through the engine andto the excess of the jet velocity over the flight velocity. The specific thrust of an engineis defined as the ratio of the engine thrust to its mass flow rate. From Equation (5.48)

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the specific thrust is

F/m = (V5 � Va) + ( p5 � pa)A5/m [lbf -s/lbm | kN-s/kg] (5.50)

Because the engine mass flow rate is proportional to its exit area, as seen in Equation(5.49), A5/m depends only on design nozzle exit conditions. As a consequence, F/m isindependent of mass flow rate and depends only on flight velocity and altitude. Assigning an engine design thrust then determines the required engine-mass flow rateand nozzle exit area and thus the engine diameter. Thus the specific thrust, F/m, is an important engine design parameter for scaling engine size with required thrust at givenflight conditions.

Another important engine design parameter is the thrust specific fuel consumption,TSFC, the ratio of the mass rate of fuel consumption to the engine thrust

TSFC = mf /F [lbm / lbf-s | kg / kN-s ] (5.51)

Low values of TSFC, of course, are favorable. The distance an aircraft can fly withoutrefueling, called its range, is inversely proportional to the TSFC of its engines. Thefollowing example demonstrates the evaluation of these parameters.

EXAMPLE 5.6

An aircraft flies at a speed of 250 m/s at an altitude of 5000 m. The engines operate ata compressor pressure ratio of 8, with a turbine inlet temperature of 1200K. Thecompressor and turbine efficiencies are 0.9 and 0.87, respectively, and there is a 4%pressure loss in the combustion chamber. The inlet total pressure recovery is 0.97, and the engine-mass flow rate is 100 kg/s. Use an engine mechanical efficiency of 0.99 anda fuel heating value of 43,000 kJ/kg. Assume that the engine has a convergent,isentropic, nozzle flow. Determine the nozzle exit area, the engine thrust, specificthrust, fuel flow rate, and thrust specific fuel consumption.

SolutionThe solution details are presented in Table 5.5 in spreadsheet form. At 5000 m

altitude, the ambient static temperature and pressure are determined from standard-atmosphere tables such as those given in Appendix H. The ambient stagnation pressureand temperature are then determined for the given flight speed. The stagnationtemperature at the compressor entrance is the same as the free-stream value for anadiabatic inlet. The inlet pressure recovery determines the total pressure at the com-pressor entrance. Using the notation of Figures 5.21 and 5.22, and given the com-pressor pressure ratio and turbine inlet temperature, the stagnation conditions at thecompressor, combustor, and turbine exits may be determined in the same way as thestatic conditions were determined earlier for stationary two-shaft gas turbines.

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The calculated available nozzle pressure ratio po4/pa = 4.118 is then compared withthe critical pressure ratio p04/pc = 1.852, which indicates that the convergent nozzle ischoked; i.e., sonic velocity exists at the throat. The nozzle exit plane pressure mustthen be given by p04 divided by the critical pressure ratio. Equation (5.45) for a soniccondition then determines the nozzle exit plane temperature, and the exit plane densityfollows from the ideal gas law. Because the exit is choked, the exit plane temperaturedetermines the exit velocity through the sonic velocity relation.

The nozzle exit area may then be determined by using the given mass flow rate and

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the exit velocity and density from Equation (5.49). The fuel-air ratio for the combustoris estimated from a simple application of the steady-flow energy equation to thecombustor, which neglects the fuel sensible heat with respect to its heating value andassumes hot-air properties for the combustion products. The thrust, specific thrust, andTSFC are then found from Equations (5.48), (5.50), and (5.51)._____________________________________________________________________

The spreadsheet in Table 5.5 (available from Spreadsheet Examples as Example5.6) is set up with conditional statements that treat the convergent nozzle for both thechoked and subsonic exit conditions. The calculations of T5, V5, and p5 depend onwhether the throat is choked or not. The format for the spreadsheet conditionalstatements is:

Conditional test, Result if true, Result if not true

The low value of fuel-air ratio, f = 0.0174, obtained in this example is typical of mostgas turbines and jet engines. In comparison with the stoichiometric value of 0.068, itcorresponds to an equivalence ratio of 0.256.

Modern Jet Engines

Full and cutaway views of a small modern jet engine used in business aircraft are shownin Figure 5.23. The engine is a turbofan engine, a type of gas turbine engine that isused in all large commercial aircraft and is gaining popularity in the business jet market. The fan referred to in the turbofan name is seen at the left in Figure 5.23(b). Theincoming air splits after passing through the fan, with the flow through the outerannulus passing to a nozzle that expels it without heating. The inner-core flow leavingthe fan passes through an axial flow compressor stage and a centrifugal compressorbefore entering the combustor, turbine stages, and exit nozzle. Thus the core flowprovides the turbine power to drive the fan and the compressors. The thrust specificfuel consumption and the thrust/weight ratio of turbofans is superior to those ofconventional jet engines, because a larger mass flow rate of air is processed and exits athigh velocity. The lower average exit velocity of the turbofan engine (compared toturbojets) is secondary in importance to the increased total mass flow rate through theengine.

A large turbofan engine designed to power Boeing 747 and 767, Airbus A310 andA300, and McDonnell-Douglas MD-11 aircraft is shown in Figure 5.24. In largeturbofans and most large jet engines, axial compressors and turbines are used ratherthan centrifugal compressors. The axial compressors are capable of much higherpressure ratios, allow compression without turning the flow through a large angle, andhave somewhat higher efficiencies than centrifugals. Large axial compressors havemany axial stages and are capable of overall pressure ratios in excess of 30. Turbofansare studied in more detail in Chapter 9.

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Afterburning

The exhaust of a jet engine contains a large amount of unused oxygen because of thehigh air-fuel ratios necessary to limit the gas stagnation temperature to which the

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turbine blading is exposed. This excess oxygen at the turbine exit makes it possible toburn additional fuel downstream and thereby to increase the nozzle exit temperatureand jet velocity. By extending the interface between the turbine and the nozzle in a jetengine and by adding fuel spray bars to create a large combustion chamber called anafterburner, it is possible to dramatically increase the thrust of a jet engine. Muchhigher afterburner stagnation temperatures are allowed than those leaving thecombustor, because (a) there is no highly stressed rotating machinery downstream ofthe afterburner, and (b) afterburner operating periods are usually limited to durations ofa few minutes. Afterburners are used for thrust augmentation of jet aircraft to assist intakeoff and climb and to provide a brief high-speed-dash capability and increasedmaneuver thrust in military aircraft. However, the substantial fuel consumption penaltyof afterburning restricts its use to brief periods of time when it is badly needed.

The T-s diagram for a jet engine with afterburner seen in Figure 5.25 shows thatafterburning is analogous to reheating in an open-cycle stationary gas turbine, as shownin Figure 5.16. The energy release in the afterburner at approximately constantstagnation pressure shifts the nozzle expansion process to the right on the T-s diagram.As a result, the nozzle enthalpy and temperature differences between T-s diagram

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constant-pressure lines increase as the nozzle inlet stagnation temperature increases.This produces higher jet velocities. Using Equation (5.47) and the notation of Figure 5.16, we get for the ratio of fully expanded nozzle isentropic jet velocity withafterburning to that without:

V/Vwo = [(Tom* � T4)/(Tom � T4*)]½

= (Tom* /Tom)½ [(1 � T4 /Tom*)/(1 � T4*/Tom)]½

= (Tom* /Tom)½

where the static-to-stagnation temperature ratios are eliminated using thecorresponding equal isentropic pressure ratios. Thus, for example, the jet velocity ratio,and therefore the thrust ratio, for a reheat temperature ratio of 4 is about 2. Theanalysis of a jet engine with afterburner is illustrated in the following example.

EXAMPLE 5.7

Consider the performance of the engine analyzed in Example 5.6 when an afterburner isadded. Assume that heat addition in the afterburner raises the nozzle entrancestagnation temperature to 2000 K, with a 5% stagnation pressure loss in theafterburner. What is the increase in nozzle exit temperature, jet velocity, and thrust?

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SolutionTable 5.6 is from an adaptation of the spreadsheet (Table 5.5) used for Example

5.6. The calculations are identical up to the turbine exit at station 4. Heat addition inthe afterburner raises the stagnation temperature at the nozzle entrance, Tob, to 2000K,while the 5% pressure loss drops the stagnation pressure to 2.115 bar. Comparing theapplied nozzle pressure ratio (pob/pa) with the critical pressure ratio (pob/pc) shows that

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the nozzle remains choked. The remainder of the calculation follows Example 5.6except that the fuel consumption associated with the afterburner temperature rise istaking into account in the fuel-air ratio, fuel flow rate, and specific fuel consumption._____________________________________________________________________

Table 5.7 compares some of the performance parameters calculated in Examples 5.6and 5.7. The table shows clearly the striking gain in thrust provided by the high nozzleexit temperature produced by afterburning and the accompanying high penalty in fuelconsumption. Note, however, that even with afterburning the overall equivalence ratio0.0448/0.068 = 0.659 is well below stoichiometric.

Table 5.7 Examples 5.6 and 5.7 Compared

Parameter Without Afterburner With Afterburner

Temperature, K

Nozzle entrance 970.2 2000.0

Nozzle exit 831.6 1714.3

Nozzle exit velocity, m/s 564.1 809.9

Thrust, kN 54.7 87.99

Fuel-air Ratio 0.0174 0.0448

TSFC, kg/kN-s 0.032 0.051

Bibliography and References

1. Cohen, H., Rogers, G. F. C., and Saravanamuttoo, H. I. H., Gas Turbine Theory,3rd Edition. New York: Longman Scientific and Technical, 1987. 2. Potter, J. H., �The Gas Turbine Cycle." ASME paper presented at the Gas TurbineForum Dinner, ASME Annual Meeting, New York, N.Y., November 27, 1972.

3. Bathie, William W., Fundamentals of Gas Turbines. New York: Wiley, 1984. 4. Whittle, Sir Frank, Gas Turbine Aerothermodynamics. New York: Pergamon, 1981.

5. Wilson, David Gordon, The Design of High-Efficiency Turbomachinery and GasTurbines. Cambridge, Mass.: MIT Press, 1984.

6. Chapman, Alan J., and Walker, William F., Introductory Gas Dynamics. New York:Holt, Rinehart and Winston, 1971.

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7. Oates, Gordon C., Aerothermodynamics of Gas Turbine and Rocket Propulsion.Washington, D. C.: American Institute of Aerodynamics and Astronautics, 1988.

8. Bammert, K., and Deuster, G., �Layout and Present Status of the Closed-CycleHelium Turbine Plant Oberhausen.� ASME paper 74-GT-132, 1974.

9. Bammert, Karl, �The Oberhausen Heat and Power Station with Helium Turbine.�Address on the Inauguration of the Helium Turbine Power Plant of EVO atOberhausen-Sterkrade, Westdeutschen Allgemeinen Zeitung, WAZ, December 20,1974.

10. Zenker, P., �The Oberhausen 50-MW Helium Turbine Plant,� Combustion, April1976, pp. 21-25.

11. Weston, Kenneth C., �Gas Turbine Analysis and Design Using InteractiveComputer Graphics.� Proceedings of Symposium on Applications of ComputerMethods in Engineering, University of Southern California, August 23-26, 1977.

12. United States Standard Atmosphere, 1976, NOAA, NASA, and USAF, October1976.

13. Weston, Kenneth C., �Turbofan Engine Analysis and Optimization UsingSpreadsheets.� ASME Computers in Engineering Conference, Anaheim, California,July 30-August 3, 1989.

14. Kerrebrock, Jack K., Aircraft Engines and Gas Turbines. Cambridge, Mass.: MITPress, 1983.

15. Dixon, S. L., Fluid Mechanics, Thermodynamics of Turbomachinery. New York:Pergamon Press, 1978.

16. Bammert, K., Rurik, J., and Griepentrog, H., �Highlights and Future Developmentof Closed-Cycle Gas Turbines.� ASME paper 74-GT-7, 1974.

EXERCISES

5.1 For the Air Standard Brayton cycle, express the net work in terms of thecompressor pressure ratio, r, and the turbine-to-compressor inlet temperature ratio,T3/T1. Nondimensionalize the net work with cpT1, and derive an expression for thepressure ratio that maximizes the net work for a given value of T3/T1. 5.2 For a Brayton Air Standard cycle, work out an expression for the maximumpossible compressor pressure ratio for a given turbine-to-compressor inlet temperature

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ratio. Draw and label the cycle on a T-s diagram. What is the magnitude of the network for this cycle? Explain.

5.3 For a calorically perfect gas, write an expression for the temperature difference,T2 � T1, on an isentrope between two lines of constant pressure in terms of the initialtemperature T1 and the pressure ratio p2/p1. Sketch a T-s diagram showing two differentisentropes between the two pressure levels. Explain how your expression demonstratesthat the work of an isentropic turbomachine operating between given pressure levelsincreases with temperature.

5.4 Derive an expression for the enthalpy difference, h2 � h1, along a calorically perfect gas isentrope spanning two fixed pressure levels, p2 and p1, in terms of the dischargetemperature T2. Note that as T2 increases, the enthalpy difference also increases.

5.5 Derive Equation (5.9).

5.6 Derive Equations (5.24) and (5.25).

5.7 A simple-cycle stationary gas turbine has compressor and turbine efficiencies of0.85 and 0.9, respectively, and a compressor pressure ratio of 20. Determine the workof the compressor and the turbine, the net work, the turbine exit temperature, and thethermal efficiency for 80°F ambient and 1900°F turbine inlet temperatures.

5.8 A simple-cycle stationary gas turbine has compressor and turbine efficiencies of0.85 and 0.9, respectively, and a compressor pressure ratio of 20. Determine the workof the compressor and the turbine, the net work, the turbine exit temperature, and thethermal efficiency for 20°C ambient and 1200°C turbine inlet temperatures.

5.9 A regenerative-cycle stationary gas turbine has compressor and turbine isentropicefficiencies of 0.85 and 0.9, respectively, a regenerator effectiveness of 0.8, and a compressor pressure ratio of 5. Determine the work of the compressor and the turbine,the net work, the turbine and regenerator exit temperatures, and the thermal efficiencyfor 80°F ambient and 1900°F turbine inlet temperatures. Compare the efficiency of thecycle with the corresponding simple-cycle efficiency.

5.10 A regenerative-cycle stationary gas turbine has compressor and turbine isentropicefficiencies of 0.85 and 0.9, respectively, a regenerator effectiveness of 0.8, and a compressor pressure ratio of 5. Determine the work of the compressor and the turbine,the net work, the turbine and regenerator exit temperatures, and the thermal efficiencyfor 20°C ambient and 1200°C turbine inlet temperatures. Compare the efficiency of thecycle with the corresponding simple-cycle efficiency.

5.11 A two-shaft stationary gas turbine has isentropic efficiencies of 0.85, 0.88, and 0.9

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for the compressor, gas generator turbine, and power turbine, respectively, and a compressor pressure ratio of 20.

(a) Determine the compressor work and net work, the gas generator turbine exittemperature, and the thermal efficiency for 80°F ambient and 1900°F compressor-turbine inlet temperatures.

(b) Calculate and discuss the effects of adding reheat to 1900°F ahead of thepower turbine.

5.12 A two-shaft stationary gas turbine has isentropic efficiencies of 0.85, 0.88, and0.9 for the compressor, gas generator turbine, and power turbine, respectively, and a compressor pressure ratio of 20.

(a) Determine the compressor work and net work, the gas generator turbine exittemperature, and the thermal efficiency for 20°C ambient and 1200°C compressor-turbine inlet temperatures.

(b) Calculate and discuss the effects of adding reheat to 1200°C ahead of thepower turbine.

5.13 A two-shaft stationary gas turbine with an intercooler and reheater hasefficiencies of 0.85, 0.88, and 0.9 for the compressor, gas-generator turbine, and powerturbine, respectively, and a compressor pressure ratio of 5.

(a) Determine the compressor work and net work, the gas generator turbine exittemperature, and the thermal efficiency for 80°F ambient and 1900°F turbine inlettemperatures.

(b) Calculate and discuss the effect on thermal efficiency, exhaust temperature, andnet work of adding a regenerator with an effectiveness of 75%.

5.14 A two-shaft stationary gas turbine with an intercooler and reheater hasefficiencies of 0.85, 0.88, and 0.9 for the compressor, gas-generator turbine, and powerturbine, respectively, and a compressor pressure ratio of 5.

(a) Determine the compressor work and net work, the gas generator turbine exittemperature, and the thermal efficiency for 20°C ambient and 1200°C turbine inlettemperatures.

(b) Calculate and discuss the effect on thermal efficiency, exhaust temperature, andnet work of adding a regenerator with an effectiveness of 75%.

5.15 Consider a pulverized-coal-burning, single-shaft gas turbine in which thecombustion chamber is downstream of the turbine to avoid turbine blade erosion andcorrosion. The combustion gases leaving the burner heat the compressor discharge airthrough the intervening walls of a high temperature ceramic heat exchanger.

(a) Sketch the flow and T-s diagrams for this gas turbine, showing the influence ofpressure drops through the combustor and the heat exchanger. The ambient, turbineinlet, and combustor exhaust temperatures are 80°F, 1900°F, and 3000°F, respectively. The compressor pressure ratio is 5. Assume perfect turbomachinery.

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(b) For zero pressure drops, determine the net work, the thermal efficiency, andthe heat exchanger exhaust temperature.

(c) If the coal has a heating value of 14,000 Btu/lbm, what is the coal consumptionrate, in tons per hour, for a 50-MW plant?

5.16 Consider a pulverized-coal-burning, single-shaft gas turbine in which thecombustion chamber is downstream of the turbine to avoid turbine blade erosion andcorrosion. The combustion gases leaving the burner heat the compressor discharge airthrough the intervening walls of a high-temperature ceramic heat exchanger.

(a) Sketch the flow and T-s diagrams for this gas turbine, showing the influence ofpressure drops through the combustor and the heat exchanger. The ambient, turbineinlet, and combustor exhaust temperatures are 20°C, 1200°C, and 2000°C, respectively. The compressor pressure ratio is 5. Assume perfect turbomachinery.

(b) For zero pressure drops, determine the net work, the thermal efficiency, andthe heat exchanger exhaust temperature.

(c) If the coal has a heating value of 25,000 kJ/kg, what is the coal consumptionrate, in tons per hour, for a 50-MW plant?

5.17 A stationary gas turbine used to supply compressed air to a factory operates withzero external shaft load. Derive an equation for the fraction of the compressor inlet airthat can be extracted ahead of the combustion chamber for process use in terms of thecompressor pressure ratio, the ratio of turbine-to-compressor inlet temperatures, andthe turbomachinery efficiencies. Plot the compressor mass extraction ratio as afunction of compressor pressure ratio for temperature ratios of 3 and 5, perfectturbomachinery, and identical high- and low-temperature heat capacities.

5.18 A stationary gas turbine used to supply compressed air to a factory operates withzero external shaft load. Derive an equation for the fraction of the inlet air that can beextracted ahead of the combustion chamber for process use in terms of the compressorpressure ratio, the ratio of turbine-to-compressor inlet temperatures, and the turbo-machinery efficiencies. What is the extraction mass flow for a machine that has acompressor pressure ratio of 10, turbomachine inlet temperatures of 1800°F and 80°F,turbomachine efficiencies of 90%, and a compressor inlet mass flow rate of 15 lbm/s?

5.19 A stationary gas turbine used to supply compressed air to a factory operates withzero external shaft load. Derive an equation for the fraction of the inlet air that can beextracted ahead of the combustion chamber for process use in terms of the compressorpressure ratio, the ratio of turbine-to-compressor inlet temperatures, and the turbo-machinery efficiencies. What is the extraction mass flow for a machine that has acompressor pressure ratio of 8, turbomachine inlet temperatures of 1000°C and 25°C,turbomachine efficiencies of 90%, and a compressor inlet mass flow rate of 10 kg /s?

5.20 Do Exercise 5.18, but account for combustion chamber fractional pressure drops

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of 3% and 5% of the burner inlet pressure. How does increased pressure loss influenceprocess mass flow?

5.21 Do Exercise 5.19, but account for combustion chamber fractional pressure dropsof 3% and 5% of the burner inlet pressure. How does increased pressure loss influenceprocess mass flow?

5.22 A gas-turbine-driven car requires a maximum of 240 shaft horsepower. Theengine is a two-shaft regenerative gas turbine with compressor, gas generator turbine,and power turbine efficiencies of 0.86, 0.9, and 0.87, respectively, and a regeneratoreffectiveness of 0.72. The compressor pressure ratio is 3.7, and the turbine andcompressor inlet temperatures are 1800°F and 90°F, respectively. What air flow ratedoes the engine require? What is the automobile exhaust temperature? What are theengine fuel-air ratio and specific fuel consumption if the engine burns gaseous methane?

5.23 A gas-turbine-driven car requires a maximum of 150kW of shaft power. Theengine is a two-shaft regenerative gas turbine with compressor, gas generator turbine,and power turbine efficiencies of 0.84, 0.87 and 0.9, respectively, and a regeneratoreffectiveness of 0.75. The compressor pressure ratio is 4.3, and the turbine andcompressor inlet temperatures are 1250°C and 20°C, respectively. What air flow ratedoes the engine require? What is the automobile exhaust temperature? What are theengine fuel-air ratio and specific fuel consumption if the engine burns gaseoushydrogen?

5.24 A simple-cycle gas turbine is designed for a turbine inlet temperature of 1450°F, a compressor pressure ratio of 12, and compressor and turbine efficiencies of 84% and88%, respectively. Ambient conditions are 85°F and 14.5 psia.

(a) Draw and label a T-s diagram for this engine.(b) Determine the compressor, turbine, and net work for this cycle.(c) Determine the engine thermal efficiency.(d) Your supervisor has requested that you study the influence of replacing the

compressor with dual compressors and an intercooler. Assume that the newcompressors are identical to each other and have the same efficiencies and combinedoverall pressure ratio as the original compressor. Assume intercooling to 85°F with nointercooler pressure losses. Show clearly the T-s diagram for the modified systemsuperimposed on your original diagram. Calculate the revised system net work andthermal efficiency.

5.25 Determine the air and kerosene flow rates for a 100-MW regenerative gas turbinewith 1800K turbine inlet temperature, compressor pressure ratio of 5, and 1 atm. and 300K ambient conditions. The compressor and turbine efficiencies are 81% and 88%,respectively, and the heat exchanger effectiveness is 75%. Use a heating value forkerosene of 45,840 kJ/kg. What is the engine specific fuel consumption?

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5.26 A regenerative gas turbine has compressor and turbine discharge temperatures of350K and 700K, respectively. Draw and label a T-s diagram showing the relevantstates. If the regenerator has an effectiveness of 70%, what are the combustor inlettemperature and engine exhaust temperature?

5.27 A gas turbine has a turbine inlet temperature of 1100K, a turbine pressure ratio of6, and a turbine efficiency of 90%. What are the turbine exit temperature and theturbine work?

5.28 A two-shaft gas turbine with reheat has turbine inlet temperatures of 1500°F, a compressor pressure ratio of 16, and turbomachine efficiencies of 88% each. Thecompressor inlet conditions are 80°F and 1 atm. Assume that all heat capacities are0.24 Btu/lbm-R and k = 1.4.

(a) Draw T-s and flow diagrams.(b) Make a table of temperatures and pressures for all real states, in °R and atm.(c) What are the compressor work and power turbine work?(d) What is the power-turbine-to-compressor work ratio?(e) What is the cycle thermal efficiency?(f) Evaluate the recommendation to add a regenerator to the system. If a 4-count

(0.04) increase in thermal efficiency can be achieved, the addition of the regenerator isconsidered economically feasible. Give your recommendation, supporting arguments,and substantiating quantitative data.

5.29 Consider a gas turbine with compressor and turbine inlet temperatures of 80°Fand 1200°F, respectively. The turbine efficiency is 85%, and compressor pressureratio is 8.

(a) Draw coordinated T-s and plant diagrams.(b) What is the turbine work?(c) What is the minimum compressor efficiency required for the gas turbine toproduce a net power output?(d) What is the thermal efficiency if the compressor efficiency is raised to 85%?

5.30 The first closed-cycle gas turbine power plant in the world using helium as a working fluid is a 50-MW plant located in Oberhausen, Germany (ref. 8). It wasdesigned as an operating power plant and as a research facility to study aspects ofcomponent design and performance with helium as a working fluid. It has twocompressors, with intercooling, connected directly to a high-pressure turbine. The high-pressure turbine is in turn connected through a gearbox to a low-pressure turbine withno reheat. Helium is heated first by regenerator, followed by a specially designedheater that burns coke-oven gas. A water-cooled pre-cooler returns the helium to thelow-pressure compressor inlet conditions. The high-pressure turbine mass flow rate is84.4 kg/s, and the heater efficiency is 92.2%. The following design data are given inthe reference:

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Temperature, °C Pressure, Bar

1. Low-pressure compressor inlet 25 10.5

2. Intercooler inlet 83 15.5

3. High-pressure compressor inlet 25 15.4

4. Regenerator inlet, high-pressure side 125 28.7

5. Heater inlet 417 28.2

6. High-pressure turbine inlet 750 27.0

7. Low-pressure turbine inlet 580 16.5

8. Regenerator inlet, low-pressure side 460 10.8

9. Precooler inlet 169 10.6

In the following, assume k = 1.67 and cp = 5.197 kJ/kg-K for helium.(a) Sketch and label flow and T-s diagrams for the plant.(b) What is the overall engine pressure ratio of the gas turbine?(c) Estimate the mechanical power output and plant thermal efficiency. Reference

8 gives 50-MW and 31.3% as net electrical output and efficiency, respectively.Evaluate your calculations with these data.

5.31 Determine the thrust for a turbojet engine flying at 200 m/s with a compressorinlet temperature of 27°C, a compressor pressure ratio of 11, a turbine inlettemperature of 1400K, and compressor and turbine efficiencies of 0.85 and 0.9,respectively. The engine mass flow rate is 20 kg/s. You may use k = 1.4 and cp =1.005 kJ/kg-K throughout and neglect the differences between static and stagnationproperties in the turbomachinery. Assume an ambient pressure of 0.2615 atmospheres.

5.32 The gas generator of a two-shaft gas turbine has a compressor pressure ratio of 5and compressor and turbine inlet temperatures of 80°F and 2000°F, respectively, at sea level. All turbomachines have efficiencies of 90%, and the inlet air flow is 50 lbm/s.

(a) What are the net work, pressure ratio, and horsepower of the power turbineand the cycle efficiency?

(b) Suppose the power turbine is removed and the gas generator exhaust gas flowsisentropically through a convergent-divergent propulsion nozzle that is fully expanded(exit pressure is ambient). What are the nozzle exhaust velocity and the static thrust?

(c) Repeat part (b) for a choked conversion nozzle.

5.33 A gas turbine with reheat has two turbines with efficiencies e1 and e2. Deriverelations for the turbine pressure ratios r1 and r2 that maximize the total turbine workfor a given overall turbine pressure ratio, r, if both turbines have the same inlettemperature. How do the pressure ratios compare if the turbine efficiencies are equal?

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If the efficiency of one turbine is 50% higher than the other, what is the optimumpressure ratio?

5.34* For a simple-cycle gas turbine, develop a multicolumn spreadsheet thattabulates and plots (a) the net work, nondimensionalized by using the product of thecompressor constant-pressure heat capacity and the compressor inlet temperature, and(b) the thermal efficiency, both as a function of compressor pressure ratio (only onegraph with two curves). Use 80°F and 2000°F as compressor and turbine inlettemperatures, respectively, and 0.85 and 0.90 as compressor and turbine efficiencies,respectively. Use appropriate constant heat capacities. Each of the input values shouldbe entered in separate cells in each column so that the spreadsheet may be used forstudies with other parametric values. Use your plot and table to determine the pressureratio that yields the maximum net work. Compare with your theoretical expectation.

5.35* Solve Exercise 5.34 for a regenerative cycle. Use a nominal value of 0.8 forregenerator effectiveness.

5.36* Solve Exercise 5.34 for a two-shaft regenerative cycle. Use a nominal value of0.8 for regenerator effectiveness and 0.88 for the power turbine efficiency. Accountalso for 3% pressure losses in both sides of the regenerator.

5.37 For the conditions of Exercise 5.31, but using more realistic properties in theengine hot sections, plot curves of thrust and specific fuel consumption as a function ofgaseous hydrogen air-fuel ratio. Determine the maximum thrust corresponding to thestoichiometric limit for gaseous hydrogen fuel.

5.38* Use the spreadsheet corresponding to Table 5.4 to plot a graph showing theinfluence of turbine inlet temperature on net work and thermal efficiency for two-shaftgas turbines, with and without regeneration, for a compressor pressure ratio of 4.

5.39 Extend Example 5.6 by using hand calculations to evaluate the thrust and thethrust specific fuel consumption for the engine if it were fitted with a fully expanded(exit pressure equal to ambient pressure) convergent-divergent isentropic nozzle.

5.40* Modify the spreadsheet corresponding to Table 5.6 to evaluate the thrust and thethrust specific fuel consumption for the engine if it were fitted with a fully expanded(exit pressure equal to ambient pressure) convergent-divergent isentropic nozzle.

5.41 A gas turbine engine is being designed to provide work and a hot, high-velocityexhaust flow. The compressor will have a pressure ratio of 4 and an isentropicefficiency of 90% at the design point. The compressor and load are driven by separate ___________________* Exercise numbers with asterisks involve computer usage.

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turbines, but the overall expansion pressure ratio across the turbines will be 3 to 1, andthe efficiency of each turbine will be 90%. The exhaust of the low-pressure turbine isexpanded through a convergent nozzle to provide the high-velocity exhaust. At thedesign point the turbine inflow air temperature will be held to 1140°F, and the air flowrate will be 300,000 pounds mass per hour. Ambient conditions are 60°F and 14.7 psia. Calculate the brake power of the engine, in kW, and the temperatures at the entranceand the exit of nozzle.

5.42 A ramjet is a jet engine that flies at speeds high enough that the pressure riseproduced by ram effect in the inlet makes a compressor and turbine unnecessary. At50,000 feet and Mach 3, the inlet has a stagnation pressure recovery of 85%. Combustion raises the air temperature to 2500K. What is the thrust per unit mass flowrate of air and the exit velocity of the engine if the nozzle is (a) convergent, and (b)fully expanded (exit pressure equal to ambient pressure) convergent-divergent?

5.43 Using the First Law of Thermodynamics, derive an equation for the work ofcompression in a reversible steady flow in terms of volume and pressure. Use theequation to derive an expression for the reversible isothermal work of compression of a calorically perfect gas, with compressor pressure ratio as an independent variable.

5.44 Sketch a pressure-volume diagram comparing isothermal and isentropiccompressions starting at the same state and having the same pressure ratio. Show for athermally perfect gas that, at a given state, the isentrope has a steeper (negative) slopethan an isotherm. Use your diagram to prove that the isothermal work of compressionis less than the isentropic work.

5.45 A supersonic aircraft flies at Mach 2 at an altitude of 40,000 feet. Its engineshave a compressor pressure ratio of 20 and a turbine inlet temperature of 2000°F. Theinlets have total pressure recoveries of 89%, the compressors and turbines all haveefficiencies of 90%, and the nozzles are convergent and isentropic. Determine thenozzle exit velocity and thrust, and estimate the thrust specific fuel consumption ofeach engine. The mass flow rate of air of each engine is 750 lbm/s. Use a fuel heatingvalue of 18,533 Btu/lbm.

5.46 A supersonic aircraft flies at Mach 2 at an altitude of 13,000 m. Its engines havea compressor pressure ratio of 20 and a turbine inlet temperature of 1500K. The inletshave total pressure recoveries of 89%, the compressors and turbines all haveefficiencies of 90%, and the nozzles are convergent and isentropic. Determine thenozzle exit velocity and thrust, and estimate the thrust specific fuel consumption of an engine for an engine air mass flow rate of 100 kg /s. Use a fuel heating value of 43,100 kJ/kg.

5.47 A supersonic aircraft flies at Mach 2 at an altitude of 40,000 feet. Its engines

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have a compressor pressure ratio of 20 and a turbine inlet temperature of 2000°F. Theinlets have total pressure recoveries of 89%, and the compressors and turbines all haveefficiencies of 90%. The engines have afterburners that raise the temperature of the gasentering the nozzles to 3000°F. The nozzles are convergent and isentropic. Compare the design thrust and thrust specific fuel consumption with the afterburner on and off for an engine air mass flow rate of 100 lbm/s.

5.48* For a simple-cycle gas turbine, develop a multicolumn spreadsheet that tabulatesand plots (a) the net work, nondimensionalized by using the compressor constant-pressure heat capacity and the compressor inlet temperature, and (b) the thermalefficiency, as a function of compressor pressure ratio (only one graph with two curves). Use 30° C and 1500°C as compressor and turbine inlet temperatures, respectively, and0.85 and 0.90 as compressor and turbine efficiencies, respectively. Assume appropriateconstant heat capacities. Each of the input values should be entered in separate cells ineach column, so that the spreadsheet may be used for studies with other parametricvalues. Use your plot and table to determine the pressure ratio that yields themaximum net work. Compare with your theoretical expectation.

5.49* Solve Exercise 5.48 for a regenerative cycle. Use a nominal value of 0.85 forregenerator effectiveness.

5.50 Methane is burned in an adiabatic gas turbine combustor. The fuel enters thecombustor at the reference temperature for the JANAF tables and mixes with aircompressed from 80°F through a pressure ratio of 16 with a compressor efficiency of90.5%. Determine the equivalence ratio that limits the turbine inlet temperature to2060°F by:

(a) Using the JANAF tables.(b) Using an energy balance on the combustion chamber and the lower heating

value for methane.

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C H A P T E R 6

RECIPROCATING INTERNAL COMBUSTION ENGINES

6.1 Introduction

Perhaps the best-known engine in the world is the reciprocating internal combustion(IC) engine. Virtually every person who has driven an automobile or pushed a powerlawnmower has used one. By far the most widely used IC engine is the spark-ignitiongasoline engine, which takes us to school and work and on pleasure jaunts. Althoughothers had made significant contributions, Niklaus Otto is generally credited with theinvention of the engine and with the statement of its theoretical cycle. Another important engine is the reciprocating engine that made the name of RudolfDiesel famous. The Diesel engine, the workhorse of the heavy truck industry, is widelyused in industrial power and marine applications. It replaced the reciprocating steamengine in railroad locomotives about fifty years ago and remains dominant in that roletoday. The piston, cylinder, crank, and connecting rod provide the geometric basis of thereciprocating engine. While two-stroke-cycle engines are in use and of continuinginterest, the discussion here will emphasize the more widely applied four-stroke-cycleengine. In this engine the piston undergoes two mechanical cycles for eachthermodynamic cycle. The intake and compression processes occur in the first twostrokes, and the power and exhaust processes in the last two. These processes are madepossible by the crank-slider mechanism, discussed next.

6.2 The Crank-Slider Mechanism

Common to most reciprocating engines is a linkage known as a crank-slider mechan-ism. Diagramed in Figure 6.1, this mechanism is one of several capable of producingthe straight-line, backward-and-forward motion known as reciprocating. Fundament-ally, the crank-slider converts rotational motion into linear motion, or vice-versa. Witha piston as the slider moving inside a fixed cylinder, the mechanism provides the vitalcapability of a gas engine: the ability to compress and expand a gas. Before delving intothis aspect of the engine, however, let us examine the crank-slider mechanism moreclosely.

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It is evident from Figure 6.2 that, while the crank arm rotates through 180°, thepiston moves from the position known as top-center (TC) to the other extreme, calledbottom-center (BC). During this period the piston travels a distance, S, called thestroke, that is twice the length of the crank.

For an angular velocity of the crank, �, the crank pin A has a tangential velocitycomponent �S/2. It is evident that, at TC and at BC, the crank pin velocity componentin the piston direction, and hence the piston velocity, is zero. At these points,corresponding to crank angle � = 0° and 180°, the piston reverses direction. Thus as �varies from 0° to 180°, the piston velocity accelerates from 0 to a maximum and thenreturns to 0. A similar behavior exists between 180° and 360°.

The connecting rod is a two-force member; hence it is evident that there are bothaxial and lateral forces on the piston at crank angles other than 0° and 180°. Theselateral forces are, of course, opposed by the cylinder walls. The resulting lateral forcecomponent normal to the cylinder wall gives rise to frictional forces between the pistonrings and cylinder. It is evident that the normal force, and thus the frictional force,alternates from one side of the piston to the other during each cycle. Thus the pistonmotion presents a challenging lubrication problem for the control and reduction of bothwear and energy loss.

The position of the piston with respect to the crank centerline is given by

x = (S/2)cos� + Lcos� [ft | m] (6.1)

where yA = (S/2)sin� = Lsin� can be used to eliminate � to obtain

x/L = (S/2L)cos� + [1� (S/2L)2 sin2 � ]½ [dl] (6.2)

Thus, while the axial component of the motion of the crank pin is simple harmonic, xA = (S/2)cos�, the motion of the piston and piston pin is more complex. It may be

226

seen from Equation (6.2), however, that as S/L becomes small, the piston motionapproaches simple harmonic. This becomes physically evident when it is recognizedthat, in this limit, the connecting rod angle, � , approaches 0 and the piston motionapproaches the axial motion of the crank pin. Equations (6.1) and (6.2) may be used topredict component velocities, accelerations, and forces in the engine.

The volume swept by the piston as it passes from TC to BC is called the pistondisplacement, disp. Engine displacement, DISP, is then the product of the pistondisplacement and the number of cylinders, DISP = (n)(disp). The piston displacement isthe product of the piston cross-sectional area and the stroke. The cylinder insidediameter (and, approximately, also the piston diameter) is called its bore. Cylinder bore,stroke, and number of cylinders are usually quoted in engine specifications along withor instead of engine displacement. It will be seen later that the power output of areciprocating engine is proportional to its displacement. An engine of historical interestthat also used the crank-slider mechanism is discussed in the next section.

6.3 The Lenoir Cycle

An early form of the reciprocating internal combustion engine is credited to EtienneLenoir. His engine, introduced in 1860, used a crank-slider-piston-cylinder arrangement

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in which a combustible mixture confined between the piston and cylinder is ignited afterTC. The resulting combustion gas pressure forces acting on the piston deliver work byway of the connecting rod to the rotating crank. When the piston is at BC, combustiongases are allowed to escape. The rotational momentum of the crank system drives thepiston toward TC, expelling additional gases as it goes. A fresh combustible mixture isagain admitted to the combustion chamber (cylinder) and the cycle is repeated.

The theoretical Lenoir cycle, shown in Figure 6.3 on a pressure-volume diagram, consists of the intake of the working fluid (a combustible mixture) from state 0 to state1, a constant-volume temperature and pressure rise from state 1 to state 2, approxim-ating the combustion process, an isentropic expansion of the combustion gases to state3, and a constant-pressure expulsion of residual gases back to state 0. Note that aportion of the piston displacement, from state 0 to state 1, is used to take in thecombustible mixture and does not participate in the power stroke from state 2 to state3. The engine has been called an explosion engine because the power delivered is dueonly to the extremely rapid combustion pressure rise or explosion of the mixture in theconfined space of the cylinder. Hundreds of Lenoir engines were used in the nineteenth century, but the engine isquite inefficient by todays standards. In 1862, Beau de Rochas pointed out that the

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efficiency of internal combustion could be markedly improved in reciprocating enginesby compression of the air-fuel mixture prior to combustion. In 1876 Niklaus Otto (whois thought to have been unaware of Rochas� suggestion) demonstrated an engine that incorporated this important feature, as described next.

6.4 The Otto Cycle

The Otto cycle is the theoretical cycle commonly used to represent the processes inthe spark ignition (SI) internal combustion engine. It is assumed that a fixed mass ofworking fluid is confined in the cylinder by a piston that moves from BC to TC andback, as shown in Figure 6.4. The cycle consists of isentropic compression of anair-fuel mixture from state 1 to state 2, constant-volume combustion to state 3,isentropic expansion of the combustion gases to state 4, and a constant-volume heatrejection back to state 1. The constant-volume heat rejection is a simple expedient toclose the cycle. It obviates the need to represent the complex expansion and outflow of

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combustion gases from the cylinder at the end of the cycle. Note that the Otto cycle isnot concerned with the induction of the air-fuel mixture or with the expulsion ofresidual combustion gases. Thus only two mechanical strokes of the crank-slider areneeded in the Otto cycle, even when it is used to represent an ideal four-stroke-cycleOtto engine. In this case the remaining strokes are used to execute the necessary intakeand exhaust functions. Because it involves only two strokes, the Otto cycle may alsorepresent a two-stroke-cycle engine. The two-stroke-cycle engine is in principlecapable of as much work in one rotation of the crank as the four-stroke engine is intwo. However, it is difficult to implement because of the necessity of making theintake and exhaust functions a part of those two strokes. It is therefore not as highlydeveloped or widely used as the four-stroke-cycle engine. We will focus on the four-stroke-cycle here.

The simplest analysis of the Otto cycle assumes calorically perfect air as the work-ing fluid in what is called the Air Standard cycle analysis. Following the notation ofFigure 6.4, the compression process can be represented by the isentropic relation for acalorically perfect gas, Equation (1.21), as

p2/p1 = (V1/V2)k [dl] (6.3)

where the compression ratio, CR = V1/V2, is a fundamental parameter of all recipro-cating engines. The diagram shows that the expansion ratio for the engine, V4 /V3, hasthe same value, V1/V2. The clearance volume, V2, is the volume enclosed between the cylinder head and the piston at TC. Thus the compression ratio may be expressed as theratio of the sum of the clearance and displacement volumes to the clearance volume:

CR = [V2 + (V1 � V2)]/V2

Thus, for a given displacement, the compression ratio may be increased by reducing theclearance volume.

The efficiency of the cycle can be most easily determined by considering constant-volume-process heat transfers and the First Law cyclic integral relation, Equation (1.3).The heat transferred in the processes 2�3 and 4�1 are

q2�3 = cv (T3 � T2) [Btu/lbm | kj/kg] (6.4)

and

q4�1 = cv (T1 � T4) [Btu/lbm | kJ/kg] (6.5)

Both the expansion process, 3�4, and the compression process, 1�2, are assumed tobe isentropic. Thus, by definition, they are both adiabatic. From the cyclic integral, thenet work per unit mass is then:

w = q2�3 + q4�1 = cv (T3 � T2 + T1 � T4) [Btu/lbm | kJ/kg] (6.6)

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As before, the cycle thermal efficiency is the ratio of the net work to the external heatsupplied:

�Otto = w/q2�3 = cv (T3 � T2 + T1 � T4) / [cv (T3 � T2)]

= 1 + (T1 � T4) / (T3 � T2)

= 1 � T1/T2 = 1 � 1 / CR k-1 [dl] (6.7)

where Equation (1.20) has been used to eliminate the temperatures. Equation (6.7)shows that increasing compression ratio increases the cycle thermal efficiency. This istrue for real engines as well as for the idealized Otto engine. The ways in which realspark ignition engine cycles deviate from the theoretical Otto cycle are discussed later.

EXAMPLE 6.1

An Otto engine takes in an air-fuel mixture at 80°F and standard atmosphere presssure.It has a compression ratio of 8. Using Air Standard cycle analysis, a heating value of20,425 Btu/lbm, and A/F = 15, determine:

(a) The temperature and pressure at the end of compression, after combustion, andat the end of the power stroke.

(b) The net work per pound of working fluid.(c) The thermal efficiency.

SolutionWe use the notation of Figure 6.4:

(a) p2 = p1(V1/V2)k = 1(8)1.4 = 18.38 atm

T2 = T1(V1/V2)k � 1 = (540)(8)0.4 = 1240.6°R

T3 = T2 + qa /cv = T2 + (F/A)(HV)k/cp = 1240.6 + 1.4�20,425/15�0.24 = 9184°R

p3 = p2T3 /T2 = 18.38(9184/1240.6) = 136.1 atm

T4 = T3 /CRk�1 = 9184/ 80.4 = 3997.2°R

p4 = p3 /CRk = 136.1/81.4 = 7.4 atm

(b) The constant-volume heat addition is governed by the fuel-air ratio and the fuelheating value:

qa = HV(F/A) = 20,425/15 = 1361.7 Btu/lbm of air

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qr = cv (T1 � T4) = (0.24/1.4)( 540 � 3997.4) = � 592.7 Btu/lbm

w = qa + qr = 1361.7 + ( � 592.7) = 769 Btu/lbm

(c) The cycle termal efficiency may then be determined from the definition of theheat engine thermal efficiency or Equation (6.7):

�th = w/qa = 769/1361.7 = 0.565

�th = 1 � 1/80.4 = 0.565_____________________________________________________________________

In view of the discussion of gas properties and dissociation in Chapter 3, the valuesof T3 and T4 in Example 6.1 are unrealistically high. Much of the energy released by thefuel would go into vibration and dissociation of the gas molecules rather than into thetranslational and rotational degrees of freedom represented by the temperature. As aresult, significantly lower temperatures would be obtained. Thus, while the analysis isformally correct, the use of constant-low-temperature heat capacities in the AirStandard cycle makes it a poor model for predicting temperature extremes when highenergy releases occur. Some improvement is achieved by using constant-high-temperature heat capacities, but the best results would be achieved by the use of realgas properties, as discussed in several of the references.

6.5 Combustion in a Reciprocating Engine

The constant-volume heat transfer process at TC in the Otto cycle is an artifice toavoid the difficulties of modeling the complex processes that take place in thecombustion chamber of the SI engine. These processes, in reality, take place over acrank angle span of 30° or more around TC. Let us consider aspects of these processesand their implementation in more detail.

Normally, the mixture in the combustion chamber must have an air-fuel ratio in theneighborhood of the stoichiometric value for satisfactory combustion. A more or lesshomogeneous mixture may be produced outside the cylinder in a carburetor, byinjection into the intake manifold, or by throttle-body injection into a header servingseveral intake manifolds. In the case of the carburetor, fuel is drawn into the enginefrom the carburetor by the low pressure created in a venturi through which thecombustion air flows. As a result, increased air flow causes lower venturi pressure andhence increased fuel flow. The fuel system thus serves to provide an air-fuel mixturethat remains close to the stoichiometric ratio for a range of air flow rates. Variousdevices designed into the carburetor further adjust the fuel flow for the specialoperating conditions encountered, such as idling and rapid acceleration.

Maximum fuel economy is usually attained with excess air to ensure that all of thefuel is burned. A mixture with excess air is called a lean mixture. The carburetor

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usually produces this condition in automobiles during normal constant-speed driving.On the other hand, maximum power is achieved with excess fuel to assure that all

of the oxygen in the air in the combustion chamber is reacted. It is a matter of exploit-ing the full power-producing capability of the displacement volume. A mixture withexcess fuel is called a rich mixture. The automotive carburetor produces a rich mixtureduring acceleration by supplying extra fuel to the air entering the intake manifold.

The equivalence ratio is sometimes used to characterize the mixture ratio, whetherrich or lean. The equivalence ratio, �, is defined as the ratio of the actual fuel-air ratioto the stoichiometric fuel-air ratio. Thus � > 1 represents a rich mixture and � < 1represents a lean mixture. In terms of air-fuel ratio, � = (A/F)stoich /(A/F).

Homogeneous air-fuel mixtures close to stoichiometric may ignite spontaneously(that is, without a spark or other local energy source) if the mixture temperatureexceeds a temperature called the autoignition temperature. If the mixture is brought toand held at a temperature higher than the autoignition temperature, there is a period of delay before spontaneous ignition or autoignition This time interval is called theignition delay, or ignition lag. The ignition delay depends on the characteristics of thefuel and the equivalence ratio and usually decreases with increasing temperature.

In spark-ignition engines, compression ratios and therefore the temperatures at theend of compression are low enough that the air-fuel mixture is ignited by the spark plugbefore spontaneous ignition can occur. SI engines are designed so that a flame frontwill propagate smoothly from the spark plug into the unburned mixture until all of themixture has been ignitied. However, as the flame front progresses, the temperature andpressure of the combustion gases behind it rise due to the release of the chemicalenergy of the fuel. As the front propagates, it compresses and heats the unburnedmixture, sometimes termed the end-gas. Combustion is completed as planned when thefront smoothly passes completely through the end-gas without autoignition. However,if the end-gas autoignites, a pinging or low-pitched sound called knock is heard.

The avoidance of knock due to autoignition of the end-gas is a major constraint onthe design compression ratio of an SI engine. If hot spots or thermally inducedcompression of the end-gas ignite it before the flame front does, there is a more rapidrelease of chemical energy from the end-gas than during normal combustion. Knock issometimes thought of as an explosion of the end gas that creates an abrupt pulse andpressure waves that race back and forth across the cylinder at high speed, producingthe familiar pinging or low-pitched sound associated with knock. Knock not onlyreduces engine performance but produces rapid wear and objectionable noise in theengine. Thus it is important for a SI engine fuel to have a high autoignitiontemperature. It is therefore important for SI engine fuel to have a high autoignitiontemperature. Thus the knock characteristics of commercially available fuels limit themaximum allowable design compression ratio for SI engines and hence limit their bestefficiency.

The octane number is a measure of a gasoline's ability to avoid knock. Additivessuch as tetraethyl lead have been used in the past to suppress engine knock. However,the accumulation of lead in the environment and its penetration into the food cycle has

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resulted in the phaseout of lead additives. Instead refineries now use appropriate blendsof hydrocarbons as a substitute for lead additives in unleaded fuels.

The octane number of a fuel is measured in a special variable-compression-ratioengine called a CFR (Cooperative Fuels Research) engine. The octane rating of a fuel isdetermined by comparison of its knocking characteristics with those of different mixtures of isooctane, C8H18, and n-heptane, C7H16. One hundred percent isooctane isdefined as having an octane number of 100 because it had the highest resistance toknock at the time the rating system was devised. On the other hand, n-heptane isassigned a value of 0 on the octane number scale because of its very poor knockresistance. If a gasoline tested in the CFR engine has the same knock threshold as ablend of 90% isooctane and 10% n-heptane, the fuel is assigned an octane rating of 90.

In combustion chamber design, the designer attempts to balance many factors toachieve good performance. Design considerations include locating intake valves awayfrom and exhaust valves near spark plugs, to keep end-gas in a relatively cool area ofthe combustion chamber and thereby suppress hot-surface-induced autoignitiontendencies. Valves are, of course, designed as large as possible to reduce induction andexhaust flow restrictions. More than one intake and one exhaust valve per cylinder arenow used in some engines to improve �engine breathing.� In some engines, four valvesin a single cylinder are employed for this purpose. The valves are also designed toinduce swirl and turbulence to promote mixing of fuel and air and to improvecombustion stability and burning rate.

Pollution and fuel economy considerations have in recent years profoundlyinfluenced overall engine and combustion chamber design. Stratified-charge engines,for example, attempt to provide a locally rich combustion region to control peaktemperatures and thus suppress NOx formation. The resulting combustion gasescontaining unburned fuel then mix with surrounding lean mixture to complete thecombustion process, thus eliminating CO and unburned hydrocarbons from the exhaust.These processes occur at lower temperatures than in conventional combustion chamberdesigns and therefore prevent significant nitrogen reactions.

6.6 Representing Reciprocating Engine Perfomance

In an earlier section, the theoretical work per unit mass of working fluid of the Ottoengine was evaluated for a single cycle of the engine, using the cyclic integral of theFirst Law of Thermodynamics. The work done by pressure forces acting on a pistoncan also be evaluated as the integral of pdV. It is evident therefore that the work doneduring a single engine cycle is the area enclosed by the cycle process curves on thepressure-volume diagram. Thus, instead of using the cyclic integral or evaluating pdVfor each process of the cycle, the work of a reciprocating engine can be found bydrawing the theoretical process curves on the p�V diagram and graphically integratingthem. Such a plot of pressure versus volume for any reciprocating engine, real ortheoretical, is called an indicator diagram.

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In the nineteenth and early twentieth centuries a mechanical device known as anengine indicator was used to produce indicator cards or diagrams to determine thework per cycle for slow-running steam and gas reciprocating engines. The indicatorcard was attached to a cylinder that rotated back and forth on its axis as the pistonoscillated, thus generating a piston position (volume) coordinate. At the same time apen driven by a pressure signal from the engine cylinder moved parallel to the cylinderaxis, scribing the p-V diagram over and over on the card. The work of high speedengines is still evaluated from traces of pressure obtained with electronic sensors anddisplayed on electronic monitors and through digital techniques.

The work done per cycle (from an indicator card, for instance) can be representedas an average pressure times a volume. Because the displacement volumes of enginesare usually known, an engine performance parameter known as the mean effectivepressure, MEP, is defined in terms of the piston displacement. The mean effectivepressure is defined as the value of the pressure obtained by dividing the net work percylinder per cycle at a given operating condition by the piston displacement volume:

MEP = W/disp [lbf/ft2 | kPa] (6.8)

Thus the MEP is a measure of the effectiveness of a given displacement volume inproducing net work.

The power output of an engine with identical cylinders may be represented as theproduct of the work per cycle and the number of cycles executed per unit time by theengine. Thus if the engine has n cylinders, each executing N identical thermodynamiccycles per unit time, and delivering W work units per cylinder, with a pistondisplacement, disp, the power output is given by

P = n�N�W = n�N �MEP � disp [ft-lbf /min | kW] (6.9)

Expressed for the entire engine, the engine displacement is DISP = n�disp and theengine work is MEP �DISP. Hence the engine power is:

P = N �MEP�DISP [ft-lbf /min | kW] (6.10)

where N, the number of thermodynamic cycles of a cylinder per unit time, is the numberof crank-shaft revolutions per unit time for a two-stroke-cycle engine and one-half ofthe revolutions per unit time for a four-stroke-cycle engine. The factor of ½ for thefour-stroke-cycle engine arises because one thermodynamic cycle is executed each timethe crank rotates through two revolutions.

EXAMPLE 6.2

What is the displacement of an engine that develops 60 horsepower at 2500 rpm in afour-stroke-cycle engine having an MEP of 120 psi?

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SolutionFrom Equation (6.10), the displacement of the engine is

DISP = P/(N �MEP) = (60)(33,000)(12)/[(2500/2)(120)] = 158.4 in3

Checking units: (HP)(ft-lbf/HP-min)(in/ft)/[(cycles/min)(lbf/in2)] = in3

_____________________________________________________________________

If the work is evaluated from an indicator diagram the work is called indicatedwork; the MEP is called the indicated mean effective pressure, IMEP; and the power isindicated power, IP. Note that the indicated work and power, being associated with thework done by the combustion chamber gases on the piston, do not account forfrictional or mechanical losses in the engine, such as piston-cylinder friction or the dragof moving parts (like connecting rods) as they move through air or lubricating oil.

Brake Performance Parameters

Another way of evaluating engine performance is to attach the engine output shaftto a device known as a dynamometer, or brake. The dynamometer measures thetorque, T, applied by the engine at a given rotational speed. The power is thencalculated from the relation

P = 2��rpm �T [ft-lbf /min | N-m/min] (6.11)

A simple device called a prony brake, which was used in the past, demonstrates theconcept for the measurement of the shaft torque of engines. Figure 6.5 shows the pronybrake configuration in which a stationary metal band wrapped around the rotatingflywheel of the engine resists the torque transmitted to it by friction. The product of theforce measured by a spring scale, w, and the moment arm, d , gives the resisting torque.The power dissipated is then given by 2�(rpm)w �d.

Modern devices such as water brakes and electrical dynamometers long agoreplaced the prony brake. The water brake is like a centrifugal water pump with nooutflow, mounted on low-friction bearings, and driven by the test engine. As with theprony brake, the force required to resist turning of the brake (pump) housing providesthe torque data. This, together with speed measurement, yields the power output fromEquation (6.11). The power dissipated appears as increased temperature of the waterin the brake and heat transfer from the brake. Cool water is circulated slowly throughthe brake to maintain a steady operating condition. The torque measured in this way iscalled the brake torque, BT, and the resulting power is called the brake power, BP. Tosummarize: while indicated parameters relate to gas forces in the cylinder, brakeparameters deal with output shaft forces.

Thus the brake power differs from the indicated power in that it accounts for theeffect of all of the energy losses in the engine. The difference between the two isreferred to as the friction power, FP. Thus FP = IP � BP.

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Friction power varies with engine speed and is difficult to measure directly. Anengine is sometimes driven without fuel by a motor-dynamometer to evaluate frictionpower. An alternative to using friction power to relate brake and indicated power isthrough the engine mechanical efficiency, �m:

�m = BP/IP [dl] (6.12)

Because of friction, the brake power of an engine is always less than the indicatedpower; hence the engine mechanical efficiency must be less than 1. Clearly, mechanicalefficiencies as close to 1 as possible are desired.

The engine indicated power can also be expressed in terms of torque, throughEquation (6.11). Thus an indicated torque, IT, can be defined. Similarly, a brake meaneffective pressure, BMEP, may be defined that, when multiplied by the engine displace-ment and speed, yields the brake power, analogous to Equation (6.10). Table 6.1summarizes these and other performance parameters and relations.

The thermal efficiency, as for other engines, is a measure of the fuel economy of areciprocating engine. It tells the amount of power output that can be achieved for agiven rate of heat release from the fuel. The rate of energy release is, in turn, theproduct of the rate of fuel flow and the fuel heating value. Thus, for a given thermalefficiency, power output can be increased by employing a high fuel flow rate and/orselecting a fuel with a high heat of combustion.

If the thermal efficiency is evaluated using the brake power, it is called the brakethermal efficiency, BTE. If the evaluation uses the indicated power, it is called theindicated thermal efficiency, ITE.

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It is common practice in the reciprocating engine field to report engine fueleconomy in terms of a parameter called the specific fuel consumption, SFC, analogousto the thrust specific fuel consumption used to describe jet engine performance. Thespecific fuel consumption is defined as the ratio of the fuel-mass flow rate to the poweroutput. Typical units are pounds per horsepower-hour or kilograms per kilowatt-hour.Obviously, good fuel economy is indicated by low values of SFC. The SFC is calledbrake specific fuel consumption, BSFC, if it is defined using brake power or indicatedspecific fuel consumption, ISFC, when based on indicated power. The SFC for areciprocating engine is analogous to the heat rate for a steam power plant in that bothare measures of the rate of energy supplied per unit of power output, and in that lowvalues of both are desirable.

Volumetric Efficiency

The theoretical energy released during the combustion process is the product of themass of fuel contained in the combustion chamber and its heating value if the fuel iscompletely reacted. The more air that can be packed into the combustion chamber, the

Table 6.1 Engine Performance Parameters

Indicated Brake Friction

Mean effective pressure IMEP BMEP FMEP = IMEP – BMEP �m = BMEP / IMEP

Power IP BP FP = IP – BP �m = BHP / IHP

Torque IT BT FT = IT – BT �m = BT / IT

Thermal efficiency ITE BTE �m = BTE / ITE

Specific fuel consumption ISFC BSFC �m = ISFC / BSFC

more fuel that can be burned with it. Thus a measure of the efficiency of the inductionsystem is of great importance. The volumetric efficiency, �v, is the ratio of the actualmass of mixture in the combustion chamber to the mass of mixture that the displace-ment volume could hold if the mixture were at ambient (free-air) density. Thus theaverage mass-flow rate of air through a cylinder is �v (disp) �aN. Pressure lossesacross intake and exhaust valves, combustion-chamber clearance volume, the influenceof hot cylinder walls on mixture density, valve timing, and gas inertia effects allinfluence the volumetric efficiency.

EXAMPLE 6.3

A six-cylinder, four-stroke-cycle SI engine operates at 3000 rpm with an indicatedmean effective pressure of five atmospheres using octane fuel with an equivalence ratio

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of 0.9. The brake torque at this condition is 250 lbf�ft., and the volumetric efficiency is85%. Each cylinder has a five inch bore and 6 inch stroke. Ambient conditions are14.7 psia and 40°F. What is the indicated horsepower, brake horsepower, and frictionhorsepower; the mechanical efficiency; the fuel flow rate; and the BSFC?

SolutionThe six cylinders have a total displacement of

DISP = 6�×52×6/4 = 706.86 in3

Then the indicated horsepower is

IP = MEP×DISP×N /[12×33,000] [lbf /in2][in3][cycles/min]/[in/ft][ft-lbf /HP-min]

= (5)(14.7)(706.86)(3000/2)/[12×33,000] = 196.8 horsepower

The brake horsepower, from Equation (6.11), is:

BP = 2� × 3000 × 250 / 33,000 = 142.8 horsepower

Then the friction power is the difference between the indicated and brake power:

FP = 196.8 � 142.8 = 54 horsepower

and the mechanical efficiency is

�m = 142.8/196.8 = 0.726

The ambient density is

�a = 14.7 × 144/ [53.3 × 500] = 0.0794 lbm /ft3

and the mass flow rate of air to the engine is

ma = 0.85×0.0794×706.86×(3000/2)/1728 = 41.4 lbm /min

For octane the stoichiometric reaction equation is

C8H18 + 12.5O2 + (12.5×3.76)N2 � 8CO2 + 9H2O + (12.5×3.76)N2

The fuel-air ratio is then

F/A = 0.9×[(8×12) + (18×1)]/[12.5(32 + 3.76×28)] = 0.0598 lbm-fuel /lbm-air

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The fuel flow rate is

mf = ma (F/A) = 41.4 × 0.0598 = 2.474 lbm /min

The brake specific fuel consumption is

BSFC = 60 mf /BHP = 60×2.474/142.8 = 1.04 lbm /BHP-hr____________________________________________________________________

6.7 Spark-Ignition Engine Performance

A typical indicator diagram showing intake and exhaust processes, valve actuation,and spark timing for a four-stroke-cycle SI engine is shown in Figure 6.6. It is assumedthat an appropriate air-fuel mixture is supplied from a carburetor through an intakemanifold to an intake valve, IV, and that the combustion gas is discharged through anexhaust valve, EV, into an exhaust manifold.

The induction of the air-fuel mixture starts with the opening of the intake valve atpoint A just before TC. As the piston sweeps to the right, the mixture is drawn into thecylinder through the IV. The pressure in the cylinder is somewhat below that in theintake manifold due to the pressure losses across the intake valve. In order to use themomentum of the mixture inflow through the valve at the end of the intake stroke toimprove the volumetric efficiency, intake valve closure is delayed to shortly after BC atpoint B. Power supplied from inertia of a flywheel (and the other rotating masses in theengine) drives the piston to the left, compressing and raising the temperature of thetrapped mixture.

The combustion process in a properly operating SI engine is progressive in that thereaction starts at the spark plug and progresses into the unburned mixture at a finitespeed. Thus the combustion process takes time and cannot be executed instantaneouslyas implied by the theoretical cycle. In order for the process to take place as near to TCas possible, the spark plug is fired at point S. The number of degrees of crank rotationbefore TC at which the spark occurs is called the ignition advance. Advances of 10° to30° are common, depending on speed and load. The spark advance may be controlledby devices that sense engine speed and intake manifold pressure. Microprocessors arenow used to control spark advance and other functions, based on almost instantaneousengine performance measurements.

Recalling the slider-crank analysis, we observ that the piston velocity at top centeris momentarily zero as the piston changes direction. Therefore no work can be done atthis point, regardless of the magnitude of the pressure force. Thus, to maximize thework output, it is desired to have the maximum cylinder pressure occur at about 20°after TC. Adjustment of the spark advance (in degrees before TC) allows some controlof the combustion process and the timing of peak pressure. For a fixed combustionduration, the combustion crank-angle interval must increase with engine speed. As aconsequence, the ignition advance must increase with increasing engine speed to

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maintain optimum timing of the peak pressure. Following combustion, the piston continues toward bottom center as the highpressure gases expand and do work on the piston during the power stroke. As thepiston approaches BC, the gases do little work on the piston as its velocity againapproaches zero. As a result, not much work is lost by early opening of the exhaustvalve before BC (at point E) to start the blowdown portion of the exhaust process. It isexpedient to sacrifice a little work during the end of the power stroke in order toreduce the work needed to overcome an otherwise-high exhaust stroke cylinderpressure. Inertia of the gas in the cylinder and resistance to flow through the exhaustvalve opening slow the drop of gas pressure in the cylinder after the valve opens. Thus the gases at point E are at a pressure above the exhaust manifold pressure and,during blowdown, rush out through the EV at high speed. Following blowdown, gasesremaining in the cylinder are then expelled as the piston returns to TC. They remainabove exhaust manifold pressure until reaching TC because of the flow resistance of theexhaust valve. The EV closes shortly after TC at point C, terminating the exhaustprocess. The period of overlap at TC between the intake valve opening at point A and exhaust valve closing at point C in Figure 6.6 allows more time for the intake andexhaust processes at high engine speeds, when about 10 milliseconds may be availablefor these processes. At low engine speed and at idling there may be some mixture lossthrough the exhaust valve and discharge into the intake manifold during this valveoverlap period.

The combined exhaust and induction processes are seen to form a �pumping loop�that traverses the p-V diagram in a counterclockwise direction and therefore

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represents work input rather than work production. The higher the exhaust strokepressure and the lower the intake stroke pressure, the greater the area of the pumpingloop and hence the greater the work that must be supplied by the power loop(clockwise) to compensate. Great attention is therefore paid to valve design and otherengine characteristics that influence the exhaust and induction processes. Volumetricefficiency is a major parameter that indicates the degree of success of these efforts.

Performance Characteristics

A given ideal Otto-cycle engine produces a certain amount of work per cycle. For sucha cycle, MEP = W/disp is a constant. Equating the power equations (6.9) and (6.11)shows that the average torque is proportional to MEP and independent of engineengine speed. Therefore power output for the ideal engine is directly proportional tothe number of cycles executed per unit time, or to engine speed. Thus an Otto enginehas ideal torque and power characteristics, as shown by the solid lines in Figure 6.7.

The characteristics of real engines (represented by the dashed lines) tend to besimilar in nature to the ideal characteristics but suffer from speed-sensitive effects,particularly at low or high speeds. Torque and power characteristics for a 3.1 liter V6engine (ref. 9) are shown by the solid lines in Figure 6.8. Note the flatness of thetorque-speed curve and the expected peaking of the power curve at higher speed thanthe torque curve. Rather than present graphical characteristics such as this in their

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brochures, automobile manufacturers usually present only values for the maximumpower and torque and the speeds at which they occur. Engine characteristics such asthose shown in the figure are invaluable to application engineers seeking a suitableengine for use in a product.

6.8 The Compression-Ignition or Diesel Cycle

The ideal Diesel cycle differs from the Otto cycle in that combustion is at constantpressure rather than constant volume. The ideal cycle, shown in Figure 6.9, iscommonly implemented in a reciprocating engine in which air is compressed withoutfuel from state 1 to state 2. With a typically high compression ratio, state 2 is at atemperature high enough that fuel will ignite spontaneously when sprayed directly intothe air in the combustion chamber from a high-pressure fuel injection system.

By controlling the fuel injection rate and thus the rate of chemical energy release inrelation to the rate of expansion of the combustion gases after state 2, a constant-

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pressure process or other energy release pattern may be achieved as in Figure 6.9. Forexample, if the energy release rate is high, then pressure may rise, as from 2 to 3', andif low may fall to 3''. Thus constant-pressure combustion made possible by controllingthe rate of fuel injection into the cyclinder implies the use of a precision fuel injectionsystem.

Instead of injecting fuel into the high-temperature compressed air, the cycle mightbe executed by compression of an air-fuel mixture, with ignition occurring eitherspontaneously or at a hot spot in the cylinder near the end of the compression process.Inconsistency and unpredictability of the start of combustion in this approach, due tovariations in fuel and operating conditions, and to lack of control of the rate of heatrelease with the possibility of severe knock, makes the operation of such an engineunreliable, at the least, and also limits the maximum compression ratio. The Dieselengine therefore usually employs fuel injection into compressed air rather thancarbureted mixture formation.

In the Air Standard cycle analysis of the Diesel cycle, the heat addition process is atconstant pressure:

q2�3 = cp(T3 � T2) [Btu/lbm | kJ/kg] (6.13)

and, as with the Otto cycle, the closing process is at constant volume:

q4�1 = cv(T1 � T4) [Btu/lbm | kJ/kg] (6.14)

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The net work and thermal efficiency are then:

w = q2�3 + q4�1

= cp(T3 � T2) + cv(T1 � T4)

= cvT1[k(T3/T1 � T2/T1) + 1 � T4/T1] [Btu/lbm | kJ/kg] (6.15)

�Diesel = w/q2�3 = 1 + q4-�1/q2�3 = 1 + (cv/cp)(T1 � T4)/(T3 � T2)

= 1 � (1/k)(T1/T2)(T4/T1 � 1)/(T3/T2 � 1) [dl] (6.16)

The expressions for the net work and cycle efficiency may be expressed in termstwo parameters, the compression ratio, CR = V1/V2 (as defined earlier in treating theOtto cycle) and the cutoff ratio, COR = V3/V2. The temperature ratios in Equations(6.15) and (6.16) may be replaced by these parameters using, for the constant-pressureprocess,

COR = V3/V2 = T3/T2

and by expanding the following identity:

T4 /T1 = (T4/T3)(T3/T2)(T2 /T1)

= (V3 /V4)k-1(V3/V2)(V1/V2)k-1

= [(V3/V4)(V1/V2)]k-1COR = (COR)k-1COR

= CORk

where the product of the volume ratios was simplified by recognizing that V4 = V1.Thus the nondimensionalized net work and Diesel-cycle thermal efficiency are given by

w /cvT1 = kCRk-1(COR � 1) + (1 � CORk) [dl] (6.17)

and

�Diesel = 1 � (1/k)[(CORk � 1)/(COR � 1)]/CRk-1 [dl] (6.18)

where the cutoff ratio, COR, is the ratio of the volume at the end of combustion, V3, tothat at the start of combustion, V2. Thus the cutoff ratio may be thought of as ameasure of the duration of fuel injection, with higher cutoff ratios corresponding tolonger combustion durations.

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Diesel-cycle net work increases with both compression ratio and cutoff ratio. This isreadily seen graphically from Figure 6.9 in terms of p-V diagram area. As with theOtto cycle, increasing compression ratio increases the Diesel-cycle thermal efficiency.Increasing cutoff ratio, however, decreases thermal efficiency. This may be rationalizedby observing from the p-V diagram that much of the additional heat supplied wheninjection is continued is rejected at increasingly higher temperatures. Another view isthat heat added late in the expansion process can produce work only over the remainingpart of the stroke and thus adds less to net work than to heat rejection.

EXAMPLE 6.4

A Diesel engine has a compression ratio of 20 and a peak temperature of 3000K. Usingan Air Standard cycle analysis, estimate the work per unit mass of air, the thermalefficiency, the combustion pressure, and the cutoff ratio.

SolutionAssuming an ambient temperature and pressure of 300K and 1 atmosphere, thetemperature at the end of the compression stroke is

T2 = (300)(20)1.4 � 1 = 994.3K

and the combustion pressure is

p2 = (1)(20)1.4 = 66.3 atm

Then the cutoff ratio is

V3/V2 = T3/T2 = 3000/994.3 = 3.02

The expansion ratio is calculated as follows:

V4 /V3 = (V1/V2)/(V3 /V2) = 20/3.02 = 6.62

T4 = T3 (V3 /V4)1.4 � 1 = 3000/6.620.4 = 1409K

w = 1.005(3000 � 994.3) + (1.005/1.4)(300 � 1409) = 1219.6 kJ/kg

qa = 1.005(3000 � 994.3) = 2015.7 kJ/kg

�th = w/qa = 1219.6/2015.6 = 0.605, or 60.5%_____________________________________________________________________

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6.9 Comparing Otto-Cycle and Diesel-Cycle Efficiencies

A reasonable question at this point is: Which cycle is more efficient, the Otto cycle orthe Diesel cycle? Figure 6.10 assists in examining this question. In general notation,the cycle efficiency may be written as

�th = wnet /qin = wnet /(wnet + |qout|)

= 1 /(1 + |qout| /wnet) [dl] (6.19) Comparing the Otto cycle 1�2�3�4 and the Diesel cycle with the same compressionratio 1�2�3'�4, we see that both have the same heat rejection but that the Otto cyclehas the higher net work. Equation (6.19) then shows that, for the same compressionratio, the Otto cycle has the higher efficiency.

It has been observed that Diesel-cycle efficiency decreases with increasing cutoffratio for a given compression ratio. Let us examine the limit of the Diesel-cycleefficiency for constant CR as COR approaches its minimum value, 1. We may writeEquation (6.18) as

�Diesel = 1 � 1 /(kCRk-1) f (COR)

where f(COR) = (CORk � 1)/(COR � 1). Applying L'Hospital's rule, with primes

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designating differentiation with respect to COR, to the limit of f(COR) as COR �1,yields

lim f(COR) = lim (CORk � 1)'/ Lim (COR� 1)' = lim kCORk � 1 = kCOR�1 COR�1 COR�1

and

lim�Diesel = 1 � 1 /CRk � 1 = �OttoCOR�1

Thus the limit of the Diesel-cycle efficiency as COR approaches 1 is the Otto cycleefficiency. Hence Equation (6.18) shows that the efficiency of the Diesel cycle must beless than or equal to the Otto-cycle efficiency if both engines have the samecompression ratio, the same conclusion we reached by examination of the p-V diagram. Suppose, however, that the compression ratios are not the same. Compare the Ottocycle 1�2'�3'�4 with the Diesel cycle 1�2�3'�4 having the same maximum temperaturein Figure 6.10. The Otto cycle has a smaller area, and therefore less work, than theDiesel cycle, but the same heat rejection. Equation (6.19) demonstrates that the Ottocycle has a lower thermal efficiency than the Diesel cycle with the same maximumtemperature.

The conclusion that must be drawn from the above comparisons is quite clear. As inmost comparative engineering studies, the result depends on the ground rules whichwere adopted at the start of the study. The Otto cycle is more efficient if thecompression ratio is the same or greater than that of the competing Diesel cycle. Butknock in spark-ignition (Otto) engines limits their compression ratios to about 12,while Diesel-engine compression ratios may exceed 20. Thus, with these highercompression ratios, the Air Standard Diesel-cycle efficiency can exceed that of the Ottocycle. In practice, Diesel engines tend to have higher efficiencies than SI enginesbecause of higher compression ratios.

6.10 Diesel-Engine Performance

In 1897, five years after Rudolph Diesel's first patents and twenty-one years afterOtto's introduction of the spark-ignition engine, Diesel's compression-ignition enginewas proven to develop 13.1 kilowatts of power with an unprecedented brake thermalefficiency of 26.2% (ref. 7). At that time, most steam engines operated at thermalefficiencies below 10 %; and the best gas engines did not perform much better than thesteam machines.

Diesel claimed (and was widely believed) to have developed his engine from theprinciples expounded by Carnot. He had developed "the rational engine." Whether hisclaims were exaggerated or not, Diesel's acclaim was well deserved. He had developedan engine that operated at unprecedented temperatures and pressures, had proven hisconcept of ignition of fuel by injection into the compressed high-temperature air, andhad overcome the formidable problems of injecting a variety of fuels in appropriate

248

amounts with the precise timing required for satisfactory combustion. His is a fascinating story of a brilliant and dedicated engineer (refs. 7, 8).

In the Diesel engine, the high air temperatures and pressures prior to combustionare attributable to the compression of air alone rather than an air-fuel mixture.Compression of air alone eliminates the possibility of autiognition during compressionand makes high compression ratios possible. However, because of the high pressuresand temperatures, Diesel engines must be designed to be structurally more rugged.Therefore, they tend to be heavier than SI engines with the same brake power.

The energy release process in the Diesel engine is controlled by the rate of injectionof fuel. After a brief ignition lag, the first fuel injected into the combustion chamberautoignites and the resulting high gas temperature sustains the combustion of theremainder of the fuel stream as it enters the combustion chamber. Thus it is evident thatthe favorable fuel characteristic of high autoignition temperature for an SI engine is anunfavorable characteristic for a Diesel engine.

In the Diesel engine, a low autoignition temperature and a short ignition delay aredesirable. Knock is possible in the Diesel engine, but it is due to an entirely differentcause than knock in a spark-ignition engine. If fuel is ignited and burns as rapidly as itis injected, then smooth, knock-free combustion occurs. If, on the other hand, fuelaccumulates in the cylinder before ignition due to a long ignition lag, an explosion ordetonation occurs, producing a loud Diesel knock. The cetane number is the parameterthat identifies the ignition lag characteristic of a fuel.

The cetane number, like the octane number, is determined by testing in a CFRengine. The ignition lag of the test fuel is compared with that of a mixture of n-cetane,C16H34, and heptamethylnonane, HMN (ref. 10). Cetane, which has good ignitionqualities, is assigned a value of 100; and HMN, which has poor knock behavior, a valueof 15. The cetane number is then given by the sum of the percentage of n-cetane and0.15 times the percentage of HMN in the knock-comparison mixture. A cetane numberof 40 is the minimum allowed for a Diesel fuel.

6.11 Superchargers and Turbochargers

The importance of the volumetric efficiency, representing the efficiency of inductionof the air-fuel mixture into the reciprocating-engine cylinders, was discussed earlier.Clearly, the more mixture mass in the displacement volume, the more chemical energycan be released and the more power will be delivered from that volume.

During the Second World War, the mechanical supercharger was sometimes usedwith SI aircraft engines to increase the power and operational ceiling of Americanairplanes. Today supercharging is used with both Diesel engines and SI engines. Thesupercharger is a compressor that supplies air to the cylinder at high pressure so thatthe gas density in the cylinder at the start of compression is well above the free-airdensity. The piston exhaust gases are allowed to expand freely to the atmospherethrough the exhaust manifold and tailpipe. The supercharger is usually driven by a beltor gear train from the engine crank shaft.

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Figure 6.11 shows a modification of the theoretical Otto cycle to accommodatemechanical supercharging. The supercharger supplies air to the engine cyclinders atpressure p7 in the intake process 7 � 1. The processes 4 � 5 � 6 purge most of thecombustion gas from the cylinder. The most striking change in the cycle is that theinduction-exhaust loop is now traversed counterclockwise, indicating that the cylinderis delivering net work during these processes as well as during the compression-expansion loop. It should be remembered, however, that part of the cycle indicatedpower must be used to drive the external supercharger.

The turbosupercharger or turbocharger, for short, is a supercharger driven by aturbine using the exhaust gas of the reciprocating engine, as shown schematically inFigure 6.12. A cutaway view of a turbocharger is shown in Figure 6.13(a). Figure6.13(b) presents a diagram for the turbocharger. Compact turbochargers commonlyincrease the brake power of an engine by 30% or more, as shown in Figure 6.8, wherethe performance of an engine with and without turbocharging is compared. There, asubstantial increase in peak torque and flattening of the torque-speed curve due toturbocharging is evident.

For a supercharged engine, the brake power, BP, is the indicated power (as inFigure 6.11) less the engine friction power and the supercharger shaft power:

BP = DISP � IMEP � N � Pm � FP [ft-lbf /min | kJ/s] (6.15)

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where Pm is the supercharger-shaft mechanical power supplied by the engine (0 for aturbocharger). The IMEP includes the positive work contribution of the exhaust loop.The exhaust back pressure of the reciprocating engine is higher with a turbochargerthan for a naturally aspirated or mechanically supercharged engine because of the dropin exhaust gas pressure through the turbine. The engine brake power increasesprimarily because of a higher IMEP due to the added mass of fuel and air in thecylinder during combustion. Intercooling between the compressor and the intakemanifold may be used to further increase the cylinder charge density. Turbochargingmay increase engine efficiency, but its primary benefit is a substantial increase in brakepower.

In a turbocharged engine, a wastegate may be required to bypass engine exhaustgas around the turbine at high engine speeds. This becomes necessary when thecompressor raises the intake manifold pressure to excessively high levels, causingengine knock or threatening component damage. Thirty to forty percent of the exhaustflow may be bypassed around the turbine at maximum speed and load (ref. 1).

251

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6.12 The Automobile Engine and Air Pollution

Since the Second World War, concern for environmental pollution has grown fromacceptance of the status quo to recognition and militance of national and internationalscope. Among other sources, causes of the well-known Los Angeles smog problemwere identified as hydrocarbons (HC) and oxides of nitrogen (NOx) in exhaustemissions from motor vehicle reciprocating engines. As a result, national andCalifornia automobile air pollution limits for automobiles have been established andtoughened. Prior to the Clean Air Act of 1990, the U.S. federal exhaust-gas emissionsstandards limited unburned hydrocarbons, carbon monoxide, and oxides of nitrogen to0.41, 3.4, and 1.0 g/mile, respectively. According to reference 12, today it takes 25autos to emit as much CO and unburned hydrocarbons and 4 to emit as much NOx as asingle car in 1960. The reference anticipated that, led by existing California law andother factors, future engine designs should be targeted toward satisfying a tailpipestandard of 0.25, 3.4, 0.4 g/mile. Indeed, the 1990 Clean Air Act (refs. 15,16)specified these limits for the first 50,000 miles or five years of operation for allpassenger cars manufactured after 1995. In addition to the regulations on gaseousemissions, the Clean Air Act of 1990 adopted the California standard for particulatematter of 0.08 g/mile for passenger cars. The standards on particulates are particularlydifficult for the Diesel engine, because of its of soot-producing tendency.

The automobile air pollution problem arises in part because the reactions in theexhaust system are not in chemical equilibrium as the gas temperature drops. Oxides ofnitrogen, once formed in the cylinder at high temperature, do not return to equilibriumconcentrations of nitrogen and oxygen in the cooling exhaust products. Likewise, COformed with rich mixtures or by dissociation of CO2 in the cylinder at high temperaturedoes not respond rapidly to an infusion of air as its temperature drops in the exhaustsystem. Their concentrations may be thought of as constant or frozen. Unburnedhydrocarbons are produced not only by rich combustion but also by unburned mixturelurking in crevices (such as between piston and cylinder above the top piston ring), bylubricating oil on cylinder walls and the cylinder head that absorbs and desorbs hydrocarbons before and after combustion, and by transient operating conditions.

Starting in 1963, positive crankcase ventilation was used in all new cars to ductfuel-rich crankcase gas previously vented to the atmosphere back into the engine intakesystem. Later in the �60s, various fixes were adopted to comply with regulation oftailpipe unburned hydrocarbons and CO, including lowering compression ratios.

In 1973, NOx became federally regulated, and exhaust gas recirculation (EGR) wasemployed to reduce NOx formation through reduced combustion temperatures. At thesame time, HC and CO standards were reduced further, leading to the use of theoxidizing catalytic converter. Introduction of air pumped into the tailpipe providedadditional oxygen to assist in completion of the oxidation reactions. In 1981, areducing catalytic converter came into use to reduce NOx further. This device does notperform well in an oxidizing atmosphere. As a result, two-stage catalytic converterswere applied, with the first stage reducing NOx in a near-stoichiometric mixture and the

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second oxidizing the combustibles remaining in the exhaust with the help of airintroduced between the stages. This fresh air does not the increase NOx significantly,because of the relatively low temperature of the exhaust. The three-way catalyticconverter using several exotic metal catalysts to reduce all three of the gaseouspollutants was also introduced.

The use of catalytic converters to deal with all three pollutants brought aboutsignificant simultaneous reductions in the three major gaseous pollutants fromautomobiles. This allowed fuel-economy-reducing modifications that had beenintroduced earlier to satisfy emission reduction demands to be eliminated or relaxed,leading to further improvements in fuel economy.

Catalytic converters, however, require precise control of exhaust gas oxygen tonear-stoichiometric mixtures. The on-board computer has made possible control ofmixture ratio and spark timing in response to censor outputs of intake manifoldpressure, exhaust gas oxygen, engine speed, air flow, and incipient knock. The oxygen,or lambda, censor located in the exhaust pipe upstream of the three-way converteror between the two-stage converters is very sensitive to transition from rich to leanexhaust and allows close computer control of the mixture ratio to ensure properoperation of the catalytic converter. Computer control of carburetors or fuel injectionas well as other engine functions has allowed simultaneous improvement in fueleconomy and emissions in recent years. Thus, while emissions have been drasticallyreduced since 1974, according to reference 11 the EPA composite fuel economy of theaverage U.S. passenger car has nearly doubled; although this improvement has notcome from the engine alone. Despite the hard-won gains in emissions control and fueleconomy, further progress may be expected.

EXAMPLE 6.5

The 1990 NOx emissions standard is 0.4 grams per mile. For an automobile burningstoichiometric octane with a fuel mileage of 30 mpg, what is the maximum tailpipeconcentration of NOx in parts per million? Assume that NOx is represented by NO2 andthat the fuel density is 692 kilograms per cubic meter.

SolutionFor the stoichiometric combustion of octane, C8H18, the air-fuel ratio is 15.05 and

the molecular weight of combustion products is 28.6. The consumption of octane is

mf = (692)(1000)(3.79×10-3)/ 30 = 87.4 g/mile

[Note: (kg/m3)(g/kg)(m3/gal)/(mile/gal) = g/mile.] The concentration of NOx is the ratioof the number of moles of NOx to moles of combustion gas products:

mole Nox /mole cg = (mNOx /mf)(mf / mcg)(Mcg /MNOx)

= (0.4/87.4)(28.6/46)/ (15.05 + 1) = 0.0001773

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or 177.3 parts per million (ppm)._____________________________________________________________________

Bibliography and References

1. Heywood, John B., Internal Combustion Engine Fundamentals. New York:McGraw-Hill, 1988.

2. Ferguson, Colin R., Internal Combustion Engines. New York: Wiley, 1986.

3. Adler, U., et al., Automotive Handbook, 2nd ed. Warrendale, Pa.: Society ofAutomotive Engineers., 1986.

4. Lichty, Lester C., Internal Combustion Engines. New York: McGraw Hill, 1951.

5. Crouse, William H., Automotive Engine Design. New York: McGraw-Hill, 1970.

6. Obert, Edward, Internal Combustion Engines, Analysis and Practice. Scranton,Pa.: International Textbook Co., 1944.

7. Grosser, Morton, Diesel: The Man and the Engine. New York: Atheneum, 1978.

8. Nitske, W. Robert, and Wilson, Charles Morrow, Rudolph Diesel: Pioneer of theAge of Power. Norman, Okla.: University of Oklahoma Press, 1965.

9. Demmler, Albert W. Jr., et al., �1989 Technical Highlights of Big-three U.S.Manufacturers,� Automotive Engineering. Vol. 96, No. 10, October 1988, p. 81.

10. Anon., �Ignition Quality of Diesel Fuels by the Cetane Method,� ASTM D 613-84,1985 Annual Book of ASTM Standards, Section 5.

11. Amann, Charles A., �The Automotive Spark Ignition Engine-A HistoricalPerspective,� American Society of Mechanical Engineers, ICE-Vol. 8, Book No.100294, 1989.

12. Amann, Charles A., �The Automotive Spark-Ignition Engine-A FuturePerspective,� Society of Automotive Engineers Paper 891666, 1989.

13. Amann, Charles A., �The Passenger Car and the Greenhouse Effect,� Society ofAutomotive Engineers Paper, 1990.

14. Taylor, Charles Fayette, The Internal Combustion Engine in Theory and Practice,2nd ed., revised. Cambridge, Mass.: MIT Press, 1985.

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15. Public Law 101-549, �An Act to Amend the Clean Air Act to Provide forAttainment and Maintenance of Health, Protection, National Air Quality Standards,and Other Purposes,� November 15, 1990.

16. Anon., �Provisions�Clean Air Amendments,� Congressional Quarterly, November 24, 1990.

EXERCISES

6.1 Plot dimensionless piston position against crank angle for S/2L = 0.5, 0.4, 0.3,and 0.2.

6.2* Obtain expressions for the piston velocity and acceleration as a function of thecrank angle, constant angular velocity, and S/2L ratio. Use a spreadsheet to calculate and plot velocity and acceleration against crank angle for S/2L = 0.5,0.4, 0.3, and 0.2.

6.3 Determine the equation for the piston motion for a scotch yoke mechanism interms of crank angle. Obtain an equation for the piston velocity for a crank thatturns with a given angular velocity, �.

6.4 Derive an equation for the Otto-engine net work by integration of pdV for the AirStandard cycle. Compare with Equation (6.6).

6.5* Use a spreadsheet to calculate and plot cycle efficiency as a function ofcompression ratio for the Diesel cycle for cutoff ratios of 1, 2, and 3. Indentifythe Otto-cycle efficiency on the plot. Explain and show graphically from the plothow a Diesel engine can be more efficient than an Otto engine.

6.6 A single-cylinder Air Standard Otto engine has a compression ratio of 8.5 and apeak temperature of 3500°F at ambient conditions of 80°F and one atmosphere. Determine the cycle efficiency, maximum cylinder pressure, and mean effectivepressure.

6.7 A six-cylinder engine with a compression ratio of 11 runs at 2800 rpm at 80°F and 14.7 psia. Each cylinder has a bore and stroke of three inches and avolumetric efficiency of 0.82. Assume an Air Standard, four-stroke Otto cycle

_______________________* Exercise numbers with an asterisk indicate that they involve computer usage.

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with stoichiometric octane as fuel. Assume that the energy release from thefuel is equally divided between internal energy increase in cylinder gases andcylinder wall heat loss. What are the cylinder mean effective pressures and theengine horsepower and specific fuel consumption?

6.8 A single-cylinder four-stroke-cycle spark-ignition engine has a BSFC of 0.4kg/kW-hr and a volumetric efficiency of 78% at a speed of 45 rps. The bore is 6cm and the stroke is 8.5 cm. What is the fuel flow rate, fuel-air ratio, and braketorque if the brake power output is 6 kW with ambient conditions of 100 kPa and 22°C?

6.9 A single-cylinder four-stroke-cycle spark-ignition engine operating at 3500 rpmhas a brake mean effective pressure of 1800 kPa and a displacement of 400 cm3. Atmospheric conditions are 101kPa and 27°C.

(a) If the stroke is 6 cm, what is the bore?(b) What is the brake power?(c) If the mass air-fuel ratio is 16 and the fuel flow rate is 0.00065 kg/s, whatis the volumetric efficiency?(d) Compare your results with the performance of a two-cylinder engine withthe same overall geometric characteristics.

6.10 A four-cylinder four-stroke-cycle spark-ignition engine operating at 3500 rpm hasa brake mean effective pressure of 80 psi and a displacement of 400 cm3. Atmospheric conditions are one bar and 80°F.

(a) If the stroke is 3 inches, what is the bore?(b) What is the brake power?(c) If the mass air-fuel ratio is 16 and the fuel flow rate is 0.00065 kg/s, whatis the volumetric efficiency?

6.11 A four-cylinder four-stroke-cycle spark-ignition engine with 200 cm3

displacement and operating in air at 27°C and 110 kPa has a friction power of 27kW and a brake power output of 136 kW at 3600 rpm.

(a) What is the mechanical efficiency?(b) If it has a volumetric efficiency of 74% and burns liquid methanol with15% excess air, what is the brake specific fuel consumption?

6.12 An eight-cylinder four-stroke-cycle engine has a bore of three inches and a strokeof 4 inches. At a shaft speed of 3000 rpm, the brake horsepower is 325 and the mechanical efficiency is 88%. Fuel with a heating value of 19,000 Btu/lbm issupplied at a rate of 80 lbm/hr. What are the engine displacement, BMEP, brake

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torque, and indicated specific fuel consumption in lbm/HP-hr?

6.13 An eight-cylinder four-stroke-cycle engine has a bore of 10 cm and a stroke of 12cm. At a shaft speed of 53 rps, the brake power is 300kW and the mechanicalefficiency is 85%. Fuel with a heating value of 40,000 kJ/kg is supplied at a rateof 40 kg/hr. What are the engine displacement, BMEP, brake torque, andindicated specific fuel consumption in kg/kW-hr?

6.14 Consider a naturally aspirated eight-cylinder four-stroke-cycle Diesel engine witha compression ratio of 20 and cutoff ratio of 2.5. Air is inducted into the cylinderfrom the atmosphere at 14.5 psia and 80°F. Assume an Air Standard cycle.

(a) Determine the temperatures and pressures immediately before and aftercombustion.(b) What is the heat added in the combustion process, in Btu/lbm?(c) What is the net work, in Btu/lbm, and the thermal efficiency?(d) If the volumetric efficiency is 85%, the engine displacement is 300 in3, and theengine speed is 2000 rpm, what is the mass flow rate of air through the engine inlbm/min?(e) What is the engine horsepower?(f) Assuming that losses through the valves cause a 20-psi pressure differential

between the pressures during the exhaust and intake strokes, estimate theactual and fractional losses, in horsepower, due to these processes. Sketchthe appropriate p-V diagram.

6.15 A two-cylinder four-stroke-cycle engine produces 30 brake horsepower at a brakethermal efficiency of 20% at 2600 rpm. The fuel is methane burning in air with anequivalence ratio of 0.8 and a heating value of 21,560 Btu/lbm. Ambientconditions are 520°R and 14.7 psia. The engine mechanical efficiency is 88%,and the volumetric efficiency is 92%. What are the fuel flow rate, thedisplacement volume per cylinder, and the brake specific fuel consumption? Whatis the bore if the bore and stroke or equal?

6.16 Sketch carefully a single p�V diagram showing Otto and Diesel cycles having thesame minimum and maximum temperatures. Shade the area representing thedifference in net work between the cycles. Repeat for cycles having the samecompression ratio. Discuss the implications of these sketches.

6.17 An eight-cylinder reciprocating engine has a 3-in. bore and a 4-in. stroke and runs at 1000 cycles per minute. If the brake horsepower is 120 and the mechanicalefficiency is 80%, estimate the indicated mean effective pressure.

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6.18 Consider a naturally aspirated eight-cylinder two-stroke-cycle Diesel engine witha compression ratio of 20 and a cutoff ratio of 2.5. Air is inducted into thecylinder at 1 atm and 23°C. Assume an Air Standard cycle.

(a) Determine the temperatures and pressures immediately before and aftercombustion.(b) What is the heat added in the combustion process, in kJ/kg?(c) What is the net work, in kJ/kg, and the thermal efficiency?(d) If the volumetric efficiency is 85%, the engine displacement is 2500 cc, and

the engine speed is 2200 rpm, what is the mass flow rate of air through theengine, in lbm per minute?

(e) What is the engine horsepower?(f) If losses through the valves cause a 120-kPa pressure differential between the

pressures during the exhaust and intake strokes, estimate the actual andfractional losses, in horsepower, due to these processes. Sketch theappropriate p�V diagram.

6.19 A twelve-cylinder four-stroke-cycle Diesel engine has a 4-in. bore, a 4.5-in.stroke, and a compression ratio of 20. The mechanical efficiency is 89%, thecutoff ratios is 2, and the engine speed is 1200 rpm. The air entering the cylinderis at 14.5 psia and 60° F. Assuming Air Standard cycle performance, determinethe cycle temperatures, indicated power, IMEP, and engine brake horsepower.

6.20 A hypothetical engine cycle consists of an isentropic compression, a constant-pressure heat addition, and a constant-volume blowdown, consecutively.

(a) Draw and label a p�V diagram for the cycle.(b) Use the cyclic integral of the First Law to derive an equation for the cycle net

work in terms of the temperature.(c) Use the definition of mechanical work to derive an equation for the cycle net

work also. Show that your equation is equivalent to the result obtained inpart (b) using the cyclic integral.

(d) Express an equation for the cycle thermal efficiency in terms of cycletemperature ratios and k.

(e) If T1 = 60°F and the volume ratio is 10, determine the other cycletemperatures; and compare the cycle efficiency with the efficiency the Ottocycle having the same compression ratio.

6.21 A slightly more complex model of a reciprocating engine cycle than those discussed combines constant-pressure and constant-volume heat additions in asingle Air Standard cycle.

(a) Sketch and label a p�V diagram for this cycle that consists of the following

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consecutive processes: isentropic compression, constant-volume heat addition,constant-pressure heat addition, isentropic expansion, and constant-volumeblowdown.(b)The engine may be characterized by three parameters: the compression ratio;the Diesel engine cutoff ratio; and a third parameter, the ratio of the pressureafter to that before the constant-volume combustion. Define these parameters interms of the symbols in your sketch and derive an equation for the thermalefficiency of the cycle.(c) Show how varying the parameters appropriately reduces your efficiencyequation to the equations for the Otto and Diesel cycles.

6.22 As a plant engineer you must recommend whether electric power for a plantexpansion (2.5 MW continuous generation requirement) will be purchased from apublic utility or generated using a fully attended Diesel-engine-driven generator oran automatic remotely controlled gas turbine generator set. The price ofelectricity is 4.8 cents per kW-hr, and the price of natural gas is 60 cents perthousand cubic feet. Both Diesel engine and gas turbine are to be natural-gasfired. The gas turbine has a heat rate of 11,500 Btu/kW-hr, and the Diesel engine13,200 Btu/kW-hr. The initial costs of Diesel engine and gas turbine are$750,000 and $850,000, respectively. Control equipment for the gas turbinecosts an additional $150,000. The engines and control equipment are estimatedto have a useful life of 20 years. The annual wages and benefits for a Dieselengine operating engineer working eight-hour daily shifts is $36,000. Assume a10% per annum interest-rate. Evaluate the alternatives for a natural gas heatingvalue of 1000 Btu/ft3, and present a table of their costs, in cents per kW-hr. Discuss your recommendation.

6.23 Evaluate the alternatives in Exercise 6.22 based on the present-worth method.

6.24 In terms of the notation of Figure 6.3, what are the piston displacement,compression ratio, and expansion ratio for the Lenoir cycle?

6.25 What are the fuel and air flow rates and brake specific fuel consumption for aneight-cylinder engine with a 3.75-in. bore and 3.5-in. stroke delivering 212horsepower at 3600 rpm with a brake thermal efficiency of 25%? The fuel isC8H18, and the equivalence ratio is 1.2. What is the power per cubic inch ofengine displacement?

6.26* Construct a spreadsheet to perform an Air Standard cycle analysis for a Diesel engine with a compression ratio of 20 and a range of peak temperatures from1000K to 3000K, in 500° increments. Use it to tabulate and plot both the net workper unit mass of air and the thermal efficiency against the cutoff ratio.

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6.27 Determine the maximum tailpipe concentrations of the three federally regulatedgaseous pollutants based on the existing standards for an automobile that achieves28 mile/gal of iso-octane. Assume that the engine mixture equivalence ratio is0.9, that NOx is represented by NO2 and unburned hydrocarbons by monatomiccarbon, and that the fuel density is 700 kg/m3.

6.28 A single-cylinder Air Standard Otto engine has a compression ratio of 9.0 and apeak temperature of 3000°F at 80°F and one atmosphere ambient conditions. Determine the net work, cycle efficiency, maximum cylinder pressure, and meaneffective pressure.

6.29 A six-cylinder engine with a compression ratio of 11 runs at 3200 rpm and 80°Fand 14.7 psia. Each cylinder has a bore of 3 inches, a stroke of 3.25 inches, and a volumetric efficiency of 0.85. Assume an Air Standard four-stroke Otto cyclewith stoichiometric octane as fuel. Assume that the energy release from the fuelis equally divided between internal energy increase in cylinder gases and cylinderwall heat loss. What are the cylinder mean effective pressures and the enginehorsepower and specific fuel consumption? Assume a heating value of 20,600Btu/lbm.

6.30 An eight-cylinder four-stroke-cycle compression-ignition engine operates with afuel-air ratio of 0.03 at 2400 rpm. It has a turbocharger and intercooler, asdiagrammed nearby, with compressor pressure ratio of 1.7 and intercooler exittemperature of 320K. The engine bore and stroke are 10 cm and 12 cm,respectively. The compressor efficiency, turbine efficiency, and volumetricefficiency are 70%, 75%, and 87%, respectively. The entrance temperature of theturbine gases is 1000K. What are the compressor power, the turbine pressureratio, and the engine power, in kW and in horsepower? Assume that the engine isconstructed of ceramic components that minimize engine heat losses so that theymay be neglected�and ideal �adiabatic� engine.

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C H A P T E R 7

THE WANKEL ROTARY ENGINE

7.1 A Different Approach to the Spark-Ignition Engine

The reciprocating internal combustion engine has served mankind for over a century,and will continue to do so for the foreseeable future. The Wankel rotary engine, amuch more recent development, is said to have been conceived in its present form in1954 (ref. 2). An implementation of the rotary engine used in the 1990 Mazda RX-7automobile and its turbocharger are shown in Figures 7.1(a) and 7.1(b). As of 1987,over 1.5 million Wankel engines had been used in Mazda automobiles (ref. 6).

The rotary engine has a host of advantages that make it a formidable contender forsome of the tasks currently performed by reciprocating engines. The piston in a four-stroke-cycle reciprocating engine must momentarily come to rest four times per cycleas its direction of motion changes. In contrast, the moving parts in a rotary engine arein continuous unidirectional motion. Higher operating speeds, ease of balancing, andabsence of vibration are a few of the benefits. The high operating speeds allow theengine to produce twice as much power as a reciprocating engine of the same weight. Ithas significantly fewer parts and occupies less volume than a reciprocating engine ofcomparable power.

With all these advantages, why are there so few Wankel engines in service? Part ofthe answer lies in the reciprocating engine�s remarkable success in so many applicationsand its continuing improvement with research. Why change a good thing?Manufacturing techniques for reciprocating engines are well known and established,whereas production of rotary engines requires significantly different tooling.

It must be admitted, however, that the rotary engine has some drawbacks. A majorproblem of the Wankel automobile engine is that it does not quite measure up to thefuel economy of some automotive reciprocating SI engines. It is the judgment of someauthorities that it does not offer as great a potential for improvement in fuel economyand emissions reduction as reciprocating and gas turbine engines. However, althoughthe rotary engine may never dominate the automotive industry, it is likely to findapplications where low weight and volume are critical, such as in sports cars, generalaviation, and motorcycles.

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While the rotary engine may not enjoy the great success of reciprocating engines, itis worthy of study as an unusual and analytically interesting implementation of thefamiliar Otto cycle. Even the present success of this latter-day Otto engine should serveas an inspiration to those who search for novel ways of doing things. This chapter is atribute to Felix Wankel and those who are helping to develop this remarkable engine.

7.2 Rotary Engine Operation

Figure 7.2 shows a cross-section of a rotary engine. The stationary housing encloses amoving triangular rotor that rotates with its apexes in constant contact with the housinginner surface. Air and combustion gases are transported in the spaces between therotor and the housing. The rotor rides on an eccentric that is an integral part of a shaft,as shown in the dual rotor crank shaft of Figure 7.3(a). The housing and rotor of a rotary engine designed for aircraft application are shown in Figure 7.3(b).

The operation of the Wankel engine as an Otto-cycle engine may be understood byfollowing in Figure 7.4 the events associated with the counterclockwise movement of agas volume isolated between the housing and one of the rotor flanks. The air-fuelmixture may be supplied, by a conventional carburetor, through the intake port labeledI in Figure 7.4(a). As the shaft and rotor turn, the intake port is covered, trapping afixed mass of air and fuel (assuming no leakage). This is analogous to the gas masscaptured within the cylinder-piston volume by closure of a reciprocating engine intakevalve. As the rotor continues to turn, the captured (crosshatched) volume contained

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between the rotor and housing decreases, compressing the air-fuel mixture [part (b)]. When it reaches the minor diameter, the active mixture volume is a minimumcorresponding to the volume at top center in the reciprocating engine. One or morespark plugs, as indicated at the top of each housing, initiate combustion, causing rapidrises in pressure and temperature [part (c)]. The hot, high-pressure combustion gas[part (d)] transmits a force to the eccentric through the rotor. Note that, during the

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power phase, the line of action of the force F provides a counterclockwise torqueacting about the shaft axis. As the rotation proceeds, the expanding gases drive therotor until the exhaust port is exposed, releasing them [part (e)]. The exhaust processcontinues as the intake port opens to begin a new cycle. This port overlap is apparentin the lower volume shown in part (b). In summary, each flank of the rotor is seen toundergo the same intake, compression, ignition, power, and exhaust processes as in afour-stroke-cycle reciprocating Otto engine.

All three flanks of the rotor execute the same processes at equally spaced intervalsduring one rotor rotation. Hence three power pulses are delivered per rotation of therotor. Because there are three shaft rotations per rotor rotation, the Wankel engine hasone power pulse per shaft rotation. Thus it has twice as many power pulses as a single-cylinder four-stroke-cycle reciprocating engine operating at the same speed, a clearadvantage in smoothness of operation. This feature of one power pulse per shaftrotation causes many people to compare the Wankel engine with the two-stroke-cyclereciprocating engine.

7.3 Rotary Engine Geometry

The major elements of the rotary engine�the housing and the rotor� are shown incross-section in Figure 7.2. The housing inner surface has a mathematical form knownas a trochoid or epitrochoid. A single-rotor engine housing may be thought of as twoparallel planes separated by a cylinder of epitrochoidal cross-section. Following thenotation of Figure 7.5, the parametric form of the epitrochoid is given by

x = e cos 3� + Rcos� [ft | m] (7.1a)

y = e sin 3� + R sin � [ft | m] (7.1b)

where e is the eccentricity and R is the rotor center-to-tip distance. For given values ofe and R, Equations (7.1) give the x and y coordinates defining the housing shape when� is varied from 0 to 360 degrees.

The rotor shape may be thought of as an equilateral triangle, as shown in Figures 7.2 and 7.4 (flank rounding and other refinements are discussed later in the chapter).Because the rotor moves inside the housing in such a way that its three apexes are inconstant contact with the housing periphery, the positions of the tips are also given byequations of the form of Equations (7.1):

x = e cos 3� + R cos(� + 2n�/3) [ft | m] (7.2a)

y = e sin 3� + R sin(� + 2n�) [ft | m] (7.2b)

where n = 0, 1, or 2, the three values identifying the positions of the three rotor tips,each separated by 120°. Because R represents the rotor center-to-tip distance, the

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motion of the center of the rotor can be obtained from Equations (7.2) by setting R = 0.The equations and Figure 7.5 indicate that the path of the rotor center is a circle ofradius e.

Note that Equations (7.1) and (7.2) can be nondimensionalized by dividing throughby R. This yields a single geometric parameter governing the equations, e/R, known asthe eccentricity ratio. It will be seen that this parameter is critical to successfulperformance of the rotary engine.

The power from the engine is delivered to an external load by a cylindrical shaft.The shaft axis coincides with the axis of the housing, as seen in Figure 7.2. A secondcircular cylinder, the eccentric, is rigidly attached to the shaft and is offset from theshaft axis by a distance, e, the eccentricity. The rotor slides on the eccentric. Note thatthe axes of the rotor and the eccentric coincide. Gas forces exerted on the rotor aretransmitted to the eccentric to provide the driving torque to the engine shaft and to theexternal load.

The motion of the rotor may now be understood in terms of the notation of Figure7.5. The line labeled e rotates with the shaft and eccentric through an angle 3�, whilethe line labeled R is fixed to the rotor and turns with it through an angle � about themoving eccentric center. Thus the entire engine motion is related to the motion of thesetwo lines. Clearly, the rotor (and thus line R) rotates at one-third of the speed of theshaft, and there are three shaft rotations for each rotor revolution.

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EXAMPLE 7.1

Derive expressions for the major (largest) and minor (smallest) diameters of anepitrochoid in terms the notation of Figure 7.5.

SolutionThe major diameter is defined by adding the lengths of the lines representing the

eccentricity and the rotor radius when they are horizontal and colinear or by usingEquation 7.1(a). Thus the major diameter at y = 0 corresponds to � = 0° and 180°, forwhich x = e + R and x = � e � R, respectively. The distance between these x coordinatesis the length of the major diameter 2(e + R).

The minor diameter is similarly defined along x = 0, but with e and R linesoppositely directed. The two cases correspond to � = 90° and 270°. For � = 90°, the eline is directed downward and the R line upward in Figure 7.5. This yields y = R � eand, by symmetry, the minor diameter is 2(R � e). Hence

Major diameter = 2(R + e)

Minor diameter = 2(R � e)_____________________________________________________________________

7.4 A Simple Model for a Rotary Engine

Additional important features of the rotary engine can be easily studied by consideringan engine with an equilateral triangular rotor. Figure 7.6 shows the rotor in the positionwhere a rotor flank defines the minimum volume. We will call this position top center,TC, by analogy to the reciprocating engine. The rotor housing clearance parameter, d,is the difference between the housing minor radius, R � e, and the distance from thehousing axis to mid-flank, e + R cos 60 = e + R/2:

d = (R � e) � (e + R/2) = R/2 � 2e [ft | m] (7.3)

Setting the clearance to zero establishes an upper limiting value for the eccentricityratio: (e/R)crit = 1/4. Study of Equations (7.1), at the other extreme, shows that, for e/R= 0, the epitrochoid degenerates to a circle. In this case the rotor would spin with noeccentricity and thus produce no compression and no torque. Thus, for the flat-flankedrotor, it is clear that usable values of e/R lie between 0 and 0.25.

Now let's examine some other fundamental parameters of the flat-flanked enginemodel. Consider the maximum mixture volume shown in Figure 7.7. For a given rotorwidth w, the maximum volume can be determined by calculating the area between thehousing and the flank of the rotor. Using Equations (7.1), the differential area 2y dx canbe written as:

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dAmax = 2y dx

= 2(e sin3� + R sin�) d(e cos3� + R cos�) [ft2 | m2] (7.4)

Dividing by R2 and differentiating on the right-hand side, we obtain an equation forthe dimensionless area in terms of the eccentricity ratio and the angle �:

Amax/R2 = � 2 [(e/R)sin3� + sin�][3(e/R)sin3� + sin�]d� [dl] (7.5)

In order for the differential area to sweep over the maximum trapped volume in

Figure 7.7, the limits on the angle � must vary from 0° to 60°. Thus integration ofEquation (7.5) with these limits and using standard integrals yields

Amax/R2 = � [(e/R)2 + 1/3] � 31/2/4[1 � 6(e/R)] [dl] (7.6)

Similarly, using Figure 7.6 and the differential volume shown there, the nondimension-alized minimum area can be written as:

Amin/R2 = � [(e/R)2 + 1/3] � 31/2/4 [1 + 6(e/R)] [dl] (7.7)

These maximum and minimum volumes (area-rotor width products) are analogousto the volumes trapped between the piston and cylinder at BC and TC in thefour-stroke reciprocating engine. In that engine the difference between the volumes atBC and TC is the displacement volume, and their ratio is the compression ratio. A littlethought should convince the reader that the analogy holds quantitatively for thedisplacement and compression ratio of the rotary engine. Therefore, subtractingEquation (7.7) from Equation (7.6) gives the displacement for a rotor width w for oneflank of the flat-flanked engine as

disp = 3 �31/2 wR2(e/R) [ft3 | m3] (7.8)

and forming their ratio yields the compression ratio as

Amax/R2 � [(e/R)2 + 1/3] � 31/2/4[1 � 6(e/R)]CR = --------- = ------------------------------------------- [dl] (7.9)

Amin/R2 � [(e/R)2 + 1/3] � 31/2/4 [1 + 6(e/R)]

Thus the displacement increases with increases in rotor width, the square of therotor radius, and with the eccentricity ratio, whereas the compression ratio isindependent of size but increases with increase in eccentricity ratio.

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EXAMPLE 7.2

What are the displacement and the compression ratio for a flat-flanked rotary enginewith a rotor radius of 10 cm, an eccentricity of 1.5 cm, and a rotor width of 2.5 cm?

SolutionFor this engine, e/R = 1.5/10 = 0.15. Equation (7.8) then yields the displacement:

3(3)0.5(0.15)(10)2(2.5) = 194.9 cm3 or (194.9)(0.0610) = 11.89 in.3

Equation (7.9) can be written as

CR = (a + b)/(a � b)

where a = (3.14159)[(0.15)2 + 1/3] � 31/2/4 = 0.6849, and b = 3� 31/2(0.15)/2 = 0.3897.Then

CR = (0.6849 + 0.3897)/(0.6849 - 0.3897) = 3.64_____________________________________________________________________

The very low compression ratio of Example 7.2 would yield a poor Otto-cyclethermal efficiency. The compression ratio could be increased by increasing e/R, but itwould still be low for most applications. It is therefore important to consider thefavorable influence of flank rounding on rotary engine performance.

7.5 The Circular-Arc-Flank Model

While the triangular rotor model represents a possible engine and is useful as a learningtool, such an engine would perform poorly compared with one having a rotor withrounded flanks. A more realistic model is one in which the triangular rotor isaugmented with circular-arc flanks, as shown in Figure 7.8. The radius of curvature, r,of a flank could vary from infinity, corresponding to a flat flank, to a value for whichthe arc touches the minor axis of the epitrochoid. Note that the center of curvature ofan arc terminated by two flank apexes depends on the value of r. It can also be seenfrom Figure 7.8 that r is related to the angle, �, subtended by the flank arc by

r sin(�/2) = R sin(�/3) = 31/2R/2 [ft | m]or

r/R = 31/2/[2sin(�/2)] [dl] (7.10)

Thus either the included angle, �, or the radius of curvature, r, may be used to definethe degree of flank rounding for a given rotor radius R.

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Clearance with Flank Rounding

The additional area obtained by capping a side of a triangle with a circular arc is calleda segment. The segment height, h, shown in Figure 7.8, is the difference between r andthe projection of r on the axis of symmetry:

h/R = (r/R)[1 � cos(� /2)] [dl] (7.11)

Substitution of Equation (7.10) in Equation (7.11) yields

h/R = 31/2 [1 � cos(� /2)] / [2sin(� /2)] [dl] (7.12)

It is evident from the figure that the clearance for the rotor with circular arc flanks isthe difference between the clearance of the flat-flanked rotor and h. Thus, usingEquation (7.3), the clearance is given by

d/R = 1/2 � 2(e/R) � 31/2 [1 � cos(� /2)] / [2sin(� /2)] [dl] (7.13)

In a real engine, of course, the clearance must be non-negative.

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Added Volume per Flank Due to Rounding

The segment area is the difference between the pie-shaped area of the sector subtendedby its included angle, �, and the enclosed triangular area. The sector area, or volumeper unit rotor width, is the fraction of the area of a circle of radius, r, subtended by theangle �; i.e., � r2 (� /2� ) = r2�/2. Thus using Equation (7.10), the dimensionlesssegment volume is

As /R2 = (Asec � Atri ) /R2 = (r/R)2(� � sin �) /2

= (3/8)(� � sin � ) /sin2�/2 [dl] (7.14)

Displacement and Compression Ratio

It was pointed out earlier that the displacement of the flat-flanked engine is the differ-ence between the maximum and minimum capture volumes, and is given by Equation(7.8). This is true also for the engine with rounded flanks. The additional volume addedto the rotor by flank rounding subtracts from both of the flat-flanked maximum andminimum capture volumes, leaving the difference unchanged. Thus the displacement ofone flank of a rounded-flank engine is

disp = 3 �31/2 wR2 (e/R) [ft3 | m3] (7.15)

Likewise, the ratio of the maximum and minimum capture volumes given byEquations (7.6) and (7.7), corrected for the segment volume from Equation (7.14)provides a relation for the rounded-flank engine compression ratio:

� [(e/R)2 + 1/3] � 31/2/4 [1 � 6(e/R)] � As /R2

CR = ----------------------------------------------------- [dl] (7.16) � [(e/R)2 + 1/3] � 31/2/4 [1 + 6(e/R)] � As /R2

The added rotor volume due to rounding subtracts from the flat-flanked capturevolumes and therefore reduces the denominator of Equation (7.16) more than thenumerator. As a result, the compression ratio is greater for rounded-flank than for flat-flanked engines. Rotary engines usually have the maximum rounding possible consistentwith adequate engineering clearances.

Effect of the Recess Volume

Equation (7.16) accounts for flank rounding but not for the recess usually found inrotor flanks. The additional capture volume associated with the recess is seen in Figure7.9. Its influence on the displacement and compression ratio may be reasoned in the

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same way as with the segment volume. The recess increases both minimum andmaximum mixture volumes by the same amount. It therefore has no effect ondisplacement and it decreases the compression ratio.

Figure 7.10 shows the influence of flank rounding and recession on clearance andcompression ratio. While flank recession reduces the compression ratio for given valuesof � and e/R, it improves the shape of the long, narrow combustion pocket forming theminimum capture volume. Rotary engines usually have more than one spark plug, tohelp overcome the combustion problems associated with this elongated shape.

EXAMPLE 7.3

Rework Example 7.2, taking into account a flank-arc included angle of 0.65 radians.What is the flank clearance for this engine?

SolutionBecause flank rounding does not influence it, the displacement is still 194.9 cm3.

Equation (7.16) rewritten using the notation of Example 7.2 becomes

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CR = (a + b � As /R2) / (a � b � As /R2)

where As /R2 = (3/8)[0.65 � sin(0.65)]/sin2(0.65/2) = 0.1648. Then the compressionratio is

CR = [0.6849 + 0.3897 � 0.1648] / [0.6849 � 0.3897 � 0.1648] = 6.98

This represents a significant improvement over the value of 3.64 for the flat-flankedrotor.

The flank clearance is given by Equation (7.13):

d = 10{0.5 � 2(0.15) � 31/2[1 � cos(0.65/2)]} / [2sin(0.65/2)] = 0.58 cm. _____________________________________________________________________

We have already noted that the displacement volume associated with one flank ofthe rotary engine produces one power stroke during each rotor revolution and duringthree shaft rotations. Because there are three flanks per rotor, a rotor executes onecomplete thermodynamic cycle per shaft rotation. Thus the power produced by a singlerotor is determined by the displacement volume of a single flank and the rotationalspeed:

(disp [cm3/Rev])(MEP [kN/cm2])(N [Rev/min])Power = ------------------------------------------------------- [kW]

(60 [sec/min])(100 [cm/m])

or

(disp [in3/Rev])(MEP [lb/in2])(N [Rev/min])Power = --------------------------------------------------- [HP]

(12 [in/ft])(33,000 [ft-lb/HP-min])

7.6 Design and Performance of the Wankel Engine

It is evident from Figure 7.4 that, in the Wankel engine, the opening and closing of theintake and exhaust ports by the motion of the rotor apexes serves a function equivalentto that of mechanical valves in reciprocating engines. This simple operation in theWankel engine eliminates the need for many of the moving parts required by thereciprocating engine, such as cams, camshafts, tappets, valves, and lifters. There are, infact, many more parts in a reciprocating engine than in a comparable rotary engine.

However, sealing at the apexes and sides of the rotor is critical for efficientoperation of the rotary engine. Significant pressure differences between the three activemixture volumes of a rotor in different phases of the Otto cycle require efficient seals

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analogous to piston rings in the reciprocating engine. These are needed to avoidleakage between adjacent volumes, which causes a loss of compression and power. Sealfriction has been estimated to account for about 25% of rotary engine friction. Springloaded, self-lubricating apex seals, as shown in Figure 7.11, allow for sliding with lowfriction over the treated-chrome-alloy-plated housing inner surface.

The figure shows improvements in apex seal design (ref. 6). The three-piece sealdesign, with two leaf springs rather than one, decreases seal mass through reducedthickness, and offers a configuration that promotes area contact rather than line contactbetween seal and rotor. Side seals are also important to maintain pressure integrity ofeach flank mixture pocket. Reductions in the thickness of both apex and side seals havedecreased friction with the housing by reducing the seal area producing the friction-

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causing normal force on the housing. Oil seals, also on the rotor sides, are used tocontrol oil consumption.

Though the peripheral intake port shown in Figures 7.4 and 7.9 provides betterperformance under heavy loads than a single side port, its associated intake-exhaustport overlap may allow excessive flow of exhaust gas into the fresh mixture, causingunreliable combustion in low-speed operation. Consequently, one or more side intakeports, in addition to or instead of a peripheral intake port, are sometimes used. Sideports, of course, are also opened and closed by rotor motion. In addition to reducingintake-exhaust overlap at light loads, side intake ports also induce combustion-enhancing swirl in the air-fuel mixture.

It is evident that the moving combustion volume at the time of ignition has a longand narrow flame propagation path. Rounded rotor flanks are usually recessed toprovide a wider flame front path between the two lobes of the active volume. In high-speed operation, the brief time for combustion may dictate additional design features.Multiple spark plugs, swirl induced by side intake ports and multiple ports, the "squish"produced by the the relative motion of the walls of the active volume, fuel injection,and stratified-charge design all can contribute to improvement of the combustionprocess.

It may be noted in Figures 7.3 and 7.9 that an internal ring gear is attached to therotor. This gear meshes with a stationary gear attached to the engine housing. Thefunction of this gearset is to position the rotor as the shaft turns�not to transmittorque. Engine torque, as indicated earlier, is transmitted by direct contact of surfaceforces between the rotor and the eccentric.

Stratified-Charge Rotary Engine

Reference 7 discusses the design and performance of stratified-charge rotary enginesdeveloped for commercial aviation propulsion and APU (auxiliary power unit)application as well as for marine, industrial, and military requirements. Figure 7.12shows a direct fuel injection configuration that has performed well under a wide rangeof speed, load, and environmental conditions and with a variety of liquid fuels. Thereference reports a lack of octane and cetane sensitivities, so that diesel, gasoline, andjet fuel can all be used with this configuration.

As air in the rotor recess passes below, the spark plug ignites a locally rich pilotstream that in turn ignites the fuel from the main injector. The net fuel-air ratio is lean,resulting in improved fuel economy over normal carburetion. Figure 7.13 presents datafor full-load brake horsepower and specific fuel consumption obtained with Jet-A fuelfor the twin-rotor 2034R engine. The maximum takeoff power at 5800 rpm was 430horsepower, with a brake specific fuel consumption (BSFC) of 0.44 lbm/BHP-hr. Throughout a range of loads and altitude conditions the engine operates with a fuel-airratio between 0.035 and 0.037, well below the stoichiometric value. The referencereports a best thermal efficiency of 35.8% (BSFC = 0.387 lbm/BHP-hr) at 3500 rpmand 225-horsepower output.

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Closure

Continued engineering research on the rotary engine has resulted in performanceimprovements through improved seals, lean-burn combustion, fuel injection, integralelectronic control, improved intake design, weight reduction, and turbocharging.Despite vehicle weight increases, the Mazda RX-7 with a two-rotor 80-in.3 -displace-ment engine improved 9.4% in fuel consumption and 8% in power output between1984 and 1987 (ref. 6). During this time period, the addition to the engine of aturbocharger with intercooling increased its power output by 35%.

Reference 8 reports that the Mazda RX-Evolv, a year-2000 concept car, has anaturally-aspirated rotary engine called �RENESIS.� The two-rotor, side intake andexhaust engine is reported to have reduced emissions and improved fuel economy andto have attained 280 horsepower at 9000 rpm and 226 N-m torque at 8000 rpm.

EXAMPLE 7.4

If the BMEP of the 11.89-in3-diplacement engine in Example 7.2 is 150 psi at 4000rpm, what is the brake horsepower?

SolutionThe brake horsepower is

BHP = (150)(4000)(11.89)/(12)(33000) = 18 horsepoweror

BHP = (18)(0.746) =13.44kW_____________________________________________________________________ Bibliography and References

1. Cole, David E. "The Wankel Engine," Scientific American, Vol. 227, No. 2, (August1972): 14�23.

2. Ansdale, R. F., The Wankel R C Engine. South Brunswick, N.J.: A. S. Barnes, 1969.

3. Yamamoto, Kenichi, Rotary Engine, Tokyo: Sankaido Co., 1981.

4. Weston, Kenneth C. "Computer Simulation of a Wankel Rotary Engine�Analysisand Graphics." Proceedings of the Conference of the Society for Computer Simulation,July 1986, pp. 213-216.

5. Weston, Kenneth C., "Computerized Instruction in the Design of the Wankel RotaryEngine." ASEE Annual Conference Proceedings, June 1988.

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6. Fujimoto, Y. et al., "Present and Prospective Technologies of Rotary Engine."Society of Automotive Engineers Paper 870446, 1987.

7. Mount, Robert E., and LaBouff, Gary A., �Advanced Stratified-Charge RotaryEngine Design.� Society of Automotive Engineers Paper 890324 (also in SAE SP-768,Rotary Engine Design; Analysis and Developments), 1989.

8. Jost, Kevin, (ed.), �Global Concepts�Mazda RX�Evolv,� Automotive EngineeringInternational, Vol 8, No.8, (August 2000), p 59.

EXERCISES

7.1 Using graph paper, plot, on a single sheet, epitrochoids for e/R = 0, 0.15, 0.2, 0.25,0.3, and 0.4. On a separate sheet, draw the epitrochiod and the triangular rotor forthree rotor positions (separated by 30°) for the case of e/R = 0.15.

7.2 Verify the results of Example 7.1 by specializing Equations (7.1) for the appropriatevalues of �.

7.3 Derive Equation (7.5) using a differential area given by (x � xf) dy, where xf is theconstant x-coordinate of the flank.

7.4 Following the approach in the derivation of Equations (7.3)�(7.6), and using thenotation of Figure 7.6, derive Equation (7.7).

7.5 Derive Equation (7.7) using a differential area (y � yf ) dx, where yf is the constanty-coordinate of the flank.

7.6 Derive Equation (7.8).

7.7 Derive Equation (7.9).

7.8 Show that the radius of curvature for a circular-arc flank that touches theepitrochoid at its midpoint is given by

r/R = 1 � e/R + 3(e/R)/(1 � 4e/R)

7.9 Use Equation (7.13) to derive an expression for the limiting value of e/R as afunction of the flank included angle. Plot the limiting value of e/R as a function ofthe included angle.

7.10 Solve Example 7.3, accounting for a rotor flank recess of 3% of R2w.

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7.11 If combustion takes place in an engine rotor rotation interval of 40° in an engineoperating at 8000 rpm, how much time is available for the combustion process?

7.12* Develop a single-column spreadsheet that determines the compression ratio,clearance ratio, and nondimensional displacement for a given value of e/R andflank-rounding angle. Use a copy command to replicate the column, forming a tableof alternative design characteristics, for a reasonable range of rounding angles.

7.13* Use the spreadsheet graphics option to develop plots of compression ratio andclearance ratio, as seen in Figure 7.10.

7.14 A snowmobile single-rotor Wankel engine developed 6.65 brake horsepower at

5500 rpm with a displacement volume of 108 cc and a compression ratio of 8.5.The fuel consumption was 2.77 lbm/hr. Determine the BMEP (in psi), the BSFC(in lbm/hp-min), the brake torque (in lbf-ft), and the brake thermal efficiency,assuming a fuel heating value of 18,900 Btu/lbm.

7.15 A rotary engine mounted on a dynamometer develops 23 lbf-ft of torque at 5000rpm. When driven by a motor-generator at the same speed, a torque of 7 lbf-ft isrequired. Determine the brake and indicated horsepower and the enginemechanical efficiency. What additional information is needed to determine theindicated mean effective pressure?

7.16 A rotary engine has an eccentricity of 2 in. and an equilateral triangular rotor witha tip radius of 10 in.

(a) Determine the major and minor diameters of the epitrochoidal housing.(b) Sketch the housing and its axes of symmetry and the rotor when it is inthe nominal spark-plug firing position.(c) For the configuaration of part (b), determine the minimum rotorclearance.(d) Write equations for the relations between the shaft speed (rpm), the sparkplug firing rate (FR), and the rotor speed (RS). Identify any new symbolsused.

7.17 A rotary engine has an eccentricity of 3 cm and an equilateral triangular rotorwith a tip radius of 13 cm.

(a) Determine the major and minor diameters of the epitrochoidal housing.(b) Sketch the housing and its axes of symmetry and the rotor when it is inthe nominal spark-plug firing position.(c) For the configuaration of part (b), determine the minimum rotorclearance.

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(d) Write equations for the relations between the shaft speed (rpm), the sparkplug firing rate (FR), and the rotor speed (RS). Identify any new symbolsused.

7.18 A rotary engine with a flat-flanked rotor has a ratio of maximum to minimumhousing inside diameters of 1.4. What is the engine compression ratio?

7.19 The major diameter of the epitrochoidal housing of a flat-flanked single-rotorindustrial rotary engine is 39.6 inches. The engine turns at 1000 rpm whiledelivering 500 brake horsepower at a BMEP of 79.2 psi. If the eccentricity ratiois 0.14, what are the minor diameter, the rotor thickness, and the rotordisplacement, [in.3]?

7.20 A flat-flanked dual-rotor industrial rotary engine has a 60-cm minor diameter.Theengine delivers 800kW from an IMEP of 700kPa at a shaft speed of 20 rps. Themechanical efficiency is 89%, and the eccentricity ratio is 0.16. Determine themajor diameter and the thickness and displacement of each rotor.

7.21 A Wankel rotary engine has an eccentricity of 2.2 in. and a major diameter of 28in. It has a compression ratio of 9.5 and a 600 in.3 displacement. Determine therotor width and the rotor sector included angle if the rotor flanks are circular andhave no indentations.

7.22 A Wankel rotary engine has an eccentricity of 2.5 cm and a major diameter of 32cm. The engine compression ratio is 9.0, and the displacement is 540 cm3. Whatare the eccentricity ratio, the rotor width, and the rotor sector included angle ifthe rotor flanks are circular and have no indentations?

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C H A P T E R 8

REFRIGERATION AND AIR CONDITIONING

8.1 Introduction

Up to this point we have considered fossil-fueled heat engines that are currently in use. Thesedevices have provided society's answers to the thermodynamic question: How can the chemicalenergy of fossil fuels be converted into mechanical work and motive power? Let us now turn ourattention to the another great thermodynamic question: How can thermal energy be transferredfrom cold to warmer regions?

The well-known Clausius statement of the Second Law of Thermodynamics asserts: It isimpossible to construct a device that, operating in a cycle, has no effect other than the transferof heat from a cooler to a hotter body. Thus the Clausius statement tells us that energy (heat) willnot flow from cold to hot regions without outside assistance. The devices that provide this helpare called refrigeration units and heat pumps. Both types of devices satisfy the Clausiusrequirement of external action through the application of mechanical power or natural transfers ofheat (more on this later).

The distinction between refrigerator and heat pump is one of purpose more than technique.The refrigeration unit transfers energy (heat) from cold to hot regions for the purpose of coolingthe cold region while the heat pump does the same thing with the intent of heating the hot region.The following will focus on refrigeration and make the distinction between refrigeration and heatpumps only when it is essential to the discussion.

It should be pointed out that any heat engine cycle, when reversed, becomes a refrigerationcycle, becuase the cyclic integral of the heat transfer, and thus the net work, becomes negative.This implies heat rejection at higher than the lowest cycle temperature.

The vapor compression refrigeration system is the mainstay of the refrigeration and airconditioning industry. Absorption refrigeration provides an alternative to the vapor compressionapproach, particularly in applications where a heat source is economical and readily available. Thischapter considers both of these system types in turn, and closes with a discussion of moist airbehavior and its influence on air conditioning system design.

Numerous other specialized refrigeration systems exist. As just stated, in principle any heatengine cycle, when reversed, becomes a refrigeration cycle. Such cycles are usually discussed inthermodynamics texts. Other systems that are occasionally used in special applications includethermoelectric coolers, which employ electrical work, and Hilsch or vortex tubes, which employcompressed gas as an energy source.

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8.2 Vapor Compression Refrigeration

Consider an insulated cold region of temperature TL as shown in Figure 8.1. Heat leakage fromthe surroundings to the system tends to increase the system�s temperature. In order to keep thecold region at temperature TL, the conservation of energy requires the removal of an amount ofheat equal to the energy inflow. This is done by a cold region heat exchanger that has an evencolder liquid flowing through it to carry away the heat. If the fluid is a saturated liquid, it willevaporate and absorb energy from the cold region in its heat of vaporization. Such a heatexchanger is called an evaporator. Thus the basic problem of refrigeration may be reduced to one

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of providing a mechanism to supply cool saturated liquid or a mixture of liquid and vapor, therefrigerant, to an evaporator.

Vapor compression refrigeration, as the name suggests, employs a compression process toraise the pressure of a refrigerant vapor flowing from an evaporator at pressure p1 to p2, as shownin Figure 8.2. The refrigerant then flows through a heat exchanger called a condenser at the highpressure, p2 = p3, through a throttling device, and back to the low pressure, p1, in the evaporator.The pressures p2 = p3 and p4 = p1 correspond to refrigerant saturation temperatures, T3 and T1 =T4, respectively. These temperatures allow natural heat exchange with adjacent hot and coldregions from high temperature to low. That is, T1 is less than TL; so that heat, QL, will flow fromthe cold region into the evaporator to vaporize the working fluid. Similarly, the temperature T3

allows heat, QH, to be transferred from the working fluid in the condenser to the hot region at TH.This is indicated by the arrows of Figure 8.2.

Thus the resulting device is one in which heat is transferred from a low temperature, TL, to ahigh temperature, TH, using a compressor that receives work from the surroundings, thereinsatisfying the Clausius statement.

The throttling device, as shown in Figure 8.2, restrains the flow of refrigerant from thecondenser to the evaporator. Its primary purpose is to provide the flow resistance necessary tomaintain the pressure difference between the two heat exchangers. It also serves to control therate of flow from condenser to evaporator. The throttling device may be a thermostatic expansionvalve (TEV) controlled by evaporator exit temperature or a long, fine-bore pipe called a capillarytube.

For an adiabatic throttling device, the First Law of Thermodynamics requires that h3 = h4 forthe irreversible process, because Q and W are zero and kinetic energy change is negligible. Thussaturated liquid at T3 flashes to a mixture of liquid and vapor at the evaporator inlet at theenthalpy h4 = h3 and pressure p4 = p1. Also the evaporator entrance has the quality x4 andtemperature T4 = T1. Heat from the cold source at TL > T4 boils the mixture in the evaporator to asaturated or slightly superheated vapor that passes to the suction side of the compressor.

The compressor in small and medium-sized refrigeration units is usually a reciprocating orother positive-displacement type, but centrifugal compressors often are used in large systemsdesigned for commercial and industrial service.

It may be noted from the T-s diagram in Figure 8.2 that the vapor compression cycle is areversed Rankine cycle, except that the pressure drop occurs through a throttling device ratherthan a turbine. In principle, a turbine or expansion device of some kind could be used tosimultaneously lower the refrigerant pressure and produce work to reduce the net work requiredto operate the compressor. This is very unlikely because of the difficulty of deriving work from amixture of liquid and vapor and because of the low cost and simplicity of refrigeration throttlingdevices.

An exploded view of a through-the-wall type room air conditioner commonly used in motelsand businesses is shown in Figure 8.3. A fan coil unit on the space side is the evaporator. Athermally insulating barrier separates a hermetically sealed, electric-motor-driven positivedisplacement compressor unit and a finned-tube heat exchanger condenser from the room on theoutdoor side.

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Figure 8.4 shows a packaged air-conditioning unit designed for in-space use or for a nearbyspace with short duct runs. Units sometimes are designed to operate with either one or twocompressors, coils, and fans to better accommodate varying cooling demands. A unit with water-cooled condensers such as that shown requires an external heat sink, usually provided by a nearbyground-level or rooftop evaporative cooling tower.

Rather than being combined in a single enclosure, refrigeration units frequently are installedas split systems. Figure 8.5 shows an uncovered rooftop condensing unit that contains acompressor and air cooled condenser. Such units are, of course, covered to resist the outdoorenvironment over many years. Cooled refrigerant is piped in a closed circuit to remote airdistribution units that contain cooling coils (evaporators) and throttling devices. Figure 8.6shows a skid-mounted air-cooled condensing unit also designed to function with remoteevaporators in applications such as walk-in coolers.

Refrigerants

Refrigerants are specially selected substances that have certain important characteristics includinggood refrigeration performance, low flammability and toxicity, compatibility with compressorlubricating oils and metals, and good heat transfer characteristics. They are usually identified by anumber that relates to their molecular composition.

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The ASHRAE Handbook (ref. 1) identifies a large number of refrigerants by number, asshown in Table 8.1. Inorganic refrigerants are designated by 700, plus their molecular weight. Forhydrocarbon and halocarbon refrigerants, the number scheme XYZ works as follows: (1) Z, on theright is the number of fluorine atoms; (2) Y is the number of hydrogen atoms plus one; and (3) theleftmost digit, X, is one less than the number of carbon atoms in the compound.

Two important examples are refrigerants R-12 and R-22. R-12, dichlorodifluoromethane, hastwo fluorine, one carbon, and two chlorine atoms in a methane-type structure. Thus the halogens,

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chlorine and fluorine, replace hydrogen atoms in the CH4 molecular structure as shown in Figure8.7. R-22, monochlorodifluoromethane, has a similar structure to R-12, except for a singlehydrogen atom replacing a chlorine atom. Charts of the thermodynamic properties of theserefrigerants are given in Appendix F.

The commonly used chlorofluorocarbon (CFC) refrigerants are a cause of great concern,because their accumulation in the upper atmosphere creates a �hole� in the ozone layer thatnormally shields the earth from solar ultraviolet radiation (refs. 8 and 9). In 1987, more than 35countries, including the United States, signed the Montreal Protocol on Substances that Depletethe Ozone Layer. The Montreal Protocol called for a freeze in 1989 and reductions in the 1990son the production levels of R-11, R-12, R-113, R-114, and R-115. The halocarbon refrigerants,some of which are also widely used as aerosol propellants, foams, and solvents, are nowcategorized as chlorofluorocarbons (CFCs), hydrochlorofluorocarbons (HCFCs), orhydrofluorocarbons (HFCs). The HFCs, lacking chlorine, are no threat to the ozone layer but arenot in common usage as refrigerants. CFCs, which contain more chlorine than do HCFCs, are themost serious offenders, are very stable, and do not break down rapidly in the lower atmosphere.The Clean Air Act of 1990 (ref. 15) mandated termination of production in the United States ofall CFCs such as R-12 by the year 2000. Government data indicate that, because of the structuraldifference between them, R-12 has twenty times the ozone-depletion potential in the upperatmosphere of R-22. Nevertheless, R-22 and other HCFCs are also scheduled by the law forphaseout of production by the year 2030.

Thus, the search for alternate refrigerants to replace those used in existing applications(worth hundreds of billions of dollars) has assumed enormous importance. It is a difficult,expensive, and continuing task to which industry is vigorously applying its efforts. Charts ofthermodynamic properties for two of the newer refrigerants, R-123 and R134a, are given inAppendix F.

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Vapor-Compression Cycle Analysis

A vapor-compression cycle was shown in Figure 8.2, The work required by the refrigerationcompressor, assuming adiabatic compression, is given by the First Law of Thermodynamics:

w = h1 � h2 [Btu/lbm | kJ/kg] (8.1)

where the usual thermodynamic sign convention has been employed. The enthalpies h1 and h2

usually are related to the temperatures and pressures of the cycle through the use of charts ofrefrigerant thermodynamic properties such as those given in Appendix F.

In the ideal vapor compression cycle, the compressor suction state 1 is assumed to be asaturated vapor. The state is determined when the evaporator temperature or pressure is given.For the ideal cycle, for which compression is isentropic, and for cycles for which the compressionis determined using a compressor efficiency, state 2 may be defined from state 1 and thecondensing temperature or pressure by using the chart of refrigerant thermodynamic properties.Assuming no heat exchanger pressure losses, the evaporator and condenser heat transfers areeasily determined per unit mass of refrigerant by application of the First Law of Thermodynamics:

qL = h1 � h4 [Btu/lbm | kJ/kg] (8.2)

qH = h3 � h2 [Btu/lbm | kJ/kg] (8.3)

The evaporator heat transferred, qL, is commonly referred to as the refrigeration effect, RE.The product of the refrigerant mass flow rate and RE, the rate of cooling produced by the unit, is called the refrigeration capacity [Btu/hr | kW].

Applying the First Law to the refrigerant in the system as a whole, we find that the work andheat-transfer terms are related by

qL + qH = w [Btu/lbm | kJ/kg] (8.4)

where qH and w are negative for both refrigerators and heat pumps. Hence

qL + |w| = |qH| [Btu/lbm | kJ/kg] (8.5)

Equation (8.5) is written here with absolute values to show that the sum of the compressor workand the heat from the low-temperature source is the energy transferred by the condenser to thehigh-temperature region. This may be seen graphically by addition of the enthalpy incrementsrepresenting Equations (8.1) to (8.3) in the pressure-enthalpy diagram shown in Figure 8.8. Thep�h diagram is applied often in refrigeration work because of its ease of use in dealing withenthalpy differences and constant-pressure processes.

EXAMPLE 8.1

Derive Equation (8.1) by using Equations (8.2) to (8.4).

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Solution

From Equation (8.4),

w = qL + qH = (h1 � h4) + (h3 � h2)

But for the adiabatic throttling valve, h3 = h4. Hence,

w = h1 � h2

as in Equation (8.1)._____________________________________________________________________________

System Performance

As mentioned in Chapter 1, a measure of the efficiency of a refrigerator or heat pump is thecoefficient of performance, COP. The COP for a refrigerator is defined as the ratio of the useful

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effect or desired energy transfer accomplished by the evaporator (RE) to the energy cost toachieve the effect (the compressor work), in equivalent units:

COPr = qL/|w| [dl] (8.6)

Like a refrigerator, a heat pump may be a vapor compression system; but unlike therefrigerator it takes energy from a cold source and transfers it to a hot region for the purpose ofraising the temperature of the region or compensating for heat losses from the region. Somecommercial heat pumps are systems that combine both cooling and heating actions in a singlepackage. These systems are discussed more fully later. For a heat pump, the useful effect or desired energy transfer is the passage of energy from thecondenser to the high-temperature region at TH. Thus the COP for a heat pump is

COPhp = |qH| / |w| [dl] (8.7)

It is evident from Equations (8.6) and (8.7) that high values of COP are desired and may beachieved by minimizing the compressor work input for given values of heat transfer. It may beparticularly advantageous to apply a vapor compression system in situations and applications inwhich simultaneous heating and cooling functions are required�to cool a computer room with theevaporator while heating rooms on the cold side of the building with the condenser, for example.

EXAMPLE 8.2

Show that the COP for a heat pump exceeds 1.

SolutionCombining Equations (8.7) and (8.5) yields

COPhp = |qH| / |w| = (qL + |w|) / |w|

= 1 + qL / |w| = 1 + COPr > 1

Thus, because qL > 0 and COPr >0 , the heat pump COP must always exceed 1. This is alsoevident directly from the definition Equation (8.7) because the First Law [Equation (8.5)] requiresthat |qH| > |w|. Note the relationship of refrigerator and heat pump COPs for the same cycle._____________________________________________________________________________

The T-s diagram in Figure 8.2 shows that condenser and evaporator pressures of both refrigerators and heat pumps are determined largely by the source and sink (cold-region and hot-region) temperatures, TL and TH, respectively. Thus, the greater the difference between the hot-region and cold-region temperatures (the application temperature difference), the higher thecompressor pressure ratio and the higher the compressor work. For low application temperaturedifferences, |w| can be significantly less than qL; and thus COPr may also exceed 1. Conversely,large application temperature differences lead to low coefficients of performance.

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Prior to the twentieth century, the cooling of both people and perishable goods was achievedby using ice formed in the winter by nature from liquid water. Ice from lakes and ponds innorthern climates was cut, hauled to ice storage houses, and delivered to customers as needed.

As late as the 1930s and �40s, the ice vendor would make a daily delivery of ice so thatperishable goods could be kept fresh in insulated ice storage boxes, the predecessors of themodern refrigerator. By then, ice was being made by freezing water via large refrigeration unitswith ammonia compressors; and comfort air cooling was being installed in the best theaters andwealthiest homes in the United States.

It is not surprising that a measure of rate of cooling, or cooling capacity, is often related tothe cooling capability of a ton of ice. A ton of refrigeration is defined as the rate of coolingproduced by melting a ton of ice in 24 hours. Based on this definition, and on the fact that thelatent heat of fusion of water is 144 Btu/lbm, it may be deduced that a ton of refrigeration is a rateof cooling of 200 Btu/min, or 12,000 Btu/hour (see Exercise 8.3). Thus a 3-ton air conditioner isone that nominally removes 36,000 Btu/hr from the cooled space.

When shopping for an air conditioner, one encounters labels that give the product's energyefficiency ratio (EER), that is, the ratio of the cooling capacity of the unit, measured in Btu/hr, tothe power required to operate it, in watts. It is evident that the EER is proportional to the COP,neglecting fan power. Applying the conversion factor between Btus and watt-hours one sees thatEER = 3.413�COP.

EXAMPLE 8.3

For an ideal vapor compression refrigeration system operating with refrigerant 22 at anevaporator temperature of 0°F and condensing at 100°F, find the following: the compressorsuction and discharge pressures, enthalpies, and specific volumes; the condenser dischargepressure and enthalpy; the refrigeration COP; and the refrigerant mass flow rate and powerrequirement for a 10-ton refrigeration unit.

SolutionFollowing the notation of Figures 8.2 and 8.8, from the chart (Appendix F) for refrigerant 22

at T1 = 0°F, the other properties at state 1 are p1 = 38 psia, h1 = 104 Btu/lbm, v1 = 1.4 ft3/lbm, and s1 = 0.229 Btu/lbm-R.

The saturated-liquid condenser discharge properties at T3 = 100°F are p3 = 210 psia and h3 =39 Btu/lbm.

The compressor discharge-state properties at s2 = s1 and p2 = p3 = 210 psia are h2 = 123Btu/lbm, T2 = 155°F, and v2 = 0.31 ft3/lbm.

The evaporator inlet enthalpy is the same as that at condenser discharge, h4 = h3 = 39Btu/lbm.

The refrigeration effect and the compressor work are then

RE = h1 � h4 = 104 � 39 = 65 Btu/lbm

w = h2 � h1 = 123 - 104 = 19 Btu/lbm

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ThusCOPr = RE /w = 65/19 = 3.42.

The rate of cooling, or cooling capacity, for a 10-ton unit is 10×200 = 2000 Btu/min. Therefrigerant mass flow rate is the capacity divided by the refrigeration effect = 2000/65 = 30.8lbm/min.

The power required by the compressor is the product of the mass flow rate and thecompressor work = 30.8×19 = 585.2 Btu/min, or 585.2×60/ 3.413 = 10,290 W, or 10.29 kW.

The ideal EER may then be calculated from the capacity and power as 2000×60/10,290 =11.7 Btu/Watt-hr, or from the COP as 3.413×3.42 = 11.7 Btu/Watt-hr._____________________________________________________________________________

Compressors

While most small- and medium-capacity refrigeration systems use hermetically sealed, electric-motor-driven compressor units or open (externally powered) reciprocating compressors,centrifugal compressors are frequently found in large units for cooling buildings and for industrialapplications.

The reciprocating compressor has much in common geometrically with a simple two-strokereciprocating engine with intake and exhaust valves. As in that case, the compressor clearancevolume Vc is the volume at top center, and the piston sweeps out the displacement volume, asindicated in Figure 8.9. The processes 1-2-3-4-1 on the idealized pressure-volume diagramrepresent the following:

� 1�2 Both valves are closed. Compression of the maximum cylinder volume V1 = Vc + Vd

of refrigerant vapor through the pressure ratio p2/p1 to a volume V2.

� 2�3 Exhaust valve is open. Discharge of refrigerant through the exhaust valve atcondenser pressure p3 until only the clearance volume V3 = Vc remains when the piston isat top center.

� 3�4 Both valves are closed. Expansion of the clearance gas with both valves closed fromV3 to V4. Note that the inlet valve cannot open until the cylinder pressure drops to p4 = p1

without discharging refrigerant back into the evaporator.

� 4�1 Intake valve is open. Refrigerant is drawn from the evaporator into the cylinder atconstant pressure p1 through an intake valve by the motion of piston. Refrigerant in theamount V1 � V4 is processed per cycle.

Assuming polytropic compression and expansion processes with the same exponent k:

V4 = V3( p3/p4)1/ k = Vc( p2/p1)

1/ k

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Then the volume of refrigerant vapor processed per cycle (or per shaft revolution)

V1 � V4 = Vd + Vc � Vc ( p2/p1)1/ k = Vd � Vc [( p2/p1)

1/ k � 1]

is less than the displacement volume and depends on the compressor pressure ratio. Neglectingthe difference between the refrigerant density leaving the evaporator and that in the compressorcylinder just before compression, we may write the compressor volumetric efficiency as the ratioof V1 � V4 to the displacement:

�v = (V1 � V4)/Vd = 1 � (Vc /Vd) [( p2/p1)1/ k � 1]

Examination of the compressor processes for different pressure ratios, as in Figure 8.9 ( p2' /p1,for example), shows that the refrigerant volume processed per cycle, and thus the volumetric

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efficiency, decreases with increasing pressure ratio. It is also evident that the clearance volumemust be kept small to attain high volumetric efficiency. It is clear that, for a given positive-displacement compressor, the volumetric efficiency limits the usable pressure ratio and thus thedifference between the condensing and evaporating temperatures.

EXAMPLE 8.4

What is the volumetric efficiency of the reciprocating compressor used with the refrigerationsystem of Example 8.3 if the clearance volume is 3% of the displacement volume and thepolytropic exponent is 1.16?

SolutionThe compressor pressure ratio is p2/p1 = 210/38 = 5.526. From the preceding equation, the

volumetric efficiency is then

�v = 1 � (0.03) [(5.526)1/1.16 � 1] = 0.899_____________________________________________________________________________

The centrifugal compressor used in refrigeration is similar in principle to that introducedearlier in connection with the turbocharger. It is fundamentally different than the positive-displacement reciprocating compressor in that it operates on a fluid dynamic principal in whichkinetic energy imparted to the refrigerant by an impeller is converted to pressure analogous to thatindicated by the Bernoulli equation. The high-speed rotor accelerates and thus increases thekinetic energy of the refrigerant, which is then converted to pressure in diffuser passages. Because the efficiency of small centrifugal compressors is limited, centrifugals are used primarilyin large-scale applications with cooling loads in tens or hundreds of tons of refrigeration.

The scroll compressor, shown in Figure 8.10, though based on a century-old concept, is arecent development in rotary compressors. In it, an electric-motor-driven orbiting scroll mesheswith a stationary scroll. Six crescent-shaped pockets formed between the involute scrolls trap,transport, and compress low-pressure refrigerant from the outside of the scrolls to the dischargepassage at the scroll center. This process is more easily visualized by following the shaded pocketof refrigerant in Figure 8.11. Some of the advantages attributed to the scroll compressor are:continuous and smooth compression with 100% volumetric efficiency, no valves, physicalseparation of suction and discharge lines (which avoids the temperature cycling of reciprocatingcompressors), and higher compression efficiency than the best reciprocating units (ref. 10).Seasonal energy efficiency ratios (EERs averaged over a cooling season) in excess of 10 arereadily attained with this technology. An animation of a scroll compressor may be seen on theinternet at www.copeland-corp.com.

Suction and Subcooling Considerations

Let's examine two items of concern with respect to some vapor compression systems. In systemswith reciprocating compressors, there is a danger that, due to changing cooling loads, that theliquid refrigerant in the evaporator may not be completely vaporized, causing slugs of liquid to

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enter the compressor. Because liquids are essentially incompressible, positive-displacement compressors with fixed clearance volumes can be damaged when such "slugging" occurs.

The use of a thermostatic expansion valve (TEV) that responds to change in the degree ofsuperheat in the suction line provides one solution to this problem. A bulb filled with refrigerantattached to the suction line, when heated by superheated vapor, transmits an increasing pressuresignal to a diaphragm in the TEV, which adjusts the valve flow area and in turn changes the mass

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flow rate of refrigerant. This control is usually set to maintain a minimum of about five degrees ofsuperheat to avoid liquid slugs entering the compressor inlet.

A second concern is the possibility of entry of vapor into the throttling valve if the refrigerantat the condenser exit is not completely condensed. Because vapor occupies much more space thanliquid, the throttling valve will not function properly if vapor can enter from the condenser. Oneapproach to dealing with this is to locate a liquid receiver downstream of the condenser to assurethe availability of liquid to the expansion device.

Both of the above concerns may be dealt with simultaneously by the addition of a suction-lineheat exchanger that superheats the evaporator discharge about five degrees, ensuring that onlyvapor enters the compressor. The heat exchanger that provides suction superheat from state 6 tostate 1 in Figure 8.12 may be set up to receive heat from the subcooling of the condenserdischarge from state 3 to state 4. This ensures the absence both of vapor entering the throttlingvalve and of liquid slugs entering the compressor. Note that the subcooling also tends to increasethe refrigeration effect over that of the ideal cycle by decreasing the enthalpy entering the

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evaporator. This effect may be accomplished by running the compressor suction line next to thecondenser discharge, allowing some thermal contact.

Combining Heating and Cooling in a Single System

It is possible to combine both heating and cooling functions in a single system by providing heatexchangers that can operate as both evaporator and condenser and a control system that canreroute the flow of refrigerant when switching functions is required. Figure 8-13 presents a

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schematic diagram for such a system, commonly called a heat pump (context usually determineswhether the term �heat pump� refers to a device that heats only or that combines heating andcooling functions). The key component in a commercial heat pump is a reversing valve. With thevalve shown in the figure, rotation through an angle of 90° reroutes the flow of refrigerant fromthe indoor coil to the outdoor coil, and vice versa. As a result of this change, the indoor coil thatserved as a condenser in the winter becomes an evaporator in the summer. The outdoor coilchanges accordingly. Separate throttling devices may be used to accommodate differing loadconditions in winter and summer. One-way check valves ensure that refrigerant flow is throughthe appropriate throttling device during each season.

8.3 Absorption Refrigeration

Example 8.3 shows that vapor compression refrigeration requires a significant supply of workfrom an electric motor or other source of mechanical power. Absorption refrigeration is analternate approach to cooling that is largely thermally driven and requires little external work.This form of refrigeration is growing in importance as energy conservation considerations demandcloser scrutiny of the disposition of heat rejection from thermal processes. Absorptionrefrigeration provides a constructive means of utilizing waste heat or heat from inexpensivesources at a temperature of a few hundred degrees, as well as directly from fossil fuels. Theeventual abolition of the use of CFCs may also boost absorption refrigeration technology.

This system relies on the fact that certain refrigerant vapors may be dissolved in liquids calledabsorbents. For instance, water vapor is a refrigerant that tends to dissolve in liquid lithiumbromide, an absorbent. Just as when they condense, vapors release heat when they go intosolution. This heat must be removed from the system in order to maintain a constant temperature.Thus, cooling causes vapor to be absorbed in absorbents, just as cooling causes vapor tocondense. On the other hand, heating tends to drive vapor out of solution just as it turns liquid tovapor. This solution phenomenon and the fact that pumping liquid requires a relatively smallamount of work compared with that required to compress a gas are the secrets of absorptionrefrigeration.

Consider the schematic diagram in Figure 8.14, which shows a basic absorption refrigerationunit. The condenser / throttling valve / evaporator subsystem is essentially the same as in thevapor compression system diagram of Figure 8.2. The major difference is the replacement of thecompressor with a different form of pressurization system. This system consists primarily of anabsorber at the pressure of the evaporator, a vapor generator at the pressure of the condenser,and a solution pump. A second throttling valve maintains the pressure difference between theabsorber and the generator.

The system operates as follows: Refrigerant vapor from the evaporator flows into theabsorber, where it mixes with the absorbent. The mixture is cooled by heat transfer QA to air orwater at the temperature of the environment, causing the vapor to go into solution. Therefrigerant-absorbent solution flows to the solution pump where it is pressurized to the pressurelevel of the generator and condenser. Heat from an energy source, QG, then drives the vapor fromthe cold liquid solution. The vapor flows into the condenser while the heated liquid in the

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generator passes back to the absorber through an absorbent throttling valve, thus completing theabsorbent loop.

As in vapor compression systems, the condensed vapor from the condenser passes throughthe refrigerant throttling valve into the evaporator as a chilled refrigerant liquid-vapor mixture,where its vaporization creates the desired refrigeration effect. The evaporator vapor then flowsback to the absorber to again go into solution and repeat the process. It should be noted that theonly work required is a relatively small amount needed to operate the solution pump.

Thus the major energy requirement of an absorption refrigeration system is a transfer of heatQG from a source of moderate temperature, such as an engine combustion-gas exhaust, steamfrom a low-pressure boiler, or perhaps a heat transfer fluid from solar collectors. The coefficientof performance of an absorption system is therefore defined as the ratio of the refrigerationcapacity to the rate of heat addition in the generator. This definition is fundamentally different

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from that of the COP of a vapor compression system, because the energy source is in the form ofheat rather than work. Therefore, it should not be surprising that absorption system COPs aremuch lower than those for vapor compression systems. 8.4 Comfort Air Conditioning

Professionals define air conditioning to include heating, humidifying, dehumidifying, circulating,cleaning, and those other operations involving air (in addition to cooling) that provide thermalcomfort to people and effectiveness to processes. Studies of what constitutes thermal comfort toindividuals have identified air temperature, humidity, ventilation, air movement, and air cleanlinessas important control parameters. The systems needed to control all of these parameters arediscussed in many references, including refs. 1 through 5. This chapter is concerned primarily withtemperature and humidity control.

Space cooling and space heating are the primary functions that rely on vapor compressionand absorption systems. In specifying systems to perform these functions, it is necessary to definethe winter heating loads and summer cooling loads [Btu/hr | kW] for the space to be heated orcooled. Most HVAC (heating, ventilating, and air conditioning) professionals now have computerprograms that assist in defining these loads.

The determination of cooling loads involves delimiting the space to be conditioned and defining the appropriate indoor and outdoor design conditions. For example, for someapplications the indoor design temperature and humidity might be specified as 74°F and 50%relative humidity based on thermal comfort data and experience. For a given locale, the ASHRAEHandbook gives outdoor temperatures that are not exceeded during a given fraction of thesummer season. For example, the Handbook indicates that the temperature in Tulsa, Oklahoma isusually not greater than 98°F during 97.5% of the four-month period from June to September.This, therefore, might be selected as a suitable summer outdoor design temperature.

A thermal cooling load analysis is performed with these design conditions and data on thestructure and contents of the enclosure of the conditioned space for times when the load isexpected to be greatest. The analysis involves heat transfer through walls, roofs, doors, andwindows; and internal heat generation due to electrical devices, gas burning appliances, humanoccupants, and other sources. At such times, solar radiation is usually a strong influence. In mostapplications, additions of water vapor to air in the space must be considered also. The results ofthese calculations yield the design rates at which heat and moisture must be removed from thespace. The air conditioning evaporator cooling capacity and the fan air circulation rate are thensized to provide for these near-extreme rates.

Cooling and heating loads at a given locale may be expected to vary approximately linearlywith the deviation of the ambient temperature from a break-even temperature of about 65°F. Atthis outdoor temperature, no mechanical heating or cooling is required for thermal comfort. Thebreak-even ambient temperature is usually below the indoor design temperature because of thenecessity of a temperature difference for the rejection of heat through the walls required tobalance the energy generation from sources in the space. Figure 8.15 shows typical performancebehavior of vapor compression heating and cooling equipment superimposed on correspondingload curves. The equipment is necessarily oversized for mild temperatures. In the case of the heat

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pump, there is an ambient temperature at which the equipment capacity matches the heating load,the balance point. At lower temperatures the heat pump is unable to maintain the design indoortemperature, a heating deficit. This situation is usually handled by the use of electrical resistanceheaters to supplement the heat pump output. The electric resistance heaters usually are switchedon in banks as the ambient temperature decreases further below the break-even point. While theheat pump itself may be quite efficient, pure resistance heating with COP = 1.0 results indegradation in overall efficiency and economy while the resistance heaters are in operation.

Operationally, for the frequent periods when capacity exceeds load, a thermostat is usuallyused to cycle the system on and off to provide for indoor comfort in a limited range oftemperature around the indoor design value. More sophisticated systems resulting from advancesin motor control, compressors, and fans may allow the system to operate at part load continuous-ly, to avoid the discomfort and inefficiency of temperature cycling associated with simplethermostat operation.

The dual-fuel heat pump, a new approach to dealing with the weaknesses of the heat pumpat low temperatures, has been proposed and studied by the Electric Power Research Institute, anequipment manufacturer, and several electric utilities (ref. 14). Instead of expensive and energy-wasteful resistance heating making up the heating deficit when ambient temperatures requireoperation below the heat pump balance point, a gas furnace built into the heat pump packagecomes online and economically satisfies the need for additional heat. The combined electric heatpump and gas unit operate together, unless the ambient temperature drops to still lower

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temperature levels at which the cost of gas operation falls below the cost of electricity. Belowthis break-even point electric heat pump component shuts down and gas heat is used exclusively.

The break-even point may be adjusted when energy prices change. According to thereference, in regions where gas prices are high and electricity low, dual-fuel heat pump users willbenefit from substantial reductions in operating costs and the utilities from increased off-peaksales in the winter months. However, the purchase price of the dual-fuel heat pump unit isexpected to be higher than combined gas heating and electric cooling systems now used. Thiscreates a marketing problem, because consumers often are more impressed by low purchase pricethan by low life-cycle costs resulting from low operating costs.

8.5 Moist Air and Air Conditioning System Design

Psychrometrics

Although the mass fraction of water vapor in ambient air usually is numerically small, water vaporcan have a significant influence on an individual's thermal comfort and on the design of airconditioning equipment. Most people can recall instances of discomfort associated with bothexcessively dry or moist air. Numerous industrial processes also require moisture control tomaintain high standards of quality and productivity. The engineering fundamentals of dealing withmoist air and the application of those fundamentals are treated here.

The study of the behavior of moist air is called psychrometrics. We will consider moist air asa two-component mixture of dry air and water vapor, and treat both components and the mixtureas ideal gases in the context of the mixture theory discussed in Chapter 3.

Humidity Ratio

Properties of moist air are commonly referred to the mass of dry air mda, apart from the watervapor, contained in the air. For example, the humidity ratio is defined as the ratio of the mass ofwater vapor to the mass of dry air present in the moist air:

W = mwv/mda [dl] (8.8)

Together, the humidity ratio and the ambient temperature and pressure define thethermodynamic state of moist air, and therefore may be used to define other properties of moistair. Using the ideal gas law, the humidity ratio may be written as

W = ( pwvV/RwvT)/( pdaV/RdaT) = ( pwv/pda)(Rda/Rwv)

= (pwv/pda)(Mwv/Mda)

= 0.622( pwv/pda) [dl] (8.9)

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where we have used the facts that the water vapor and dry air occupy the same space andtherefore have the same volume, and that both components are in thermal equilibrium andtherefore are at the same temperature. The ratio of gas constants in Equation (8.9), expressed interms of the Universal Gas Constant, reduces to the inverse ratio of molecular weights Mwv /Mda =18/28.9 = 0.622.

Thus the humidity ratio is equivalent to the ratio of partial pressures of water vapor and dryair. Going one step further, consider the total air pressure P, the pressure that we measure with abarometer, as the sum of partial pressures of dry air and water vapor according to Dalton's Law:

P = pda + pwv [lbf/ft2 | kPa] (8.10)

The humidity ratio then may be written as:

W = 0.622pwv/(P � pwv) [dl] (8.11)

Thus, for a given ambient pressure, the humidity ratio is determined by the existing partialpressure of water vapor. It is evident that because pwv << P, the humidity ratio variesapproximately linearly with the partial pressure of water vapor in the air. Humidity ratio issometimes expressed in grains of water vapor per pound of dry air where there are 7000 grainsper pound.

Saturated Air

Moist air is said to be saturated when its humidity ratio is a maximum for the existing temperatureand total pressure. The humidity ratio for saturated air may be determined using Equation (8.11)with the saturation pressure of water vapor obtained from saturated steam tables at the knowntemperature.

EXAMPLE 8.5

Determine the partial pressures of water vapor and dry air and the humidity ratio in saturatedmoist air for an ambient total pressure and temperature of 14.7 psia and 80°F, respectively.

SolutionFrom the saturated-steam tables at 80°F, the partial pressure of water vapor is 0.50683 psia.

The partial pressure of the dry air component is then 14.7 � 0.50683 = 14.2 psia, and the humidityratio is

W = 0.622(0.50683)/(14.7 � 0.50683) = 0.0222

Thus, at 80°F, the moist air may have no more than 0.0222 lb (or 155.4 grains) of moisture perpound of dry air._____________________________________________________________________________

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While air at a given temperature and pressure can have no more water vapor than indicatedby the humidity ratio for saturated air, the air can be drier. This is, of course, indicated by lowervalues of humidity ratio and also by relative humidity, as discussed next.

Relative Humidity and Dew Point

An alternative parameter for indicating the moisture content of air is the relative humidity, �. Therelative humidity is defined as the ratio of the mole fraction of water vapor in unsaturated air tothe mole fraction of water vapor in air that is saturated at the same temperature and totalpressure:

� = xwv /xwv,sat [dl] (8.12)

It has been shown in Chapter 3 that the mole fraction of a component of an ideal gas is the ratioof the partial pressure of that component to the total pressure of the gas. Thus the relativehumidity may be written as

� = (pwv /P)/(pwv /P)sat = pwv /pwv,sat [dl] (8.13)

where the partial pressures are evaluated at the same temperature and total pressure. The relativehumidity of the air may be related to the p-V diagram for its water component in Figure 8-16. Thediagram shows two isotherms where T1 = T2 > T3. Considering the top isotherm, the relativehumidity of state 1 is the ratio of the pressure of water vapor at state 1 to that at the saturationvapor state at the same temperature, state 2. The figure shows that the maximum pressure ofwater vapor possible at a given temperature corresponds to the saturation pressure of water atthat temperature. Thus increasing amounts of water vapor in air at a given ambient temperatureare indicated by higher relative humidity, higher humidity ratio, higher water vapor partialpressure, and by vapor states closer to the saturated-vapor line for water. Thus the term"saturated air" implies that the water vapor in the air is a saturated vapor. These two uses of theterm "saturated" should be distinguished to avoid confusion.

When a given mass of moist air cools slowly, the mass of water vapor in the air remainsconstant until the air becomes saturated. Thus the humidity ratio and the partial pressure of thewater vapor are also fixed. This process is depicted by the line of constant pressure from 1 to 3 to4 in Figure 8.16. Note that, once state 4 is reached, no further cooling can occur withoutcondensation of water vapor. This condition, at which liquid water first appears, is known as thedew point. The formation of dew is a familiar early morning occurrence in which water vaporcondenses, leaving drops of liquid on grass and other surfaces that are cooler than thesurrounding air. The temperature at which water vapor begins to condense is therefore called thedew point temperature. It is evident from Figure 8.16 that as the air temperature at state 1 drops,approaching the dew point temperature, the relative humidity increases until it finally reaches100% at state 4. Thus the dew point temperature for any moist air state may be obtained from thesaturated-steam tables, if the partial pressure of the water vapor is known.

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EXAMPLE 8.6

In ambient moist air, the temperature and pressure are 70°F and 14.7 psia, respectively, and therelative humidity is 50%. What are the partial pressures of dry air and water vapor and thehumidity ratio and dew point?

SolutionAt 70°F, the steam tables give the saturation pressure of water as 0.36292 psia. This is the

partial pressure of water vapor in saturated air. From Equation (8.13), the partial pressure ofwater vapor in the ambient air is then (0.5)(0.36292) = 0.181, and the corresponding partialpressure of dry air is 14.7 � 0.181 = 14.52 psia. Equation (8.11) then gives the humidity ratio:

W = 0.622(0.181)/14.52 = 0.0078 lb

of water vapor per pound of dry air.

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The tables for saturated steam give the dew point temperature of about 51°F for the vaporpartial pressure of 0.181 psia._____________________________________________________________________________

Enthalpy of Moist Air

Since any extensive property of a mixture of ideal gases is the sum of the correspondingproperties of the components, the enthalpy of moist air is the sum of the enthalpies of dry air andwater vapor. Hence

H = Hda + Hwv = mdahda + mwvhwv [Btu | kJ]

where the mass of the mixture is mda + mwv. In order to define an enthalpy per unit mass for themixture, it must be decided which mass to select as a reference for the specific enthalpy of themixture. As with the humidity ratio, the usual choice in psychrometrics is the mass of dry air.Thus the specific enthalpy of moist air is given by h = H /mda:

h = hda + (mwv/mda)hwv = hda + Whwv [Btu/lbm | kJ/kg] (8.14)

Because the temperature range of most air conditioning operations is limited, it is usuallysatisfactory to use constant heat capacities in dealing with enthalpy calculations. Selecting thereference temperatures for enthalpy as 0° on both scales, the enthalpies of the dry air, expressed interms of Fahrenheit and Celsius temperatures, may be written as

hda = cpdaT [F] = 0.24T [F] [Btu/lbm]

and

hda = cpdaT [C] = 1.004T [C] [kJ/kg]

The enthalpy of water vapor is the sum of the sensible and latent enthalpies:

hwv = hs + hfg = cpwvT + hfg

= 0.446T [F] + 1061.2 [Btu/lbm] (8.15a)

= 1.867 T [C] + 2501.3 [kJ/kg] (8.15b)

EXAMPLE 8.7

Compare the enthalpy of water vapor from Equation (8.15a) with that from the steam tables for32°, 100°, 150°, and 200°F.

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SolutionFor 32°F, the enthalpy is

hwv = 0.446(32) + 1061.2 = 1075.47 Btu/lbm

This compares with 1075.5 Btu/lbm from the steam tables. The comparison for other temperaturesis as follows:

Temperature, °F Enthalpy, Btu/lbm � Equation (8.15a)

Enthalpy, Btu/lbm � Steam Tables

32 1075.47 1075.5

100 1105.8 1105.1

150 1128.1 1126.1

200 1150.4 1146

Thus the approximation is quite accurate in and beyond the air conditioning range, but it decreases in accuracy as temperature increases._____________________________________________________________________________

Combining the above relations, we may then write the enthalpies of moist air as

h = cpdaT [F] + W{cpwvT [F] + hfg}

= 0.24T [F] + W{0.446T [F] + 1061.2} [Btu/lbm] (8.16a)

h = cpdaT [C] + W{cpwvT [C] + hfg}

= 1.004T [C] + W{1.867T [C] + 2501.3} [kJ/kg] (8.16b)

Adiabatic Saturation

Consider a steady-flow process in which air flows through a liquid water spray in a duct where theair entrains moisture until it becomes saturated. If the duct is insulated so that the flow isadiabatic, and the water spray and the moist air leaving the duct are at the same temperature, theprocess is referred to as an adiabatic saturation process. It is shown shortly that measurement ofthe temperature and pressure of the air entering the duct and the final moist air temperature,called the adiabatic saturation temperature, determines the state of the moist air entering theduct. Thus two temperature measurements and a pressure measurement are sufficient todetermine the state of moist air.

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Figure 8.17 shows a schematic of an adiabatic saturation device in which liquid water at theadiabatic saturation temperature T2 is evaporated at a rate mw into a moist air stream flowing at arate of mda(1 + W) through the duct. Thus conservation of mass applied separately to both dry airand water components yields

mda2 = mda1 = mda [lbm/s | kg/s]

mdaW1 + mw = mdaW2 [lbm/s | kg/s]

Combining these equations, we find that the ratio of the mass flow rate of entrained water ormakeup water to the mass of dry air is

mw/mda = (W2 � W1) [dl] (8.17)

The steady flow form of the First Law yields

mda 1(hda 1 + W1hwv 1) + mw hf = mda 2(hda 2 + W2hwv 2)

where hf is the enthalpy of the saturated liquid at T2. Substituting Equation (8.17) into this

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equation yields

(hda1 + W1hwv1) + (W2 � W1)hf = (hda 2 + W2hwv 2)

or

W1(hwv 1 � hf) = cpda(T2 � T1) + W2(hwv 2 � hf) [Btu/lbm | kJ/kg] or

W1 ={cpda(T2 � T1) + W2(hwv 2 � hf)}/(hwv 1 � hf) [dl] (8.18)

Given the initial temperature T1, hwv 1 is determined from Equation (8.15). If T2 and P1 = P2 arealso measured, the saturation partial pressure, the humidity ratio W2, the vapor enthalpy hwv 2, andthe liquid-water enthalpy hf = hf(T2) may also be determined using the steam tables and Equations(8.11) and (8.15). Equation (8.18) may then be solved for the upstream (ambient) humidity ratio.The following example demonstrates the use of the adiabatic saturation concept in thedetermination of the ambient moist air state.

EXAMPLE 8.8

Ambient air at 14.7 psia and 70°F drops to 56°F in passing through an adiabatic saturator. What isthe ambient humidity ratio and the relative humidity?

SolutionSaturated air at 56°F yields a partial pressure of 0.22183 psia, a humidity ratio

W2 = 0.622(0.22183)/(14.7 � 0.22183) = 0.00953

a water-vapor enthalpy hwv 2 = 1086 Btu/lbm, and makeup-water enthalpy hf = 24.059 Btu/lbm.At the inlet where T1 = 70°F, the water-vapor enthalpy hwv 1 = 1092.1 Btu/lbm. Thus, from

Equation (8.18)

W1 = [0.24(56 �70) + 0.00953(1086 � 24.059)] /(1092.1 � 24.059) = 0.00633 lbof water vapor per pound of dry air.

Solving Equation (8.11) for the partial pressure of water vapor gives

pwv 1 = (W1)(P)/(0.622 + W1) = (0.00633)(14.7)/(0.622 + 0.00633) = 0.1481 psia

Because the saturation partial pressure of water vapor at 70°F is pwv 1, sat = 0.36292 psia, therelative humidity at state 1 is

� = 0.1481/0.36292 = 0.408, or 40.8%. _____________________________________________________________________________

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We may summarize the adiabatic saturation process as one in which liquid waterevaporates into the air flow, increasing the mass of water vapor in the air and thereforeincreasing the humidity ratio and, according to Equation (8.11), increasing the partial pressureof the vapor also. Because the flow is adiabatic, the heat of vaporization for the water mustcome from the flow of moist air, decreasing the air temperature. The increase in vapor partialpressure implies that the air dew point temperature increases as the flow passes through thesaturator, ending at the adiabatic saturation temperature.

In order to distinguish among the various temperatures dealt with in psychrometrics, thetemperature of moist ambient air, as measured by a mercury-in-glass thermometer, athermocouple, or an equivalent device, is known as the dry bulb temperature.

The sling psychrometer, a more convenient and inexpensive device for determining moistair states than the adiabatic saturator, uses two mercury-in-glass thermometers rigidlyattached to a frame with a pivoted handle as shown in Figure (8.18). The bulb of onethermometer, called the wet bulb, is wrapped in moistened wicking, and the other is left bareto measure the dry-bulb temperature. In operation, the user twirls the psychrometer for aminute or two until the readings of the thermometers stabilize. Evaporation of water from thewicking cools the wet bulb in proportion to the dryness of the ambient air. The temperaturedifference between the bulbs is commonly called the wet-bulb depression. Tables aresometimes provided to give relative humidity in terms of wet-bulb depression, but apsychrometric chart may be conveniently used to identify the moist air properties, as discussedin the next section.

The wet-bulb temperature, as an approximation to the adiabatic saturation temperature,together with the dry-bulb temperature and barometric pressure, determines the state of themoist ambient air. While the wet-bulb process differs fundamentally from the adiabaticsaturation process, in practice it provides a reliable approximation to the wet-bulbtemperature. Thus the adiabatic saturation temperature is sometimes referred to as thethermodynamic wet-bulb temperature.

Devices that give direct readings of humidity are called hygrometers. These devices areusually more complicated and expensive and less reliable than the psychrometer. They

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commonly make use of properties of materials that are influenced by the presence of watervapor, such as elongation and shrinkage. Human hair, paper, and nylon are examples of suchmaterials. More information on hygrometers is available in reference 2. The Psychrometric Chart

The properties and processes described above are conveniently related on a graph commonlyknown as a psychrometric chart. Although close examination of the skeleton chart of Figure8.19 and the detailed chart in the Appendix shows that it is not quite a rectangular coordinateplot, for most purposes it may be thought of as such, with humidity ratio as ordinate and drybulb temperature as abscissa. A psychrometric chart is constructed for a specified value ofbarometric pressure. Thus charts for different barometric pressures should be selected foranalyses for high altitudes, such as at Denver�s, where atmospheric pressure is low, and thosenear sea level, such as Houston�s, where it is normally high.

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While most psychrometric charts are based on the precise methods of reference 7, it isinstructive to consider how a chart could be constructed based on the ideal gas theorydiscussed earlier and on the steam tables. The curve labeled "saturation" on the psychrometricchart corresponds to states of air containing saturated water vapor at each dry-bulbtemperature. For each value of dry-bulb temperature, the saturation partial pressure obtainedfrom the steam tables determines the saturation humidity ratio for a specified atmosphericpressure [Equation (8.11)]. Thus the saturation curve relates the dew point temperature(measured on the abscissa) to the humidity ratio and to the corresponding partial pressure ofwater vapor. The saturation line also defines the conditions for 100% relative humidity.

For a given dry-bulb temperature, Equation (8.13) shows that the vapor partial pressurevaries linearly with �:

pwv = ��pwv, sat

For a given value of �, the humidity ratio can then be established using Equation (8.11).Hence the value of � and the dry-bulb temperature determine the value of the ordinate W.Thus a line of constant relative humidity � lies between the horizontal axis (pwv = 0 or W = 0)and the chart saturation curve at a vertical position in proportion of the magnitude of �.

Lines of constant volume of moist air per unit mass of dry air may be approximated usingthe ideal gas law applied to the dry air component:

v = RdaT/pda = RdaT/(P � pwv) [ft3/lbm | m3/kg] (8.19)

where Rda is the gas constant for dry air. Because dry air and water vapor occupy the samevolume independently, the specific volume of moist air based on unit mass of dry air is thesame as that for dry air. For zero humidity ratio, the air is dry and pwv = 0. Then the dry-bulbtemperature and ambient pressure alone determine the specific volume for each point on thehorizontal axis of the chart. As the humidity ratio and hence the vapor partial pressureincrease at constant ambient pressure, Equation (8.19) shows that the dry-bulb temperaturemust decrease to hold v constant. This establishes the negative slope of constant-specific-volume lines on the psychrometric chart. It is also easily seen from Equation (8.19) that, fordry air along the horizontal axis, lines of higher specific volume are located at higher dry-bulbtemperatures. The same conclusion follows for any constant value of humidity ratio.

Similar arguments may be used, together with Equation (8.14), to explain the slope andvariation of lines of constant enthalpy with dry-bulb temperature on the psychrometric chart.

Processes on the Psychrometric Chart

Consider point A on the psychrometric chart in Figure 8.19. The processes of heating andcooling to points B and C, respectively, are called sensible heating and sensible cooling. Theterm �sensible� implies change in dry-bulb temperature without change in moisture content.These processes occur at constant humidity ratio and thus at constant dew point temperature,Tdp.

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Next consider processes in which either moistening or drying of air takes place. Anexample of a humidification, or air-moistening, process already considered is the adiabaticsaturation process. It is represented by movement along a line of constant wet-bulbtemperature toward the saturation curve, such as the line AD on the figure.

A pure humidification process is one in which the humidity ratio increases at constantdry-bulb temperature. The line AE in Figure 8.19 represents this process. The process requiresthe addition of water vapor to the air, with a consequent energy increase corresponding to theheat of vaporization of water, (WE � WA)hfg. The enthalpy increase between points A and Eindicated by the chart is equal to this quantity because no sensible heating has occurred. Theprocess is said to involve latent heat as opposed to sensible heat. Thus an isothermal changein humidity ratio represents a latent heat load associated with the moist air mass and theequipment dealing with it. The reverse process EA may be thought of as puredehumidification. The modifier "pure" is meant to suggest that there are other, more commonand easily executed, humidification and dehumidification processes in which dry bulbtemperature change is involved. Examples are air washer processes like the adiabatic saturatorand desiccant drying (to be discussed later).

Let us now consider the steady-flow, adiabatic mixing of two moist air streams at states iand o to form a single stream at state m as shown in Figure 8.20. Conservation of mass of dryair and water vapor yields

mi + mo = mm [lbm/s | kg/s] (8.20)

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andmiWi + moWo = mmWm

Dividing through by the mixture mass flow rate, the latter equation becomes

Wm = (mi /mm)Wi + (mo /mm)Wo [dl] (8.21)

Thus the mixture humidity ratio is the mass-weighted average of the humidity ratios of thetwo streams. Applying conservation of energy to the adiabatic mixing process, we obtain

0 = mmhm � mihi � moho

which may be written as

hm = (mi/mm)hi + (mo/mm)ho [Btu/lbm | kJ/kg] (8.22)

This equation is of the same form as Equation (8.21).Thus, given the flow rates and properties of the two streams, the mass flow of the

mixture is the sum of the masses of the two inflowing streams and the humidity ratio andenthalpy of the mixture may be obtained by calculating the weighted means of the humidityratios and enthalpies from Equations (8.21) and (8.22).

A graphical solution for the mixture state can be easily obtained on the psychrometricchart. For fixed states i and o, Equations (8.21) to (8.22) depend on a single parameter,mi/mm, or

mo/mm = mo/(mo + mi) = 1 � mi /mm

Rearranging Equations (8.21) and (8.22) they become

(Wm � Wi)/(Wo � Wi) = mo/mm [dl] (8.23a)

(hm � hi)/(ho � hi) = mo/mm [dl] (8.23b)

Eliminating the mass ratio between these equations defines a linear relation between themixture humidity ratio Wm and enthalpy hm:

Wm = Wi + (hm � hi)(Wo � Wi)/(ho � hi) [dl]

as shown in Figure 8.20. The explicit linear dependence of Wm and hm on mo/mm is shown byrearranging Equations (8.23a and b) into

Wm = Wi + (Wo � Wi)(mo/mm) [dl] (8.23c)

hm = hi + (ho � hi)(mo/mm) [Btu/lbm | kJ/kg] (8.23d)

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When the flow rates and properties of states i and o are known, the latter two equations definethe mixture state.

These equations are the basis for a graphical method to determine the mixture state, assuggested earlier. A straight line connecting states i and o is drawn on the psychrometricchart, and the mass fraction mo/mm in Equations (8.23) is laid out as the fraction of thedistance between i and o measured from state i. The resulting point defines the properties ofthe mixture state.

EXAMPLE 8.9

Ten lbm/sec of outdoor air at 100°F dry-bulb and 60% relative humidity at sea level is mixedwith 20 lbm/sec of indoor air at 75°F dry-bulb and 40% relative humidity. Determine the moistair properties of the mixture.

SolutionFrom the psychrometric chart for the outdoor air state: ho = 51.8 Btu/lbm and Wo = 0.0252 lbm

water vapor per lbm dry air. For indoor air: hi = 26 Btu/lbm and Wi = 0.0074 lbm of water vaporper lbm dry air. From Equations (8.20) and (8.21), the humidity ratio is then:

Wm = (10/30)(0.0252) + (20/30)(.0074) = 0.01333 lbm

water vapor per lbm dry air, and the enthalpy is

hm = (10/30)(51.8) + (20/30)(26) = 34.6 Btu/lbm

Other mixture properties from the psychrometric chart are then 83.4°F dry-bulb, 65°Fdew point, and 70.8°F wet-bulb temperatures, 54% relative humidity, and 13.97 ft3/lbm

specific volume.__________________________________________________________________________

Psychrometric Analysis of Space Cooling

An important application of psychrometrics is in the analysis and design of air conditioningsystems. In comfort cooling, a control volume (see Figure 8.21), usually referred to as a spaceor zone, is identified as a room or group of rooms that have common thermal conditions andare to be put under the control of a single thermostat or control. Those conditions are usuallydefined by the atmospheric pressure, dry-bulb temperature, a humidity parameter such ashumidity ratio, and specified sensible and latent cooling loads, qs and qL. The design sensibleand latent loads are defined based on the characteristics of the boundaries of the zone and onits external and desired internal environments, as discussed earlier. The sensible heat factor,

SHF = qs /qt = qs /(qs + qL) [dl] (8.24)defines the fraction of the total load qt due to the sensible load. The sensible heat factor, SHF,

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together with the sensible load, is frequently used in defining the conditions for cooling azone.

Consider the conditioned space in Figure 8.21 as having specified latent and sensibleloads. It is desired to hold the dry-bulb temperature of the indoor space at TDBi and relativehumidity, �i (the wet-bulb temperature TWBi or humidity ratio Wi could be used instead). Thedesign outdoor environment is represented by design dry-bulb and wet-bulb temperatures,TDBe and TWBe, respectively, as shown on the psychrometric chart in Figure 8.22. A system isto be designed to supply cool, dry air to offset the sensible and latent loads imposed on thezone. The system is represented by a frictionless adiabatic ducting system that suppliesconditioned air, the supply air (state su on the chart), to replace an equal mass of room air atthe space conditions (state i). Part of the air leaving the zone is exhausted to the environment;and the balance, the return air, is mixed with fresh outdoor ventilation air (state e). Themixture of return air and ventilation air (state m) is cooled and dehumidified by a cooling coiland circulated back to the space by a supply-air fan. It is desired to determine suitable valuesfor supply-air temperature, coil cooling rate, and fan volume flow rate.

Assuming that air leaves the space at temperature TDBi, the steady-flow form of the FirstLaw applied to the space yields

qt = qs + qL = msu(hi � hsu) [Btu/hr | kW] (8.25)

The enthalpy difference in this equation can be written as

hi � hsu = hi � h3 + h3 � hsu [Btu/lbm | kJ/kg]

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This decomposition is seen graphically in the psychrometric chart in terms of the triangle i �su �3. Thus

qt = msu(hi � h3 + h3 � hsu) [Btu/hr | kW] (8.26)

Because the sensible heat, reflecting the supply-air dry-bulb temperature rise, may be writtenas

qs = msu(h3 � hsu) [Btu/hr | kW] (8.27a)

it is clear from Equation (8.26) that the latent heat is given by

qL = msu(hi � h3) [Btu/hr | kW] (8.27b) Equations (8.27) show that both sensible and latent loads are proportional to the supply-airmass flow rate. Graphically, Equations (8.27) indicate that the line segments su �3 and 3 �irepresent sensible and latent zone heat loads, respectively.

The sensible heat factor may be expressed as

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SHF = qs /qt = [(hi � h3)/(h3 � hsu) + 1]-1 [dl] (8.28)

It is seen from Equation (8.28) and Figure 8.22 that the space sensible heat factor depends onthe slope of the room condition line. The ASHRAE psychrometric chart includes a"protractor" that relates the condition line slope to the sensible heat factor. Thus the designindoor state and the cooling loads determine the room condition line.

Selection of the supply-air dry-bulb temperature together with the room condition linedetermines the supply-air state. The supply-air temperature must be chosen within thecapabilities of available refrigeration cooling coils. Note that the coil cooling load

qc = msu(hm � hsu) [Btu/hr | kW]

exceeds the space-cooling load whenever warm ventilation air is provided.The fan mass flow rate requirement may be evaluated from Equation (8.25) or (8.27)

when the supply-air state is known. The fan volume flow rate is then msuvm [ft3/hr | m3/s].However, it is usually quoted in ft3/min or CFM as

CFM = msuvm/60 [ft3/min]

where the mass flow rate is in lbm/hr of dry air. There is a significant tradeoff here which may be considered with the help of the

psychrometric chart. Higher supply-air dry-bulb temperatures allow higher evaporator coiltemperatures, which improve the system cooling COP. However, the resulting lower supply-air to room-air temperature difference [according to Equation (8.25)] implies a larger air flowrate and thus a larger fan and larger ducts.

A moisture mass balance on the cooling coil gives the mass rate of condensation of waterfrom the supply air:

mw = msu(Wm � Wsu) [lbm/hr | kg/s]

It should be recognized that this is usually greater than the rate of removal of water vaporfrom the zone because of the higher vapor content of the ventilation air.

EXAMPLE 8.10

A single zone is to be maintained at 70°F dry-bulb and 50 % relative humidity by arefrigeration coil in a forced-air system. The total cooling load on the zone is 55,000 Btu/hrand the latent load is 11,000 Btu/hr. Outdoor air at 98°F dry-bulb and 80°F wet-bulb isprovided at a mass rate that is 25% of the supply-air rate. Air leaving the coil is at 50°F dry-bulb.

(a) Show the states of the indoor and outdoor air on a psychrometric chart and identifytheir enthalpy, wet-bulb temperature, humidity ratio, specific volume, and relative humidity.

(b) Show the mixture line and state on a psychrometric chart.(c) Show the space and coil condition lines on a psychrometric chart. What are the

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enthalpy, wet-bulb temperature, humidity ratio, specific volume, and relative humidity of thesupply air?

(d) What are the mass and volume flow rates of the outdoor and supply air?(e) What is the total coil cooling load, in Btu/hr and in tons?

Solution(a) From the psychrometric chart, the zone and outdoor properties are:

TDBi = 70°F TDBe = 98°F�i = 50% TWBe = 80°FWi = 0.0078 We = 0.018hi = 25.3 Btu/lbmda he = 43.4 Btu/lbmdavi = 13.51 ft3/lbmda ve = 14.47 ft3/lbmdaTWBi = 58°F �e = 47%

(b) For the properties listed above and for mo/mm = 0.25 in Equation (8.23), the mixtureenthalpy and humidity ratio are

hm = 25.3 + 0.25(43.4 � 25.3) = 29.83 Btu/lbmda

Wm = 0.0078 + 0.25(0.018 � 0.0078) = 0.01035 lbm/lbda

It is seen in Figure 8.22 that these properties verify that the mixture state lies on a straight linebetween the outdoor and indoor states. Thus the mixture line and either hm or Wm could havebeen used to identify state m. The mixture dry-bulb temperature read from the chart is 77°Fand the specific volume is 13.7 ft3/lbmda.

(c) The sensible heat factor is SHF = 44,000/55,000 = 0.8. A line parallel to the roomcondition line is obtained using the chart protractor and the space sensible heat factor. Usingtwo triangles, a parallel line through state i gives the space (room) condition line shown inFigure 8.22. From the intersection of the space condition line and the 50°F dry-bulbtemperature line, the supply-air properties are:

hsu = 19.2 Btu/lbmdaWsu = 0.0066vsu = 12.95 ft3/lbmda�su = 86%

The coil condition line is drawn to connect states m and su.

(d) The enthalpy of the supply air rises as it mixes with room air. The rise rate must equal thetotal space load in order for the space temperature to remain unchanged. Thus

msu = (qs + qL)/(hi � hsu) = 55,000/(25.3 � 19.2) = 9016.4 lbmda/hr

Using the vsu, the supply air volume flow rate is 9016.4(12.95)/60 = 1946 cfm. Similarly, the

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on-coil volume flow rate is 9016.4(13.7)/60 = 2059 cfm. The outdoor-air mass and volumeflow rates are then, respectively, 9016.4/4 = 2254.1 lbmda/hr and 2254.1(14.45)/60 = 543 cfm.Note that, because of the different specific volumes, the outdoor-air volume flow rate is notone-fourth of the supply-air rate; and the supply-air and on-coil volume flow rates are notequal.

(e) The mixed ventilation and return air are cooled and dehumidified by heat transfer to therefrigeration coil. The coil load is therefore

qcoil = msu(hm � hsu) = 9016.4(29.83 � 19.2) = 95,844 Btu/hr, or 95,844/12,000 = 8 tons

The coil load is greater than the space-cooling load because of the cooling anddehumidification of the outdoor ventilation air. __________________________________________________________________________

Dehumidification Without Cooling Coils

It was shown in Example 8.10 that dehumidification usually accompanies cooling in typicalspace-cooling systems using refrigerant coils. In industrial processes and other applicationswhere precise humidity control is important, special equipment often is used. Substancesknown as desiccants, which absorb water vapor, may be employed in such dehumidificationapplications, alone or in combination with refrigerant coil cooling and dehumidification.Examples of solid desiccants are silica gel, lithium chloride, and activated alumina. You mayhave observed that cartons which contain merchandise that must be kept dry are often shippedwith small packets of a desiccant such as silica gel.

The difference in water vapor pressure between neighboring regions is a driving force forwater vapor transport by diffusion from the high-vapor-pressure region to the lower-pressureregion. A desiccant absorbs water vapor from air adjacent to its surface, lowering the local vapor pressure and thereby attracting an inflow of additional vapor from the surroundings.

As a desiccant absorbs water vapor, however, the water vapor pressure at its surfacegradually increases. The material will continue to accept water vapor until its vapor pressureequals or exceeds that of the surrounding air. Thus the material must be reactivated tocontinue the drying process. Since vapor pressure increases with temperature, a desiccant canbe reactivated by heating it with hot air. At high temperatures the desiccant vapor pressureexceeds that of the surrounding air, and water vapor then passes from the desiccant to thesurrounding air.

Commercial dehumidifiers may use liquid or solid desiccants. Figure 8.23(a) shows aschematic of a compact desiccant-wheel dehumidification system that has a wide range ofindustrial applications and that allows the drying of process air to a very low dew point. In thedehumidifier, the process air passes through a duct leading to a sector of a rotating,fiberglass-composite honeycomb wheel having many small passages impregnated with lithiumchloride dessicant. The water-laden sector rotates slowly into another duct, sealed off from

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the process air, through which a separate air flow passes at a temperature from 250°� 300° F.This hot air reactivates the desiccant in preparation for its next exposure to the process air.Though heat (not work) is required to reactivate the desiccant, the process requires lessenergy than other dehumidification processes. A desiccant-wheel storage room installation isshown in Figure 8.23(b). As in the figure, heated outdoor air is usually provided forreactivation of the desiccant. The reader is referred to references 11 and 12 for furtherinformation on dessicant dehumidification.

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Bibliography and References

1. Parsons, Robert A. (Ed.), ASHRAE Handbook of Fundamentals. Atlanta: AmericanSociety of Heating, Refrigerating, and Air-Conditioning Engineers, 1989, p. 16.3.

2. McQuiston, Faye C. and Parker, Jerald D., Heating Ventilating, and Air Conditioning,3rd ed. Wiley, New York, N.Y., 1988.

3. Stoecker, W. F. and Jones, J. W., Refrigeration and Air Conditioning, 2nd ed.McGraw-Hill, New York, N.Y., 1982.

4. Threlkeld, James L., Thermal Environmental Engineering, 2nd ed. Prentice-Hall, Inc.,

Engelwood Cliffs, N.J., 1970.

5. Jennings, Burgess H., The Thermal Environment � Conditioning and Control, Harperand Row, 1978.

6. Stoecker, W.F., Design of Thermal Systems, 3rd ed. McGraw-Hill, New York, N.Y.,1989.

7. Goff, John A. and Gratch, S., �Thermodynamic Properties of Moist Air," ASHRAEHandbook of Fundamentals, Atlanta: American Society of Heating, Refrigerating, andAir-Conditioning Engineers, 1972.

8. Derra, Skip, "CFCs No Easy Solutions," R&D Magazine, May 1990, pp. 56-66.

9. Moore, Taylor, "CFCs: The Challenge of Doing Without," EPRI Journal, Vol. 14, No. 6,September 1989.

10. Purvis, Ed, "Applications of Scroll Compressors in Heat Pumps," Energy TechnologyXVI, Government Institutes, Inc., February 1989, pp. 410-420.

11. Parsons, Robert A. (Ed.), �Sorbents and Desiccants,� ASHRAE Handbook ofFundamentals. Atlanta: American Society of Heating, Refrigerating, and Air-Conditioning Engineers, 1989, Ch 19.

12. Parsons, Robert A. (Ed.), �Sorption Dehumidification and Pressure Drying Equipment,�ASHRAE Handbook of Equipment. Atlanta: American Society of Heating, Refrigerating,and Air-Conditioning Engineers, 1988, Ch 7.

13. Althouse, Andrew D., et al., Modern Refrigeration and Air-Conditioning. SouthHolland, Ill.: Goodheart-Wilcox Co., 1975.

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14. York, Thomas, �Heat Pumps: Developing the Dual Fuel Option,� EPRI Journal,December 1990, pp. 22-27.

15. Public Law 101-549, � An Act to Amend the Clean Air Act to Provide for Attainmentand Maintenance of Health Protection, National Air Quality Standards, and OtherPurposes,� November 15, 1990.

16. Anon., Refrigeration and Ice Making. Alymer, Ontario: Ontario Arenas Association Inc.,March 1988.

EXERCISES

8.1 Derive a relationship between the coefficient of performances for the heat pump andrefrigerator using the ideal cycles of Figure 8.1.

8.2 Carefully sketch a flow diagram and T-s diagram for a refrigeration cycle that uses asingle heat exchanger for suction line heating and condenser discharge subcooling.Explain how the addition of this heat exchanger influences the cycle refrigeration effect?

8.3 Derive the conversion factors in Btu/hr and in kJ/hr, for a ton of refrigeration, based onthe latent heat of fusion of ice of 144 Btu/lb.

8.4 Derive the following convenient relation for the horsepower required to produce acooling capacity of one ton of refrigeration: horsepower/ton = 4.72/COP.

8.5 When shopping for an air conditioner, one encounters labels that give the product's EER.What is the coefficient of performance for an air conditioner with an EER of 9.3?

8.6 Determine the capacity and COP of a vapor compression chiller operating with areciprocating compressor between 35°C condensing and �10°C evaporating withrefrigerant 22. The compressor has a volumetric efficiency of 70% and operates at 1800rpm with a displacement volume of 750 cubic centimeters.

8.7 Compare the refrigeration effect, compressor work, and COPr for refrigerants 12 and 22

for a refrigeration system that has a 40°F evaporator and condenses at 140°F.

8.8 Compare the refrigeration effect, COPr, compressor work, and EER for an R-12refrigeration unit with 120°F condensing and 50°, 30°, and 10°F evaporators. Tabulatethe results.

8.9 Compare the refrigeration effect, COPr, compressor work, and EER for an R-22refrigeration unit with 120°F condensing and 50°, 30°, and 10°F evaporators. Tabulatethe results.

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8.10 Compare the refrigeration effect, COPr, compressor work, and EER for an R-22refrigeration unit with 100°, 120°, and 140°F condensing and 50°F evaporator.Tabulate the results.

8.11 Compare the refrigeration effect, COPr, compressor work, and EER for an R-12

refrigeration unit with 100°, 120°, and 140°F condensing and 50°F evaporator.Tabulate the results.

8.12 Compare the results of Exercise 8.10 with those for a similar system, accounting for a

compressor with a 75% isentropic efficiency.

8.13 Compare the results of Exercise 8.11 with those for a similar system, accounting for acompressor with a 75% isentropic efficiency.

8.14 Consider a two-stage refrigeration system that operates with a 40°F evaporator andcondenses at 140°F. Draw labeled T-s and flow diagrams for the cycle that consists oftwo ideal vapor compression cycles connected by a heat exchanger that serves as thecondenser for the low-pressure cycle and the evaporator for the high-pressure cycle.The evaporating and condensing temperature in the heat exchanger is 80°F. For R-12,determine the ratio of refrigerant flows for the two cycles and compare the refrigerationeffect, total compressor work, COPr, and EER with that for the single-stage system.

8.15 Consider a two-stage refrigeration system that operates with a 40° F evaporator and

condenses at 140°F. Draw labeled T�s and flow diagrams for the cycle that consists oftwo ideal vapor compression cycles connected by a heat exchanger that serves as thecondenser for the low-pressure cycle and the evaporator for the high-pressure cycle.The evaporating and condensing temperature in the heat exchanger is 80°F. For R-22,determine the ratio of refrigerant flows for the two cycles; and compare therefrigeration effect and the total compressor work per unit mass in the evaporator, theCOPr, and the EER with that for the single-stage system.

8.16 A Diesel engine develops 500 horsepower at an efficiency of 38%. Almost all of theheat loss from the engine is transferred through the lubricating oil, through the coolingwater, and through an exhaust gas heat exchanger to the generator of an absorptionrefrigeration system with a COP of 0.8. Derive an equation for the capacity of theevaporator in terms of the engine efficiency and power output and the COP of therefrigeration system. What is the capacity of the evaporator of the refrigeration systemin Btu/min and in tons?

8.17 The volumetric efficiency, mvs /(N�disp), of a reciprocating compressor may be crudelymodeled as:

Vol. eff. = 1 + C � Cr(1/n)

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where C is the ratio of the clearance volume to the piston displacement disp, vs is thecompressor suction specific volume, r is the compressor pressure ratio, and n is thepolytropic exponent. Taking C = 0.15 and n = 1.2, plot capacity and compressor powerversus evaporator temperature for a condensing temperature of 140°F for R-12. Thecompressor has a 3 in3 displacement and operates at 1200 rpm.

8.18 To attempt to compare COPs for vapor compression and absorption refrigerationsystems, consider a vapor compression system (with COPr) driven by a heat engine(with efficiency �E). The combined system is one that accepts heat from a high-temperature source and uses it to cool a low-temperature region. Work out an equationfor the COP for the combined system COPa,and compare it with the COPr for thecorresponding vapor compression system. What is COPa, when COPr = 6 and theengine efficincy is 25%?

8.19 Determine the COP of an ideal vapor-compression-refrigeration system using R-22with a 40°F evaporator and a condensing temperature of 110°F. Next, assume that thework for the system is provided by a Carnot engine operating between 250°F and100°F reservoirs. Diagram the system. Considering the combined system as a model ofan absorption system, compare its absorption COP with the COP of the vaporcompression system alone. Discuss the conditions under which higher-absorption COPscould be attained.

8.20 Sketch and discuss thoroughly the slope and variation of lines of constant enthalpy onthe psychrometric chart.

8.21 Compare the sea level relative humidity, enthalpy, specific volume, wet-bulbtemperature, and dew point using (a) steam tables and ideal gas mixture relations and(b) the psychrometric chart for the following conditions: 1. 60°F dry-bulb, W = 0.01. 2. 90°F dry-bulb, W = 0.02. 3. 90°F dry-bulb, W = 0.01.

8.22 Compare the sea level humidity ratio, enthalpy, specific volume, wet bulb temperatureand dew point using (a) steam tables and ideal gas relations and (b) the psychrometricchart for the following conditions: 1. 60°F dry-bulb, � = 50% 2. 90°F dry-bulb, W = 80% 3. 90°F dry-bulb, W = 40%

8.23 Compare the sea level relative humidity, humidity ratio, enthalpy, specific volume, anddew point using (a) steam tables and ideal gas relations, and (b) the psychrometric chartfor the following conditions: 1. 60°F dry-bulb, 55°F wet-bulb.

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2. 90°F dry-bulb, 55°F wet-bulb. 3. 90°F dry-bulb, 80°F wet-bulb.

8.24 Two hundered lbm/hr of moist air at 65°F dry-bulb and 50°F wet-bulb temperaturesmixes with 40 lbm/hr of air at 100°F and 85°F wet-bulb. What is the mixture dry-bulbtemperature, humidity ratio, relative humidity, and dew point? What are theseparameters if the second stream flow rate is increased to 80 lbm/hr?

8.25 One hundred lbm/hr of moist air at 70°F dry-bulb and 50% relative humidity mixes with35 lbm/hr of air at 90°F wet-bulb and 70% relative humidity. What are the mixture dry-bulb and wet-bulb temperatures, humidity ratio, relative humidity, enthalpy, specificvolume, and dew point?

8.26 Determine the sensible, latent, and total heat transfer for moist air flowing over acooling coil. The on-coil wet-bulb and dry-bulb temperatures are 85°F and 105°F,respectively, and the coil-leaving conditions are 60°F dry-bulb and 60% relativehumidity. How much water is condensed per unit mass of dry air?

8.27 Determine the sensible, latent, and total heat transfer for moist air flowing over acooling coil. The on-coil wet-bulb and dry-bulb temperatures are 75°F and 95°F,respectvely, and the coil-leaving conditions are 50°F dry-bulb and 85% relativehumidity. How much water is condensed per unit mass of dry air?

8.28 Air at sea level has a dry-bulb temperature of 80°F and an adiabatic saturationtemperature of 60°F. Evaluate the humidity ratio, relative humidity, and dew pointtemperature using (a) equations and steam tables and (b) the psychrometric chartmethod.

8.29 Air at sea level has a dry-bulb temperature of 85°F and an adiabatic saturationtemperature of 55°F. Evaluate the humidity ratio, relative humidity, and dew pointtemperature using (a) equations and steam tables and (b) the psychrometric chartmethod.

8.30 A zone is to be maintained at 72°F dry-bulb and 50% relative humidity. The zonesensible and latent loads are 150,000 Btu/hr and 50,000 Btu/hr, respectively. Thesupply-air temperature is to be 48°F. What is the supply-air mass flow rate and volumeflow rate?

8.31 Rework the parts of Example 8.10 that are changed by reducing ventilation air to 10%of supply air. Compare the coil cooling load and volume flow rate with the givenresults.

8.32 Rework the parts of Example 8.10 that are changed by increasing the supply-air dry-

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bulb temperature to 55°F. Compare the coil mass and volume flow rates and coolingload with the given results.

8.33 A room is to be maintained at 74°F dry-bulb and 62°F wet-bulb by a vapor compressionair conditioner. The total cooling load on the zone is 24,000 Btu/hr, and the latent loadis 4,000 Btu/hr. Outdoor air is provided for ventilation at 100°F dry-bulb and 60%relative humidity. The ventilation mass rate is 15% of the supply-air rate. Air leavingthe coil is at 52°F dry-bulb. (a) Show the significant states and processes on the psychrometric chart, and construct a table of wet- and dry-bulb and dew point temperatures, enthalpy, humidity ratio, specific volume and relative humidity for the space, outdoor, mixture, and supply- air conditions. (b) What are the mass and volume flow rates of return air, ventilation air, and supply air? (c) What is the coil cooling load in Btu/hr and in tons?

8.34 Resistance heaters are switched on in 5-kW increments to satisfy the demand for heatfrom the electric heat pump system characterized by Figure 8.15. At what temperaturesshould the first and second 5-kW banks be turned on?

8.35 Indicate whether the following are true or false: (a) An isotherm on a steam p-V diagram is a line of constant relative humidity for a given total moist air pressure. (b) The saturated vapor line on a steam p-V diagram is a line of constant relative humidity for a given total moist air pressure. (c) Increasing partial pressure of water vapor for fixed moist air pressure always increases relative humidity.

(d) Increasing partial pressure of water vapor for a fixed moist air pressure and fixed dry-bulb temperature always increases relative humidity. (e) Decreasing dry-bulb temperature for fixed moist air pressure and fixed water vapor pressure always increases relative humidity.

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C H A P T E R 9

ADVANCED SYSTEMS

9.1 Introduction

For decades the response to the ever-growing need for electric generation capacitywas to build a new steam power plant, one not very different from the last. Todaythe energy conversion engineer is faced with a variety of issues and emergingtechnologies and a changing social and technological climate in which a diversityof approaches is likely to be accepted. This chapter intends to indentify some ofthese concerns and opportunities. No claim of completeness is made. No chapter,or book for that matter, could thoroughly cover this domain. The reader is referredto the bibliography at the end of the chapter as a starting point for continued study.

A few characteristics of importance in new power initiatives are: low capitaland operating costs, ability to operate with a variety of fuels and with hightolerance to fuel variability, short construction time, low emission of pollutants,marketable or at least inert and easily disposable waste products, and highefficiency, maintainability, financeability, and reliability. Increasingly, the newinitiative may take the form of repowering the the old plant so as to increaseefficiency, meet pollution standards, and minimize the financial impact of meetingnew power demands.

The improvement of the efficiency of power plants using conventional cyclesis usually evolutionary in nature, by virtue of high temperature limitations andadvances in materials. Hence, only gradual improvements in efficiency can beexpected. On the other hand, significant improvements in efficiency cansometimes be obtained by combining conventional cycles in appropriate ways.Such power plants are referred to as combined-cycle plants. This chapter willexamine the characteristics of several combined-cycle plants.

It is evident from the study of the Rankine and Brayton cycles, and in fact allheat engines, that the rejection of large amounts of thermal energy to thesurroundings accompanies the production of useful power. This heat rejection canbe reduced by improving the thermal efficiency of the cycle but cannot beeliminated. If this energy is not to be wasted, it is logical to seek applicationswhere both power and rejected heat may be utilized. Power plants that producemechanical or electrical power and utilize “waste heat” for industrial processes arecalled cogeneration plants. Several examples of cogeneration are considered inthis chapter. District heating and other possible applications of waste heat are alsodiscussed.

Another key problem facing the energy conversion engineer is the anticipatedscarcity, in a few decades, of fuels such as natural gas and oil, relative to the vastresources of coal available in the United States and elsewhere. Perhaps futurepower plants should utilize this coal and nuclear energy to save the natural gas and

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petroleum for industrial feedstocks and other more critical future needs. On theother hand, serious problems exist with respect to utilization of these resources.Nuclear power, an important alternative, replete with problems, is considered inthe next chapter. Much of the readily available coal has unacceptably high sulfur,which significantly degrades the environment when released from power plantstacks in untreated combustion products. The well-known problem of acid rain hasbeen attributed to emissions from coal-burning power plants. Thus the search fortechnology to utilize medium- and high-sulfur coal and to reduce levels ofpollutant emissions of all types is an important area for research and development.

In this chapter advanced technologies that may contribute solutions to theseand other crucial problems are considered. Some recent U.S. Department ofEnergy efforts in these areas may be found in Reference 66.

9.2 Combined-Cycle Power

One of the unfavorable characteristics of the gas turbine is that the exhaust gasissuing from the turbine is at high temperature, thereby wasting much energy andcreating a local hazard. One solution to this problem was considered in Chapter 5:the addition of an exhaust gas heat exchanger to preheat the combustion air. Theresulting regenerative cycle was found to be much more efficient than thecorresponding simple cycle and to produce a lower exhaust gas temperature.

An alternative approach to dealing with the high gas turbine exhausttemperature is to provide a separate bottoming cycle to convert some of the energyof the turbine exhaust into additional power. Let’s consider the use of a Rankinecycle that uses gas turbine exhaust as its energy source. It is clear that, if theRankine cycle does not interfere with the operation of the gas turbine, thecombined cycle will produce additional power and will have a higher efficiencythan the gas turbine alone. Even if more heat is required for the Rankinebottoming cycle to produce additional work, the overall combined efficiency willincrease if the additional work is large enough and supplemental heat smallenough.

A combined gas turbine–Rankine cycle can be implemented in several ways. One method makes use of the fact that the exhaust of gas turbines usually has ahigh residual oxygen content because of the high air-fuel ratio required to limit theturbine inlet temperature when burning conventional fossil fuels. This hot,oxygen-rich exhaust gas can be used instead of air as the oxidizer in a steamgenerator as shown in Figure 9.1. For a moderate expenditure of additional naturalgas in the furnace the resulting combustion products can provide heat for a high-temperature steam cycle with conventional steam plant technology.

The Horseshoe Lake combined-cycle plant was designed in this way to yieldadditional power and high efficiency when operating in the combined mode, and tooperate with the gas turbine alone or with the steam turbine alone by direct-firing

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of the steam generator with fuel and air. The plant, about ten miles east of

Oklahoma City, was the first its kind, first producing power in 1963. It wasdesigned for a net output of 200 MW and a net heat rate of 9350 Btu/kW-hr. Theplant cross-section and photos of the turbines are seen in Figures 9.2–9.4. Note theevaporative cooler to reduce the temperature and to increase the density of the airentering the gas turbine compressor. This reduces compressor work and increasesinlet mass flow rate. Twenty years of successful operating experience with theHorseshoe Lake plant and information on similar German plants is documented inreference 46.

An open-cycle gas turbine also may be linked to a steam cycle through whatmay be considered a gas turbine exhaust heat exchanger containing an economizer,a boiler, and perhaps a superheater. This device, called a heat-recovery steamgenerator (HRSG), may be used to create and superheat steam as in a conven-tional steam cycle. A flow diagram for such a cycle is shown in Figure 9.5,together with a T-s diagram for the steam bottoming cycle. Gas turbine exhaustgas cools as it superheats, boils, and then warms liquid water in counterflow as itpasses through the HRSG.

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Combined-Cycle Analysis

Assuming a constant heat capacity for the gas turbine combustion gas, we maycompare gas temperatures with adjacent HRSG local water temperatures on the T-s diagram, Figure 9.5, by applying the steady-flow First Law of Thermodynamicsto appropriate sections of the HRSG:

ms(h12 – h11) = mgCpg(T6 – T7) = msA1 [Btu/hr | kJ/hr] (9.1)

ms(h13 – h12) = mgCpg(T5 – T6) = msA2 [Btu/hr | kJ/hr] (9.2)

ms(h8 – h13) = mgCpg(T4 – T5) = msA3 [Btu/hr | kJ/hr] (9.3)

where ms and mg are the mass flow rates for the steam and gas turbine cycles,respectively, and the Ai (i = 1, 2, 3) are the areas on the T-s diagram representingheat transfer per unit mass of steam. By adding the three equations, we obtain thesame equation as would result from application of the steady-flow First Law ofThermodynamics to the entire HRSG:

ms(h8 – h11) = mgCpg(T4 – T7) = ms ( A1 + A2 + A3) [Btu/hr | kJ/hr] (9.4)

Thus the enthalpy rise of the steam in the HRSG is controlled by the ratio of massflow rates and the hot-gas temperature drop. Expressing gas temperature in theHRSG in terms of steam enthalpy allows us to condense these equations into

T = T7 + (ms /mgCpg)(h – h11) [R | K] (9.5)

It is evident that the gas temperature is linearly proportional to the water enthalpyon a T-h diagram, as shown in Figure 9.6. The abscissa may be viewed as thecumulative heat transfer per unit mass of water, which is in turn proportional to theexhaust gas heat transfer.

The temperature difference T6 – T12, known as the pinchpoint temperaturedifference, is at a critical location in the heat recovery steam generator, because itoccurs at the point of minimum temperature difference between the two fluids. Itshould exceed some minimum design value (about 30°F) for all operatingconditions of the system to make effective use of all of the HRSG heat transfersurface. Smaller temperature differences would substantially increase the heattransfer surface area needed, while significantly larger values would necessitatereducing the boiling temperature and would adversely affect the combined-cyclethermal efficiency.

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EXAMPLE 9.1

Let us examine several possible Rankine bottoming cycles for the gas turbineconsidered earlier in Example 5.1. There a gas turbine with a compressor pressureratio of 6 and a turbine inlet temperature of 1860°R was analyzed. The turbineexhaust temperature was found to be 1273°R, or 813°F. With 813°F as the HRSGgas entrance temperature, select the steam turbine throttle temperature as 700°Fand consider Rankine cycles with a range of boiling temperatures, a condensingtemperature of 100°F, and a pinchpoint temperature difference of at least 30°F. Determine a cycle with a satisfactory boiling temperature, and compare it with

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other cycles in the family. Take the HRSG gas exit temperature as 350°F to avoidcondensation.

SolutionThe spreadsheet in Table 9.1 follows the notation of Figures 9.5 and 9.6. It

tabulates three combined-cycle cases in which the boiling temperatures, T12 = T13,were selected as 300°, 400°, and 500°F. These temperatures, together with the throttle and condensing temperatures and the turbine efficiency, determine theRankine-cycle thermodynamic conditions (neglecting pump work). Equation (9.5)relates the HRSG enthalpy changes to the corresponding gas temperatures. Thuswith a mass ratio Rmass given by (h8 – h11)/(T4 – T7), the pinchpoint temperature T6for Case 2 is

T6 = T7 + (ms /mgCpg)(h12 – h11) = T7 + (h12 – h11)/Rmass

= 350 + (375.1 – 68)/2.82 = 350 + 108.9 = 459°F

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and the pinchpoint temperature difference is

T6 – T12 = 459 – 400 = 59°F

Again, by Equation (9.5),

T5 = 350 + (1201 – 68)/2.82 = 350 + 401.8 = 752°F

With temperatures and enthalpies determined, work, heat, and efficiencyvalues may be determined as usual, observing carefully the distinction betweensteam-mass and gas-mass references. It is seen that Case 2 has a satisfactorypinchpoint temperature difference and a combined-cycle efficiency of 39%, whichis significantly greater than that of the gas turbine cycle (25%) and Rankine-cycle(29%) operating separately. The temperature distributions for this case are shownin Figure 9.7.__________________________________________________________________

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More than one gas turbine could be used in conjunction with a single steamturbine by using a larger HRSG and multiple gas turbine exhaust ducts. Alternatively, each gas turbine may have its own HRSG, with all HRSGsthermally coupled to the steam turbine through feedwater and steam line headers.

An HRSG may be designed to burn additional fuel at its inlet using oxygenremaining in the gas turbine combustion gas to raise the temperature the HRSGgas and provide higher heat transfer to the water. Such a design is called a firedHRSG, and its use is usually referred to as supplemental firing.

Figure 9.8 shows large heat-recovery steam generators built for a nominal 450-MW combined-cycle plant in Texas utilizing three 100-MW gas turbines and one140-MW steam turbine.

The Comanche power station, located near Lawton, Oklahoma, is an exampleof a combined-cycle facility that employs HRSGs. Two Westinghouse simple-cycle natural-gas-fired gas turbines, as diagramed in Figure 9.9, drive 85-MWelectrical generators and exhaust into separate HRSGs designed for supplementalfiring. Steam at 1200 psia and 950°F produced by the HRSGs is supplied to onenonreheat, single-extraction (not shown) steam turbine that drives a 120-MWelectrical generator.

The Comanche unit, first operated in 1974 (ref. 10) was upgraded in 1986 (ref.11), resulting in a measured plant heat rate of 8508 Btu/kW-hr (40.1% thermalefficiency), with a gas turbine inlet temperature of 1993°F, an HRSG gas inlet

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temperature of 1200°F, and supplemental firing. In 1986, the plan was indentifiedby the Utility Data Institute as the most efficient steam-electric generating stationin the United States, with an average net plant heat rate of 8821 Btu/kW-hr(thermal efficiency of 38.7%).

High-temperature combined-cycle plants are now achieving thermalefficiencies exceeeding 50%. A combined-cycle heat rate for a United Technol-ogies Turbo Power FT8 gas turbine is said to operate at a gross plant heat rate of6815 Btu/kW-hr (50.1%) based on lower heating value (ref. 51). Reference 50indicates that the Pegus Unit 12 combined-cycle cogeneration plant in theNetherlands produces 223.3 MW at maximum electrical output, with a netelectrical yield of 51.74% based on lower heating value. According to reference67, “Both GE and Siemens Westinghouse turbines will be able to break the 60 percent efficiency barrier in combined-cycle operation, and a 3 to 6 per cent reductionin CO2 emissions should be possible because of the increased efficiency.”

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9.3 Integrated Gasification Combined-Cycle (IGCC) Power Plants

The Cool Water IGCC Plant

One approach to the problem of clean coal utilization lies in the technologyexemplified by the Cool Water integrated gasification combined cycle (IGCC)power plant located near Barstow, California. This plant, which went intooperation in 1984, demonstrated the capability of producing power for the

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Southern California Edison System with very low levels of pollutants by using both low-sulfur and medium-sulfur coals. The Cool Water demonstration plantshown in Figure 9.10 (named after the ranch on which it is located) utilized a coalgasification technique known as the Texaco process. The medium-heating-valuesynthetic gas called syngas produced by the process has about one-third theheating value of natural gas, about 265 Btu per standard cubic foot on a dry basis.

As it is produced, the syngas is first cooled, then treated to remove pollutants,and finally burned to drive turbine generators, as seen in Figure 9.11. The initialcooling in the syngas coolers produces saturated steam,which is later superheatedby gas turbine exhaust gas in an HRSG, to power a steam turbine (path A in thediagram). The syngas emerging from the coolers (path B) is processed to removeparticulates and sulfur and to control oxides of nitrogen, and is then burned in thegas turbine to produce additional power. The high-temperature gas turbine exhaust(path C) then passes through the heat-recovery steam generator, adding energy tothe steam before it passes to the steam turbine. Thus the gasifier flows and thesteam turbine and gas turbine flows interact, hence the name “integratedgasification combined cycle,” IGCC.

The Texaco process requires oxygen of at least 95% purity to gasify the coal inthe gasifier. The Cool Water plant thus has a small, independently owned oxygen plant (seen surrounding the single tower left of center in figure 9.10) that separates

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oxygen from air to feed the gasifier. The Cool Water plant produces oxygen of99.5% purity in order to produce argon as a by-product (as mentioned earlier, theprocess otherwise requires only 95% purity). Nitrogen is also a by-product.Oxygen is produced continually for gasifier use, but oxygen is also stored so thatthe IGCC plant can continue to operate if the oxygen plant is shut down. A coal slurry, a mixture of nominally 60% coal and 40% water produced in awet grinding process, is introduced with oxygen into the Texaco gasifier (Figure9.11), where partial combustion of the coal takes place at about 600 psig and2500°F. The gasifier yields a mixture of mainly carbon dioxide, carbon monoxide,and hydrogen gases, with sulfur primarily in the form of hydrogen sulfide. Arelatively inert slag containing most of the mineral matter of the coal passes fromthe gasifier into a pool of water in the bottom of the radiant cooler. The slag istaken out periodically through a lockhopper process. The seal at the bottom of theradiant cooler is maintained by the water that is recycled.

The syngas, after cooling in the radiant and convective coolers, passes througha carbon scrubber, where a water spray removes most of the particulates andfurther cools the gas. After additional cooling to ambient temperature, the gasflows to a sulfur-removal unit, where a solvent removes the hydrogen sulfide andtherefore most of the sulfur from the stream. The relatively particle-free andsulfur-free gas is then saturated with moisture to control the formation of oxides ofnitrogen (NOx) during combustion in the gas turbine. The water-quenching processsuppresses NOx formation by reducing the gas combustion temperature, and it alsoincreases the turbine power output by adding to the mass flow in the gas turbinecombustor. Combustion gases from the gas turbine then pass through the heat

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recovery steam generator, where they produce additional steam as they drop intemperature to about 400°F. The combustion products leaving the plant areremarkably pollutant free. Performance of the Cool Water Station is summarizedin Table 9.2 below.

Table 9.2 Cool Water IGCC Station Nominal Performance

Gas turbine electric generator power 65 MW

Steam turbine electric generator power +55 MW

Gross power 120 MW

Air-separation oxygen plant –17 MW

Internal plant consumption – 7 MW

Net power 96 MW

Design heat rate 11,510 Btu/kW-hr

Observed heat rate 10,950 Btu/kW-hr Source: Reference 3.

The operators of the Cool Water station have demonstrated values of criticalpollutants well below current environmental limits for both permit and regulatorylimits and New Source Performance Standards (NSPS), as indicated in Table 9.3.The Cool Water plant has operated successfully with Utah run-of-mine coal with0.4% sulfur, Illinois #6 coal with 3.1% sulfur, and Pittsburg #8 coal with 2.9%sulfur. The sulfur removal process in the plant yields about 99.6% pure elementalsulfur, which can be sold for the production of sulfuric acid and fertilizers. Theslag produced by the gasifier is considered nonhazardous and suitable for theproduction of road-making materials or cement.

Table 9.3 Emissions from the Cool Water Station HRSG (lb/million Btu)

High SulfurCoal SO2

Low SulfurCoal SO2

NOx CO ParticulateMatter

Permit and regulatory limit 0.16 0.033 0.13 0.07 0.01

Utah coal – 0.018 0.07 0.004 0.001

Illinois #6 0.068 – 0.094 0.004 0.009

Pittsburgh #8 0.122 – 0.066 <0.002 0.009

Federal NSPS 0.6 0.24 0.6 – 0.03Source: Reference 3

The net effect of the plant then is to generate power efficiently by utilizingwidely available coals with sulfur content up to 3.5% in an environmentally sound

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way while producing nonhazardous waste that offers the possibility of constructiveuse.

It should be understood that the Cool Water plant was built as a demonstrationplant to prove a technology on a large but not a full scale. Engineering studiesbased on the Cool Water operating experience (ref. 4) have indicated that a 360-MW full-scale IGCC plant using Illinois #6 coal could be built with a net heat rateof 9000 Btu/kW-hr at a capital cost of $1530/kW and with operating andmaintenance costs of 2.3 cents/kW-hr (or 5.2 cents/kW-hr, which includes all fixedcharges) (ref. 3). These costs are competitive with those for conventional newcoal-burning power plants using flue gas desulfurization.

An important feature of the IGCC plant concept is the attractiveness of phasedconstruction. A conventional coal-burning steam power plant must be constructedas a unit and takes a relatively long time to erect. On the other hand, one or moregas turbines of a planned IGCC plant may be quickly put into service using naturalgas as a fuel. Additional gas and steam turbines may be added later to transformthe plant into a combined-cycle plant, with coal gasifiers and an oxygen plantadded still later at a third stage. Because the units may be paid for as they are built,phased construction offers significant financial benefits as well as ordely growth.

Significant materials problems have been overcome in the Texaco and CoolWater plant technology. The inside of the gasifier requires refractory materials towithstand the severely corrosive high-temperature environment. Cool Waterexperience (ref . 5) indicates that 10,000–14,000 hours of refractory life isattainable. This implies that gasifier overhaul will be required at least every twoyears. Additional problems with cooler tube-wall materials, coal slurry pumps, andpiping; and other severe material operating environments offer challenges formaterials research to improve IGCC operation.

Several of the references expand on the idea of cogeneration–polygeneration–by pointing out that, based on the Cool Water technology today, a single facilitywhose only major requirements are air, water, and coal can simultaneouslyproduce electricity, steam, sulfur, inert slag for road construction, oxygen,nitrogen, hydrogen, carbon monoxide, carbon dioxide, argon, methanol, otherchemical products, and even syngas. The Dow Gasification Process

Another coal gasification process combined-cycle system (CGCC), developed byDow Chemical Co., operates in Plaquemine, Louisiana. Like the IGCC system,the Dow process reacts a coal slurry with oxygen to produce syngas, from whichmost of the sulfur is removed for by-product use. The raw syngas from thegasifier produces steam in an HRSG, is cooled, passes through pollutant removalequipment, and is burned in gas turbines. Reference 47 indicates that “of the newcoal-based technologies, the CGCC system has the highest efficiency and thelowest emission of environmental pollutants.” The 161-MW plant, built in theremarkable time of twenty-one months, was completed in 1987. The referenceindicates for CGCC plants a net heat rate based on lower heating value of 8670

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Btu/kW-hr (39.4%) and a total capital cost of $1201/kW, based on 1988 dollars. Additional information on combined-cycle coal gasification systems is givenreferences 58, 59, and 65. Tampa Electric’s 250-MW IGCC Polk Power Stationbegan commercial operation in 1996. A website (ref. 65) indicates that the stationis 10-12% more efficient than conventional coal-fired plants.Three dimensionalviews may be seen at the website.

9.4 Combustion in Fluidized Beds

The need to develop environmentally sound methods for utilizing a variety ofcoals, industrial and municipal wastes, and other solids as fuels has stimulatedresearch in a variety of areas. Another prominent and promising technologyapplicable to these goals is fluidized bed combustion (FBC). Atmosphericfluidized bed combustors (AFBCs) operate near atmospheric pressure; pressurizedfluidized combustors (PFBCS) are enclosed in a pressure vessel and operate at a pressure of about 12 atmospheres.

As the name suggests, a principal unique feature of this technology is thatcombustion takes place in a bed of solid particles supported in vigorous, turbulentmotion on an upwardly directed stream of air. The bed may consist of acombination of particles of fuel, sand, ash, waste materials, limestone, and othercompounds, depending on the function and design of the fluidized bed combustor.The key point is that these materials mingle and react in continually changingorientations, providing ample opportunity for intimate contact of fuel and oxidizerand for removal of combustion products, while supported on the fluidizing airstream. The name bubbling bed is sometimes used to decribe this action.

It has been found that horizontal water tubes located within the fluidized bed,in crossflow to the upward air stream, experience very high heat transfercoefficients in comparison with those in normal furnace convective and radiativeenvironments. It is even more important that fluidized beds containing coal andlimestone produce combustion gases with both low sulfur content and lowconcentrations of oxides of nitrogen.

Limestone reacts with sulfur in the coal to produce particles of calcium sulfatethat are removed as bed materials are renewed. More specifically, the limestone,CaCO3, reacts to form CO2 and lime, CaO:

CaCO3 => CaO + CO2

and the CaO reacts with the sulfate vapors:

H2SO4 + CaO => CaSO4 + H2O.

The calcium sulfate forms as a solid that becomes a bed material, and the watervapor passes off as a component of the flue gas.

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Oxides of nitrogen (NOx) produced by the FBC are maintained at low values,because bed and gas temperatures are well below those at which NOx forms inconventional combustors. The bed temperature is determined primarily by the ratesof air and fuel supplied. Normally, FBC bed temperatures are about 1550°F,compared with conventional furnace temperatures in the neighborhood of 3000°F.The reactivity of nitrogen is low at 1500°F, and chemical equilibrium calculationsand laboratory observations show that the nitrogen in flue gas is almost entirely inits normal, diatomic form.

The combustion of coal or other solids occurs largely at the interface betweenthe solid and the surrounding oxidizing gases. The rate at which the solid burnsdepends on the rate at which oxygen is brought to the solid interface and on therate at which combustion products are removed, as well as on the rate of chemicalreaction at the interface itself. The vigorous relative motion of the bed particlesand the intervening air flow provide an excellent mechanism for delivery ofoxygen to and the transport of combustion products from the interface.

Normally, FBC occurs with enough excess air, in the primary supporting air

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stream and possibly in a secondary over-fire air flow, that combustion is virtually complete. Flyash and other airborne particles are removed by centrifugalseparators and baghouse filters. Solids from the separators may be reinjected intothe bed to further ensure almost complete burnup of carbon.

The circulating fluidized bed (CFB) combustor (refs. 26 and 30) is one inwhich smaller solid bed materials are carried upward by the combustion air/gasstream. A return passage transports the unburnt and inert particles and part of thecombustion gas back to the main furnace, allowing the remaining flue gas to passto the heat-recovery area, as seen in Figure 9.12. The solid bed materials continueto burn as they circulate, thus maintaining an approximately uniform temperatureof about 1550°F throughout the furnace. As a result, there is a long residence timefor particles of the furnace to complete their reactions. A mechanical cycloneseparator built into the furnace helps to separate the particles from the exiting fluegas. As a result, reinjection of the unburnt carbon makes possible very highcombustion efficiencies.

According to reference 26, CFB designs achieve higher combustion efficiency,reduced NOx emissions, minimum CO formation, and reduced limestoneutilization in capturing SO2 when compared with bubbling fluidized bedcombustors. Much continues to be learned about problems and opportunitiesinherent in fluidized bed combustion as more units come into use. In a December1998 work (ref. 65), the U.S. Department of Energy (DOE) proposed a 379-MWe,pressurized circulating fluidized bed combustor combined-cycle plant with a netefficiency of 47%.

9.5 Energy Storage

It has been observed that there is no existing means of storing electrical power on alarge scale. As a consequence, power generation varies from instant to instant, tosatisfy the immediate demands of consumers. Utility generation capacity must therefore be great enough to satisfy the peak demand, or the utility must purchasepower at a premium from other utilities to make up its generation deficit.

Demand varies from place to place, seasonally, daily, and hourly. Forinstance, the loads of utilities in the southern United States are usually greatestduring hot summer days, when air conditioning and industrial demands coincide. As a result, southern utilities may have excess capacity at night and in the winter. It were possible to generate a full capacity during off-peak hours and store theenergy in excess of demand, the utilities could operate with installed capacitybelow the demand peak and operate more units as base-load plants close to theirhigh-efficiency design points.

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Pumped Storage

One approach to off-peak energy storage utilizes a high reservoir into which waterfrom a low elevation is pumped, using electricity generated during off-peak hours.Thus energy may be stored in the form of potential energy of the elevated waterfor later use during peak loads. There are about thirty such pumped-storagefacilities in the United States.

Figure 9.13 depicts the two phases of operation of a pumped-storage facility. These installations usually employ motor-generator sets driven by hydraulicturbines. Mechanically reversible pump-turbines generate electricity by usingwater from an elevated reservoir during periods of peak demand. When there is anexcess of base-load power, the motor-generators can be reversed to drive thepump-turbines as pumps for filling the reservoir.

It is clear that net energy is lost in the use of pumped storage. Its success relieson the availability of cheap power during off-peak hours and consistent demandfor electricity, with its associated high price during peak hours. Pumped storageallows utilities to generate more electricity with their most efficient base-loadplants instead of handling peaks with less efficient equipment.

Figure 9.14 is a photograph of the Salina pumped-storage facility (ref. 34) ofthe Grand River Dam Authority, located about 50 miles from Tulsa, Oklahoma. The facility was designed for gradual expansion, in three steps, from the 130-MWconfiguration that went online in 1968, to the current 260-MW facility shown in

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the figure, to a 520-MW final modification. The eventual completion of theproject will require the development of an additional upper reservoir as well asconstruction of six additional penstocks and the installation of their hydroelectricpump-turbine motor-generator sets. The present plant is capable of producingpower at full-rated load for eight hours a day or at part load for twelve hours a day. The reservoir is 251 feet above the lower lake. The penstocks are 14 feet indiameter and 720 feet long.

Reference 36 identifies an early pumped-storage facility named Rocky River atNew Milford, Conn., which was in operation in 1928. The limited number of highhead surface sites for such facilities, the high capital cost of building a dam, andthe large land area impacted by these facilities make the future use of pumpedstorage questionable. Reference 37, however, predicted continued growth inpumped storage capacity. One possibility that would support such growth is tohave the upper pool at ground level and to use underground mines for the lowerpool (ref. 38). Another is the development of low head facilities.

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Compressed-Air Energy Storage

A relatively new approach to energy storage is compressed-air energy storage(CAES), which employs underground caverns for storage of air under pressure(refs. 25 and 49). Air is compressed into the cavern using off-peak power and lateris released to oxidize fuel in a gas turbine combustor to generate electrical powerduring hours of peak demand. The first CAES plant, a 290-MW unit, firstoperated at Huntorf, West Germany, in 1978. A second CAES plant located inMcIntosh, Ala, with phased construction of two 110-MW units, first went on-linein May, 1991 (ref. 64).

CAES plants are located in the vicinity of underground caverns. The cavernsmay be natural or may be created and/or enlarged by solution-mining of under-ground salt domes. In solution-mining, water is pumped into a salt formation, thewater dissolves salt locally, enlarging the cavern, and the resulting brine is pumpedto the surface, where the salt is driven from the solution and the water reused.

Figure 9.15 shows a schematic of the McIntosh plant. During off-peak hours amotor-generator, powered by electricity produced elsewhere in the utility systemand with turbine-expander clutch disengaged, drives the compressor set that packsair into the salt cavern. Later, air is allowed to escape from the cavern to oxidizefuel at the high cavern pressure, forming combustion gas to pass through the gasturbines (expanders) that drive the motor-generator in the generation mode with

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the compressor clutch disengaged. The generator thus is able to return electricity tothe utility system during periods of peak demand.

The CAES plant is essentially a gas turbine in which the compression processis decoupled from the power delivery process. During daily cyclical operation, themass of air supplied to the cavern at night is utilized during the day. Since thecharging period may be different than the utilization period, the average mass ratesof storage and delivery may differ. The McIntosh plant is designed for twenty-sixhours of storage capacity and therefore does not operate on a strict daily cycle butrather on a weekly cycle that takes advantage of less expensive weekend power.

The presence of the recuperator in the McIntosh plant, the first in use in aCAES facility, provides regeneration, which, as discussed in Chapter 5,substantially reduces gas turbine fuel consumption and thus improves plantefficiency. Regeneration may also be accomplished by absorption and storage ofthe heat of compression of the air in an aftercooling, high heat capacity heatexchanger such as a pebble bed recuperator. The stored energy is later returned tothe air discharged from the cavern as it flows to the turbines.

The success of a CAES plant, like that of pumped storage, depends on theavailability of cheap, off-peak power. The objective is to drive the compressortrain during these periods so that high-priced power may be produced later inperiods of high demand. It should be noted that the compression process is notonly decoupled mechanically from the power delivery process, but the compress-ion process may take place over a longer period of time than the period of CAESpower production. This allows the use of smaller compressors than would beneeded for a conventional peaking gas turbine. According to reference 25, theHuntorf plant compresses for 4 hours, and the McIntosh plant for 1.7 hours, foreach hour of power production. CAES may become a more viable option in theUnited States than surface pumped storage because of the existence of morepotential sites and lower land surface area requirements for CAES.

9.6 District Heating and Cooling and Cogeneration

Comfort heating and cooling in homes, businesses, and industry consumes largequantities of energy. Much of this low-temperature-energy use is accomplished bythe direct or indirect burning of fossil fuels and high temperature. Electricalresistance heating, especially, and heat pumps may be included in this because the electricity they consume is produced largely from high-temperature sources. Manybelieve that these are inappropriate uses of fossil fuels, in a conservation sense,because of the unnecessary loss of the availability of high-temperature energy todo work. It is simply a reflection of the desirability of using high temperatureswhere needed and low-temperature sources for low-temperature functions. Forinstance, it is no revelation that some of the obligatory heat rejection by modernheat engines is at a high enough temperature to supply energy for comfort heatingand cooling.

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District heating, the provision of heat to a populated area by a nearby centralheating plant, has been in use for one hundred years or more. More recently theaddition of cooling to such plants has become widespread. Now, colleges anduniversities frequently have such a facility. Shopping malls and blocks of businessdistricts in many cities also take advantage of the economic benefits provided by acentral plant. This is usually accomplished through the use of boilers that producehot water or steam for heating, and by vapor compression or steam-drivenabsorption refrigeration machines that produce chilled water for cooling. Water,steam, or brine is usually used to deliver the energy to the user.

In municipal facilities, the steam, water, or brine is metered and circulated in insulated pipes under the streets to air-handling units and other point-of-usedevices in the customers’ buildings, and thence to return pipes that bring the fluidback to the central plant or, in single-pipe one-way systems, to sewer mains.

Until recently, most of the district heating facilities the United States did notfind significant advantage in producing electrical power and using the waste heatfor district heating, cooling, or industrial process energy. With Europe’s morelimited energy resources, combined heat and power (CHP), or cogeneration, thesynergistic generation of electric power and heat, found more extensive use therethan in United States following the Second World War. In fact, a number of steamturbines and closed-cycle gas turbines burning a variety fuels were developed forsimultaneous electric power generation and district heating or industrialcogeneration purposes (refs. 12 and 15–20). These activities are closely related to total energy systems, which seek to utilize natural gas for other purposes whilegenerating electricity (ref. 17). In the United States this term has been used in thepast for the promotion of natural-gas-burning systems that provided heating,cooling, and electricity for shopping malls, colleges, and similar customers.

Some of the possible cogeneration steams include:

1. Steam turbine power with condenser heat rejection for low-temperatureprocesses, facility heating, or district heating.

2. Steam turbine power with steam extraction or use of a back-pressure turbinefor process or district heating use.

3. Steam turbine power with exhaust steam or steam extraction heat transfer toabsorption refrigeration system generators for chilling processes or districtcooling in summer.

4. Closed-cycle gas turbine power with coolers (intercoolers and pre-coolers)used for district heating.

5. Closed-cycle gas turbine power with coolers used with absorptionrefrigeration system for chilling processes or district cooling in summer.

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6. Open-cycle gas turbine power with exhaust heat recovery for process use ordistrict heating.

7. Diesel or gas reciprocating engine power with water-jacket cooling, oilcooler, and/or exhaust gas heat recovery used for process or district heating.

A measure of the efficiency of energy utilization in a cogeneration plant is theenergy utilization factor, EUF, which is the sum of the net work, w, and the usefulheat produced, qu, divided by the energy supplied to achieve the combined heatand work, qin, (ref. 18):

EUF = (w + qu )/qin [dl] (9.6)

Care should be taken when considering the EUF and the thermal efficiency ofa plant. The EUF is not restricted by the Carnot efficiency and can thereforeapproach 100%. For instance, when there is no useful heat transfer, the EUF is theplant thermal efficiency. At the other extreme, when no power is produced theplant can utilize almost all the energy supplied by the fuel for useful heat; andtherefore the EUF could approach 100%. Thus the EUF is a measure of the extentof productive use of the energy source, with no consideration for work–useful heatproportions. An EUF in excess of 80 percent is possible for a CHP plant.

Other efficiencies for CHP plants may be defined. For instance, a weightingfactor might be used in the numerator of the EUF to attempt to give appropriateweight to both heat and work. For example, since conventional power plantsconvert heat to work with about 33% efficiency, one might define a special EUF*as (w + 0.33qu )/qin for comparison with thermal efficiencies. While such factorsmay be useful in evaluating the design of a plant, they should be applied with care.Reference 18 discusses alternative definitions more fully.

9.7 Electricity Generation and Legislation

Historically, the electric power industry in the United States developed byrecognizing the economic advantages of scale of large central plants that usedextensive power transmission and distribution systems, following the lead of thegreat governmental hydroelectric power projects. They also recognized theenhanced growth potential inherent in the society’s becoming “totally electric.”The productive disposition of condensor rejected heat had no place as a revenueproducer in this scheme, for the important reasons that it would reduce plantefficiency and power output and could not reach across the miles as powertransmission lines could. Individual consumers as well as industry happilyaccepted this approach to the electrification of America, as electricity pricesdropped decade after decade. Natural-gas and fuel oil companies were also thereto satisfy the vast needs for heat.

More recently, energy-consuming industries came to recognize the possibilitiesof simultaneous power and heat generation to satisfy their energy requirements. At

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the same time, a consciousness grew that there are limits to the world’s energyresources and that a more thoughtful stewardship of them would be prudent, whenit is economically attractive. However, in the early 1970s the OPEC oil embargobrought these notions into clearer focus, and federal legislation in 1978 broughtcogeneration to everyone’s attention with the enactment of PURPA, the PublicUtility Regulatory Policies Act (ref. 14).

PURPA, a part of the National Energy Act, was intended to bring competitioninto the electric power generation business and to increase the nationaleffectiveness of energy utilization by making cogeneration more economicallyattractive. It required that utilities purchase power from qualified cogenerators at areasonable price, created tax incentives for developers, and established the FederalEnergy Regulatory Commission, FERC, to regulate and administer the activity. Prospective cogenerators must apply to the FERC for certification as a qualifiedfacility, QF. The FERC, however, left the deliberation of what constitutes a fairprice to state public utility commissions. As a result, industrial cogenerationthrives in some states and is virtually nonexistent and others. PURPA requiresutilities to buy electricity from QFs at the marginal cost of new generation, that is,the cost of electricity generated by new power plant if it were built, the so-calledavoided cost.

The growth of cogeneration, state and national actions to change the regulatorystructure of the electric utility industry, the difficulties of acquiring capital forlarge, long-term projects and other uncertainties are significantly changing theoutlook and structure of the utility industry (ref. 21). Utilities now appear to belooking toward incremental and modular growth and the avoidance of long-termcommitment, which is reflected in their reluctance or inability to undertake theconstruction of large, capital-intensive base-load plants. Legislative changes in the1990's and beyond are bringing substantial deregulation to the industry, openaccess to power transmission systems, and the introduction of merchant plantscreated to offer electricity to the highest bidder in the new free-enterprise climate(ref. 66).

9.8 Steam-Injected Gas Turbines

Water injection has been used for many years for brief augmentation of the thrustof jet engines. More recently, liquid water injection and steam injection have beenused to control the formation of NOx in gas turbines. Injection of water or steaminto the combustion chamber reduces the combustion temperature, which in turnsuppresses the formation of NOx caused by high temperature. Power output is maintained or increased because the injection increases both the turbine mass flowand the energy extraction by the turbine. The latter is possible because the heatcapacity of steam is almost twice that of normal combustion products. Thus theenthalpy change of steam for a given temperature drop is about double that of airor combustion gas. If water is injected as a liquid, additional energy must beextracted from the combustion gas to vaporize the water.

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As a consequence of these considerations, an important cogeneration technol-ogy is emerging. The steam-injected gas turbine, STIGTM* or SIGT, uses the gasturbine exhaust flow through an HRSG to produce steam that is partially orentirely injected into the gas turbine combustion chamber and possibly thecompressor and/or turbine, as indicated in Figure 9.16, resulting in augmented gasturbine power output. Reduced steam requirements for process use in an industrialcogeneration plant may be used to increase electrical power generation, which inturn may be sold to a local utility under PURPA if it is not needed for other on-siteuses.

References 13, 22, 23, and 31 show increased power output, increased thermalefficiency, and reduced NOx as benefits of steam injection. Table 9.4, for example,lists performance measurements of the effects of steam injection into twoindustrial gas turbines. The table shows that as the cogeneration process steampercentage is reduced and the steam flow to the gas turbine increases accordingly,substantial increases in power output and thermal efficiency are consistentlyattained. Thus the steam-injected gas turbine holds considerable promise forcogeneration and process applications.

A cross-section of the General Electric LM 5000 gas turbine without steaminjection is shown in Figure 9.17; a photo of the same is presented in Figure 9.18.The LM 5000 is a compact, high-performance, aeroderivative gas turbine intendedfor marine and industrial applications. It is derived from the CF6 family of high-bypass-ratio turbofan engines but burns either distillates or natural gas fuels. TheLM 5000 has a dual-rotor gas generator and a three-stage power turbine. Themanufacturer quotes a power output of 46,200 shaft horsepower (34,451 kW) anda heat rate of 9160 Btu/kW-hr at a power turbine speed of 3600 rpm for the LM5000 without STIG.____________________*STIG is a trademark of General Electric Co., U.S.A.

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Table 9.4 Performance of Steam-Injected Gas Turbines with Unfired HRSG

Cogeneration SteamPercentage

Power Output (kW) Thermal Efficiency

Allison 501-KH

100% 3500 0.24

50% 4750 0.30

0% 6000 0.35

General Electric LM5000

100% 33000 0.33 (0.36)

50% 40000 0.36 (0.38)

0% 47000 0.38 (0.42)

Adapted from reference 22. Data in parentheses from references 56 and 57.

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A schematic of the LM 5000 STIG configuration is shown in Figure 9.19; a photoof the same is presented in Figure 9.20. The steam manifolds and injection linesmay be seen in both figures. Table 9.4 shows the performance of this configur-ation without supplemental firing of the HRSG. With full STIG and supplementalfiring, the power output increases to 72,100 shaft horsepower (53,765 kW) and aheat rate of 7580 Btu/kW-hr (or a thermal efficiency of 45%) is attained.

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Figure 9.21 shows the average steam generation capability of the engine exhaustwith an unfired HRSG having a 30°F pinchpoint temperature difference. It is seenthat, as expected, higher steam flow rates are obtained with reduced steam temp-erature and pressure for given HRSG exhaust and turbine exhaust temperatures.

A study the economic advantages of steam-injected gas turbines for utility userelative to the combined-cycle power plant (ref. 33) has shown combined cycles tobe superior in unit sizes above 50 MW. However, utilities are becomingincreasingly interested in plants of 50 MW or less because of their small size andquick availability.

Intercooled Steam-Injected Gas Turbines

A modification of the steam-injected gas turbine is ISTIG, or, intercooled STIG, inwhich steam is injected into compressor bleed air for turbine cooling, together withsteam injection into the combustor and into one or more turbine stages (ref. 22). The enhanced blade cooling allows increased turbine inlet temperature and furtherpower and efficiency increases. The reference predicts efficiencies for ISTIGturbines better than for existing combined cycles and comparable to advancedcombined cycles. Thus STIG and ISTIG show great promise for cogenerationapplications and are likely to find their way into future power generation plans.

Steam-Injected Gas Turbine Analysis

The influence of steam injection into the combustor can be analyzed with a modelthat accounts for the major effects on the engine performance: the added mass and

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heat capacity of steam in the turbine flow. The model assumes the following:

�Downstream steam injection has no effect on the compressor.

�The properties of steam may be represented by constant values of the heatcapacity Cps: 1.86 kJ/kg-K and 0.444 Btu / lbm -R.

�The heat capacity of the mixture of combustion gas and steam may be repre-sented by a temperature-independent mass-weighted heat capacity given by

Cpm = Cpg + (ms /ma )Cps [Btu/lbm-R | kJ/kg-K]

where ms /ma is the steam-air mass ratio. Here, as before, the mass of fuel isneglected with respect to the mass of air.

�The isentropic exponent of the mixture, km , remains the isentropic exponentof combustion gas with no water injection, km = kg = 4/3.

�The fuel control system maintains the turbine inlet temperature at a constantvalue, irrespective of steam injection rate.

EXAMPLE 9.2

Consider the injection of steam in the combustion chamber of the single-shaft gasturbine studied in Example 5.1. Five pounds of steam are injected for everyhundred pounds of compressor discharge air. The fuel flow rate is adjusted tomaintain the turbine inlet temperature at a constant 1400°F. Compare the poweroutput for a compressor flow rate of 100 lbm /s, the thermal efficiency, the workratio, and the fuel-air ratio with like parameters for the machine with no injection. Assume that the steam injected is saturated at the combustor pressure level and isheated by the turbine exhaust from pressurized feedwater at 70°F in a heat-recovery steam generator as shown in Figure 9.22.

SolutionTable 9.5 presents the spreadsheet solution as well as the no-injection

reference solution, which is repeated there for convenience.The algorithm is based on the assumptions just enumerated and uses the HRSG

analysis techniques discussed in section 9.2 in connection with the combined-cyclesteam generator study. Steam conditions entering the combustor are obtained fromthe saturated-vapor tables, assuming constant pressure mixing at the known com-bustor pressure level; feed water conditions are for saturated liquid water at 70°F.

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A 3-point increase in thermal efficiency and a substantial increase in poweroutput due to steam injection are evident.

The steam generator easily provides the required steam mass with a largepinchpoint temperature difference. This indicates that the turbine exhaust gas iscapable of producing a still larger steam fraction with a smaller pinchpointtemperature difference while maintaining acceptable exhaust gas temperature withrespect to the exhaust gas dew point.__________________________________________________________________

By expanding the spreadsheet, with mass rate as a parameter, we present theinfluences of increased steam mass rate on performance and on steam generatorsteam-air temperature differences in Figures 9.23 and 9.24, respectively. Thefigures indicate that steam-air mass ratios up to about 0.18 are possible with thisconfiguration and that substantial performance benefits are the result.

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9.9 Resource Recovery

The disposal of municipal and industrial solid waste has become a nationalconcern with significant energy conversion aspects. It has been estimated that50% of sanitary landfills that are convenient or accessible to urban areas would befully utilized in the 1990s and that satisfactory new sites will be progressivelymore difficult to find. This has led to the rapid growth of the resource recoveryindustry.

Resource recovery deals with the environmentally sound disposal of municipaland industrial waste. It starts, at one extreme, with garbage landfills andincineration and extends to the ultimate conversion of components of waste totheir useful constituent elements, with recovery of available energy in the process. Vigorous activity in this area is producing a variety of approaches to the problem. Many of the solutions focus on the development of a central waste disposal facilitythat receives and prepares waste for efficient landfill disposal. This sectionconsiders a modern facility that reduces the volume of about 1125 tons per day ofsolid waste by over 90% by burning, to produce steam for industrial use and togenerate electricity when steam is not needed.

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Figure 9.25 shows a modern facility designed to serve the waste disposal needsof almost 500,000 Tulsa-area residents. The plant is similar in many respects tothe power plant steam generators discussed earlier. Major exceptions are themodifications required to deal with the unique problems of handling municipalsolid waste (MSW). The right foreground of Figure 9.25 shows where wastehaulers’ trucks are weighed on outside scales and enter the facility’s tipping floor. Large noncombustible materials and other refuse that cannot be burned in thefacility, such as steel bars and tires, are left outside the plant by the hauler forseparate disposal. The trucks dump the MSW in or near a large refuge storage pitadjacent to the tipping floor and leave from the opposite side of the tipping floorfrom which they entered.

Haulers are held responsible for the waste that they deliver, although overheadcrane operators who feed refuse to the three boilers also attempt to sort the refuse,to help achieve uniform heat release in the furnace and to keep large, noncombust-ible, hazardous, and otherwise inappropriate materials from entering the furnaces.

Figure 9.26 shows how refuse handled by cranes is hoisted from the holding pitand lowered through near-vertical feed chutes (4) to inclined reciprocating-gratefurnaces (5). There, rugged hydraulically operated rams meter the flow of refuseas they push it onto the grates. The combination of underfire air from a forced-draft fan (6) and vertically reciprocating grates keeps the refuse in continuousmotion. Overfire air assures 98% burnup of combustible materials. Supply air forthe forced-draft fans is drawn from the tipping-floor enclosure (1), to retain odorswithin the plant.

At the end of the grate, a variable-speed ash discharge roller controls the rateof discharge and hence the depth of the ash bed at the end the grate. The ash falls

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from the roller into a water-bath ash discharger that cools the residue. The wateralso seals the bottom of the furnace, which is slightly below atmospheric pressure.The cooling ash residue is pushed by a ram out of the water bath, up an inclinedsurface, and into a compartment where water is allowed to drain off and evaporatefrom the residue for fifteen minutes. The ash residue then falls to a conveyor,where it is transported to the ash house. There, salvageable metals are separatedand recycled. The remaining ash is then trucked to a sanitary landfill. The plant achieves about a 90% reduction in refuse volume delivered to the landfill.

In the furnaces, the combustion gas from the burning MSW on the grates heatsthe water in the welded-membrane water walls of the furnaces and the varioussteam generator tube banks (8-12). Each of the steam generators produces 88,500lbm/hr of steam at 680 psia and 700°F. The entire steam production passes in a 12-inch underground steam line to a refinery about a mile away, for process use. Infigure 9.25, the refinery process plant utilizing the steam is located just beyond the tank farm in the upper left. A steam turbine and generator set rated at 16 MW isavailable for electrical generation as an alternative to refinery use of steamproduction. The electricity may be used on site or sold to the local electric utility.

After leaving the economizer (12), combustion gas passes into an electrostaticprecipitator (14), where most of the fly ash remaining from the numerous passesthrough the boiler is collected. The precipitators have automatic rapping systemsthat free the collected particles, allowing them to drop into flyash hoppers to betransported to the residue conveyor. Induced-draft fans (16) transport the cleanedcombustion gas from the precipitators to the stacks.

Tulsa’s Walter B. Hall Resource Recovery Facility, described here, is anenvironmentally sound example of an increasing number of facilities operating orunder construction. These facilities typically are externally neat and are suited foroperation in industrial and some commercial locations. Massive reductions inwaste volumes are achieved in these facilities, with the possibility of generatingsteam for process use, district heating, and steam turbine generation of electricity. In resource recovery facilities of differing design, fluidized bed combustors mightbe employed, and the released refuse heat might instead be used in connectionwith closed-cycle gas turbines or other heat-driven devices.

9.10 Polytropic Efficiency

To this point the performance of turbomachinery has been represented byisentropic efficiencies. In comparisons of turbomachines with differing pressureratios, the use of the isentropic efficiency gives an undeserved advantage to somemachines over others with different pressure ratios. Another approach toefficiency, called the small-stage efficiency or polytropic efficiency, is consideredhere as an alternative and, under certain circumstances, a more consistent way ofrepresenting the quality of turbine and compressor performance.

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Axial compressors and turbines consist of alternating rows of stationary, orstator, blades and rotating, or rotor, blades, with the rotor rows firmly attached toa rotating shaft. In a turbine, the stationary blades act as nozzles to increase flowvelocity, and the rotor blades downstream turn and decelerate (change themomentum of) the flow. The reaction to the momentum change is a force with alarge component in the direction of blade rotation. This blade force deliverstorque and power about the rotor axis. The combination of the stator row and arotor row is called a stage.

Instead of thinking in terms of efficiency for the entire machine, we focus ourattention on the efficiency of a single turbine stage. The efficiency of a stage maybe defined in terms analogous to the definition of the isentropic efficiency. Consider the T-s diagram of Figure 9.27, which shows the temperature drop of a calorically perfect gas in a stage of a multistage turbine. We assume here that allstages have identical pressure ratios. In the notation of the figure, the turbineisentropic efficiency is given by (T1 – T2)/(T1 – T2i), and by analogy the stageisentropic efficiency is �T/ �Ti.

The expansion process in the turbine may be thought of as a stairstep sequenceof expansions through individual small stages, each having its own efficiency. Afew such steps are indicated in the figure. It should be observed that, in eachsuccessive stage, the isentropic temperature drop �Ti moves to the right on thediagram and therefore is larger than the corresponding drop between the samepressure levels on the expansion line from 1 to 2i; i.e., �Ti > �Ts. Thus theirreversibility of the expansions through earlier stages results in a sum of stageisentropic temperature drops greater than the overall isentropic temperature dropand hence greater work-producing capability for successive stages.

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Let us now imagine the turbine as comprised of an infinite number of stageswith infinitesimal pressure drops of equal efficiency. Thus the �Ts become dTsand the small-stage efficiency becomes

�s = dT/dTi [dl] (9.7)

The isentropic relation for a calorically perfect gas, Equation (1.19), may bewritten in the form T/p (k – 1)/k = a constant. Using differentiation by parts yields

dTi – [(k – 1)/k]Tp – 1dp = 0

which combined with Equation (9.7), gives

dT/T = [�s (k – 1)/k] p – 1dp [dl] (9.8a)

If it is assumed that the stage efficiency is constant, integration of Equation(9.8a) between the turbine inlet and exit states gives

T1/T2 = (p1/p2)�s (k– 1)/k [dl] (9.8b)

Thus the requirement that all stages have equal efficiency yields a temperaturepressure relationship of the same form as Equation (1.19), except for the exponent. Such relationships, of the form T1/T2 = (p1/p2) (n – 1)/n are called polytropic. Forexample, the isentropic Equation (1.19) is polytropic with n = k. Because the useof constant small-stage efficiency yields a pressure-temperature relation ofpolytropic form, �s is often called the polytropic efficiency. The exponent for theturbine expansion with constant stage efficiency is then given by

(n – 1)/n = �s(k –1)/k [dl] (9.9)

If the polytropic efficiency is unity, n becomes k, and Equations (9.8) become theusual isentropic relation. Values of �s less than 1 reduce the turbine temperatureratio, T1/T2, for a given turbine pressure ratio, p1/p2, below the isentropic value inqualitatively the same way that decreasing turbine isentropic efficiency does. A similar statement applies to turbine work.

The turbine isentropic efficiency, �t, can be expressed in terms of thepolytropic efficiency by substitution in the isentropic efficiency definition:

�t = (T2/T1 – 1)/(T2i /T1 – 1)

= (1/r �s(k –1)/k – 1)/ (1/r (k –1)/k – 1) [dl] (9.10)

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where r is the turbine pressure ratio, p1/p2.If we define a compressor polytropic efficiency as a small-stage isentropic

efficiency, a similar analysis yields relations analogous to Equations (9.8) – (9.10):

T2/T1 = ( p2/p1) (k– 1)/(k�s) [dl] (9.11)

(n – 1)/n = (k –1)/(k�s) [dl] (9.12)

�c = (r(k –1)/k – 1)/ (r (k –1)/(k�s) – 1) [dl] (9.13)

where �c and r are the compressor isentropic efficiency and pressure ratio,respectively. Figure 9.28 shows isentropic efficiencies for both compressors andturbines as functions of pressure ratio for two values of polytropic efficiencies.Applying L’Hospital’s rule to Equations (9.10) and (9.13), we can show that theisentropic efficiency of both turbines and compressors approaches the polytropicefficiency in the limit as the pressure ratio approaches 1. This fact is evident in thefigure, and it is apparent that as pressure ratio increases, isentropic efficienciesincrease for turbines and decrease for compressors for fixed values of thepolytropic efficiencies.

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The polytropic efficiencies may be regarded as measures of the internal qualityof turbomachines, that is, superior internal blade passage design. For example, fortwo turbines having the same internal aerothermodynamic quality, Figure 9.28indicates that a machine with a high pressure ratio will have a higher overallisentropic efficiency than one with a low pressure ratio.

As another example, the compressor curves imply that, in a comparison of two compressors having the same isentropic efficiency, the one with the higherpressure ratio has a superior aerothermodynamic quality. This suggests thatparametric studies involving varying compressor pressure ratio should use aconstant value of polytropic efficiency rather than constant isentropic efficiency torepresent comparable compressor quality.

9.11 Turbofan Engines

The turbofan engine, ducted fan, or fanjet, discussed briefly in Chapter 5, is thedominant gas turbine engine in commercial aircraft and is extensively employed inmilitary aircraft is well. Its primary feature is a large fan that accelerates a largemass of unheated air in an annular duct surrounding the central core engine, as in Figures 9.29 and 9.30, which show, respectively, a cutaway diagram and a photo-graph of the General Electric CF6-80C2 high-bypass-ratio engine. The large fandiameter produces a large jet exhaust consisting of a cylindrical wake of hot combustion gas surrounded by an annular flow of slower-moving warm air.

The bypass ratio, B, is the ratio of the mass flow rate through the outer coolerduct, mc, to the flow rate of the hot core engine, mh:

B = mc/mh [dl] (9.14)

Bypass ratios range from 0 for the pure turbojet engine studied in Chapter 5 to values in the neighborhood of 10. The bypass ratio is a design parameter that isprimarily determined by the mission of the aircraft. High-bypass-ratio engines aredesirable for long-range commercial aircraft because of their excellent fueleconomy. The CF6-80C2 engine has a bypass ratio of 5.05 and a total airflow of1769 lbm/s (802 kg/s).

The bypass air may have its own nozzle, separate from the core engine as inthe CF6 engine, or the core and bypass flows may be mixed in a specially designednozzle. The mixing nozzle helps to reduce jet noise by transferring momentumfrom the fast-moving core gas to the slower-moving bypass air, thereby reducing the wake shear noise source. The mixing process, however, involves a thrust-losspenalty.

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Figure 9.31 presents some of the nomenclature and notation that will be usedin discussing the turbofan engine. For the configuration shown in the figure, thefan pressurizes the compressor inlet air as well as delivering the power needed toaccelerate the bypass air through its nozzle downstream. While other fan configur-ations are possible, this frequently used arrangement is the only one analyzed here.

In addition to the bypass ratio, a second important design parameter is the fan pressure ratio, the ratio of the stagnation pressure downstream to that upstream ofthe fan:

FPR = po2/po1 [dl] (9.15)

The fan pressure ratio, together with the bypass ratio, determines the powertransferred between the hot core engine and the bypass flow. For a given coreengine configuration, higher bypass ratios and fan pressure ratios cause morepower to be extracted from the turbine and passed to the bypass air by the fan. Thisproduces higher bypass duct thrust. However, as more power is extracted from thecore flow, the core nozzle velocity and core engine thrust are reduced. Thedetermination of the design values of these parameters therefore involves acomplex tradeoff with numerous other design factors.

The combination of a compressor and a turbine joined by a shaft is sometimesreferred to as a spool. High-pressure-ratio engines are frequently arranged in atwin-spool or even a three-spool configuration. In a twin-spool engine, a high-pressure turbine drives a high-pressure compressor with a hollow shaft, and a low-pressure turbine delivers power to a low-pressure compressor and/or a fan bymeans of a shaft that turns inside of the high-pressure-spool shaft. The turbofanconfiguration in Figure 9.31 has the low-pressure turbine driving both the fan and

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the low-pressure compressor, as does the PW 4000 fanjet shown in Figure 5.24. Inanother arrangement, the low pressure turbine powers the fan only, and the entirecompressor is driven by the high-pressure turbine. Regardless of engine layout, theobjective of the fanjet is to accelerate a large mass of air and thereby increase the propulsive efficiency of the engine.

Let us now consider the analysis of the turbofan configuration shown in Figure9.31. The following parameters are assumed to be specified:

� Ambient conditions at a specified altitude and flight speed or Mach number, pa, ta, ca, or Ma

� Inlet stagnation-pressure recovery, IPR

� High-pressure-turbine inlet stagnation temperature, To5

� Fan pressure ratio, FPR = po2/po1

� Bypass ratio, B

� Low-pressure-compressor pressure ratio, LPCPR = po3/po2

� High-pressure-compressor pressure ratio, HPCPR = po4/po3

� Low- and high-pressure compressor efficiencies, �LPC, �HPC

� Low- and high-pressure turbine efficiencies, �LPT, �HPT

� Fan efficiency, �F (defined analogous to compressor efficiency)

The inlet is assumed adiabatic, the nozzles are assumed to be isentropic, and allmechanical efficencies are taken to be unity.

The calculation procedure parallels that of the turbojet analysis of Chapter 5.Free-stream stagnation conditions are determined from the ambient conditions andflight speed or Mach number. For an adiabatic inlet, the stagnation temperature atthe fan face, To1, is the same as the free-stream value, Toa, and the fan face stag-nation pressure, po1, is given by the product of the inlet stagnation-pressurerecovery and the free-stream stagnation pressure: po1 = IPR�poa.

The conditions immediately downstream of the fan, assumed to be the same forboth hot and cold paths, are given by

po2 = po1�FPR [lbf /in2 | kPa] (9.16)

and

To2 = To1 + To1[FPR(k – 1)/k – 1]/�F [R | K] (9.17)

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The pressurized flow in the bypass duct then accelerates from po2 and To2 throughthe convergent nozzle to a high velocity c9 with static pressure p9, contributingm9(c9 – ca) + A9( p9 – pa) to the engine thrust. Here, p9 = pa if the nozzle pressureratio, po2/p9, is less than or equal to the critical pressure ratio; otherwise it is equalto the critical pressure ratio.

The stagnation conditions downstream of both low- and high-pressurecompressors are determined in the same way as for the fan. For instance, for thelow-pressure compressor,

po3 = po2�LPCPR [lbf /in2 | kPa] (9.18)

and

To3 = To2 + To2[LPCPR(k – 1)/k – 1]/�LPC [R | K] (9.19)

Similarly, for the high-pressure compressor,

po4 = po3�HPCPR [lbf /in2 | kPa] (9.20)

and

To4 = To3 + To3[HPCPR(k – 1)/k – 1]/�HPC [R | K] (9.21)

Neglecting combustor pressure losses, the high-pressure-turbine inlet pressure is po5 = po4 and the known turbine inlet temperature is To5. Recognizing that, apartfrom assumed-small frictional losses, the power delivered by the high-pressureturbine is delivered to the high-pressure compressor; the steady-flow form of theFirst Law of Thermodynamics applied to the high-pressure spool yields thestagnation temperature upstream of the low-pressure turbine:

m8Cpa(To4 – To3 ) = m8Cpg(To5 – To6)

To6 = To5 – (Cpa /Cpg)(To4 – To3 ) [R | K] (9.22)

The turbine isentropic efficiency definition then gives the isentropic dischargetemperature, To6s, which in turn yields the high-pressure-turbine pressure ratio.

A similar procedure for the low-pressure turbine spool results in the energyrate balance:

(m9 + m8)Cpa(To2 – To1) + m8Cpa(To3 – To2) = m8Cpg(To6 – To7) [Btu/hr | kW]

which, after dividing by m8Cpa and using the bypass ratio equation, yields

(B + 1)(To2 – To1) + (To3 – To2) = (Cpg / Cpa)(To6 – To7) [R | K]

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This equation may be solved for the nozzle entrance stagnation temperature:

To7 = To6 – (Cpa /Cpg)[(B + 1)(To2 – To1) + (To3 – To2 )] [R | K] (9.23)

The turbine pressure ratio follows from the application of the isentropic turbineefficiency definition and the isentropic temperature-pressure relation, as in theearlier high-pressure-turbine analysis.

Alternatively, polytropic efficiencies may be used in the solutions. This is donefor the fan and compressors by replacing Equations (9.16) to (9.21) by theirpolytropic equivalents:

po2 = po1�FPR [lbf /in2 | kPa] (9.24)

To2 = To1 + To1[FPR(n – 1)/n – 1] [R | K] (9.25)

po3 = po2�LPCPR [lbf /in2 | kPa] (9.26)

To3 = To2 + To2[LPCPR(n – 1)/n – 1] [R | K] (9.27)

po4 = po3�HPCPR [lbf /in2 | kPa] (9.28)

To4 = To3 + To3[HPCPR(n – 1)/n – 1] [R | K] (9.29)

where (n – 1)/n is given by Equation (9.12) for the appropriate compressorpoltytropic efficiency.

Equations (9.22) and (9.23) are still applicable for the analysis of the turbines.However, polytropic equations, using exponents given by Equation (9.9):

n /(n – 1) = k /(k – 1)� s

where � s is the appropriate turbine polytropic efficiency, are required to determinethe turbine pressure ratios, namely

po6/po5 = (To6/To5)n/(n – 1) [dl] (9.30)

and

po7/po6 = (To7/To6)n/(n – 1) [dl] (9.31)

Once po7 and To7 are known, the core nozzle may be treated in the same way as theturbojet nozzle in Chapter 5. The engine thrust is then the sum of the thrustsproduced by the core and bypass flows.

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EXAMPLE 9.3

A turbofan engine has a fan and a low-pressure compressor driven by the low- pressure turbine, as shown in Figure 9.31. It operates at a flight velocity of 200 m/sat 12,000 meters altitude, where the ambient temperature and pressure are 216.65Kand 0.1933 bar, respectively. The overall pressure ratio the engine is 19, and thefan and low-pressure-compressor pressure ratios are 1.65 in 2.5, respectively. Thefan, compressors, and turbines have 90% polytropic efficiencies. Assume anisentropic inlet and separate isentropic convergent nozzles, a fuel heating value of43,000 kJ/kg, and a combustor total pressure loss of 6.5%.

Determine the core engine and bypass duct exit velocities, the engine thrustspecific fuel consumption, thrust, and specific thrust for an engine with an inlet airflow of 100 kg/s and a bypass ratio of 3.0. Determine whether the nozzles arechoked.

Solution

Table 9.6 tabulates, in spreadsheet format, the design data for the example and systematically calculates the engine parameters. After computing the fan and LP-turbine exit total pressures, the nozzles may be checked for choking at the throats.The applied pressure ratios are compared with the critical pressure ratios for coldbypass air (k = 1.4) and for hot gas (k = 4/3), respectively. Branching in thecomputation required by the presence or absence of choking is easily handled bythe @IF function of popular spreadsheets, which allows a conditional selectionbetween specified alternatives, as discussed previously in connection withExample 5.6.

TABLE 9.6 Spreadsheet Solution to Example 9.3

AEROTHERMODYNAMIC ANALYSIS OF A FANJET

Ca 200 m/s Flight velocityTa 216.65 K Ambient temperaturePa 0.1933 Bar Ambient pressureFuel HV 43000 kJ/kg Heating valueCpa 1.005 kJ/kg Air heat capacityCpa/Cpg 0.8754 Air/gas heat capacity ratioOPR 19.0000 Overall pressure ratioFPR 1.6500 Fan pressure ratioLPCPR 2.5000 Low pressure compressor pressure ratioHPCPR 4.6100 HPCPR = OPR/[(FPR)(LPCPR)]B 3.0000 Bypass Ratiomair 100 kg/s Inlet total mass flow ratecpeta 0.9 Compressor and fan polytropic efficiencies(n-1)/n l comp 0.3175 (n-1)/n = (k-1)/[k�cpeta] (for fan & comp.)

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tpeta 0.9 Turbine polytropic efficiencyn/(n-1) l turbine 4.4400 Turbine polytropic factor, k/[(k-1)�tpeta](dp/p) l combust 0.065 Combustor fractional pressure dropTo5 1027 C High pressure turbine inlet temperatureTo5 1300 K To5 (K) = To5 (C) + 273To1=Toa 236.550 K Toa=Ta+Ca^2/(2�Cpa*1000)Poa = Po1 0.263 Bar Po1 = Poa = Pa�(Toa/Ta)^3.5Po2 0.434 Bar Po2=FPR�Po1Po3 1.085 Bar Po3=(LPCPR)Po2Po4 4.995 Bar Po4=Po3(HPCPR)Po5 4.671 Bar Po5=Po4�[1 - (dp/p)] | combustTo2 277.31 K To2=To1�FPR^[(n-1)/n] I compTo3 370.93 K To3=To2�(LPCPR)^[(n-1)/n] | compTo4 602.39 K To4=To3�HPCPR^[(n-1)/n] | compTo6 1097.38 K To6=To5-(Cpa/Cpg)�(To4-To3)To7 872.68 K To7=To6-(Cpa/Cpg)[(B+1)(To2-To1)+(To3-To2)]Po6 2.199 Bar Po6=Po5�(To6/To5)^[n/(n-1)] | turbinePo7 0.795 Bar Po7=Po6�(To7/To6)^[n/{n-1)] | turbinePo7/Pc8 1.8530 Crit. press. ratio: Po7/Pc8=[(4/3+1)/2]^4Po7/Pa 4.1100 Core flow is chokedPo2/Pc9 1.8930 Crit. Press. ratio: Po2/Pc9=[(7/5+1)/2]^3.5Po2/Pa 2.2440 Bypass flow is chokedT8 748.02 K IF (Po7/Pc8) < (Po7/Pa) THEN T8=2�To7/(4/3+1)

ELSE T8=To7/(Po7/Pa)^0.25C8 535.02 m/s IF (Po7/Pc8)<(Po7/Pa) THEN C8=(287�T8�4/3)^0.5

ELSE C8 = [(To7-T8)�2000�Cpg)]^0.5P8 0.429 Bar IF Pc8>Pa THEN P8=Po7/(Po7/Pc8) ELSE P8=PaRho8 0.200 kg/m^3 Rho8=100�P8/(0.287*T8)A8/m8 0.0094 m^2-s/kg A8/m8 = 1 / (C8�Rho8)T9 231.09 K IF Pc9>Pa THEN T9=2To2/(7/5+1)

ELSE T9=To2/(Po2/Pa)^(1/3.5)C9 304.72 m/s IF Pc9>Pa THEN C9=(287�T8�7/5)^0.5

ELSE C=[(To2-T9)�2000�Cpa]^0.5P9 0.229 Bar IF Pc9>Pa THEN P9=Po2/(Po2/Pc9) ELSE P9=Parho9 0.346 kg/m^3 Rho9=100�P9/(0.287�T9)A9/m9 0.009 m^2-s/kg A9/m9=1 / (C9�Rho9)Specific Thrust, ST 242.9 N-s/kg ST = C8+C9�B-Ca�(B+1)+[(P8-Pa)�(A8/m8)

+(P9-Pa)�(A9/m9)�B]�10^5/(B+1)Thrust 24294.4 N Total engine thrust= mair(specific thrust)f/a 0.0186 f/a=Cpg(To5-To4)/HVmfuel 0.466 kg/s Fuel flow rate=mair�(f/a)/(B+1)TSFC 0.000019 kg/N-s TSFC=(f/a)/[spec. thrust)(B+1)]TSFC' 0.069 kg/N-hr TSFC'=3600�TSFCST | core m 971.8 N-s/kg ST based on m8 = (B+1)�ST

The specific thrust is presented in two forms, one based on the total mass flowrate to the engine and the other based on the mass flow rate to the core engine. Theformer, ST, allows easy comparison with engines of comparable frontal area,while the latter, ST | core m, is useful in showing the thrust increase due to adding

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a fan to a given core engine. Both definitions reduce to the common turbojetdefinition for B = 0, but deviate progressively with increasing bypass ratio.__________________________________________________________________

It is of interest to compare the performance of a turbofan and twin-spool turbojetin flight at the same altitude and velocity based on our computations. This is easilyaccomplished with the spreadsheet of Table 9.6 by setting FPR = 1.0 and B = 0 forthe turbojet engine. Table 9.7 shows some of the parameters resulting from thiscomparison.

Table 9.7 Turbofan–Turbojet Comparison

Turbofan Turbojet

Core jet velocity, m/s 535.02 566.88

Bypass jet velocity, m/s 304.72 ---------

Thrust, N 24,294.4 67,745

Specific thrust based on total mass, N-s/kg 242.9 677.5

Specific thrust based on core mass, N-s/kg 971.8 677.5

Fuel mass flow rate, kg/s 0.466 1.862

TSFC, kg/N-hr 0.069 0.099

It is evident that the extraction of power to drive the fan reduces the core jetvelocity. The lower jet velocities substantially reduce the turbofan engine thrustfor the same size inlet and total engine mass flow rate. Thus the high-bypass-ratioturbofan is less likely to be used for military applications requiring high speed andtherefore high thrust per unit of frontal area. This is reflected in the specific thrustbased on total engine mass flow. On the other hand, the specific thrust based oncore mass flow shows that a significant increase in thrust can be achieved byadding a fan to an existing turbojet design. The major advantage of the fanjet isshown in the fuel flow rate and TSFC comparisons, where the superior fueleconomy of the fanjet appears.

The success of the turbofan or ducted-fan engine has made it clear that furtheradvances in jet engine fuel economy are possible with higher bypass ratios. Thelarge-diameter cowlings necessary for drastic increases in turbofan bypass ratio,however, appear impractical. Still, advances in propeller technology now makeflight at high subsonic Mach numbers possible with gas-turbine-driven unductedfans.

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Figures 9.32 and 9.33 show a UDF®* engine featuring contrarotating fansinstalled in an engine nacelle. A twin-spool gas generator upstream of the fanssupplies combustion gas to the cantilevered fan section. In this design the forward fan blades are coupled directly to the outer fan-turbine blades, as shown in Figure 9.34, and rotate clockwise. The free-wheeling aft fan section, which includes the_________* UDF® is a registered trademark of General Electric Co., U.S.A..

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inner fan-turbine blades, rear fairing, and tailcone assembly, is cantilevered at therear of the engine and rotates counterclockwise.

With unducted-fan engines, very high bypass ratios are possible. The UDFengine, with an overall pressure ratio of approximately 42 at top of climb, hasdemonstrated substantial SFC reductions over modern turbo fans, approximately20% lower than the best of the current turbofan engines. In-cabin noise, originallyexpected to be a problem with unducted fans, was very low–equal to or less thanthat of today’s turbofan-powered aircraft (ref. 55).

Despite the attractiveness of the unducted fan concept, it appears that in thenext decade large transport aircraft will use ducted fans that incorporate moremodest increases in bypass ratio together with increases in overall pressure ratiofor improved performance. Reference 63, for example, anticipated the family of75,000 to 95,000 pounds-thrust turbofans coming into use around 1995. They wereexpected to have a bypass ratio of 9 and overall pressure ratio about 45 to give a9% improvement in SFC over then-existing engines. The reference anticipated thatthe engines would have a low speed, low pressure ratio fan for low noise. The fanswere expected to be made of composite materials to save about 25% weight overcompeting metal fans.

Bibliography and References

1. Curran, Paul F., "Clean Power at Cool Water," Mechanical Engineering, Vol.109, No. 8 (August 1987): 68-71.

2. Grover, Ralph W., et al.; "Preliminary Environmental Monitoring Results forThe Cool Water Coal Gasification Program." Proceedings of the EPRI SixthAnnual Conference on Coal Gasification, Palo Alto, Calif., October 15–16, 1986.

3. Watts, Donald H., Dinkel, Paul W., and McDaniel, John E., "Cool Water IGCCPerformance to Date and its Future in the Electric Utility Industry," U.S. UtilitySymposium, Palm Springs, Calif., October 1987. 4. Gluckman, Michael J., et al., "A Reconciliation of the Cool Water IGCC PlantPerformance and Cost Data with Equivalent Projections for a Mature 360-MWCommercial IGCC Facility." Cool Water Gasification Program, October 1987.

5. Fahrion, M. E., "Materials Development Experience at Cool Water CoalGasification Plant." Cool Water Coal Gasification Program, October 1987.

6. Anon., "Boilers and Auxiliary Equipment," Power, June 1988: 91–229.

7. Ehrlich, Shelton, "Fluidized-Combustion Boiler Technology–1988 Update."Missouri Valley Electric Association Engineering Conference, Kansas City, Mo.,March 1988.

8. Ehrlich, Shelton, "Fluidized Combustion: Is It Achieving Its Promise?" ElectricPower Research Institute, Palo Alto, Calif.

385

9. Pollak, Robert, “Status of First U.S. CAES Plant,” EPRI Journal, December1988: 49–52.

10. Allison, P. R., and Berman, P. A., “PACE 260 at Comanche, the First TwoYears.” ASME Gas Turbine Conference, ASME Paper 76-GT-109, 1976.

11. Anon., “Upgraded Combined Cycle Will Save Over $6 Million per Year inFuel,” Gas Turbine World, May–June 1968: 13–18.

12. Bammert, K. and Groschup, G., “Status Report on Closed-Cycle Gas Turbinesin the Federal Republic of Germany.” ASME Gas Turbine Conference, ASMEpaper 76-GT-54, 1976.

13. Williams, Robert H., and Larsen, Eric D., “Aeroderivative Gas Turbines forStationary Power,” Annual Review of Energy, Vol. 13 (1988): 429-489.

14. Anon., “The Purpose of PURPA,” Mechanical Engineering, August 1988: 56.

15. Henneforth, James C., and Todd, Douglas M., “Cogeneration on a LargeScale,” Mechanical Engineering, August 1988: 51-53.

16. Anon., “Modular Cogeneration: A Capital Idea,” Mechanical Engineering,August 1988: 54-57.

17. Diamant, R. M. E., Total Energy. New York: Pergamon Press, 1970.

18. Horlock, J. H., Cogeneration: Combined Heat and Power. New York:Pergamon Press, 1987.

19. Williams, R. H., “Industrial Cogeneration,” Annual Review of Energy, Vol. 3,1978: 313-356.

20. Clarke, E. L., “Cogeneration, Efficient Energy Source,” Annual Review ofEnergy, Vol. 11, 1986: 275-294.

21. Joskow, P. L., “The Evolution of Competition in the Electric Power Industry,”Annual Review of Energy, Vol. 13, 1988: 215-238.

22. Larson, E. D., and Williams, R. H., “Steam-Injected Gas Turbines,” ASMEJournal of Engineering for Gas Turbines and Power, Vol. 109, January 1987: 55-63.

23. Burnham, J. B., Giuliani, M. H., and Moeller, D. J., “Development, Install-ation, and Operating Results of a Steam-Injection System (STIG) in a GeneralElectric LM 5000 Gas Generator,” ASME Journal of Engineering for Gas Turbinesand Power, Vol. 109 (July 1987): 257-262.

24. Boutacoff, David, “Pioneering CAES for Energy Storage,” EPRI Journal,January-February 1989: 31-39.

386

25. Pollak, Robert, “Status of First U.S. CAES Plant,” EPRI Journal, December1998: 49-52.

26. Anon., “The Circulating Fluidized Bed Boiler,” Heat Engineering, FosterWheeler Corp., September-December 1986: 88-89.

27. Anon., “A Combined-Cycle Cogeneration Plant,” Heat Engineering, FosterWheeler Corp., January-April 1986: 67-73.

28. Douglas. John, “Quickening the Pace of Clean Coal Technology,” EPRIJournal, January-February 1989: 5-15.

29. Clarke, Ed, “Cool Water Coal Gasification Program: An Update,” EPRIJournal, September 1988: 45-48.

30. Johnson, Gordon L., “Tailoring Controls to Accomodate Circulating FluidizedBed Boilers,” Power, May 1989:72-73.

31. Kolp, D. A., and Moeller, D. J., “World’s First Full STIG LM5000 Installed atSimpson Paper Company,” ASME Journal of Engineering for Gas Turbines andPower, Vol. 111 (April 1989): 200-210.

32. Scalzo, A. J., et al., “A New 150-MW High-Efficiency Heavy-DutyCombustion Turbine,” ASME Journal of Engineering for Gas Turbines and Power,Vol. 111 (April 1989): 211-217.

33. Cohn, Art, “Steam-Injected Gas Turbines Versus Combined Cycles,” EPRIJournal, October-November 1988: 40-43.

34. Holway, Donal K., and Moody, Joe R., Jr., “Salina Pumped Storage–Designand Operating History,” Waterpower ‘83, Proceedings of the InternationalConference on Hydropower, Vol. 2, 861-870, 1983.

35. Anon., “Pumped Storage,” Engineering Foundation Conference, AmericanSociety of Civil Engineers, 1975.

36. Barrows, H. K., Water Power Engineering. New York: McGraw-Hill, 1934.

37. Douglas, John, “Tools of the Trade for Hydro Relicensing,” EPRI Journal,July-August 1988: 4-13.

38. Holway, Donal K., Private Communication, June 21, 1989.

39. Makansi, Jason., “Ultralow Emissions, Performance Tied to ProcessKnowhow,” Power, June 1989: 64-66.

40. Bretz, Elizabeth A., “Examine Full Impact of Injecting Steam into Gas TurbineSystems,” Power, June 1989: 53-61.

387

41. Brunner, Calvin R., Handbook of Hazardous Waste Incineration. Blue RidgeSummit, Pa: Tab Books, 1989.

42. Spiewak, Scott A., Cogeneration and Small Power Production Manual, 3rd ed.Lilburn, Ga.: Fairmont Press, 1991.

43. Weston, Kenneth C., “Turbofan Engine Analysis and Optimization UsingSpreadsheets,” Proceedings of the ASME Computers in Engineering Conference,August 1, 1989.

44. Cohen, H., Rogers, G. F. C., and Saravanamuttoo, H. I. H., Gas TurbineTheory, 3rd ed. New York: Longman, 1987.

45. Wilson, David Gordon, The Design of High-Efficiency Turbomachinery andGas Turbines. Cambridge, Mass.: MIT Press, 1984.

46. Chenoweth, C. V., Schreiber, H., and Dolbec, A. C., “20 Years of Operation atHorseshoe Lake Station–Oklahoma Gas and Electric Company,” Procedings of theAmerican Power Conference, Vol. 96, 1984.

47. Roll, Mark W., “The Dow Syngas Project,” Turbomachinery International,Vol. 32, No. 1, January-February 1991: 28-32.

48. Larson, E. D., and Williams, R. H., “Biomass-Gasifier Steam-Injected GasTurbine Cogeneration,” ASME Journal of Engineering for Gas Turbines andPower, Vol 112, April 1990: 157-163.

49. Zaugg, P., and Stys, Z. S., “Air-Storage Power Plants with SpecialConsideration of USA Conditions,” Brown Boveri Review, Vol. 67 (December1990): 723-733.

50. Kehlhofer, Rolf, Combined-Cycle Gas and Steam Turbine Power Plants.Lilburn, Ga: Fairview Press, 1991.

51. United Technologies Turbo Power and Marine Systems, Inc. FT8 Gas TurbineBrochure. 52. Goldstein, Gina, “Rechanneling the Waste Stream,” Mechanical Engineering,Vol. 111 (August 1989): 44-50.

53. Butler, Arthur J., and Hufnagel, Karl R., “Skagit County Resource RecoveryFacility Design of a 2500-kW Waste-to-Energy Plant,” ASME ‘89–JPGC/Pwr-12,October 1989.

54. Douglas, John, “Beyond Steam,” EPRI Journal, December 1990: 4-11.

55., G.E. Aircraft Engines source, personal communication.

56. Riley, Shaun T., (manager of LM 5000 and Industrial programs, G. E. Marineand Industrial Engines & Service Div.), personal communication.

388

57. Riley, Shaun T., “STIG Increases Output from Aeroderivative Gas Turbines,”Modern Power Systems, June 1990: 41-47.

58. Stewart, Norman, “Shell Coal Gasification,” EPRI Journal, June 1989: 38-41. 59. Sundstrum, David G., “Status of the Dow Syngas Project: Three Years ofExperience at the World’s Largest Single Train Coal Gasification Plant,”Proceedings of the American Power Conference, Vol. 52, 1990, pp. 512-515.

60. Anon., “Gasifier Demo Heralds New Era for Gas Turbines,” Power, October1989: S-24–S-28.

61. Anon., “Cogen System Balances Steam Cycle Demand,” Power, October1989: S-42–S-44.

62. Ullman, Ronald R., and Duffy, Thomas J., “Integrated Material and EnergyRecovery from Municipal Solid Waste,” Proceedings of the American PowerConference, Vol. 52, 1990, pp. 570-577.

63. Smith Jeffrey J., “ The GE90 ... When is Big’ Big Enough?” ASME TurboExpo–Land, Sea, and Air, Orlando, Fla., June 6, 1991.

64. Lamarre, Leslie, “Alabama Cooperative Generates Power from Air,” EPRIJournal, December 1991: 12-19.

65. United States Department of Energy website, www.fe.doe.gov.

66. Schroeter, Jeffrey W., “The Merchant Revolution,” Mechanical EngineeringPower, May, 2000. 67. Lane, Jon, “More Power, Fewer Players,” Mechanical Engineering Power,May, 2000.

EXERCISES

9.1 A 10-MW gas turbine operating at the conditions in Example 5.1 exhauststhrough a process heat exchanger, with the combustion gas leaving at 200°F.What is the rate of heat transfer to the process? If the heat transfer is to liquidwater, what flow rate of water can be increased in temperature from 50°F to160°F in the heat exchanger? How many homes with an average heating demand of 60,000 Btu/hr could be serviced by the gas turbine in a districtheating application where there is a 20% energy loss in the system distribut-ing heat to the customers? What is the unweighted system energy utilizationfactor?

9.2 Design the combustion gas heat exchanger required in Exercise 9.1. Indicatethe type selected, the required surface area, the geometric configuration, andoverall size.

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9.3 The oxygen remaining in the exhaust gas of a 5-MW gas turbine supplies theoxidizer for a process heater that burns methane completely. The gas turbineoperates at the conditions of Example 5.1. What is the nominal heat transferrating if the heater is designed for an exit temperature of 300°F? What is therated of fuel consumption and the unweighted energy utilization factor of thiscombined heat and power system? What would the maximum heater outputbe if the gas turbine were shut down and atmospheric air were supplied to theheater with the same supplemental fuel firing rate?

9.4 Compare the unweighted energy utilization factors for the simple-cycle and regenerative gas turbines of Examples 5.1 and 5.2, assuming their exhaustsserve unfired process heat exchangers with 250°F exit gas temperatures.Compare the energy utilization factors, assuming the useful heat in each isweighted by the cycle’s thermal efficiency.

9.5 The exhaust of a 2-MW gas turbine, operating at the conditions of Example5.1, transfers heat to an absorption chiller with a COP of 0.78. The exhaustgas leaves the chiller at a temperature of 220°F. How many tons ofrefrigeration can be produced by the chiller? What is the unweighted EUFbased on shaft power and chiller rate of cooling?

9.6 A 20-MW simple-cycle gas turbine operates with compressor inletconditions of 101 kPa and 15°C, a turbine inlet temperature of 1200°C, and acompressor pressure ratio of 12. The compressor and turbine isentropicefficiencies are 84% and 88%, respectively. The turbine exhaust flowsthrough a process heat exchanger and exits to the atmosphere at 110°C.Determine the gas turbine cycle state properties, the thermal efficiency, andthe work ratio, accounting for an 80-kPa pressure drop on the process heatexchanger gas turbine exhaust side. What is the rate of heat transfer to theprocess through the heat exchanger? If the heat transfer is to liquid water,what flow rate of water can be boiled at atmospheric pressure if water entersthe heat exchanger at 30°C? How many homes with an average heatingrequirement of 20 kW can be heated in a district heating application wherethere is a 15% loss in the distribution of heat to customers? What is theunweighted system energy utilization factor?

9.7 Design the combustion gas heat exchanger required in Exercise 9.6. Indicatethe type selected, the required surface area, the geometric configuration,overall size, and the estimated pressure drops. Can you improve significant-ly on the assumed heat exchanger pressure drop used in Example 9.6?

9.8 The exhaust of a 5-MW gas turbine supplies the oxidizer for a heater thatburns methane completely. The gas turbine operates the 101 kPa and 15°C compressor inlet conditions and 1060K turbine inlet temperature. The enginehas compressor and turbine isentropic efficiencies of 86% and 88%, respect-ively, and a compressor pressure ratio of 10. What is the nominal heat transfer rating of the heater? What is the total rate of fuel consumption andthe unweighted energy utilization factor? Combustion gases leave the heaterat 200°C.

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9.9 The exhaust of a 2-MW gas turbine operating at 960°C turbine inlet temp-erature with a compressor pressure ratio of 9 transfers heat without loss tooperate an absorption chiller with a COP of 0.83. The exhaust gas leaves thechiller at a temperature of 120°C. The compressor inlet conditions are 105kPa and 25°C, and the compressor and turbine isentropic efficiencies are82% and 86%, respectively. How many tons of refrigeration can be producedby the chiller?

9.10*Develop a spreadsheet like that in Table 9.1, and investigate the influence onthe combined-cycle efficiency of varyingeach of the following:

(a) HRSG exit temperature.(b) Steam turbine throttle temperature.(c) Gas turbine inlet temperature.(d) Compressor pressure ratio.

9.11 A closed-cycle gas turbine using air as a working fluid has an intercooler, areheater, and a recuperator. The overall pressure ratio is 6. The low-pressurecompressor has a pressure ratio of 2. Both compressors are driven by thehigh-pressure turbine. All turbomachines are 85% efficient. Compressorinlet temperatures are 80°F, and turbine inlet temperatures are 1500°F.Regenerator effectiveness is 75%.

(a) Draw T-s and flow diagrams, and label both compatibly.(b) Identify actual temperatures, in degrees Rankine, at all stations of sig-nificance.(c) If the low-pressure compressor inlet pressure is 6 atm, what are theintercooler and reheater gas pressures?(d) Calculate the total compressor work, and compare with the work for asingle compressor with 85% efficiency without intercooling.(e) What is the gas turbine net work?(f) What is the total external heat addition?(g) How much heat is available from the precooler for district heating ?(h)What is the plant thermal efficiency?(i) What is the plant energy utilization factor if 80% of the precooler heatrejection is used for district heating?(j) If the turbines deliver 100MW of power, what is the air flow rate and(k) the rate of consumption of coal, in tons per hour (12,000 Btu/lbmheating value and 90% combustion efficiency)?

______________________* Exercise numbers with an asterisk involve computer usage. 9.12 A closed-cycle gas turbine using air as a working fluid has an intercooler, a

reheater, and a recuperator. The overall pressure ratio is 6. The low-pressurecompressor has a pressure ratio of 2. Both compressors are driven by thehigh-pressure turbine. All turbomachines are 85% efficient. Compressorinlet temperatures are 15°C, and turbine inlet temperatures are 1000°C.Regenerator effectivenesss is 75%.

(a) Draw T-s and flow diagrams, and label both compatibly.(b) Identify actual temperatures, in degrees Kelvin, at all stations of signif-icance.

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(c) If the low-pressure compressor inlet pressure is 6 atm, what are theintercooler and reheater gas pressures?(d) Calculate the total compressor work, and compare with the work for a single compressor with 85% efficiency without intercooling.(e) What is the gas turbine net work? (f) What is the total external heat addition?(g) How much heat is available from the precooler for district heating? (h) What is the plant thermal efficiency?(i) What is the plant energy utilization factor if 80% of the precooler heatrejection is used for district heating?( j) If the turbine delivers 100 MW of power, what is the air flow rate and(k) the rate of consumption of coal in kilograms per second (25,000 kJ/kg heating value and 90% combustion efficiency)?

9.13 Starting with the selected combined-cycle design for Example 9.1, consider amodification to the heat-recovery steam generator that allows saturated liquidwater at state 12 to be extracted for industrial process use. Assume that the water is returned to the condenser as a saturated liquid at the condensing temperature. Assume that process use dictates the mass flow rate extracted,with the balance of the water going as steam to the steam turbine. Calculateand tabulate the heat supplied per unit of gas turbine mass flow as a functionof the process liquid mass fraction of the total water flow entering the steamgenerator. Also calculate and plot the energy utilization factor as a functionof the process mass fraction. If the gas turbine power output is 6 MW, whatis the maximum saturated-liquid-water process heat transfer rate?

9.14* Starting with the selected combined-cycle design for Example 9.1, considersuperheated steam at state 8 to be extracted for industrial process use.Assume that the water is returned to the condenser as a saturated liquid atthe condensing temperature. Assume that process use dictates the massflow rate extracted, with the balance of the steam created going to thesteam turbine. Calculate and tabulate the process heat supplied per unit ofgas turbine mass flow as a function of the process steam mass fraction ofthe water flow through the steam generator. Also calculate and plot theenergy utilization factor as a function of the process mass fraction. If thegas turbine power output is 6 MW, what is the maximum superheated-steam process heat transfer rate?

9.15 Extend the analysis of Table 9.1 to include supplemental firing of the HRSG, to provide an inlet gas temperature of 1500°F and a steam turbine throttletemperature of 1000°F. Determine the influence of boiling temperature onthe pinchpoint temperature difference, and on the net work per pound of gasturbine flow; and compare the combined-cycle thermal efficiency with theefficiencies of the individual cycles. Discuss the results of the analysis.

9.16 Extend the analysis of Table 9.1, to consider a modification of the heat-recovery steam generator that allows saturated water vapor at state 13 to beextracted for industrial process use. Assume that the water is returned to thecondenser as a saturated liquid at the condensing temperature. Assume thatprocess use of the steam has priority, with the balance of the steam going to

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the steam turbine. Calculate and tabulate the process heat supplied per unit ofgas turbine mass flow as a function of the process water mass fraction of thewater flow through the steam generator. Also calculate and plot the energyutilization factor as a function of the process mass fraction. It the total poweroutput of the combined cycle is 6 MW, what is the maximum process heattransfer rate?

9.17 Extend Exercise 9.15 to consider a modification of the heat-recovery steamgenerator that allows superheated steam at state 8 to be extracted forindustrial process use. Assume that the water is returned to the condenser asa saturated liquid at the condensing temperature. Assume that process use ofthe steam has priority, with the balance of the steam going to the steamturbine. Calculate and tabulate the process heat supplied per unit of gasturbine mass flow as a function of the process steam mass fraction of thewater flow through the steam generator for a boiling temperature of 600°F.Also calculate and plot the energy utilization factor as a function of processmass fraction. If the total power output of the combined cycle is 20 MW withno process heat, what is the maximum process heat transfer rate?

9.18 Design an open-cycle regenerative gas turbine cogeneration system in whicha fraction of the turbine exhaust gas can bypass the heat exchanger forprocess use. Prepare a report stating your design criteria, defining youranalysis methodology, and presenting performance data for the nominaldesign condition that you selected.

9.19 Perform the design required in Exercise 9.18, and include consideration ofperformance for a range of off-design process heat transfer requirements thatare lower than the design value.

9.20 A 20-MW electric motor in a simple compressed-air-storage plant drives acompressor with a pressure ratio of 10 and an efficiency of 85% for six hoursnightly. What power output can be obtained with a turbine inlet temperatureof 1600°F if the plant operates for four hours during the day at constantpower output? The turbine efficiency is 90%. What is the air-fuel ratio ifmethane is the fuel used? What is the net generation efficiency, consideringonly the fuel consumption of the turbine? Assume a daily cycle, that cavernpressure changes are negligible, and that the heat of compression isdissipated before generation begins.

9.21* An ideal steam turbine operates with 1000°F, 2000-psia throttle, and 1-psiacondenser and produces 15 MW without extraction. When steam is extractedfor process use at 500 psia, after use it is condensed to a saturated liquid atthat pressure and throttled to the condenser. Tabulate and plot the processheating rate and the EUF as a function of extraction mass fraction.

9.22* A 25 MW steam turbine operates in 1000°F and 2000 psia with an efficiencyof 87%. Eighty percent of the condenser heat transfer is used for an industrial process. Tabulate and plot the process heating rate and the EUF asa function of condenser pressure between 1 psia and 2 atmospheres.

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9.23* Develop the equations and an algorithm for the analysis of the steam-injected gas turbine with an unfired steam generator producing superheatedsteam at the combustion chamber pressure and using methane as a fuel . Usethe JANAF tables for thermodynamic properties of steam in the gas turbine.State clearly the assumptions made. Write a computer programmingimplementing the algorithm.

9.24 Consider a compressor operating at a pressure ratio of 20 and a polytropic efficiency of 86% that compresses ambient air at 101 kPa and 15°C into acavern. Assume that heat losses from the cavern maintain the air at 15°C andconstant pressure during the filling period from midnight to 6 am daily. Thecompressor is driven by a 20 MW electric motor. What is the daily massaddition to the cavern? From 2 pm to 6 pm daily the same mass of air thatwas added to the cavern during the night is heated to 1200K and allowed toescape to the atmosphere at a constant flow rate through a turbine expanderwith a 90% isentropic efficiency. What is the expander power output? Whatis the daytime energy output? What is the fractional fuel consumptionreduction if a regenerator with 80% effectiveness is added to the system?

9.25 Resolve Example 9.2 for superheated steam injection at 400°, 500°, 600°, and 700°F and the combustor pressure level. Write a brief report on yourfindings on the influence of temperature of injected steam on STIG perform-ance.

9.26* Use the STIG spreadsheet shown in Table 9.5 to verify the performancecalculations of Figures 9.23 and 9.24.

9.27* Investigate the influence of compressor pressure ratio variation on STIG performance for the model of Example 9.2, and prepare a memo reportingyour results.

9.28* Evaluate the separate influences of steam heat capacity and added mass onthe thermal efficiency, power output, and work ratio for the model ofExample 9.2.

9.29 Consider a two-shaft gas turbine to be modified for steam injection. Thecompressor pressure ratio is 9.3, and the turbine inlet temperature is 982°C.The isentropic efficiencies of the compressor and turbines are 83% and 90%,respectively. The gas generator mechanical efficiency is 99%, and the powerturbine drives an electrical generator that has a 93% efficiency. Accountingfor a 4% pressure loss in the combustor and a fuel heating value of 43,000kJ/ kg, compare the electrical power output, specific fuel consumption,thermal efficiency, and fuel-air ratios for 0.0 and 0.05 steam-air ratios. Briefly described your selection of steam system design conditions.

9.30 It has been decided that the heat-recovery steam generator for a steam-injected gas turbine must be retubed to continue running it in the steaminjection mode. The expected cost of retubing is $230,000. Steam injectionproduces an additional 4000 MW-hr per year, adding two cents per kW-hr to

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revenue. What is the break-even operating time to recover the cost of thismaintenance operation?

9.31 Compare the isentropic efficiencies of two compressors, each having apolytropic efficiency of 87% and pressure ratios of 6 and 18.

9.32 What is the polytropic efficiency of a turbine, with a pressure ratio 30, thathas the same isentropic efficiency as a turbine having a polytropic efficiencyof 92% at a pressure ratio of 12?

9.33* Compare the performance of a twin-spool turbofan engine with separate fanand compressor to that of a turbojet engine, both with an overall pressureratio of 30 and a turbine inlet temperature of 1300K, and designed for analtitude of 15,000m and a flight speed of 275 m/s. The bypass ratio is 6 andthe fan pressure ratio is 1.6. Assume turbine, fan, and compressor polytropicefficiencies of 90% and a combustor pressure loss of 3% of the compressorexit total pressure. Compare specific thrusts for engines built from the samecore engine. Compare, also, specific fuel consumption and jet velocities.

9.34* Build a multicase spreadsheet for the conditions of Exercise 9.33, and use itto plot specific thrust and TSFC as a function of fan pressure ratio. Use thespreadsheet to explore further the influence of varying bypass ratio for avalue of fan pressure ratio suggested by your plot. Write a memo discussingbriefly the results of your study.

9.35 Develop an analysis for a turbofan engine, sometimes called an aftfanengine, in which the fan is located at the same axial station and directlyattached to a low pressure turbine dedicated to driving the fan. The high-pressure turbine drives the compressor, and is located upstream of the fan.The compressor and fan each have their own separate inlets.

9.36 Express the work of a compressor in terms of its inlet temperature andpressure ratio using (a) the isentropic efficiency, and (b) the polytropicefficiency. Equate the relations, and solve for the compressor isentropicefficiency. Compare your result with Equation (9.13).

9.37 Express the work of a turbine in terms of its inlet temperature and pressureratio using (a) the isentropic efficiency, and (b) the polytropic efficiency.Equate the relations, and solve for the turbine isentropic efficiency. Compareyour result with Equation (9.10).

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C H A P T E R 10

Nuclear Power Plants*

10.1 Introduction

Nuclear power is universally controversial. Many would say that it is also universallyneeded�as an alternative or supplement to power generated by fossil fuels.

The combustion of fossil fuels produces carbon dioxide, now notorious for the threat of global warming. Nuclear power plants produce neither carbon dioxide noroxides of sulfur and nitrogen, as does the burning of fossil fuels. Thus nuclear powerreduces the global production of carbon dioxide and other pollutants, and helps toalleviate many of the pervasive problems of fossil fuel supply.

Petroleum is least available in regions of widest use; natural gas is, for the timebeing, plentiful and sought after by all; and widely abundant coal has come to beregarded as the great Satan of air pollution. Water power is important, but it offerslimited possibility for growth. Solar energy, while promising, is far from being amainstay of the world�s energy supply. Thus sources other than fossil fuels and nuclearpower offer little hope to become major suppliers during our lifetimes.

Nuclear power, in stasis for many years, may make a comeback. Engineers havebeen quietly working on new and safer designs for nuclear power plants, and thepolitical climate may be swinging slowly back in favor of nuclear power. According toreferences 31 and 34, there were 434 operating nuclear plants producing 17% (350,000megawatts) of the world�s electricity in 1998. Regardless of one�s position towards it,nuclear power is a major factor in world power production.

Knowledge of nuclear power is not American, French, Indian, or British; it isvirtually universal. Nuclear power plants such as those shown in Figure 10.1 areproducing power in many nations around the world. Blockage of the growth of nuclearpower in the United States did not prevent the development and extensive use ofnuclear-generated electricity in France or Canada. Should developing countries withminimal fossil resources not use nuclear power? Should a country that has seen theterror wrought by nuclear weapons be denied the benefits of electricity from thenucleus? Can attempts to halt the growth of nuclear power stop the proliferation ofnuclear weapons? These and many other issues around nuclear power, which obviouslyextend far beyond the bounds of engineering, are much too broad to be pursued here. Nuclear power is controversial, but it is here; it is important and likely to remain so.

*Thanks to Dr. Andrew A. Dykes for his valuable inputs and comments on thischapter.

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This chapter is concerned largely with nuclear fission power reactors. Most nuclearpower plants use steam cycles that differ little from those of fossil fuel plants except forthe source of heat for the steam generator and for steam supply conditions. Steamturbine cycles were considered in some detail in Chapter 2. Thus this chapter will focusprimarily on (a) the characteristics of nuclear reactor steam supply systems and, to alesser extent, on (b) the climate for future nuclear development. In preparation for thisstudy let us first review relevant aspects of atomic and nuclear structure.

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10.2 Review of Atomic Structure

Atoms and Molecules

Somewhat more than a hundred elements are known and are thought to be the buildingblocks of everything in the universe. The atom is the basic unit of structure for eachelement. An important connection between the microscopic world of the atom and themacroscopic world of experience is given by Avogadro's number. A gram-mole of anyelement has Avogadro's number (6.023 x 1023) of atoms.

The atom may be considered as consisting of a positively charged nucleus at itscenter and one or more negative charges around the nucleus called electrons that makethe atom electrically neutral. The electron is the fundamental unit of negative charge. Itmay be viewed as a particle which is much smaller than the nucleus and which orbitsaround the nucleus as a planet orbits the sun, or it may be viewed as a diffuse electroncloud around the nucleus. Still another concept is that of a particle whose location isnot known but which is more likely to be in some places than others, its position beinggiven by a probability distribution. We will not concern ourselves with the rationalesfor these views.

Thus atoms consist of nuclei surrounded by electrons. The sizes of atoms areconveniently measured in Angstroms (10-8 cm). The nucleus typically is of the order of10 �5 Angstroms. Thus the volume of the atom is largely due to the size of the outerelectron�s orbit or to the atom�s electron cloud.

Molecules are collections of atoms held together by electromagnetic forces betweenthe nuclei and the electrons. Atoms and molecules can exist in a variety of energy statesassociated with their electron distributions. These microscopic states and theirmacroscopic influences are dealt with theoretically in the fields of quantum mechanicsand statistical thermodynamics.

Molecules and atoms can interact with each other to form different molecules inways that are controlled by their electron structures. These interactions, calledchemical reactions, have little to do with the nucleus. In Chapter 3, we consideredaspects of these reactions that are relevant to combustion. The magnitudes of theenergy associated with these chemical changes, while of great importance in thermalengineer-ing, are so small that they have no significant influence on the nuclei of thereacting atoms. Thus the nuclei may be thought of as merely going along for the ridewhen a chemical reaction occurs.

An electrically neutral particle, however, can penetrate an atom's electromagneticfield and approach the nucleus, where it interacts via short range but powerful nuclearforces. Then the electrical forces holding the nucleus together may be overcome,resulting in changes in the nucleus. In these cases the interatomic forces are largelyirrelevant and are overpowered by nuclear events. It is these changes that are theconcern of this chapter.

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The Nucleus

The nucleus, for our purposes, may be thought of as being made up of integral numbersof protons and neutrons. The proton is a particle with a positive charge of the samemagnitude as that of the electron, so that pairing a proton with an electron producesexact electrical neutrality. Thus protons account for the charge of the nucleus, and alike number of electrons ensures the electrical neutrality of the atom. Compared withthe electron, the proton is a massive particle, having a mass which is about 1800 timesthe mass of the electron. Thus the hydrogen atom, which consists of one proton in thenucleus and a single electron in orbit around the nucleus, is electronically neutral andhas a mass only slightly larger than that of the nucleus. Atoms larger than the hydrogen atom have more than one proton in the nucleus andhave one or more neutrons as well. A neutron, as the name suggests, is an electricallyneutral particle with a mass only slightly larger than that of the proton. As componentsof the nucleus, protons and neutrons are called nucleons, and are thought ofinterchangeably with respect to mass because their masses differ so little from eachother.

The number of protons in an atom of an element is called the atomic number of theelement. Thus hydrogen has an atomic number of 1. The atomic number of a givenelement is unique to that element. Thus we could identify the elements by their atomicnumbers rather than by their names if we wished. Elements are ordered in the periodictable in part by their atomic numbers.

The mass number of an element is the number of nucleons in an atom of thatelement and is therefore the sum of the number of protons and neutrons in the nucleus.Atoms of a given element that have differing mass numbers are called isotopes of theelement. A given isotope of an element is sometimes designated by a notation thatincludes the element's chemical symbol, its mass number, and its atomic number. Forexample, the most common isotope of uranium is denoted as 92U238, where 92 is theatomic number of the element uranium and 238 is the sum of the number of protonsand neutrons in the isotope nucleus. The isotopes are also sometimes simply identifiedby their name or symbol and mass number, such as U-235 or Uranium-235. Othersignificant examples are the isotopes of hydrogen, deuterium, 1H2, and tritium, 1H3,which have, respectively, one and two neutrons accompanying the proton. Theseisotopes are sometimes written 1D2 and 1T3 to reflect their commonly used names. Theform of water, H2O, in which the isotope deuterium replaces hydrogen is commonlycalled heavy water, D2O, because of the added mass of the extra neutron in eachnucleus. It will be seen later that heavy water has characteristics that are advantageousin some nuclear processes.

10.3 Nuclear Reactions

Just as chemical fuels undergo chemical reactions that release energy, nuclei may alsoparticipate in energy-releasing nuclear reactions. When this happens atoms of the

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reacting elements are converted to atoms of other elements, the sort of transmutationsought by the alchemists of the past.

Mass-Energy Equivalence

Two nuclear processes are known to be capable of releasing energy on a scale largeenough to influence the personal and business lives of humans. The reality of bothprocesses has been amply demonstrated by the production and detonation of the atomicand hydrogen bombs. Both processes owe their energy release to the annihilation ofmatter consistent with the famous Einstein formula, E = mc2, which asserts theconvertibility of mass to energy. Because the speed of light is so large, the equationshows that the annihilation of a small amount of mass yields a large quantity of energy.These energy releases are usually measured in MEV, millions of electron-volts. Theelectron-volt, EV, is defined as the energy required for an electron to pass through apotential difference of one volt. The reactions of individual nuclei typically produceparticles with energies measured in MEV. On the other hand, as indicated earlier, themost energetic of chemical reactions releases much less energy, only a few EV permolecule.

As a result of our encounter with Einstein's energy-mass relation we must adjustour philosophical position on the conservation laws of mass and energy and thinkinstead in terms of conservation of mass-energy. Mass and energy may be thought of asdifferent forms of the same thing. Any change in the mass of an isolated system must beaccompanied by a corresponding change in system energy. This in no way influencesthe discussions in previous chapters, because changes in mass are entirely insignificantin chemical and other nonnuclear processes.

Fission and Fusion

The process known as nuclear fusion occurs in nature in the stars, including our ownsun. Since the Second World War, scientists have been attempting to achieve theconditions for fusion in the laboratory. Because it can use heavy water from the sea asa fuel, controlled thermonuclear fusion offers the hope of vast quantities of power formany centuries in the future.

Fusion occurs when light atoms interact to form a heavier atom in reactions such as

1D2 + 1D2 � 2He3 + 0n1 + 3.2 MEV.

Here, two deuterium atoms collide to form helium-3 and a neutron while releasing 3.2MEV of energy. Other fusion reactions exist that provide comparable amounts ofenergy. Using precise atomic masses measured with mass spectrometers, we candetermine the differences of the masses of the reactants and products in this reaction.Application of the Einstein relation to the mass loss yields the same energy release (inthis case 3.2 MEV) as is obtained by energy measurements. Thus the energy yield of

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known nuclear reactions may be determined with only mass measurements. For over fifty years, researchers have pursued the goal of achieving controlled

thermonuclear fusion on a scale suited for commercial power production. Since thereactants are two positively charged nuclei, they must have high kinetic energies toovercome their mutual repulsion. These high energies imply a gaseous state withenormously high temperature, a condition known as a plasma. Because solid materialscannot exist at plasma conditions and plasmas would be cooled by the presence ofsolids, magnetic confinement of plasmas has been one approach to achieving athermonuclear plasma. Large experimental devices called Stellerators, Tokamaks, andmirror machines have been built to help solve the problems inherent in achieving large- scale fusion reactions and in stably confining the associated thermonuclear plasma.While progress continues, controlled thermonuclear fusion remains, and will likelycontinue, in the research stage for many years. It will therefore not be consideredfurther here.

Whereas nuclear fusion annihilates mass by forming larger atoms from light atoms,fission is a process of breaking massive atoms into two large, more-or-less equal-sizedatoms, with an accompanying mass loss and energy release. While controlled fusionremains elusive, nuclear fission has been producing electrical power on a commercialscale for decades. The remainder of this chapter therefore deals with the fundamentalsand commercial use of nuclear fission.

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10.4 Fundamentals of Nuclear Fission

Nuclear fission involves the breakup of certain massive elements resulting from col-lisions with neutrons. Uranium U-235, for instance, the isotope of uranium with 235nucleons, forms an highly excited isotope with 236 nucleons upon capture of a neutron,as in Figure 10.2(a). These excited U-236 atoms are unstable and break up into avariety of pairs of large atoms shortly after they are created. The many such fissionsoccurring in a reactor may be expressed as an average reaction:

92U235 + 0n1 � 92U236� F1 + F2 + 2.47 0n1 + 203 MEV

where F1 and F2 represent fission fragment elements, as in Figure 10.2(b). On theaverage, 2.47 neutrons are released in the process, one of which must initiate anotherfission to sustain a chain reaction, as illustrated schematically in Figure 10.3. Inaddition alpha, beta, and gamma radiation is released. One of the many reactions thatparticipate in the average reaction above creates xenon and strontium fission fragmentsand two neutrons:

92U235 + 0n1 � 92U236 � 54Xe139 + 38Sr95 + 2 0n1 + energy

U-235 reactions exemplified by the Xe-Sr reaction create diverse fission fragments andsmall integral numbers of neutrons that average to 2.47 and release energies thataverage to 203 MEV.

Over 80% of the 203 MEV of energy released by the average U-235 fissionreaction is the kinetic energy of the fission fragments associated with their large massand high velocity. Because of their high energy, the fission fragments become ionizedand ionize nearby atoms as they tear their way through surrounding materials. Thefission fragments, however, travel only a very short distance, for interactions causethem to lose much of their energy to the surrounding solid. Thus most of the fission

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energy appears immediately as internal energy and therefore locally high temperature ofthe surrounding materials. It is, therefore, necessary that the fuel be cooledcontinuously to counter the fission heat generation to avoid temperature buildup andpossible melting. The heated coolant is then provides the energy input for athermodynamic (usually steam) cycle.

As the fission fragments come to rest, their presence changes the character of thereactor materials. Non-fissionable material exists where fissionable and other atomsonce resided. In time these materials may decay radioactively, forming still differentsubstances and releasing additional heat. But, importantly, they absorb neutronswithout offering the possibility of fission. Thus the fission fragments are said to poisonthe reactor.

Fission Reactor Design Considerations

While most of the fission heat generation is due to the fission fragments, reactor designfocuses on the actions of neutrons. This is because neutron captures produce fissions,fissions produce more neutrons, and, as we will see later, fissile material may also beproduced using neutrons. Thus neutrons are the currency of the reactor and may beused constructively or wastefully. The manner in which the neutrons in a reactor areused is called the neutron economy. The reactor designer pays close attention to all ofthe details of the neutron economy. One of these details is the distribution of the kineticenergy of the neutrons, that is, the fractions of the neutrons that lie in given energyranges.

Almost all commercial power reactors (as opposed to experimental reactors) arethermal reactors in which fission is caused by thermal neutrons. A thermal neutron is aneutron that is in thermal equilibrium with the surrounding atoms. This implies thatthey have energies on the order of 0.02 EV. Because they move relatively slowly,thermal neutrons are much more likely to cause fission of U-235 than are higher-energyneutrons; therefore, they are the choice for most power reactor designs. However, theneutrons created by the fission reaction are not thermal neutrons. The neutrons createdby fission have kinetic energies that range from about 1 to 10 MEV. They are called fast neutrons because of this high kinetic energy.

Thus, if a chain reaction is to be sustained in a thermal reactor, it is necessary forthe fast neutrons to be slowed, or thermalized, to much lower energies so that they cancause fissions before they are absorbed in nonproductive captures in reactor material orbefore they escape from the reactor. They do this by colliding with certain other nucleiin the core, put there for that purpose. These nuclei are called moderators. A goodmoderator is a light element that, on collision with a neutron, is speeded up by thecollision, and thereby extracts energy from the neutron without absorbing it.

The moderator concept may be understood by considering what happens when abilliard cue ball hits another ball head-on. The cue ball stops, and the second ball carriesaway the kinetic energy. However, with balls of differing mass, if the second ball isheavy the cue ball bounces off without losing energy, whereas light balls are propelled

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at high speed and extract significant energy from the cue ball. In neutron collisions withatoms, it is the lightest atoms that extract the most energy.

Ordinary water, heavy water, graphite, and beryllium are all used as moderatorsbecause they are light and are poor absorbers of neutrons. Hydrogen and deuterium arethe moderators in water. The oxygen in the water is not an effective moderator but haslow absorption and therefore does not interfere significantly with the moderationprocess.

It is evident that conservation of neutrons is a prime consideration in any reactordesign. The number of neutrons per unit volume, or neutron density, in a reactor is animportant design parameter. Neutrons diffuse about in the reactor as they are scatteredand slowed by moderator atoms. Because they have no charge, they are uninfluencedby electromagnetic fields and therefore may travel further than charged particles.

Four events can influence local neutron densities as they pass through thesurrounding reactor core:

1. The neutrons can be captured by a fissionable atom and produce a fission.2. They can be absorbed non-productively by fission products, structural materials,or nonfissioning fuel.3. They can escape through the walls of the reactor.4. They can be absorbed in nuclei that create more fuel.

The last possibility will be considered in Sections 10.5 and 10.8.Events 2 and 3, absorption and escape, result in nonproductive waste of neutrons in

the neutron economy. The necessity that enough events of type 1, rather than types 2and 3, occur to sustain the chain reaction suggest several important considerations forreactor design:

� The reactor should be large enough that only events of types 1, 2, and 4 takeplace and hence that escape of neutrons from the reactor is rare. Since reactor sizeis dictated primarily by the cooling requirements imposed by nuclear heatgeneration, this usually follows automatically from the design process. In addition,positioning moderating material such as water at the boundaries of the reactor as areflector, to deflect escaping neutrons back into the reactor, may allow a morecompact design, in some cases.

� Materials to be used in the reactor design are selected so that type 2 events areminimized. The gradual buildup of poisons must also be considered in designing forthe change in reactor performance with time between refuelings.

� The reactor should be designed to minimize the amount of structural materials inthe active fuel region, to reduce the frequency of type 2 events.

� Neutron-absorbing materials may be moved into and out of the reactor to changethe average neutron density for reactor control purposes.

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A reactor can function over a range of neutron densities. When the neutron densityand power levels are constant, the reactor is said to be critical. This implies that thenumber of neutrons producing fission at one instant is the same as at a later instant, asituation depicted in Figure 10.4. Since high neutron densities produce more fissions,each of which produces 203 MEV in the case of a U-235 thermal reactor, heatgeneration, and thus power output, increases with neutron density.

Let's assume that 2.5 fast neutrons are produced per fission in a critical thermalreactor and track the activities of 100 fast neutrons created in an instant. In a criticalreactor, about 40 of these fast neutrons must thermalize and undergo fission to produceanother 100 fast neutrons. Figure 10.4 shows that 2 of the original 100 fast neutronsproduce (fast) fissions and 5 more neutrons immediately so that there are 103 fastneutrons diffusing around. Of these, about 10 escape from the core or are absorbed innon-fissile materials while slowing down. Of the remaining 93 thermal neutrons, about

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3 escape from the core and about 43 are absorbed in non-U-235 materials whilediffusing around at thermal energies. The remaining 47 captures by U-235 result in 7non-fission captures and 40 fissions. These 40 fissions, in turn, produce 100 fastneutrons for the next generation in a reactor operating at critical. This sequencerepresenting a single generation in a critical reactor is repeated over and over by anenormous number of neutrons. By slight changes in the reactor configuration duringoperation (for instance, by addition or removal of a small amount of poisons), theneutron economy can be adjusted to increase or decrease the number of neutrons in thenext generation and thus change the reactor operating condition.

A parameter, k, called the multiplication factor, is defined as the ratio of thenumber of neutrons in one generation to the number in the preceding generation. Thusk = 1 for a critical reactor. If k is less than or greater than 1, the neutron density isdecreasing or building and the reactor is said to be subcritical or supercritical,respectively.

A reactor is designed to have a maximum value of k > 1 so that it may be broughtup to a desired power level and maintained there over the duration of the fuel cycle.Once the reactor approaches the desired power level, control actions are taken toadjust the value of k to 1, bringing the reactor to critical and stabilizing its operation.The control action requires the introduction of additional neutron-absorbing material orthe reduction of the amount of moderator in the reactor, to reduce the rate of buildupof the neutron density to zero. As fissionable material is depleted and poisons build up,control material is gradually withdrawn from the reactor to maintain critical operation.When the reactor can no longer maintain critical at its design power level with nocontrol material present, it must be refueled.

The presence of a nuclear heat source is the major difference between fossil fueland nuclear plants. In that connection, there are several important factors that must beconsidered in the design of a nuclear power plant that are not factors in conventionalpower plant design and operation. First, the entire amount of fuel needed to operate anuclear power plant for up to two years is loaded into the plant at one time. The rateat which power is generated must then be maintained by controlling the neutron chainreaction over the wide range of reactor operating and fuel depletion conditions that canarise between refueling operations. This calls for detailed planning and analysis, bothbefore and throughout the operating cycle.

Second, because the products of fission are highly radioactive and their rate ofdecay cannot be controlled, the heat from radioactive decay of fission products aftershutdown amounts to as much as 7% of full power output. Consequently, provisionmust be made in the thermal design to remove this heat under all credible operating andaccident conditions. (The reactor at Three Mile Island melted over an hour after thenuclear chain reaction was terminated, because the operators misinterpreted theirinstruments and turned off the emergency systems that were removing the decay heat.) The nuclear decay process is not self-limiting and has no maximum temperature, aswith chemical reactions. If the heat generated is not removed from the reactor core, thecore will melt and be destroyed.

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Third, if radioactive materials from the reactor core find their way to theenvironment, they can be hazardous to nearby life. Although the proper cooling of thecore will ensure that these materials will remain contained inside the fuel assemblies, adefense-in-depth approach to safety must be employed to make the possibility of arelease extremely remote.

Table 10.1 Nuclear Fuels

Fertile Fuels Fuels Fissionable by Thermal Neutrons

U-235

U-238 � Pu-239

Th-232 � U-233

10.5. Nuclear Fuels

Uranium-235 is the only material that is both fissionable by thermal neutrons and foundin nature in sufficient abundance for power production. Other fissile fuels are uranium-233 and plutonium-239 (Table 10.1) which are created from thorium-232 and uranium-238, respectively, by absorption of neutrons. Substances from which fissionable fuelsare created, called fertile fuels, are transmuted into fissionable fuels in a reactor byextra neutrons not needed to sustain the fission chain reaction. Fertile fuel used in thisway is said to have been converted. The resulting fissile materials may be processed tomake new fuel elements when sufficient quantities have accumulated. Some of theconverted material may be consumed directly by fissions during reactor operation.

The composition of uranium ore is about 99.3% U-238 and 0.7% U-235. Becauseof the ore�s small percentage of U-235, it is difficult to design a water-cooled, thermalreactor that uses natural uranium. Therefore the power reactors in the United Statesand most other parts of the world are thermal reactors that employ uranium enriched tobetween 2% and 5% U-235. Such reactors use ordinary (light) water for both coolingand moderation and are therefore commonly called light-water reactors.

Uranium enrichment is an expensive and difficult process because it involvesseparation of two isotopes of the same element, which rules out most chemicalmethods. Thus processes that rely on the small mass difference between U-235 andU-238 are usually used. The gaseous diffusion process involves conversion of auranium compound processed from the ore to gaseous uranium hexafluoride, UF6. Thegaseous UF6 flows in hundreds of stages of diffusion through porous walls thateventually produce separate UF6 gas flows containing enriched and depleted uranium.The enriched UF6 then is processed to UO2 powder which is sintered into hard ceramicfuel pellets such as those shown in Figure 10.5. The pellets are sealed in long cylindricalmetal tubes for use in the reactor.

Newer enrichment processes currently available or under development include: highspeed centrifugal separation; which also relies on the uranium isotopic mass difference;

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a separation process that relies on differences in chemical reactivity between theisotopes; and laser enrichment, which relies on ionization of uranium by an intense lightbeam with subsequent chemical or physical separation of the ions (ref. 8).

10.6 Light-Water Power Reactors

Power reactors active in the United States today are light-water reactors. They aredesigned so that the core is both moderated and cooled by highly purified water andtherefore must use a fuel that fissions with thermal neutrons.

Water has many advantages in thermal reactors. From a neutron point of view,H2O is an extremely efficient moderator. As we know from its extensive use inconventional power plants, water has excellent heat transfer characteristics, and thetechnologies of its use in steam power plants are well established.

Water has disadvantages as well. To maintain its excellent moderation and heattransfer capabilities, it must remain a liquid. Thus water reactors are currently limited toproducing hot liquid or steam with little superheat. Moreover, boiling temperaturessuited to an efficient plant require very high pressures, as in fossil fuel plants. Thuswater-cooled reactor cores must be encased in pressure vessels that operate with hightemperatures nearby. In addition, they must endure, for the design life of the plant, the

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severe environment resulting from the fission reactions. Finally, and perhaps mostimportantly, should reactor pressure integrity be lost while the reactor is operating, theliquid water will flash to steam, losing much of its heat transfer advantages. All ofthese factors contribute significantly to the challenges that an engineer faces in thethermal and mechanical design of light-water reactors.

There are two major types of light water reactors (Figures 10.6 and 10.7), whichare differentiated primarily by the thermodynamic conditions of the water used to cooluranium fuel elements in the reactor vessel. The boiling water reactor (BWR) operatesat a pressure that allows boiling of the coolant water adjacent to the fuel elements. Thewater in the pressurized water reactor (PWR) is at about the same temperature as inthe BWR but is at a higher pressure, so that the reactor coolant remains a liquidthroughout the reactor coolant loop. In addition to their use in utility power reactors,PWRs are used in American nuclear submarines.

Boiling Water Reactors

A schematic of the layout of the General Electric BWR/6 system is shown in Figure10.8. There the turbines, condenser, pumps, and feedwater heaters studied in Chapter 2appear in a familiar configuration. Water boils inside the reactor core, producingslightly radioactive steam that passes directly to the steam turbines. The radioactivity inthe steam, however, has a half-life of only a few seconds. The carryover of radioactivityto the turbine-feedwater system is virtually nonexistent, and experience has shown thatcomponents outside the reactor vessel (turbine, condensate pump, etc.) may beserviced essentially as in a fossil-fueled system. Some other reactor designs, such as thepressurized water reactor to be considered later, have an additional separate waterloop, as seen in Figure 10.7, that isolates the turbine steam loop from the reactorcoolant to provide further assurance that the turbine-feedwater system componentsremain free of radioactivity.

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Fuel Assemblies

Uranium appears in most boiling water reactors in the form of sintered cylindrical pellets of uranium dioxide (UO2) (Figure 10.5), about 0.4 inches in diameter and about0.4 inches long. These pellets are stacked inside of long sealed zirconium alloy(zircaloy) tubes called fuel rods. Fuel rods, in turn, are mounted in an eight-by-eight array in a fuel bundle, as seen in Figure 10.9. The fuel bundle and the fuel channel thatsurrounds the fuel rods comprise a fuel assembly, as shown in Figure 10.10.

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The lower-tie-plate nose piece seen there, together with the fuel channel, directs thecoolant water flow over the fuel rods inside of the fuel assembly.

The fuel assemblies, mounted vertically in the core, are designed to minimizeoperating stresses on the fuel rods. For example, Figure 10.10 shows that the fuel rodsare spring loaded so they are free to expand in the axial direction in response tochanges in reactor operating temperatures.

The Reactor Assembly

The core of a BWR/ 6 1220-MWe reactor (MWe stands for electrical generator poweroutput, as opposed to MWt for reactor thermal output) has 732 fuel assemblies and 177control rod assemblies in an approximately 16-ft-diameter circular array, as shown inFigure 10.11. Cooling water receives heat from the fuel rods by forced-convection and

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two-phase nucleate boiling as it flows upward through the fuel assemblies. As the watercools the fuel assemblies, it also thermalizes the fast neutrons that diffuse through thecore.

Water that bypasses the fuel assemblies and flows upward on the outside of the fuelchannels is confined by the cylindrical core shroud, as seen in Figure 10.11. This flowcools the channels and the neutron absorber control rods. The control rods aremounted in cruciform assemblies, as seen in Figure 10.12. The control rod assembliesfit vertically in a fuel module between 4 fuel assemblies in the core, as diagrammed inFigure 10.13. The control rod assemblies move vertically between the fuel bundles tochange the effective multiplication factor to compensate for changes in reactoroperating conditions due to buildup of poisons over time. The stainless-steel-cladcontrol rods contain boron carbide (B4C), which absorbs neutrons and hence tends to

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terminate fission processes when in place in the core. The control rod assemblies aremoved up and down through the core to change the rate of absorption of neutronswhen significant changes in power level or adjustments to account for fuel burnup arerequired, or in the event of an emergency shutdown.

The positions of the control rods are adjusted by hydraulic drives located below thecore. The rods are fully inserted in the core during a shut down making themultiplication factor less than 1. Most of the control rods are fully out of the coreduring critical operation. When all of the rods are out of the reactor the multiplicationfactor slightly exceeds 1, which allows the neutron density and power level to increase.The bottom-entry fuel rod drives in the GE BWR are an unusual feature in reactors.Their location below the reactor simplifies the refueling process which is carried outfrom above in most reactor designs.

The quality of the steam leaving the top of the core is approximately 11%, indicat-ing that most of the water is still liquid and must be recirculated through the core foradditional heating. The liquid water leaving the top of the core and the steamseparators flows downward outside the core shroud, driven by the jet pumps locatedbetween the shroud and the reactor vessel wall, as shown in Figure 10.14. The jetpumps, in turn, are driven by recirculation pumps located outside the reactor vessel.The jet pumps induce the downward flow outside the shroud by momentum transfer tothe slower-moving liquid.

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Vapor bubbling up through the fuel assemblies leaves the core and passes upwardwith the liquid. The flow is turned by stationary vanes in the steam separator, wherethe higher angular momentum of the liquid separates it from the vapor. The separatedliquid flows to the outside of the reactor vessel where it is recirculated outside the coreshroud.

The steam passing through the separator is further dried in the steam dryerassembly before it leaves the reactor vessel as a slightly superheated or saturated vapor.The turbines are specially designed to operate with saturated vapor at the throttle andsmall amounts of liquid within. Liquid is separated from the wet steam leaving the HPturbine. Steam tapped from the HP-turbine-throttle steam line is used to reheat the HPexit steam before its entry to the LP turbines. The low HP-turbine-throttle conditionsof 550°F and 1040 psia lead to a plant thermal efficiency of about 32%.

In the boiling water reactor, control is primarily achieved by adjustment of the rateof recirculation through the reactor by the recirculation and jet pumps shown in Figures 10.11 and 10.14. Change in the rate of water recirculating through the core changesboth the onset of boiling and the volume fraction of steam in the cooling channels, andthus the amount of moderator in the core at a given time. This allows significantadjustment of reactor power output without control rod movement. For example,increasing the recirculation and jet pump speeds sweeps bubbles away faster, increasingmoderation and raising the power level until the additional boiling restores the propervoid fraction for critical operation at a higher power level.

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As the coolant passes upward through the fuel assemblies, heat from the fuel rodsproduces vapor bubbles via nucleate-boiling heat transfer. Nucleate boiling ischaracterized by the local formation of bubbles of vapor that break away from the fuelrod surface, causing vigorous, agitated, fluid motion with resulting high heat transferrates. This may be contrasted with film boiling, in which a stable vapor layer covers thetube surface, resulting in low heat transfer rates. Film boiling occurs at high surface-to-bulk fluid temperature differences. It is crucial that a departure from nucleate boiling,DNB, be avoided, because reduced heat transfer coefficients and cooling rates producedrastic increases in tube temperatures, leading to fuel melting or zircaloy fuel rodburnout. Thus the ratio of maximum heat flux to the critical heat flux for DNB is amajor thermal design parameter for the water reactor. The reactor fuel rod heat-generation rate and flow-channel convective cooling are designed to maintaintube-to-fluid heat fluxes well below the unstable transition range between nucleate andfilm boiling. The maximum UO2 fuel temperature based on the maximum design fuelrod heat-generation rate of 13.4 kW/ft is approximately 3400°F, whereas the fuelmelting temperature is about 5100°F.

The boiling water reactor, like other reactors, has numerous active and passivesafety systems. A thick pressure vessel, for instance, surrounds the reactor core. Aneven thicker concrete containment structure surrounds the pressure vessel to confineanything that may escape from it. An emergency core cooling system (ECCS) sensesoverheating of the core and supplies a flood of water to take away the heat generatedby the fuel elements. These and other safety systems clearly reduce the danger ofaccidents but also increase the cost and complexity of plant operation.

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Pressurized Water Reactors

The pressurized water reactor, PWR, is currently the predominant reactor type in theworld. It is a light-water reactor that uses slightly enriched U-235 as fuel. Figure 10.15shows the three reactor containment buildings of the Oconee PWR Nuclear Station.

A major difference between BWRs and PWRs is that the pressure of the PWRcoolant is above the saturation pressure (it is subcooled liquid) through the entirecooling loop so that there is no possibility of bulk boiling in the core. As shown inFigure 10.7, separate steam generators receive heat from the reactor liquid coolingloop, thus preventing radioactive material from entering the turbine power loop.Another difference is that control rods are at the top of the PWR and can drop bygravity into the reactor when necessary. Figure 10.16 shows a sectional view of thereactor building of the Oconee plant. The stairs and landings give some idea of the sizeof the equipment within the containment. Figure 10.17 gives a less cluttered view of themajor components. Table 10.2 shows that, for PWRs, the turbine loop is at a lowerpressure than the liquid in the reactor loop and therefore produces an outflow of steam to the turbine throttle with about 50 Fahrenheit degrees of superheat.

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We have seen that, in contrast to the BWR, which circulates the reactor coolantthrough the turbine, the PWR uses two loops: a primary loop that cools the reactorcore, and a secondary loop, heated by the primary loop, that provides steam for theturbines, as diagrammed in Figure 10.18. The primary loop is contained completelywithin the reactor containment building (Figure 10.16) and is designed so that thewater that cools the core is completely isolated from the environment. The secondaryloop executes a steam turbine cycle similar to those of conventionally fueled powerplants, with the exceptions that the steam is generated by heat transported from thereactor and that the maximum turbine inlet temperature and superheat are limited bythe maximum temperature in the reactor core.

The Babcock & Wilcox steam generator is a counterflow shell-and-tube heatexchanger, as seen in Figure 10.19. The primary water enters the top of the unit andflows downward through thousands of small-diameter tubes to provide a large heattransfer surface. Feedwater is piped into the bottom of the shell side of the steamgenerator and is first heated, then boiled by heat from the hot tubes containing theprimary water flow. As the steam rises, it encounters the hotter portions of the primarytubing, reaching 50 Fahrenheit degrees of superheat at the top of the steam generator.

An important thermal design parameter of the steam generator is its size. A largerunit increases the heat transfer area from the primary loop, but it also increases thecapital cost of the plant, both in the cost of manufacturing the generator itself and inthe larger size required of the primary containment. This must be balanced with the costsavings, lower heat exchanger effectiveness and consequent lower thermal efficiencyachieved, if the heat transfer area is made smaller.

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The pressurizer (Figure 10.17) controls the pressure in the primary loop and servesas a surge tank to accommodate the reactor coolant water�s expansion and contraction with temperature changes. It is a large tank designed to contain saturated water andsteam, each occupying about one-half of the tank volume. As shown in Figures 10.17and 10.18, the pressurizer is connected to the primary loop at the outlet of the reactor through a single pipe. The temperature of the pressurizer contents are controlled withinternal electric heaters and water sprays. The resulting saturation temperatureestablishes the operating pressure of the reactor.

An important part of the PWR thermal design is sizing the pressurizer tank topermit the primary system to respond to all possible transients without bursting thecoolant pipes. Recalling that water is an almost incompressible fluid that expands whenheated, if the reactor power rises and heats the primary water to a higher temperature,the expanding water will flow into the pressurizer, compressing the steam bubble.Pressure sensors will detect the increased pressure and open spray valves at the top ofthe pressurizer to condense some of the steam, thus restoring a lower operatingpressure. Conversely, if the primary loop temperature declines, heaters in the bottom

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of the pressurizer will turn on, creating more steam to fill the volume formerly occupiedby the contracting water and thus increasing the reactor temperature and pressure.

Although there is no bulk boiling in a PWR core, heat transfer to the coolant is bysubcooled nucleate boiling. In subcooled nucleate boiling, steam bubbles are formedon the surface of the cladding. As the bubbles expand they become detached from thetubes and immediately collapse as they are swept into the coolant channel. The lateralmotion in the coolant achieved by this action is an extremely effective mechanism totransport the energy from the cladding into the coolant stream. With subcooled-nucleate-boiling heat transfer, the cladding surface temperature will stay within 10°F ofthe water's saturation temperature while providing the very high rates of heat transferneeded to remove the fission energy from the fuel rods.

An extremely important design limitation is the critical heat flux at which steambubbles form and grow so fast that they coalesce to form a vapor film over the clad. Atthis point the heat transfer undergoes a departure from nucleate boiling to film boiling,and the clad no longer touches liquid water. Since metal to vapor heat transfer is muchmore inefficient than nucleate boiling, the fuel rod temperature rises in order to transferthe accumulating heat that is created by fissions. Unfortunately, film boiling heattransfer coefficients are so low that the high surface temperatures needed to achieve asteady state are sufficient to melt the clad and severely damage the fuel rods. For thisreason the Nuclear Regulatory Commission has established safety factors for the ratiobetween the critical heat flux and the maximum heat flux expected in a reactor under itsmost severe overpower transient conditions. In its application for an operating license,every nuclear power plant must be able to demonstrate through engineering analysisthat it will maintain the required safety factor under all credible overpower andundercooling transients.

The Babcock and Wilcox PWR uses U-235 enriched to about 3% in the form ofUO2 fuel pellets (Figure 10.20) encased in a zircaloy-clad tube similar to the fuel rodsdescribed for the BWR. Table 10.2 shows that a 1300-MWe plant has 205 fuelassemblies with 54,120 fuel rods. Figure 10.21 shows fuel assemblies for an electric

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utility PWR and for the Nuclear Ship Savannah. Each fuel assembly for electric powergeneration is fitted for the inclusion of sixteen control rods, as seen attached to anactivating spider in the left fuel assembly photo. Positioning of the fuel assemblies inthe core and those assemblies containing control rods are shown in Figure 10.22. Thereare sixty-nine control rod assemblies in Oconee Unit 1, of which sixty-one are forcontrol of power level; the remaining eight contain poisons in the lower part of the rodsfor shaping of the reactor power distribution. The control rod neutron-absorbingmaterial, silver-indium-cadmium, is encased in stainless steel.

10.7 The CANDU Reactor

The CANadian Deuterium Uranium, CANDU, reactor is a reactor of unique design thatutilizes natural uranium as a fuel and heavy water as a moderator and coolant. These reactors produce a substantial saving due to the absence of fuel enrichment costs, but alarge chemical plant is required to supply the quantities of heavy water required. ThePickering station near Toronto, an Ontario Hydro plant shown in Figure 10.23, useseight CANDU reactors to generate 4800 MWe and has been generating power since1971.

One of the important and unique features of the CANDU reactor is that, whereaslight-water reactors are shut down for refueling annually, CANDU reactors arerefueled daily. The pressurized heavy-water-cooled fuel bundles are horizon-tallyoriented in individual fuel channels inside the unpressurized "calandria," asdiagrammed in Figure 10.24. Each bundle may be individually accessed, rearranged in

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the calandria, or replaced using special fuel handling equipment while the reactor isoperating. Heavy water at atmospheric pressure in the calandria surrounds the fuelchannels and moderates the reactor. Thus, in an emergency, the reactor can be shutdown by draining the calandria to remove the moderator, thereby depriving the fuel ofthermal neutrons. The pressurized heavy-water loop draws hot coolant from the fuelchannels through headers to supply heat to steam generators as in a PWR. A light-water loop passing through the steam generator in turn supplies steam to theturbine-feedwater loop. A turbine hall at the Pickering Station is shown in Figure 10.25.

The largest cylindrical structure seen in Figure 10.23 is a vacuum building that connects with each of the reactor buildings. In the event of an emergency, escaping steam and radioactive materials would be drawn by vacuum into the structure. A cold-water spray would condense the steam to limit any pressure buildup.

The thermal efficiency of a CANDU reactor plant is only about 29%, but theCANDU reactor uses a larger fraction of U-235 in uranium ore than other reactors andalso makes better use of the U-238 to Pu-239 conversion process to extend fuelburnup.

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Moreover, statistics show that, among large reactors, CANDU reactors haveoutstanding reliability records, with annual capacity factors (the ratio of annualelectrical energy output to maximum possible annual output) as high as 96% andcumulative capacity factors as high as 88% (ref. 14).

10.8 Fast Reactors

Reactors may be designed to fission with fast neutrons, but these fast reactors mustbe more compact than thermal reactors so that the fast neutrons may produce fissionsquickly before they are absorbed or moderated by surrounding materials. They aredesigned with structural materials that are poor absorbers and moderators of neutrons,such as stainless steel. The core of a fast reactor must contain a fissionable fuel ofabout 20% enrichment to compensate for the lowered probability of fissioning withhigh-energy neutrons.

Because of their high fuel density, fast reactors have a high power density that poses a difficult cooling problem. One solution is the use of a liquid metal as coolant.Liquid metals such as sodium and potassium have excellent heat transfer characteristicsand do not interfere significantly with neutron functions.

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The choice of fuel used for thermal and fast reactors depends on the fuel�s fissionprobability and the net number of neutrons produced per neutron absorbed. Mosteffective for thermal reactors is U-235, whereas Pu-239 is most suitable for fastreactors. In fissioning with fast neutrons, Pu-239 emits almost 20% more neutrons thandoes U-235. These additional neutrons are extremely important for making breedingpractical, as will be discussed shortly. Because Pu-239 must be created from fertile U-238, a plutonium reactor can use fuel processed from fuel produced in a uraniumthermal reactor or in another plutonium reactor. This would occur in the core of aplutonium fast reactor and in a blanket of U-238 surrounding the core, whereadditional plutonium is created using neutrons that escape from the core.

Figure 10.26 diagrams two steps in a chain reaction in a fast reactor where a plu-tonium atom is created for each one consumed. Note that each fission must produce aminimum of two neutrons for this reaction to continue. As a practical matter, morethan two neutrons are required for complete replacement because of non-productiveneutron captures.

A reactor that transmutes a fertile fuel to a fissionable fuel is called a converter. Theconversion efficiency is the ratio of the number of new fissionable atoms produced tothe number of atoms consumed in the fission. Thus the conversion efficiency of thereaction shown in Figure 10.26 is 1, because a new fissionable atom is created for eachatom consumed. In this case there is no net fissionable fuel consumption.

Fast Breeder Reactors

The cover of a 1971 U. S. Atomic Energy Commission booklet (ref. 7), (see Figure10.27) shows a notebook page that poses and answers the following question: Johnnyhad 3 truckloads of plutonium. He used 3 of them to light New York for 1 year. Howmuch plutonium did Johnny have left? Answer: 4 truckloads. This neatly emphasizesthe point that if unproductive neutrons in the process shown in Figure 10.26 were to

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transmute a U-238 atom to Pu-239 occasionally, the chain reaction would producemore fissionable fuel than it consumes. The reactor would have a conversion efficiencygreater than 1. A reactor that does this is called a breeder reactor. Thus, the breederreactor provides a means of increasing the world supply of fissionable fuel as itgenerates power. While it may appear that the breeder provides something for nothing,it actually only takes advantage of the possibility of conversion of the fertile fuelsthrough more efficient use of fission neutrons.

The breeder reactor is of great importance because it would allow the use of thevast store of U-238 in uranium ore that remains as a by-product of the U-235enrichment process to provide fuel for current LWRs. This supply of U-238 has thepotential to provide fuel for many years without further uranium mining. Fission ofU-235 is currently the only natural large-scale source of neutrons. The continued use oflow-conversion-efficiency reactors could preclude the eventual use of much of theenergy resource of the U-238 in uranium ore.

One possibility for the design of a breeder reactor is a liquid-metal fast breederreactor, LMFBR, which has the characteristics described briefly in the precedingsection. The development of such a reactor involves careful design of its neutroneconomy and the development of a system of fuel reprocessing and nuclear wastestorage. These topics will be considered in the next section.

In 1971, President Nixon set the development of a breeder reactor as a nationalgoal and established a program for the development of an LMFBR pilot plant inTennessee to be known as the Clinch River Breeder Reactor, CRBR. The vocalopposition of segments of the American public to all forms of nuclear power, coupled

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with well-known nuclear accidents, bureaucratic and construction delays, and concernfor diversion of nuclear fuels to foreign military or terrorist uses, brought thedevelopment of new nuclear plants in the United States, including the CRBR, to astandstill. While some other nations, notably Japan, France, and Russia, continue tosupport extensive nuclear establishments and research, the path to the development ofbreeder reactors has been slow. The French, in particular, have developed experiencewith their 233-MW Phénix reactor, in operation since 1974, and an advanced 1200-MW LMFBR called the Super-Phénix, which went into operation in 1986. In 1994, thestatus of Super-Phénix was changed from production to a research facility; and in 1998the French government directed that it be shut down, well before its planned 2015shutdown (refs. 24 to 26). In Japan, delayed reporting of a minor sodium leak duringcommisioning of the Monjo breeder in 1995 produced a national turmoil that hadserious repercussions for the Japanese breeder program. (Ref. 27).

10.9 Innovation in Reactor Design

During the hiatus in new domestic nuclear plant orders in the United States since thelate 1970s, the industry has been working to incorporate new ideas and recommend-ations, based on utility operating experience, into the design of plants that will be morecapable of winning public, government, and industry acceptance.

Popular opinion notwithstanding, the safety record of nuclear power plants in theUnited States has been a good one. No one has been killed in a nuclear accident in a U.S. power plant. Few other industries can approach that record. Nevertheless, seriousand expensive accidents have taken place at home and abroad; and the threat ofcatastrophic accidents, while extremely remote, remains.

Designs are being considered by industry, the U.S. Department of Energy (DOE),the Electric Power Research Institute (EPRI), and the Nuclear Regulatory Commission(NRC) that would provide built-in systems and safeguards that are passive, rather thanactive, that must be depended on to perform when a malfunction occurs (ref. 15).Providing for natural conduction/convection cooling of the core in case of an accident,rather than depending on pumps to force water through the core, is one of suchfeatures under consideration. The General Electric Company, with an internationalteam, is developing a 600 MWe simplified / small boiling water reactor (SBWR) thatincorporates this approach (ref. 28).

Nuclear plant incidents sometimes result from failure of an operator to interpretinstrument data and act thereon or from instrument failure. Redundant instrumentationis provided to avoid such problems, but operators do not always analyze the readingscorrectly or make the right decisions. If the plant is designed to minimize the necessityof such actions, the likelihood of accidents can be significantly reduced. Thus, mucheffort focuses on reducing the complexity of reactor control. Designs have beenproduced that appear to accomplish this to a significant degree. The NRC must, ofcourse, conduct hearings and evaluate the adequacy of any design before a plant isbuilt.

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One of the problems of the power industry is the lack of standardization. EveryAmerican power plant is different from every other because of the number of reactorsuppliers and power plant engineering firms and the diverse requirements of thedifferent utilities that buy them. One of the obvious reasons for the reliability of theCANDU reactors is that they are planned and constructed by Ontario Hydro andAtomic Energy of Canada, largely for Ontario Hydro. American suppliers are alsostriving for standardization, but resolving the differences between differing require-ments and standards is a formidable problem. An advanced boiling water reactor(ABWR) program with an NRC certification (ref. 29) addresses these issues byproviding a standardized design that places pumps within the reactor envelope,eliminates much external piping, and incorporates other passive safety features.Reference 23 indicates that this system benefits from a new NRC licensing process inwhich safety issues are resolved, with full public participation, before constructionbegins. ABWR planning calls for a forty-eight-month construction schedule once a sitehas been approved.

The long-term trend has been for utilities to seek larger and larger plants to takeadvantage of the economies of scale. When the annual rate of growth of the electricalpower industry dropped drastically in the 1970s to 2�3% due to conservation and otherfactors, many of the utility requirements for large reactors became less critical ordisappeared. Interest has appeared in smaller reactors that can be constructedeconomically and on a timely basis. One hope is that better quality control could beexercised, and better economics result if small reactors were built, entirely or inmodular fashion, in a factory and shipped to and installed at the power plant site, ratherthan erected there. Some of theses issues are addressed in the ABWR and SBWRprograms (ref. 23).

10.10. Nuclear Reprocessing and Waste Disposal

Spent fuel is fuel that has resided in a reactor for a year or more, has been depleted ofmuch of its fissionable material, and includes a buildup of radioactive fission products.When a reactor is shut down for refueling, spent fuel assemblies are removed andpartially spent assemblies may be repositioned in the core to obtain further fuelutilization. Ideally, the spent fuel would be reprocessed to reclaim the unused andnewly created fissionable materials for use in new fuel rods, and the high-level wastewould be isolated to minimize the volume of highly toxic wastes. Even without breederreactors, current spent fuel inventories can provide ample Pu-239 to supportreprocessing. In the United States, by law, fuel assemblies are being stored indefinitelyin reactor-fuel storage pools in nuclear power plants until legal decisions are made onwhether reprocessing will take place and on the final disposition of nuclear waste.Unfortunately, these decisions are among the most politically difficult ones of ourtimes.

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10.11 Concluding Remarks

Nuclear power engineering has become so hopelessly intertwined with politics that anyspeculation on the future of nuclear power must simultaneously consider both technicaland political realities. Even the continuation and expansion of nuclear power generationin the United States is in question.

A major concern for many years has been the availability of fissionable resourcesfor an expanding nuclear power generation industry. It has been widely recognized thatU-235, as the sole natural neutron resource, places limits on extensive nucleardevelopment. For that reason, countries with nuclear generation capability haveplanned on developing breeder reactors to greatly extend that neutron resource. It ispossible that the continuing long-term use of reactors that do not provide for efficientconversion of the fertile fuels may result in the waste of this enormous energy resource.

While the American nuclear power industry languishes, foreign development ofnuclear power continues in some areas. The eventual recovery of the American nuclearpower industry may depend on the success of international nuclear development. Theabsence of CO2 production by nuclear plants, and its global warming implications, givesan added incentive for international perpetuation and development of the industry.

Expansion of the nuclear power industry and development of a breeder reactornuclear power economy inevitably entails the development of nuclear processing andhazardous waste disposal facilities and reactor fuel recycling. These activities arouseconcerns about the possible diversion of nuclear materials for the production ofweapons, (refs. 13 and 17�19). The tradeoffs among nuclear power, coal, and otherenergy conversion alternatives are also becoming more prominent as concerns overglobal warming and atmospheric pollution intensify. Well-thought-out and consistentgovernment policies and regulation, as well as international cooperation, would givewelcome direction to the power generation industry. Reference 16 observes thatsuccess in nuclear programs in other countries seems to correlate with limiting publicintervention, and it questions whether the democratic institutions of the United Statesare consistent with the growth of the American nuclear power industry.

Bibliography and References

1. Hulme, H. R., Nuclear Fusion. London: Wykeham Publications Ltd., 1969.

2. Semat, Henry, Introduction to Atomic and Nuclear Physics. New York : Holt,Rinehart & Winston, 1959.

3. Spitzer, Lyman, Physics of Fully Ionized Gases. New York: Interscience Publishers,1956.

4. Glasstone, Samuel, Principles of Nuclear Reactor Engineering. New York: D. VanNostrand, 1960.

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5. Glasstone, S., and Sesonske, A., Nuclear Reactor Engineering. New York: VanNostrand Reinhold, 1981.

6. Steam / Its Generation and Use, 39th ed. New York: Babcock and Wilcox, 1978.

7. Mitchell, Walter III, and Turner, Stanley E., Breeder Reactors. U. S. AtomicEnergy Commission, 1971.

8. Nero, Anthony V. Jr., A Guidebook to Nuclear Reactors. Berkeley, Calif.:University of California Press, 1979.

9. Winterton, R. H. S., Thermal Design of Nuclear Reactors. New York: PergamonPress, 1981.

10. Judd, A. M., Fast Breeder Reactors, An Engineering Introduction. New York:Pergamon Press, 1981.

11. Bromberg, Joan Lisa, Fusion. Cambridge, Mass.: MIT Press, 1982.

12. Robinson, Mark Aaron, 100 Grams of Uranium Equal 290 Tons of Coal. Kelso,Wash.: R & D Engineering, 1987.

13. Nuclear Proliferation and Safeguards�Summary. Congress of the United States,Office of Technology Assessment, OTA-E-148, March 1982.

14. Nuclear Power in an Age of Uncertainty. Congress of the United States, Office ofTechnology Assessment, OTA-E-216, February 1984.

15. Catron, Jack, "New Interest in Passive Reactor Designs." EPRI Journal,April/May 1989: 5�13.

16. Campbell, Jack, Collapse of an Industry-Nuclear Power and the Contradictions ofU.S. Policy. Ithaca, New York: Cornell University Press, 1988.

17. Blocking the Spread of Nuclear Weapons. New York: Council on ForeignRelations, 1986.

18. Muller, Harald, A European Non-Proliferation Policy, New York: ClarendonPress, 1987.

19. Snyder, Jed C., and Wells, Samuel F. Jr., (Eds.) Limiting Nuclear Proliferation.Cambridge, Mass.: Ballinger, 1985.

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20. Lovins, Amory, Soft Energy Paths. New York: Harper and Row, 1979.

21. Nealey, Stanley M., Nuclear Power Development�Prospects for the 1990s.Columbus, Ohio: Battelle Press, 1990.

22. �World List of Nuclear Power Plants.� Nuclear News, February 1991.

23. Wilkins, D. R., Quirk, J. F., and McCandless, A. H., �Status of Advanced BoilingWater Reactors.� American Nuclear Society, 7th Pacific Basin Nuclear Conference,March 1990.

24. Electricité de France, �Super Phénix Shutdown.�www.info-france-usa.org/america/embassy/nuclear/facts/superpr.htm(Oct. 30, 2000).

25. Electricité de France, �Permanent Shutdown.�www.info-france-usa.org/america/embassy/nuclear/facts/superpff.htm(Oct. 30, 2000).

26. Electricité de France, �Shutdown of the Superphénix breeder reactor.�www.info-france-usa.org/america/embassy/nuclear/facts/supersum.htm(Oct. 30, 2000).

27. Japan Nuclear Cycle Development Institute, �Monju Reactor Website.�www.jnc.go.jp/zmonju/mjweb/index.htm (October 30, 2000).

28. �The Simplified Boiling Water Reactor.�www.nuc.berkeley.edu/designs/sbwr/sbwr.html (December 19, 2000).

29. The Advanced Boiling Water Reactor.�www.nuc.berkeley.edu/designs/abwr/abwr.html (December 19, 2000).

30. Davis, Mary Bird, �Nuclear France: Materials and Sites�Creys-Malville-Superphénix.� www.francenuc.org/en_sites/rhone_crey_e.htm (Oct. 30, 2000).

31. Fetter, Steve, �Energy 2050,� Bulletin of the Atomic Scientists. July/August 2000.www.bullatomsci.org/issues/2000/ja00/ja00fetter.html (October 30, 2000).

32. Nuclear Regulatory Commission, �U.S. Commercial Nuclear Power Reactors.�www.nrc.gov/AEOD/pib/states.html (November 2, 2000).

33. Nuclear Regulatory Commission, �Nuclear Reactors.�www.nrc.gov/NRC/reactors.html (November 2, 2000).

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34. Environment News Service, �World Total: 434 Nuclear Power Plants.�ens.lycos.com/ens/may99/1999L-05-06-06.html (November 4, 2000).

EXERCISES

10.1 Consider the collision of a particle of mass m and velocity v with a stationaryparticle of mass M. Write energy and momentum equations for the collision.Derive an equation for the ratio of the final to the initial kinetic energy of theoriginal moving particle in terms of the masses of both particles. Use the result toshow why light atoms are used as moderators.

10.2 Evaluate the assertion of the title of reference 12 that 100 grams of uranium equal290 tons of coal. Assume coal to be represented by pure carbon. Could you makethe title of the book more accurate? If so how?

10.3 Estimate the core thermal power and thermal efficiency of a 1220-MWe boilingwater reactor that has 46,376 fuel rods in a 150-in.-high core with a maximumfuel rod linear energy density of 13.4 kW / ft and a fuel rod peak-to-averagepower release of 2.2. If there are 748 fuel assemblies, what is the average numberof fuel rods per assembly? Estimate the number of thermally inactive rods in areactor with an eight-by-eight fuel assembly array.

10.4 Assuming the fuel temperature to be 295K, calculate the energy of a thermalneutron using 3kT/ 2 where k is the Boltzmann constant.

10.5 Study the literature and then discuss the details of the nuclear processes by whichneutrons convert U-238 to fissionable fuel. Include nuclear reaction equations.

10.6 Study the literature, then discuss the details of the nuclear processes by whichneutrons convert thorium to a fissionable fuel. Include nuclear reaction equations.

10.7 Sketch and label a PWR steam generator flow diagram and a T-s diagram for thetwo flows through the PWR steam generator, using the data from Table 10.2.

10.8 Determine the ratio of the reactor-loop flow rate to the steam flow rate from thedata for the PWR given in Table 10.2.

10.9 Sketch a T-h diagram and determine the pinchpoint temperature difference for aPWR steam generator that has, respectively, 572.5°F and 630°F reactor inlet andoutlet temperatures and steam generator inlet and outlet temperatures of 473°Fand 603°F.

10.10 Discuss the changes in the neutron economy diagram (Figure 10.4) due to (a)insertion of control rods, (b) withdrawal of control rods, (c) increase and (d)

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decrease in recirculation flow rate in the boiling water reactor.

10.11 List from BWRs, PWRs, CANDUs, and LMFBRs those power reactors that are(a) Thermal reactors(b) Fast reactors(c) Heavy-water reactors(d) Light-water reactors(e) Natural uranium reactors(f) Enriched uranium reactors(g) Plutonium reactors

10.12 Identify those power reactors that use moderators and those that do not.

10.13 Which power reactor (a) uses light water as coolant and has separate reactorcoolant and steam loops, (b) uses heavy water and natural uranium, and (c) usesplutonium as fuel and liquid metal as coolant?

10.14 Based on data in Table 10.2 for the Oconee Unit 1 PWR, estimate the turbine-loop thermodynamic conditions (temperature and pressure) and power deliveredby a simple Rankine-cycle steam turbine with 85% efficiency.

10.15 Compare the reactor temperature rise shown in Table 10.2 for the Oconee Unit 1with an analysis based on heat transfer data given in the table. Discuss your result.

10.16 Compare the reactor temperature rise shown in Table 10.2 for the 1300-MWePWR with an analysis based on heat transfer data given in the table. Discuss yourresult.

10.17 Based on the data given in Table 10.2, determine the heat transfer rating for theOconee Unit 1 PWR steam generator, and evaluate the average reactor-loop heattransfer loss rate and fraction. Discuss your result.

10.18 Based on data in Table 10.2, estimate the turbine-loop thermodynamic conditions(temperature and pressure) and power delivered by a simple Rankine-cycle steamturbine with 85% efficiency for the 1300-MW PWR.

10.19 Based on the data given in Table 10.2, determine the heat transfer rating for the1300-MWe PWR steam generator, and evaluate the average reactor-loop heattransfer loss rate and fraction. Discuss your result.

10.20 Develop a thermal design for a 2000-MWe pressurized water reactor core with0.4 in. diameter, 15-ft-long fuel rods having an average linear power output of 16kW/ft. Assume an average film coefficient of 4500 Btu/hr-ft2-R. Prepare a report.

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C H A P T E R E L E V E N

Energy System Alternatives

Part 1. Electromagnetic Principles, Batteries, and Fuel Cells

11.1 Introduction

So far we have considered energy conversion systems that are in widespread use. Inthis chapter we turn our attention to systems that, to date, have not had major impactsin power generation but appear to have the potential to profoundly influence societywithin the next few decades.

Most of the preceding study has examined power and propulsion systems that wereprimarily the outgrowth of thermodynamics and fluid mechanics. We now considersystems among which are several of a fundamentally different character. To better un-derstand these systems, we must invest a little more time in reviewing electromagneticfundamentals.

In earlier chapters we considered numerous systems that employ the energy con-version chain:

Energy source � Heat � Mechanical energy � Electricity

In this chapter we are concerned with several conversion processes in which the heat-to-mechanical energy transformation link is not essential:

Energy source � Electricity

Such processes, in which the source energy is converted directly to electricity, arecalled direct energy conversion processes. While direct energy conversion techniquesother than those discussed here exist, we will concentrate on a few that show potentialfor large-scale power production and that could reach commercialization in the nextfew decades. Specifically, we focus on fuel cells, solar photovoltaics, and magneto-hydrodynamic systems. In addition, the application of batteries and hydrogen as energystorage media are considered. Let us first develop some principles needed in theanalysis of these systems. It would be advantageous for students to relate thefundamental material reviewed in the following sections to chemistry and physics textswith which they are familiar.

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11.2 Review of Electromagnetic Principles

A few principles of electromagnetics, fundamental to alternative energy systems, arenow reviewed. We will work almost exclusively with SI units in this chapter, except fora few problems framed in the English system.

Charge

A unit of electrical charge, q, is the coulomb [C]. The electron has a negative charge of1.6×10 -19C and a mass of 9.1×10 -31kg. Even the smallest positively charged particle,the proton or hydrogen ion, has a mass more than a thousand times that of the electron.As a result, the electron is much more mobile than neighboring ions and is usually thedominant charge carrier. However, because electromagnetic theory is usually based onthe concept of the positive charge, we will think of an electric current as a flow ofpositive charge, even though motion of electrons in the opposite direction usuallytransports the charge.

Potential and Field

A positively charged particle influenced by other charged particles may be thought of asbeing in a force field created by these particles. This field, known as an electric field, may be represented by nonintersecting, usually curved surfaces of constant potential(V) in three-dimensional space, as sketched in Figure 11.1.

Consider a charged particle located between two potential surfaces of potentialdifference V2 � V1. A force is exerted on the particle in a direction perpendicular to thelocal potential surface. The larger the potential difference between the two surfaces, thelarger the force. Also, the smaller the distance between surfaces of fixed potentialdifference, the larger the force. The direction of the force on positively chargedparticles is from high potential to low, and is oppositely directed for negative particles.

It was indicated that the strength of the force acting on a charged particle betweentwo potential surfaces depends on both the magnitude of V2 � V1 and how closetogether the surfaces are. An electric field vector is defined to account for both of thesefactors. In one dimension, the electric field intensity Ex is the negative of the rate ofchange of potential with respect to distance, � dV/dx, the limit of the local potentialdifference per unit distance in the direction normal to the potential surface. Thus

Ex = � dV/dx � � lim (V2 � V1)/�x [V/m] (11.1)

For a three-dimensional potential field given by V = V(x, y, z), the electric field intensityis a vector, E (vectors are represented by boldface symbols here) which in terms ofCartesian unit vectors i, j, k is

E = Exi + Ey j + Ez k [V/m] (11.2)

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The electric field vector is related to the potential function V(x, y, z) by

E = � ����V (x, y, z) = � (i�V/�x + j�V/�y + k�V/�z) [V/m] (11.3)

where the derivatives in Equation (11.3) are partial derivatives. For a one-dimensionalpotential, V = V(x), the derivatives with respect to y and z vanish; and the remainingpartial derivative becomes the total derivative dV/dx, consistent with Equation (11.1).

The electric field vector gives the magnitiude and direction of the force on apositive charge in three-dimensional space. The force on the charge due to the electricfield is explicitly represented by the vector equation

F = qE [N] (11.4)

Equation (11.4) indicates that a force of one newton is exerted on a charge of onecoulomb in an electric field of one volt per meter. Hence one coulomb is onenewton-meter per volt, or one joule per volt [1 C = 1 N-m/V = 1 J/V].

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The work done to move a positive charge a small distance dl against the electricfield (to a higher potential) is given by the scalar product of the force �F and thedifferential distance that the charge moves, dl = dxi + dyj + dzk. Using Equation (11.3),the work is

w = �����F�dl = � q����E����dl = q (V 2 � V1) [C-V] (11.5)

Work is done on the positive charge when the differential distance (dx in Figure 11.1)has a component parallel to the force (and the electric field vector).

Consistent with the observation made in connection with the units of Equation(11.4), Equation (11.5) shows that a coulomb-volt is the same as the usual measure ofwork, the newton-meter or joule.

When the electric field is specified it becomes unnecessary to think in terms of theassociated charge distribution. The electric field may give rise to or resist motions ofpositive charges or free electrons in the space occupied by the field, just as the originalcharge distribution would have.

Current and Power

A motion of charges through an electrically conducting medium is called a current. Thecurrent is the rate at which positive charge passes through a given surface. It ismeasured in coulombs per second [C/s] or amperes [A]

I = dq/dt [A] (11.6)

A current flow produced by an electric field is in the direction of the field and thus is inthe direction of decreasing potential, as shown in Figure 11.2. Correspondingly,electrons flow from low to high potential in a direction opposite to the conventionalcurrent.

Power is the rate at which work is done. Using Equation (11.5)

Power = dw/dt = Vdq/dt = VI [W] (11.7)

where we now use V for potential difference, as is common. Here it is seen that the unitof power, the watt [W], is the equivalent of the product of amperes and volts [A-V].

Current flows in a medium because of the forces exerted on the charges. For agiven potential difference, the current is smaller or larger depending on the ease withwhich the charges pass through the medium. Thus the current is proportional to theapplied potential difference and the conductance, G, of the conducting medium. Theconductance is usually expressed in terms of its inverse, the resistance R = 1/G.Resistance is defined by Ohm's Law:

R = V / I = 1/G [ � ] (11.8)

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Ohm�s Law defines the unit of resistance, the ohm, as the ratio of a volt to an ampere� =V/A. Thus the unit of conductance is the inverse ohm or mho � �1.

EXAMPLE 11.1

A 12-V battery produces a current of 7 A in a load resistor. Estimate the power deliv-ered to the resistor by the electric field created by the terminals and the magnitude ofthe resistance. Explain the phenomena in terms of current and electron flow.

SolutionThe resistive load has a resistance of R = V/I = 12 / 7 = 1.71 �

The power output of the battery is IV =7 × 12= 84 W.

Current flows from the positive terminal through the load resistor to the negativeterminal because of work done by the electric field applied by the battery terminals. Inreality, work is done on electrons passing from the negative to the positive terminals.Chemical reactions within the battery provide the energy to move the electrons fromthe positive to the negative terminals within the battery._____________________________________________________________________

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The terms conductance and resistance refer to characteristics of a specific configur-ation of a conducting material. The terms electrical conductivity and electricalresistivity refer to intrinsic electrical properties of the material itself. We shall see thatthe latter properties often are used in dealing with currents in multidimensional fields.

The current density J [A/m2] is defined as the current per unit cross-sectional areanormal to the direction of the current. The electric field vector and the current densityare related (Figure 11.2) in a generalized form of Ohm's Law by

J = �E [A/m2] (11.9)

where � is the electrical conductivity of an isotropic conducting medium. In general,when a medium has conductivity varying in direction, the medium is called�anisotropic.� Here we need only consider the isotropic case, where the conductivity ofthe medium is the same in all directions.

Note that Equation (11.9) indicates that � must have units of (A/m2)/(V/m) or(�-m)�1. For a cylindrically-shaped conductor, Equation (11.9) may be expressed interms of the current and potential difference to show that the electrical conductivity isrelated to the resistance of the conductor by

� = J/E = (I/A)/(V/L) = L/(RA) [(�-m)-1] (11.10)

where L and A are the length and cross-sectional area of the conductor, respectively.Rewriting Equation (11.10) as R = L/(A�), we see that the resistance of a conductor isproportional to its length and inversely proportional to the electrical conductivity andcross-sectional area, as expected. The electrical resistivity � [�-m] is the inverse ofthe electrical conductivity. Thus the resistance of a cylindrical conductor may bewritten in terms of resistivity as R = L�/A.

Magnetic Field

We have seen that the electric field may be thought of as representing the effect of acharge distribution on a test charge. Now we turn our attention to a fundamentallydifferent field.

A stationary magnetic field exists between the north and south poles of a bar orhorseshoe magnet (Figure 11.3). A magnet may be thought of as a collection of tinymagnetic elements, or domains, each having its own north and south poles, which arealigned to produce the magnetic field of the magnet. Magnetic field lines are alwaysclosed or terminate on magnetic poles, as opposed to electric field lines, whichterminate on charges.

It was observed by Oersted in 1819 that a small magnet deflects when placed in theneighborhood of a wire carrying an electric current. This is explained by the existenceof circular magnetic field lines that surround the conductor, as shown in Figure 11.4.Thus not only magnets but electric currents and moving electric charges give rise to

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magnetic fields. In view of the connection of currents and electric fields, Oersted�sdiscovery indicates that electric fields may produce magnetic fields.

A magnetic field vector B, called the magnetic induction or magnetic flux density,is a measure of the strength of a magnetic field and indicates its direction at any point inspace (Figures 11.3 and 11.4). Magnetic flux density is measured in webers per squaremeter [Wb/m2]. A weber is equivalent to a volt-second. The unit of weber per squaremeter is also called a tesla.

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The magnetic field lines around a circular conductor of electricity are continuousconcentric circles, as indicated in Figure 11.4. The direction of the magnetic flux orlines of force around a current carrying conductor is given by the direction of thefingers of the right hand when they are wrapped around it with the thumb pointing inthe direction of the current flow.

Twelve years after Oersted observed that electric fields (which produce chargemotion, or electric currents) give rise to magnetic fields, Michael Faraday (1831)discovered that magnetic fields can create electric fields. Faraday observed that acurrent was induced in a stationary circuit by a changing magnetic field produced by anearby moving magnet. That changing magnetic field could have been created by (1) atransient current in a nearby stationary circuit, (2) motion of a nearby circuit with aconstant current, or (3) motion of a magnet in the vicinity. A current, and thus anelectric field, is generated in any conductor that experiences a change in its localmagnetic field, regardless of whether the observed change is due to the motion of amagnet or to motion of the conductor itself through a stationary magnetic field. Thelatter phenomenon is the basis of the operation of an electrical generator in whichstationary magnets create a current in a rotating armature.

The force exerted on a positive charge in a magnetic field is proportional to thevector product of the charge velocity and the magnetic field strength:

F = qu×B [N] (11.11)

The vector product indicates that charges in a conductor moving with a velocity uthrough a magnetic field experience a force in a direction perpendicular to the planedefined by the vectors u and B. As indicated in Figure 11.5, the sense of the force isreadily determined by rotating the four fingers of the right hand from the velocityvector to the B vector. The direction of the extended thumb during this operation givesthe direction of the force. The force exerted on a coil of an armature of an electricmotor is an example of this magnetic action.

In general, both electric and magnetic fields occupy the same space, Thus it is notunusual to speak of a single electromagnetic field represented by the electric andmagnetic field vectors. If the electric field intensity is E�and the magnetic flux density isB at a given location, the total electromagnetic force acting on a charged particle there,called the Lorentz force, is given by

F = q(E' + u×B) [N] (11.12)

Thus the charge experiences a force that is the vector sum of the electric field vectorand a vector perpendicular to both the charge velocity and magnetic flux vectors.

In a coordinate system moving with the velocity u, the Lorentz force would begiven by F = qE because the observer in this system sees no motion of the charge. Thusthe electromagnetic field felt by the charge may be expressed as an electric field:

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E = E' + u×B [N/C] (11.13)

Here E' is the field existing at the charge location in a stationary coordinate system andE is the effective electric field sensed by the charge. Thus in a moving coordinatesystem, the influence of the magnetic field appears as part of the electric field vector.

While our purpose here has been to establish the concepts and relations necessaryfor an understanding of the energy conversion systems discussed in this chapter, wewould be remiss to move on without a few closing comments. We have discussed theinteractions of two invisible fields. These spatially varying fields may be thought of asexisting in all of space. These are not merely mathematical abstractions but realphenomena. What better proofs of their reality than the electromagnetic waves that aportable radio picks out of space and converts to music to entertain us as we walk orjog, the warmth of solar radiation and the chill of its absence when passing throughshade, and the interplanetary signals from space probes that provide amazing views ofother planets?

The study of the physics and mathematics of these fields and related phenomenaculminated in the formulation of a set of four partial differential equations involving Eand B that govern electromagnetics. These famous equations known as Maxwell'sequations, were firmly established on a consistent basis by James Clerk Maxwell in

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1860. They mark one of the great triumphs of science because they are capable ofquantitatively defining details of countless phenomena in the areas of electricity, magnetism, and radiation. Their use in detailed research and development in energyconversion devices goes well beyond the brief introduction to electromagnetic theorygiven here. Interested readers may consult, among others, references 6, 26, and 64 for broader and deeper treatments of the subject. 11.3 Batteries and Fuel Cells

Batteries and fuel cells, while structurally and functionally distinct, are based on similarelectrochemical principles and technologies. For that reason we have chosen tointroduce them together to emphasize their common scientific foundations. They are,however, easily distinguishable, because batteries are devices for the storage ofelectrical energy, while fuel cells are energy converters with no inherent energy storagecapability. A battery contains a finite amount of chemicals that spontaneously react toproduce a flow of electrons when a conducting path is connected to its terminals, as

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shown in Figure 11.6. Fuel cells behave similarly, except that chemical reactants arecontinuously supplied from outside the cell and products are eliminated in continuous,steady streams.

The chemical reactions in batteries and fuel cells are oxidation-reduction reactions. A common terminology exists for both types of cells. Both contain electrode pairs incontact with an intervening electrolyte (the charge-carrying medium), as in Figure 11.6.The electrolyte is a solution through which positively charged ions pass from the anodeto the cathode. The negative electrode, called the anode, is where an oxidation reactiontakes place, delivering electrons to the external circuit. The positive electrode, wherereduction occurs, is called the cathode. Electrons flow through the external load fromthe negative to the positive electrode, while the conventional current flows in theopposite direction from high to low potential. At the cathode, electrons arriving fromthe external load are neutralized by reaction with positive ions from the electrolyte.

The use of electrical storage cells dates back to the early 1800s. In 1800, Alessan-dro Volta gave the first description of an electrochemical cell in a letter to the BritishRoyal Society. The invention of the telegraph in the 1830s provided strong motivationfor the development and manufacture of this electrical energy source. Though theelectrochemical cell provided electrical energy for the telegraph and other inventions, itwould be half a century before central plant electrical power became available. In 1839,Sir William Grove demonstrated the fuel cell concept, but it was not until the 1950sthat significant progress was made toward the development of useful fuel cells.

Batteries designed for a single discharge cycle (flashlight batteries, for example) arecalled primary cells; those that can be discharged and recharged numerous times arecalled secondary cells. Fuel cells, on the other hand, utilize chemicals that flow into thecell to provide a more or less continuous supply of direct current. Before consideringthe exciting possibilities of fuel cells, we will take a brief look at batteries, withemphasis on secondary cells because of their current and future technical importance.

Batteries

Enormous numbers of batteries are used in a variety of applications, from the tinybutton cells in wristwatches, to the cells in flashlights, to the starting, lighting, andignition (SLI) batteries in automobiles, to the large batteries found in submarines forsubmersed operations, as well as in emergency power supplies, and remote relaystations. Batteries have been considered for some time for load-leveling power plants as alternatives to pumped hydroelectric and compressed-air storage. While batteries havebeen used as propulsion power sources for road vehicles for about a century (accordingto ref. 10 an electric car held the world land speed record of 66 mph in 1899), theyhave not been competitive with the fossil-fueled internal combustion engine. However,

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a renewed interest in this application has developed as research in battery technologyand the public sensitivity to pollution by internal combustion engine-propelled vehicleshas become more advanced.

It has been pointed out that batteries are basically devices for storing charge andautomatically delivering an electrical current flow, on demand, for a limited time. Abattery consists of one or more cells connected in series to provide a particular opencircuit voltage or electromotive force (EMF). Each chemical cell typically has an EMFof about two volts, although the exact value is dependent on the specific cell reactants.

A Simple Mathematical Model of a Battery. When current is drawn from a battery,the voltage across its terminals decreases below its EMF, largely due to the internalvoltage drop associated with the battery's internal resistance. Thus a simple model of abattery involves a constant voltage source in series with a constant resistance.Following Figure 11.7, the terminal voltage for this model is given by the differencebetween the EMF and the internal resistive potential drop:

V = E � IRi [V] (11.14)

and the current is

I = E /(Ri + Ro) [A] (11.15)

EXAMPLE 11.2

The voltage of a lead-acid battery with an EMF of 12.7 V is measured as 11.1 V when it delivers a current of 50 amperes to an external resistance. What are the internal and external resistances? What is the battery voltage and the current flow through a 5-� resistor? Sketch the voltage�current characteristic.

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Solution

Solving Equation (11.14) for Ri and Equation (11.15) for Ro, we obtain the internal andexternal resistances:

Ri = (E � V)/I = (12.7 � 11.1)/50 = 0.032 �

Ro = E / I � Ri = 12.7/50 � 0.032 = 0.222 �

With the constant internal resistance established, the current flow through a 5-�external resistor is then given by Equation (11.15):

I = E/(Ri + Ro) = 12.7/(0.032 + 5) = 2.524 A

Note that if Ro were infinite in Equation (11.15), the current would vanish and the cellvoltage would be equal to the EMF, as in an open circuit.

The cell voltage with the 5-� resistor is given by equation (11.14):

V = E � IRi = 12.7 � 2.524(0.032) = 12.62 Vor

V = IRo = 2.524(5) = 12.62 V

By Equation (11.14), the voltage characteristic is a straight line (see Figure 11.8) withintercept E and slope �Ri._____________________________________________________________________

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While the linear model is useful and informative, it does not conform precisely toreal battery characteristics. As in the model, real batteries are capable of providingcurrent and power to a wide range of resistances with little voltage drop for shortperiods of time. However, the model fails for long periods of time and when largecurrents are drawn, because it does not account for the finite amount of stored chemicalenergy and reaction rate limitations encountered in real batteries. In these circumstan-ces, the terminal voltage and power output drop drastically as cell chemicals areconsumed.

Secondary cell reactions are approximately internally reversible. Thus, by applica-tion of an external direct current source, internal battery reactions may be reversed overa period of time in a process called charging. In a conventional automotive application, electrical power for lighting, ignition, radio, and so on is supplied by an engine-drivenalternator. Normally, the battery is required only for engine starting and to compensatefor possible alternator output deficits when the engine runs slowly. A voltage regulatormaintains a steady electrical system voltage and controls battery charging when excessalternator current is available and needed. According to reference 27, the regulatorlimits the charging voltage of a 12-V battery to about 14.4 V.

Battery Performance Parameters. Important parameters characterizing batteryperformance are the storage capacity, cold-cranking amperes, reserve capacity, energyefficiency, and energy density.

The storage capacity, a measure of the total electric charge of the battery, is usuallyquoted in ampere-hours rather than in coulombs ( 1 amp-hour = 3600 C). It is anindication of the capability of a battery to deliver a particular current value for a givenduration. Thus a battery that can discharge at a rate of 5 A for 20 hours has a capacityof 100 A-hr. Capacity is also sometimes indicated as energy storage capacity, inwatt-hours.

Cold-cranking-amperage, CCA, for a battery composed of nominal 2-V cells is thehighest current, in amperes, that the battery can deliver for 30 seconds at a temp-erature of 0°F and still maintain a voltage of 1.2 V per cell (ref. 27). The CCA iscommonly used in automotive applications, where the problem of high engineresistance to starting in cold winter conditions is compounded by reduced batteryperformance. At 0°F the cranking resistance of an automobile engine may be increasedmore than a factor of two over its starting power requirement at 80°F, while the batteryoutput at the lower temperature is reduced to 40% of its normal output. The reductionin battery output at low temperatures is due to the fact that chemical reaction ratesdecrease as temperature decreases.

The reserve capacity rating is the time, in minutes, that an SLI battery will deliver25 amperes at 80°F. This is an indication of how long the battery would continue tosatisfy essential automotive operating requirements if the alternator were to fail.

Other parameters evaluate batteries with respect to energy for traction or otherdeep-cycling applications where, in contrast to SLI applications, their state of chargebecomes low regularly. The energy efficiency is the ratio of the energy delivered by a

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fully charged battery to the recharging energy required to restore its original state ofcharge. For transportation purposes where weight and volume are critical, the energydensity is an important parameter. It may be represented on a mass basis, in watt-hoursper kilogram, or on a volume basis, in watt-hours per cubic meter.

Analysis of a Cell. Consider a reversible cell with a terminal potential difference Vdischarging with a current I through a variable resistance. The flow of electronsthrough the load produces the instantaneous electrical power, IV. At low currents thecell voltage is close to the cell EMF and the external work done and power output aresmall. Suppose that chemical reaction in the battery consumes reactant at an electrodeat a rate of nc moles per second. Electrons released in the reaction flow through theelectrode to the external load at a rate proportional to the rate of reaction, jnc, where jis the number of moles of electrons released per mole of reactants. It will be seen laterthat, for a lead-acid battery, two moles of electrons are freed to flow through theexternal load for each mole of lead reacted. Hence, in this case j = 2. Thus jnc is therate of flow of electrons from the cell, in moles of electrons per second. There are6.023×1023 electrons per gram-mole of electrons, and each has a charge of 1.602×10-19

C. Thus the product

F = (6.023×1023)(1.602×10-19) = 96,488 C/gram-mole

is the charge transported by a gram-mole of electrons. The constant F is called Faraday's Constant, in honor of Michael Faraday, a great pioneer of electrochemistry.The electric current from a cell may then be related to the rate of reaction in the cell as

I = jncF [A] (11.16)

and the instantaneous power delivered by the cell is:

Power = jncFV [W] (11.17) The cell electrode and electrolyte materials and cell design determine the maximum cell voltage. Equations (11.16) and (11.17) show that the nature and rate of chemicalreaction control the cell current and maximum power output of a cell. Moreover, it isclear that the store of consumable battery reactants sets a limit on battery capacity.

Lead-Acid Batteries. By far the most widely used secondary battery is thelead/sulfuric acid/lead oxide (Pb/H2SO4/PbO2), or lead-acid battery, by virtue of itswidespread application in automomobiles for starting, lighting, and ignition and in

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traction batteries used in rail and street vehicles, golf carts, and electrically poweredindustrial trucks. An automotive lead-acid battery is shown schematically in Figure11.9. Today most lead-acid batteries are 12-V designs, assembled as six 2-V cells inseries.

Each cell consists of a porous sponge lead anode and a lead dioxide cathode(usually in the form of plates) separated by porous membranes in a sulfuric acidsolution. The aqueous electrolyte contains positive hydrogen and negative sulfate ionsresulting from dissociation of the sulfuric acid in solution:

2H2SO4 => 4H+ + 2(SO4)2 �

At the anode, the lead releases two electrons to the external load as it is converted tolead sulfate:

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Pb(s) + SO4 2 � (aq) => PbSO4 (s) + 2e�

This shows that two moles of electrons are produced for each mole of sulfate ions and sponge lead reacted. Thus j = 2 for the lead-acid cell in Equations (11.16) and (11.17).

Meanwhile at the cathode, lead dioxide is converted to lead sulfate in the reductionreaction involving hydrogen and sulfate ions from the electrolyte and electrons from theexternal circuit:

PbO2 (s) + 4H+ (aq) + SO4 2 � (aq) +2e � <=> PbSO4 + 2H2O (l)

The equations show that the sulfate radical (and thus sulfuric acid) is consumed at bothelectrodes, and that hydrogen ions combine with oxygen from the cathode material toform water that is retained in the cell. Thus the water formed as the battery dischargesmakes the sulfuric acid electrolyte more dilute. This is also seen in the net cell reaction,obtained as the sum of the electrode and electrolyte reactions:

Pb(s) + PbO2 (s) + 2H2SO4 (aq) <=> 2PbSO4 (s) + 2H2O (l)

It is clear that during battery discharge the lead and lead oxide electrodes are con-verted to lead sulfate. Furthermore, sulfuric acid is consumed and water is produced.The production of water and the consumption of sulfuric acid make the electrolytemore dilute and decrease its specific gravity. The specific gravity drops from about1.265 when the battery is fully charged to about 1.12 when fully discharged (ref. 27).Thus the cell specific gravity is a measure of the state of discharge of the battery. Asimple device called a hydrometer, which temporarily withdraws electrolyte from anopen battery, gives a visual indication of the electrolyte specific gravity. The depth ofthe float in the hydrometer electrolyte indicates its specific gravity and thus the batterystate of charge. When any of the reactants is depleted, the battery ceases to function.When the battery is charged by applying an external power source, the electron flow isreversed, the previously displayed reactions also reverse, and water is consumed toproduce sulfate and hydrogen ions from the lead sulfate on the electrodes.

EXAMPLE 11.3

Consider the battery model of Example 11.2 for the case of 50-A current. Determinethe instantaneous power output, in watts, and the rate of consumption of sulfuric acidof the cell, in kilograms per hour.

Solution

Power = IV = (11.1) (50) = 555 W

Two moles of electrons are produced for each mole of sulfuric acid consumed at theanode. From Equation (11.16),

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nc = I/jF = 50/(2 � 96,487) = 0.000259 gm-moles/s

Because sulfuric acid is consumed at the cathode at the same rate as at the anode, thetotal rate of acid consumption is 0.000518 gm-moles /s, or 5.18 × 10 �7 kg-moles/s. Themolecular weight of sulfuric acid is 98, so the mass rate of acid consumption is

(5.18 × 10�7)(98) = 5.078 × 10�5 kg/s

or

(5.078 × 10 �5 )(3600) = 0.1828 kg/hr_____________________________________________________________________

Battery Applications. It is important to distinguish between SLI and tractionbatteries. The design of the SLI battery is governed primarily by the crankingrequirements of the vehicle, usually a high current flow for a brief time interval. Oncethe engine is started, the alternator provides power for vehicle operating loads and torecharge the battery. During normal driving the alternator satisfies the power needs oflighting, ignition, and other electrical functions. Thus a SLI battery is usually close tofully charged and seldom deeply discharged.

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The traction battery, on the other hand, usually has no energy converter available tocontinuously maintain the fully charged level. Thus the traction battery must be capableof operation from fully charged to almost fully discharged over a period of severalhours. At least eight hours daily are normally required to return the battery to its fullycharged level. This type of secondary cell is appropriate for vehicle propulsion andpower plant peak-shaving applications. In the latter instance, off-peak base-load poweris used to charge the battery. Later, during peak-demand periods, the battery returns alarge fraction of the charging energy to the system. Such a battery is designed forthousands of charge and discharge cycles.

According to reference 4, a 10-MW, 40 MW-hr lead-acid battery storage facilitywas built at the Southern California Edison Company at Chino. It was designed forload-leveling use for at least 2000 deep-discharge cycles and 80% energy efficiency. Sixof the cells were tested at 50°C and achieved 2300 cycles while retaining 108% of ratedcapacity (ref. 33). The facility, shown in Figure 11.10, uses strings of 1032deep-discharge, 2-V lead-acid cells connected in series to provide a nominal busvoltage of 2000 V DC. A total of 8256 2-V cells, each weighing 580 pounds, are usedto store 40 MW-hr for a 4-hr daily discharge cycle. A power conditioning system actsas both an inverter to provide AC output to the utility and as a rectifier to convertutility AC to DC for battery charging. The system is designed to utilize the responsive-ness of battery storage to deal with wide load swings. The Chino facility is said to beable to swing from 10-MW discharge to 10-MW charge in about 16 milliseconds.

According to references 4 and 11, lead-acid batteries can now deliver energydensities of more than 40 W-h/kg. This is less than the expected performance of severalother battery types. Nickel-iron batteries with an energy density of 50.4 W-hr/kg, forinstance, were among 10 types endurance-tested in an electric vehicle. A van withnickel-iron batteries logged over 44,000 miles, as compared to fewer than 27,000 milesfor the best performance by a lead-acid-battery-equipped van. Nickel-iron batteries,while noted for ruggedness and long life, suffer from a low energy efficiency of about60% (ref. 34).

Space limitations allow the present discussion to only hint at the more than acentury of research and the sophistication of the science and engineering of batteries.

Fuel Cells

Whereas batteries store energy in the form of chemical energy and subsequently trans-form it to electricity, fuel cells are capable of continuous transformation of chemicalenergy into electrical energy without intermediate thermal-to-mechanical conversion.Fuel cells may be thought of as chemical reactors similar to storage batteries, but withexternal supplies of fuel and oxidizer that react for indefinitely long time periodswithout substantial change in cell materials. Thus fuel cells do not run down or requirerecharging. As in a storage battery, reactions take place at electrodes, giving rise to a

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flow of electrons through an external circuit as depicted in Figure 11.11. An electrolyte,between the electrodes, again provides a medium for ion transfer. This allows electrons to flow through the external load while ion transport through the electrolyte maintainsoverall electrical neutrality of the cell.

An important element of fuel cell design is that, like large batteries, they are builtfrom a large number of identical unit cells. Each has an open-circuit voltage on theorder of one volt, depending on the oxidation-reduction reactions taking place. The fuelcells are usually built in sandwich-style assemblies called stacks. The schematic inFigure 11.12 shows crossflows of fuel and oxidant through a portion of a stack.

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Electrically conducting bipolar separator plates serve as direct current transmissionpaths between successive cells. This modular type of construction allows research anddevelopment of individual cells and engineering of fuel cell systems to proceed inparallel.

A photograph of a 32-kW phosphoric acid fuel cell stack is shown in Figure11.13(a). A conceptual design of a nominal 100-kW stack using three of the 32-kWstacks appears in Figure 11.13(b). Phosphoric acid fuel cells ideally operate withhydrogen as fuel and oxygen as oxidizer. To be economically practical in mostapplications, these fuel cells will probably require the use of a hydrocarbon fuel as asource of hydrogen and air as an oxidizer.

Consider the basic chemical reactions in a hydrogen-oxygen fuel cell as shown inFigure 11.11. Hydrogen molecules supplied at the anode are oxidized (lose electrons),each forming two hydrogen ions that drift through an electrolyte to the cathode.Electrons liberated from the hydrogen at the anode pass through an external circuit tothe cathode, where they combine in a reduction reaction with the H+ ions from theelectrolyte and an external supply of oxygen to form water. The electrode and overallcell reaction equations are:

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H2 � 2H+ + 2e� at the anode

2H+ + 2e� + 0.5O2 � H2O at the cathode

H2 + 0.5O2 � H2O overall cell reaction

Thus the sole reaction product of a hydrogen�oxygen fuel cell is water, an idealproduct from a pollution standpoint. In manned-space-vehicle applications the water isused for drinking and/or sanitary purposes. In addition to electrons, heat is also areaction product. This heat must be continuously removed, as it is generated, in orderto keep the cell reaction isothermal.

In the fuel cell, as in the battery, the reactions at the electrodes are surface pheno-mena. They occur at a liquid-solid or gas-solid interface and therefore proceed at a rateproportional to the exposed area of the solid. For this reason porous electrode materialsare used, frequently porous carbon impregnated or coated with a catalyst to speed thereactions. Thus, because of microscopic pores, the electrodes effectively have a surfacearea many times their visible area. Any phenomenon that prevents the gas from enteringthe pores or deactivates the catalyst must be avoided if the cell is to function effectivelyover long time periods. Because of the reaction rate�area relation, fuel cell current andpower output increase with increased cell area. The power density [W/m2] therefore isan important parameter in comparing fuel cell designs, and the power output of a fuelcell can be scaled up by increasing its surface area.

The electrolyte acts as a medium for ion transport between electrodes. The rate ofpassage of positive charge through the electrolyte must match the rate of electronarrival at the opposite electrode to satisfy the physical requirement of electrical neutral-ity of the discharge fluids. Impediments to the rate of ion transport through the electro-lyte can limit current flow and hence power output. Thus care must be taken in designto minimize the length of ion travel path and other factors that retard ion transport.

Fuel Cell Thermodynamic Analysis and Efficiencies. The theoretical maximumwork of an isothermal fuel cell (or other isothermal reversible control volume) is thedifference in Gibbs� function (or free energy) of the reactants (r) and the products (p).Because the enthalpy, entropy, and temperature are all thermodynamic properties, theGibbs� function, g = h � Ts, is also a thermodynamic property. From the First Law, thework leaving the control volume is

w = hr � hp + q [J/g-mole] (11.18)

For a reversible isothermal process, q = T(sp � sr), and the maximum work can bewritten as

wmax = hr � hp + T(sp � sr) = gr � gp [J/g-mole] (11.19)

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Thus the maximum work output of which a fuel cell is capable is given by the decreasein the Gibbs function.

But the maximum energy available from the fuel in an adiabatic steady-flow processis the difference in inlet and exit enthalpies, hr � hp. Defining the fuel cell thermalefficiency as wmax /(hr � hp), we obtain

�th = (gr � gp)/(hr � hp)

= 1 � T(sr � sp)/(hr � hp) [dl] (11.20)

The enthalpies and Gibbs function may be obtained from the JANAF tables for theappropriate substances (ref. 77).

EXAMPLE 11.4

Evaluate the open-circuit voltage, maximum work, and thermal efficiency for a directhydrogen-oxygen fuel cell at the standard reference conditions for the JANAF tables.Consider the two cases when the product water is in the liquid and vapor phases.

SolutionThe maximum work of the fuel cell is given by equation (11.19). The thermodynamicproperties needed are obtained from the formation properties of the JANAFthermodynamic tables:

hp(l) = hf [H2O(l)] = � 285,830 kJ/kg-molehp(g) = hf [H2O(g)] = � 241,826 kJ/kg-mole hr = hf [H2] + 0.5hf [O2] = 0.0 kJ/kg-molegp (l) = gf [H2O(l)] = � 237,141 kJ/kg-molegp(g) = gf [H2O(g)] = � 228,582 kj/kg-mole gr = gf [H2] + 0.5gf [O2] = 0.0 kJ/kg-mole

Thus the maximum work for liquid-water product is:

wmax(l) = gr � gp (l) = 0 � (� 237,141) = 237,141 kJ/kg-mole

and for water-vapor product is

wmax(g) = gr � gp (g) = 0 � (� 228,582) = 228,582 kJ/kg-mole

of H2 consumed or water produced in the reaction.The theoretical ideal open-circuit voltage of the cell may also be determined from

the Gibbs function using, for liquid-water product,

EMF(l) = wmax(l)/( jF) = 237,141×103/(2×9.6487×107) = 1.229 V

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For water-vapor product:

EMF(g) = wmax(g)/( jF) = 228,582x103/(2×9.6487×107) = 1.18 V

By Equation (11.20), the thermal efficiency with liquid-water product is

�th(l) = 237,141/285,830 = 0.8297

and with water-vapor product is

�th(g) = 228,582/241,826 = 0.945

Regardless of the phase of the product water, a large fraction of the available enthalpyof the reactant flow is converted to work in the ideal isothermal hydrogen-oxygen fuelcell. Because the fuel cell operates as a steady-flow isothermal process and not in acycle, the Carnot efficiency limit does not apply._____________________________________________________________________

Fuel cells, like other energy conversion systems do not function in exact conform-ance with simplistic models or without inefficiencies. Open-circuit potentials of 0.7�0.9volts are typical, and losses or inefficiencies in fuel cells, often called polarizations, arereflected in a cell voltage drop when the fuel cell is under load. Three major types ofpolarization are:

� Ohmic Polarization: The internal resistance to the motion of electrons throughelectrodes and of ions through the electrolyte.

� Concentration Polarization: Mass transport effects relating to diffusion of gasesthrough porous electrodes and to the solution and dissolution of reactants andproducts.

� Activation Polarization: Related to the activation energy barriers for the varioussteps in the oxidation-reduction reactions at the electrodes.

The net effect of these and other polarizations is a decline in terminal voltage withincreasing current drawn by the load. This voltage drop is reflected in a peak in thepower-current characteristic.

The high thermal efficiencies predicted in Example 11.4 are based on operation ofthe fuel cell as an isothermal, steady-flow process. Although heat must be rejected, the transformation of chemical energy directly into electron flow does not rely on heatrejection in a cyclic process to a sink at low temperature as in a heat engine, which iswhy the Carnot efficiency limit does not apply. However, inefficiency associated withthe various polarizations and incomplete fuel utilization reduce overall cell conversionefficiencies to well below the thermal efficiencies predicted in the example.

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Fuel cell conversion efficiency, �fc, is defined as the electrical energy output perunit mass (or mole) of fuel to the corresponding heating value of the fuel consumed.Defining the voltage efficiency, �V, as the ratio of the terminal voltage to thetheoretical EMF, V/EMF, and the current efficiency, �I , as the ratio of the cellelectrical current to the theoretical charge flow associated with the fuel consumption,we find that the conversion efficiency is related to the thermal efficiency by

�fc = (V/EMF)[I /( jFnc)]�th = �V �I �th

Thus the voltage efficiency accounts for the various cell polarizations previouslydescribed. Fuel cell conversion efficiencies in the range of 40�60% have beendemonstrated and are possible in large-scale power plants.

EXAMPLE 11.5

The DC output voltage of a hydrogen-oxygen fuel cell with liquid-water product is 0.8V. Assuming a current efficiency of 1, what is the fuel cell conversion efficiency?

SolutionFrom Example 11.4, the thermal efficiency is 0.8297 and the EMF is 1.229 V. The fuelcell conversion efficiency is then

�fc = �V �I �th = (0.8/1.229)�0.8297�1.0 = 0.54_____________________________________________________________________

If heat rejected in high-temperature fuel cells is used and not wasted, overall energyefficiencies as high as 80% may be attainable. In cells that operate at high temperatures,there is an opportunity to use the thermal energy of the cell reaction products forcogeneration purposes or in a combined cycle, with a steam turbine or a gas turbineusing the waste heat.

It is evident that the power output of fuel cells depends on the rate at which thereactants are consumed in the cell, just as the output of other steady-flow energyconversion systems depend on fuel and air supply rates. In low-temperature fuel cells,expensive catalysts (such as platinum) are required, to increase the rates of chemicalreaction at the electrodes so that the fuel cells will be small enough to achievereasonable power densities (per unit mass and volume).

Because chemical reaction rates increase with temperature, high-temperature fuelcells can achieve satisfactory reaction rates without catalysts or with reduced quantities.This advantage of high temperature may be offset by increased corrosion rates andother adverse effects of high temperature that reduce the fuel cell�s active life. Forutility applications, long, active cell life (about 40,000 hours) is considered an

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important design characteristic. It has been suggested, however, that the fuel cell stack may be a relatively small part of the cost of generation in large combined-cycle fuel cellpower plants designed for long life, so that it may be acceptable to replace fuel cellstacks after shorter operating periods than other plant components, as is done with fuelrods in nuclear reactors.

While space vehicle fuel cells successfully operate on pure hydrogen and oxygen,stationary and vehicle power plants are more likely to be viable if they consumehydrocarbon fuels and air. Thus the goals of major fuel cell research and developmentprograms focus on systems that incorporate the use of natural gas or otherhydrogen-rich fuels. A utility fuel cell system appears schematically as in Figure 11.14.A fuel processor upstream of the fuel cell stack is used to prepare the incoming fuel. Itsprimary functions are conversion of the incoming fuel to a hydrogen-rich gas andremoval of impurities such as sulphur. For natural gas, syngas, or other gaseous fuels,

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the hydrocarbons may be converted by a steam-reforming process such as that formethane:

CH4 + H2O � CO + 3H2

to produce a mixture of hydrogen and carbon monoxide. The carbon monoxide in theproducts can then be converted to CO2 and produce more hydrogen in the so-calledshift reaction:

CO + H2O � CO2 + H2

The direct current generated by a fuel cell for utility use must be converted toalternating current by a power conditioner, as the figure suggests. High-efficiency solid, state inverters are used for this function.

Following Figure 11.14, the overall fuel cell system efficiency is then the product ofa power conditioner efficiency (AC power output / DC power output), a fuelprocessing efficiency (enthalpy change of hydrogen / heating value of supply fuel), andthe ratio of the sum of the fuel cell DC output and bottoming cycle power to thehydrogen enthalpy change.

Fuel Cell Types. Table 11.1 indicates the major fuel cell types, their convenientabbreviations, and their nominal operating temperatures. Fuel cells are usually classified according to their electrolytes, as indicated by the types in the table. While thealkaline and polymer electrolyte fuel cells were successfully demonstrated in the NASAGemini (1-kW PEFC), Apollo (1.5-kW AFC), and Space Shuttle Orbiter (7-kW AFC)programs, these were applications with short design operating lives that used expensivecatalysts to attain satisfactory reaction rates and thus may be appropriate only to spaceand military applications and certain limited transportation uses.

Table 11.1 Fuel Cells of Current Technical Interest

Type Abbreviation Operating Temperature

Polymer electrolyte fuel cell PEFC 80°C

Alkaline fuel cell AFC 100°C

Phosphoric acid fuel cell PAFC 200°C

Molten carbonate fuel cell MCFC 650°C

Solid oxide fuel cell SOFC 1000°C

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The last three fuel cell types listed in the table (PAFC, MCFC, and SOFC) arewidely believed to be most likely to achieve commercialization for stationary powerproduction early in the 21st century. Figure 11.15 summarizes the reactants and flowpaths of these three types. The PAFC is most likely to be commercialized first, butsignificant progress is being made in the advanced high-temperature technologiesrepresented by the MCFC and the SOFC.

Large-scale phosphoric acid fuel cell power plants have been explored based onfuel cell stacks developed by International Fuel Cells, Inc. (IFC). A basic 10-ft2, 700-kW unit is shown in Figure 11.16, together with its pressure vessel. According toreference 36, The Tokyo Electric Power Company 11-MW Goi Station had twenty700-kW fuel cell assemblies with an expected heat rate of 8300 Btu/kW-hr usingliquefied natural gas as fuel.

More recently, the IFC website (ref. 81) states that the PC 25TM (a 200-kW unit) isone of five units comprising the �largest commercial fuel cell system� in the UnitedStates, which became operational in the year 2000 for the U.S. Postal Service inAnchorage, Alaska. The PC 25TM operates with hydrogen, natural gas, propane,butane, naptha, or waste gases as its energy source. The website states that IFCdelivered its 200th PC25 in the year 2000, with 3.5 million total operating hours. ThePC 25 originally went into production in 1991.

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Reference 82 states that the ONSI Corporation (a subsidiary of IFC) PC 25 fuel cell�delivers 35 to 40 percent electrical efficiency and up to 80 percent total efficiency withheat recovery.� A 1997 DOE report (ref. 87) explores the potential for a PC 25 fuelcell cogeneration system retrofit in a large office building.

The first molten carbonate fuel cell power plant, a 100-kW plant operated byPacific Gas and Electric, despite being conservatively designed for simplicity andreliability, was expected to have a heat rate of about 7000 Btu/kW-hr without abottoming cycle. This pilot plant, located at San Ramon, Calif., uses natural gas as fueland aimed for a cell lifetime of 25,000 hours (refs. 15 and 36). Reference 78 indicatesthat verification and durability testing began in June 1991 leading to a series of 2-MWdemonstration plants starting in 1994, with first commercial units scheduled for 1997.

The Santa Clara Fuel Cell Demonstration project, fueled by natural gas, startedconstruction in 1994, produced first power in March 1996, and concluded thedemonstration in April 1997. Reference 84 indicates that it was �the largest and mostefficient fuel cell power plant ever operated in the U.S...�

Reference 85 announces the successful completion of the grid-connected operationof a 250-kW, Fuel Cell Energy, Inc., demonstration plant as of June 2000. The plantlogged more than 11,800 hours and generated more than 1.8 million kilowatt-hours ofelectricity. The plant is to be refitted to demonstrate a very-high-efficiency combined-cycle operation with a gas turbine.

Perhaps the most promising fuel cell type is the solid oxide fuel cell. The use of asolid electrolyte, as shown in Figure 11.17, allows a radical departure from the plate orsandwich type of design. The individual cells are of a closed-end cylindrical design with

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fuel manifolded to flow over the outside electrode and air on the inside communicatingwith the inner electrode through a porous cylindrical support structure. The clever fuel,air, and exhaust manifolding and electrical connections of this design are apparent inFigures 11.17 and 11.18. The high-temperature design allows operation withoutcatalysts and provides an excellent opportunity to use a bottoming cycle.

Beyond high efficiency, there are several additional major characteristics of fuelcells that make them particularly attractive as energy conversion systems. First, fuel celloperation has been shown to occur with very low levels of environmental pollution (ref. 15). It has been projected that commercial fuel cells may attain pollution levels that arefactors of ten below those of new conventional coal-burning power plants using thebest available pollution control equipment (ref. 16).

A second important characteristic is that, because most fuel cells operate with ahydrogen-rich fuel or pure hydrogen, the fuel can be obtained from a number ofsources, such as petroleum, natural gas, naptha, methanol, and syngas made from coal.Thus a commercial stationary power plant will have a chemical processor upstream ofthe fuel cell stack as in Figure 11.14. The reforming and shift reactions likely to be usedwill produce carbon dioxide as a product. This somewhat tarnishes the environmental

467

beauty of the hydrogen-oxygen fuel cell concept. The chemical process to obtainhydrogen from some of the fuels may in some cases take place within the fuel cell stack.

Electrolysis of water may also be used to produce hydrogen and oxygen for fuel celluse. This provides an opportunity for energy storage by using electricity generated bybase-loaded nuclear or conventional fossil fuel plants operating at off-peak hours. Thehydrogen and, when appropriate, oxygen could then be reacted in fuel cells to generateelectricity during periods of peak electrical demand at the central plant site or at otherlocations.

It has been pointed out that the modular nature of fuel cells and of fuel cell stacks isan important characteristic of this technology. Because of this modularity, a given celldesign could lead to a range of power plant designs such as large utility-scale plants,smaller environmentally acceptable neighborhood plants, and industrial facilitiesproviding combined heat and power. It has been demonstrated that scaling based onfuel cell modularity does not necessarily adversely influence stack efficiency. Theircompact construction suggests that stacks and other power plant components could befactory-built and assembled in plug-in style at plant sites.

The modular nature of the fuel cell offers real possibilities for distributed powerproduction, contrary to the historic long-term trend in electrical power production.

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Because of the expected short manufacturing time for fuel cells, their apparentenvironmental acceptability, and their ability to function well in a variety of sizes, theyoffer utilities alternatives to the long construction times and the financial risks of largefossil fuel and nuclear plants that have required as much as ten years to bring intoservice. Thus significant advantages may be expected to accrue from the ability toconstruct small, high-efficiency, environmentally acceptable plants in areas wheredemand is growing.

While its expected high efficiency commends its use as a utility base-load plantswith constant power output, the fuel cell also offers the potential for serving varyingloads without substantial losses in performance. Thus fuel cells may prove useful inload-following applications as well as in base-load application.

Another potential application for fuel cells is in the large vehicle transportationsector�buses, trucks, locomotives, and the like. A likely entry for the fuel cell may bein public transportation vehicles in large cities with significant pollution problems.Reference 18 reports on studies of alternatives to battery-powered trucks in Paris,where almost 200 electric garbage trucks are currently in use. It was found that fuel-cell-powered trucks and, to a lesser degree, hybrid fuel cell/flywheel trucks (theflywheel stores extra energy for occasional vehicle accelerations while the fuel cellhandles the base-load propulsion and flywheel reenergizing) would have significantlylower operating costs per ton of garbage per day than the existing battery-poweredtrucks. Studies also showed that diesel-powered buses were only about 10% to 20%cheaper to operate in Europe than fuel-cell-powered buses.

A simulation study, reported in reference 67, considered hypothetical fuel-cell-powered bus designs operating on three urban bus routes. A bus design specification,resulting from the study, combined two 30-kW phosporic acid fuel cells with a 38-kW-hr battery to power a 50-kW electric traction motor. The motor transmits power to arear wheel-axle assembly through a conventional transmission. The specification wasfor a 23 passenger bus for use at speeds up to 45 mph.

In April 2000, Daimler-Chrysler announced (ref. 88) an offering of 20-30 busesemploying 250-kW fuel cells operating on hydrogen fuel for delivery in 2002. In thesevehicles, hydrogen would be stored in eight tanks mounted on the roof over theforward axle.

These studies suggest that fuel cell technology is maturing. In many cases theeconomics of these systems is not yet competitive with that of existing systems.However, further advances may soon make fuel cells competitive for some transport-ation and stationary applications. Reference 68 gives further evidence that the quest forcommercial fuel cell technology is truly international in scope.

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Bibliography and References

1. Angrist, Stanley W., Direct Energy Conversion, 3rd ed. Newton, Mass.: Allyn &Bacon, 1976.

2. Spitzer, Lyman, Physics of Fully Ionized Gases, New York: Wiley-Interscience,1956.

3. Appleby, A. J., "Advanced Fuel Cells and Their Future Market," Ann. Rev. Energy,13 (1988): 267�316.

4. McLarnon, Frank R., and Cairns, Elton J., "Energy Storage," Ann. Rev. Energy, 14 (1989): 241�271.

5. Mellde, Rolf W., "Advanced Automobile Engines For Fuel Economy, LowEmissions, and Multifuel Capability," Ann. Rev. Energy, 14 (1989): 425�444.

6. Jackson, John David, Classical Electrodynamics. New York: Wiley, 1962.

7. Fraas, Arthur P., Engineering Evaluation of Energy Systems. New York:McGraw-Hill, 1982.

8. Vincent, Colin A., et al., Modern Batteries. Baltimore: Edward Arnold, 1984.

9. Sperling, Daniel and DeLuchi, Mark A., �Transportation Energy Futures,� Ann. Rev.Energy, 14 ( 1989): 375�424.

10. Fickett, A. P., " Fuel Cell Power Plants," Scientific American, December 1978: 70.

11. Society of Automotive Engineers, Automotive Handbook, 4th ed., Robert BoschGmbH, 1996.

12. Mayfield, Manville J., Beyma, Edmund F., and Nelkin, Gary A., "Update on U.S.Department of Energy�s Phosphoric Acid Fuel Cell Program." Proceedings of theSixteenth Energy Technology Conference, February 28�March 2, 1989, GovernmentInstitutes, Inc., pp. 184�196.

13. Myles, K. M., and Krumpelt, M., "Status of Molten Carbonate Fuel Cell Technol-ogy." Proceedings of the Sixteenth Energy Technology Conference, February 28�March 2, 1989, Government Institutes, Inc., pp. 197-204.

14. Bates, J. Lambert, "Solid Oxide Fuel Cells: A Materials Challenge." Proceedings ofthe Sixteenth Energy Technology Conference, February 28�March 2, 1989, Govern-ment Institutes, Inc., pp. 205�219.

470

15. Gillis, Edward, "Fuel Cells," EPRI Journal, September 1989: 34�36.

16. Harder, Edwin L., Fundamentals of Energy Production. New York: Wiley, 1982.

17. Sissine, F., "Fuel Cells for Electric Power Production: Future Potential, FederalRole and Policy Options," in Fuel Cells:Trends In Research and Applications, Ed. A.J. Appleby. Washington: Hemisphere, 1987.

18. Appleby, A. J., "Phosphoric Acid Fuel Cells," in Fuel Cells:Trends In Researchand Applications, Ed. A. J. Appleby. Washington: Hemisphere, 1987.

19. Appleby, A. J. (Ed.), Fuel Cells:Trends In Research and Applications,Washington: Hemisphere, 1987.

20. Ketelaar, J. A. A., "Molten Carbonate Fuel Cells," Fuel Cells:Trends In Researchand Applications, Ed. A. J. Appleby. Washington: Hemisphere, 1987.

21. Kinoshita, K., McLarnon, R. R., and Cairns, E.J., Fuel Cells: A Handbook., U.S.Department of Energy METC-88/6096, 1988.

22. Berry, D.A., and Mayfield, M.J., Fuel Cells, Technology Status Report, U.S.Department of Energy METC-89/0266, 1988.

23. Anon., Fuel Cell Systems Program Plan, FY 1990, U.S.Department of Energy,FE-0106P, October 1989.

24. Huber, W. J., Proceedings of the First Annual Fuel Cells Contractors ReviewMeeting, U.S.Department of Energy METC-89/6105, May 1989.

25. Anon., 1988 Fuel Cell Seminar. Washington, D.C.: Courtesy Associates, 1988.

26. Feynman, Richard P., Leighton, Robert B. and Sands, Matthew, The FeynmanLectures on Physics, Vol. 2. Reading, Mass.: Addison-Wesley, 1964.

27. Anon., Battery Service Manual. Chicago: Battery Council International, 1987.

28. Kotz, John C., and Purcell, Keith F., Chemistry and Chemical Reactivity.Philadelphia: Saunders, 1987.

29. Goodman, Frank, "Power Electronics for Renewables," EPRI Journal,January/February 1988: 44�47.

471

30. Schaefer, John, "Photovoltaic Operating Experience," EPRI Journal, March 1988:40�42.

31. Peterson, Terry, "Amorphous Silicon Thin Film Solar Cells," EPRI Journal, June1988: 41�44.

32. Dostalek, Frank, "High Concentration Photovoltaics," EPRI Journal, March 1989: 46�49.

33. Morris, Douglas, "The Chino Battery Facility," EPRI Journal, March 1988: 46�50.

34. Purcell, Gary, and Driggans, Rick, "Electric Vehicle Testing at TVA," EPRIJournal, March 1988: 44�46.

35. Anon., The Storage Battery. Horsham, Pa.: Exide Corp., 1980.

36. Makanski, Jason, "Fuel Cells Extend Boundaries of Process/Power Integration,"Power, May 1990: 82�86.

37. Chapman, Alan J., Heat Transfer, New York: Macmillan, 1968.

38. Swanson, Theodore D., "Large Scale Photovoltaic System Design Considerations,"The Handbook of Photovoltaic Applications. Atlanta, Ga.: Fairmont Press, 1986, pp.19�32.

39. Van Overstraeten, R.J., and Mertens, R.P., Physics, Technology, and Use ofPhotovoltaics. Boston: Adam Hilger, 1986.

40. McDaniels, David K., The Sun: Our Future Energy Source, 2nd ed. New York:Wiley, 1984.

41. Feynman, Richard P., Leighton, Robert B. and Sands, Matthew, The FeynmanLectures on Physics, Vol. 3, Reading, Mass.: Addison-Wesley, 1965.

42. Takahashi, K. and Konagai, M., Amorphous Silicon Solar Cells. New York: Wiley, 1986.

43. Duffie, John A., and Beckman, William A., Solar Engineering of ThermalProcesses. New York: Wiley, 1980.

44. Rosa, Richard J., Magnetohydrodynamic Energy Conversion, rev. ed. Washington:Hemisphere, 1987.

472

45. Gregory, Derek P., �The Hydrogen Economy,� Scientific American, Vol. 228,No.1, (January 1973) 13�21.

46. Gregory, D. P. and Pangborn, J.B., "Hydrogen Energy", Ann. Rev. Energy, 1 (1976): 279�310.

47. Cogineni, M. Rao, Andrus, H.E. Jr., and Jones, T.J., "Advanced Energy Systems."Proceedings of the American Power Conference, Vol. 50, 1988, pp. 282�287.

48. Bajura, R. A., and Halow, J.S., " Looking Beyond the Demonstration Plants:Longer-Term, Coal-Based Technology Options." Proceedings of the American PowerConference, Vol. 50, 1988, pp. 49�56.

49. DeMeo, Edgar, et al., "Thin Films: Expanding the Solar Marketplace", EPRIJournal, March 1989: 4�15.

50. Wood, Bernard W., Applications of Thermodynamics, 2nd ed. Reading, Mass.:Addison-Wesley, 1982.

51. Womack, G. J., MHD Power Generation: Engineering Aspects. London: Chapmanand Hall, 1969.

52. Soo, S. L., Direct Energy Conversion. Engelwood Cliffs, N.J.: Prentice Hall, 1968.

53. Kettani, M. Ali, Direct Energy Conversion. Reading, Mass.: Addison Wesley,1970.

54. Sutton, George W., Direct Energy Conversion. New York: McGraw-Hill, 1966.

55. Pitts, Donald R., and Sissom, Leighton E., Heat Transfer, Shaum's Outline Series.New York: McGraw-Hill, 1977.

56. Anon., The Astronomical Almanac. Washington, D.C.: U. S. Government PrintingOffice, 1990.

57. Parsons, Robert A. (Ed.), The ASHRAE Handbook, HVAC Systems and Appli-cations Volume. Atlanta, Ga.: American Society of Heating, Refrigerating, and AirConditioning Engineers, 1987.

58. DeMeo, Edgar, "Getting Down to Business with Thin Films," EPRI Journal,March 1989: 4�15.

473

59. Taylor, Roger W., Cummings, John E., and Swanson, Richard M., "High EfficiencyPhotovoltaic Device Development: An Example of the R&D Process" Proceedings ofthe American Power Conference, Vol. 47, 1985, pp. 255�259.

60. Bockris, J. O. M., Energy The Solar-Hydrogen Alternative. New York: Wiley,1975.

61. Williams, L. O., Hydrogen Power, Elmsford, N. Y.: Pergamon Press, 1980.

62. Sze, S. M., Semiconductor Devices Physics and Technology. New York: Wiley,1985.

63. Moore, Taylor, et al., "Opening the Door for Utility Photovoltaics," EPRI Journal,January�February 1987: 5�15.

64. Griffiths, David, J., Introduction to Electrodynamics, 2nd ed. Engelwood Cliffs, N. J.: Prentice-Hall, 1989.

65. Anon., �Solar Electric Generating Stations (SEGS),� IEEE Power EngineeringReview, August, 1989: 4�8.

66. Anon., �Promise of Solar Energy Being Fulfilled in California,� Power, October1989: s32�s36.

67. Romano, Samuel, �Fuel Cells for Transportation,� Mechanical Engineering,August 1989: 74�77.

68. Hirschenhofer, J. H., �International Developments in Fuel Cells,� MechanicalEngineering, August 1989: 78�83.

69. Carlson, D. E., �Photovoltaic Technologies for Commercial Power Generation,�Annual Review of Energy, Vol. 15, 1990, pp. 85�98.

70. MacDonald, Gordon J., �The Future of Methane as an Energy Resource,� AnnualReview of Energy, Vol. 15, 1990, pp. 53�83.

71. Moore, Taylor, et al., �On-Site Utility Applications for Photovoltaics,� EPRIJournal, March 1991: 26�37.

72. Moore, Taylor, et al., �Thin Films: Expanding the Solar Marketplace,� EPRIJournal, March 1989: 4�15.

474

73. Smock, Robert W., �Second Generation Fuel Cell Technology Moves TowardDemos,� Power Engineering, June 1990: 10.

74. Sapre, Alex R., �Properties, Performance and Emissions of Medium-ConcentrationMethanol-Gasoline Blends in a Single-Cylinder, Spark-Ignition Engine,� SAE Paper 881679, October 1988.

75. Douglas, John, �Beyond Steam: Breaking Through Performance Limits,� EPRIJournal, December 1990: 5�11.

76. Graff, Eric, Texas Instruments Co. Personal Communication, May 24, 1991.

77. Chase, M. W. Jr., et al., JANAF Thermodynamic Tables, 3rd ed., J. Phys. Chem.Ref. Data 14, Supplement No. 1, 1985.

78. Douglas, John, et al., �Fuel Cells for Urban Power,� EPRI Journal, September1991: 5�11.

79. Howell, John R., Bannerot, Richard B., and Vliet, Gary C., Solar-Thermal EnergySystems. New York: McGraw-Hill, 1982.

80. Levine, Jules D., et al., �Basic Properties of the Spheral Solar Cell.� 22nd

Photovoltaic Specialists Conference, Las Vegas, Nev., 1991.

81. International Fuel Cells: Clean, Reliable Fuel Cell Energy,www.internationalfuelcells.com/index_fl1.shtml, (November 14, 2000)

82. Rulseh, Ted, �Fuel Cells: From Promise to Performance,� Grid, Spring/Summer2000: 15.

83. Santa Clara Demonstration Project, www.ttcorp.com/fccg/scdpnew1.htm,(November 14, 2000)

84. Fuel Cell Energy, Carbonate Fuel Cell Manufacturer,www.ttcorp.com/fccg/erc_abt.htm, (November 15, 2000)

85. Welcome to Fuel Cell Energy, Inc., www.fuelcellenergy.com/, What�s New?,(November 15, 2000)

86. Library of Fuel Cell Related Publications, http://216.51.18.233/biblio.html,(November 14, 2000)

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87. Archer, David H., and Wimer, John G., � A Phosphoric Acid Fuel CellCogeneration System Retrofit to a Large Office Building,� Department of EnergyFETC-97/1044, April 1997., www.fetc.doe.gov/netltv/index.html, (November 15,2000)

88. DaimlerChrysler Offers First Commercial Fuel Cell Buses to Transit Agencies,www.hfcletter.com/letter/may00/feature.html, (November 15, 2000)

EXERCISES

11.1* Derive an expression for the battery power output, in terms of E and Ri /Ro, forthe linear model discussed in connection with Example 11.2. Nondimension-alizethe power by dividing by E2/Ro. Use a spreadsheet to tabulate and plot thedimensionless power as a function of Ri /Ro.

11.2* Derive an expression for the linear-battery-model power output nondimensional-ized by E2/Ri in terms of the internal-to-external resistance ratio. Use aspreadsheet to tabulate the dimensionless power function, and plot it. Is there acondition that produces an extreme value of the dimensionless power? If so, usecalculus methods to derive the condition.

11.3 An automobile storage battery with an open-circuit voltage of 12.8V is rated at260 A-hr. The internal resistance of the battery is 0.2 �. Estimate the maximumduration of current flow and its value through an external resistance of 1.8 �.

11.4 A battery electrical storage plant is to be designed for 20-MW peak powerdelivery for a duration of four hours. The plant uses 600 A-hr batteriesoperating at 400 volts DC. Estimate the minimum number of batteries and thecurrent in each during peak operation.

11.5 A hydrogen-oxygen fuel cell operates with a voltage of 0.7v with water-vaporproduct. Calculate the work per kg-mole of hydrogen, in kJ and in kW-hr, anddetermine the cell efficiency.

11.6 A neighborhood fuel cell power plant is to be designed for an electrical poweroutput of 2000 kW with liquid-water product. Estimate the flow rates ofhydrogen and oxygen during peak power production, assuming that an 80%efficient power conditioner is used to convert DC to AC power and that the fuelcell efficiency is 55%. What is the plant heat rate?

___________________* Exercise numbers with an asterisk involve computer usage.

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11.7 A hydrogen-oxygen fuel cell has liquid water as product when it operates at 0.82volts. What is the electrical energy output, in kJ / kg-mole of hydrogen, and thecell efficiency?

11.8 Prepare a typed three-page, double-spaced memorandum outlining the design ofthe power train of a hydrogen-oxygen fuel-cell-powered automobile, giving thedesign criteria, system description, and quantitative preliminary design data onthe power and fuel supply systems.

11.9 A hydrogen-oxygen fuel cell stack produces 50 kW of DC power at an efficiencyof 60%, with water-vapor as product. What is the hydrogen mass flow rate, ing/s, and the cell voltage?

11.10 A fuel cell power plant is to be designed for an electrical power output of 200-MW with liquid-water product. Estimate the flow rates of hydrogen and oxygenduring peak power production with an 85% efficient power conditioner and fuelcell efficiency of 55%. What is the plant heat rate?

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C H A P T E R E L E V E N

Energy System Alternatives

Part 2. MHD, Solar Energy, a Hydrogen Economy and Concluding Remarks

11.4 Magnetohydrodynamic Energy Conversion

Magnetohydrodynamic energy conversion, popularly known as MHD, is another formof direct energy conversion in which electricity is produced from fossil fuels withoutfirst producing mechanical energy. The process involves the use of a powerfulmagnetic field to create an electric field normal to the flow of an electricallyconducting fluid through a channel, as suggested by Equation (11.13) and depicted inFigure (11.19). The flow velocity u is parallel to the channel axis, taken in the y-direction. The drift of electrons induced by this lateral electric field produces anelectric current, represented by the current density vector J. Electrodes in oppositeside walls of the MHD flow channel provide an interface to an external circuit.Electrons pass from the fluid at one wall to an electrode, to an external load, to theelectrode on the opposite wall, and then back to the fluid, completing a circuit. Thusthe MHD channel flow is a direct current source that can be applied directly to anexternal load or can be linked with a power-conditioning inverter to producealternating current.

MHD effects can be produced with electrons in metallic liquids such as mercuryand sodium or in hot gases containing ions and free electrons. In both cases, theelectrons are highly mobile and move readily among the atoms and ions while localnet charge neutrality is maintained. That is, while electrons may move with ease, anysmall volume of the fluid contains the same total positive charges on the ions andnegative electron charges, because any charge imbalance would produce largeelectrostatic forces to restore the balance.

Though liquid metal MHD has been demonstrated experimentally, mosttheoretical and experimental work and power plant development and applicationstudies have focussed on high-temperature ionized gas as the working fluid.Unfortunately, most common gases do not ionize significantly at temperaturesattainable with fossil fuel chemical reactions. This makes it necessary to seed the hotgas with small amounts of vapor of readily ionizable materials such as the alkalimetals. The resulting ions and electrons make the hot gas sufficiently electricallyconducting that it may be influenced by the applied magnetic field.

The ionization potentials are measures of the energy needed to free valenceelectrons from an atom. Materials such as cesium and potassium have ionizationpotentials low enough that they ionize at the temperatures attainable with combustion

478

reactions in air. Recovery and reuse of seed materials from the MHD channel exhaustare usually considered necessary from both economic and pollution standpoints.

Ionized Gases in Electromagnetic Fields

Before analyzing the MHD channel, we will consider briefly the behavior ofelectrons in an ionized gas in the presence of electromagnetic fields.

In a gas at or near equilibrium, atoms, ions, and electrons are in random motion.At any given spatial position their velocities are distributed about a mean velocity thatincreases with increase in the local temperature. Consider just one of the freeelectrons moving, without collision, in a plane normal to a uniform magnetic field, asin Figure 11.20. The electron experiences a constant force qceB normal to its pathaccording to Equation (11.11). Here, q is the charge of the electron and ce themagnitude of its velocity. Because the force is normal to its path, the electron travelswith constant velocity on a circular path around magnetic lines of force. By Newton�sSecond Law, the force on the electron is

F = mece2/r = qceB [N] (11.21)

It follows that the angular frequency of the electron about a line of force ce/r, calledits cyclotron frequency, is

� = ce/r = qB/me [s �1] (11.22)

The electron cyclotron frequency is independent of electron velocity and is dependent

479

only on the magnetic field strength and electron properties. Although the cyclotronmotions of electrons exist in gases when strong magnetic fields are present, thecircular paths of the electrons may be disrupted by collisions with other particles.

The likelihood of collisions between particles depends on their effective sizes:larger particles will collide more frequently. The probability of collision is taken asproportional to the collision cross-section Q of the particle, which may be thought ofas its area. The frequency of collision of electrons �c is given by the product of theelectron number density, ne [electrons/m3], the collision cross-section, Q [m2], and thevelocity, ce [m/s]:

�c = neQce = 1/� [collisions per s] (11.23)

Here, the mean time between collisions, � [s], is the inverse of the collisionfrequency.

The ratio of the cyclotron frequency to the collision frequency �/�c , is called theHall parameter. It indicates the relative importance of the magnetic field andcollisions in controlling electron motion in the ionized gas. The Hall parameter isrelated to the magnetic field intensity by

�/�c = �� = qB� /me = qB/nemeQce [dl] (11.24)

It is proportional to the number of cyclotron loops made per collision. A Hallparameter large compared with one indicates magnetic-field-dominated motion of

480

electrons, while a small value implies that collisions quickly break up ordered motionsproduced by the magnetic field.

At least three velocities are of importance in a conducting gas in a MHD channel.First, the velocity of the gas stream is given by u (assumed constant for the presentcase for an appropriately designed channel). Secondly, the velocities of individualelectrons ce, as just introduced, are distributed about an average value that increaseswith the local temperature. In the absence of electromagnetic fields, the average valueof ce over all electrons is the flow velocity u; i.e., on the average the electrons movewith the gas flow. When fields are present, however, there may be an average motionof electrons relative to the gas. It is the ease of this motion that determines theconductivity of the gas. The third velocity, the relative velocity of an electron we, isdefined as the vector difference of its absolute velocity and the mean fluid velocity:

we = ce � u [m/s]

The drift velocity we, is the magnitude of the average of the relative velocities of theelectrons. In the absence of fields, the average of ce is u, and thus the drift velocity iszero. When an electric field is present, however, the transport of negative charge by electrons represents a current flow in the gas.

Another important parameter, the electron mobility �, is a measure of theresponse of electrons to an electric field. It is defined as the ratio of the magnitude ofthe electron drift velocity we to the local electric field intensity:

� = we /E [m2/V-s] (11.25)

If it is assumed that an electron loses all of its drift velocity on collision, the accel-eration of the electron may be approximated by the ratio of the drift velocity to themean time between collisions. Because the force due to the electric field is given byqE, Newton�s Second Law allows the drift velocity to be expressed as

we = qE�/me [m /s] (11.26)

The electron mobility can then be written as

� = q� /me [m2/V-s] (11.27)

Using equation (11.24), the product �B becomes the Hall parameter:

�B = q� B/me = �� = �/�c [dl] (11.28)

Thus the Hall parameter is large for gases of high electron mobility in strong magneticfields. It will be seen that this can have a significant effect on MHD channel design.

Assuming electrons as the dominant charge carriers, the current density can also

481

be related to the electron mobility through the drift velocity:

J = neqwe = neq2�E/me

= �neqE [A /m2] (11.29)

The electron conductivity of a stationary gas is then given by:

� = J/E = �neq [(�-m)-1] (11.30)

Thus high electron mobility and electron number density are essential to achieve thehigh conductivity needed in an MHD generator.

Analysis of a Segmented Electrode MHD Generator

Consider the one dimensional flow of a gas in an MHD channel coupled with a simplethree-dimensional model of the electromagnetic phenomena. Rather than thecontinuous electrode configuration shown in Figure 11.19, we examine a refinedconfiguration, shown in Figure 11.21. Here the electrodes, set in opposite electrically insulated channel walls, are segmented in the streamwise direction. This eliminates areturn path along the wall for axial electrical currents in the flow.

By the same notation as in Figure 11.19, a seeded, ionized gas flows through thesegmented-electrode channel in the y-direction with a constant velocity u. A uniformmagnetic field in the z-direction exists throughout the gas in the channel. A forcegiven by qu×B, and thus an equivalent electric field u×B is imposed on the flow inthe channel. Therefore, positive ions tend to drift in the positive x-direction andelectrons drift in the negative x-direction toward the right electrodes. Because theirmobility is much greater than that of the relatively massive ions, the electrons are theprimary charge carriers. The electrons are collected at the right electrodes and flowthrough the external circuits returning to the channel at the left electrodes, as shownin Figure 11.21.

When the channel is under an electrical load, the current density vector in the x-direction induces a force on the fluid in the negative y-direction. Thus the x-comp-onent of J interacts with the magnetic field to produce the axial electric fieldcomponent Ey = � JB that opposes the flow velocity u. In order to maintain aconstant velocity in the duct, a streamwise pressure gradient, dp/dy, must balance theforce due to this axial electric field and the viscous forces. Thus, ignoring viscousresistance, the axial force on the gas per unit volume is

Fy = � |J×B| = � JB = dp/dy [N/m3] (11.31)

where the negative sign indicates that the magnetic force is directed upstream. As a

482

result dp/dy < 0, indicating that the flow pressure drops as y increases. The resultingnet pressure force in the positive y-direction balances the magnetic and viscous forcesand maintains the flow velocity constant. A compressor is therefore requiredupstream of the channel to pressurize the flow, to support the field-inducedstreamwise pressure gradient, and thus to maintain the steady flow in the channel.

With segmented electrodes there is no axial current in the channel (Jy = 0), andthus the current density component Jx = J is proportional to the net electric field inthe x- direction:

J = � (uB � Ex) [A/m2] (11.32)

The combined electrical resistance of the MHD channel flow and the external loadgoverns the available potential at the MHD electrodes. If the external circuit is open,J = 0; hence, Equation (11.32) indicates that Ex|open = uB. With a finite externalresistance, current flows and the electrode potential is reduced below the open-circuitvalue. Thus, under load, the channel voltage drops to a fraction K of the open-circuitvoltage. Hence, we may write Ex = KuB, where K is called the channel load factorand where 0 � K � 1. The current density then becomes

483

J = �uB(1 � K) [A/m2] (11.33)

The electrical power delivered to the load per unit volume of channel is then given by

Power|out = J�E = �u2B2K(1 � K) [W/m3] (11.34)

Returning now to consideration of the streamwise electrical fluid interaction, wewrite the steady-flow form of the First Law for an adiabatic control volume, includingthe work done against the body force, as

m(h1 + u12/2) = m(h2 + u2

2/2) + mw [J/s] (11.35)

where m is the channel mass flow rate. Work is positive here because it is done by thefluid in the channel to produce the electrical current flow to the external load. Forconstant velocity in the channel, this becomes

h2 = h1 � w = h1 � (Power|out)(Volume)/m

= h1 � �u2B2K(1 � K)/� [kJ/kg] (11.36)

where � is the gas density. Thus the work delivered to the load reduces the thermalenergy of the flow. We have seen that a compressor is required to pressurize the flowin the channel and that heating of the flow provides a high entrance enthalpy andwork output.

From Equations (11.31) and (11.33), the electrical retarding force on the flow is

Fy = � �uB2 (1 � K) [N/m3] (11.37)

and the fluid power to push the gas through the channel per unit volume is

Power|in = |Fyu| = �u2B2 (1 � K) [W/m3] (11.38)

The Ohmic or I2R loss is given, using Equation (11.33), by

J2/� = [�uB(1 � K)]2/� = � (uB)2(1 � 2K + K2)

= �u2B2(1 � K) � �u2B2K(1 � K) [W/m3] (11.39)

Comparing equations (11.34) and (11.38) with equation (11.39), we see that theohmic loss is the difference between the power required to push the flow through thechannel and the useful power through the load.

The efficiency of the channel is defined as the ratio of the Power|out to Power|in.By Equations (11.34) and (11.38), the MHD channel efficiency is

484

� = Power|out /Power|in = K [dl] (11.40)

Thus the electrical efficiency of the segmented electrode MHD channel is equal to thechannel load factor. Examination of Equation (11.34) shows that the power outputvanishes when K = 0 and when K = 1. Thus there must be an intermediate value of Kthat maximizes the power output. By differentiation of Equation (11.34) with respectto K, the usual methods of calculus indicate that the power is maximized when K =0.5. Thus operation at this value implies that 50% of the flow energy input to thechannel is converted to electricity, and the remainder is dissipated in the flow channel.This energy is not lost from the flow but is an irreversibility that is reflected in a lossin ability of the flow to do work.

EXAMPLE 11.6

A 10-m3 MHD generator with segmented electrodes has a short-circuit currentdensity of 12,000 amperes per square meter. The gas conductivity is 20 (Ohm-m)-1.

If flow and magnetic field conditions are unchanged when the load factor is 0.6,what is the output power? What is the actual current density in the channel? If themagnetic field is doubled in strength, by what factor would you expect the outputpower to change?

SolutionFor a short-circuit condition, the load factor is zero, and Equation (11.33) yields

Jsc = �uB = 12000 A/m2

Then

uB = Jsc/� = 12000/20 = 600 V/m.

The power output is then given by Equation (11.34):

Power = �u2B2K(1 � K)V = 20(600)2(0.6)(0.4)(10) = 17,280,000 W = 17.28 MW

The channel current density is then given by

J = �uB(1 � K) = 20(600)(0.4) = 4800 A/m2.

If the magnetic field strength is doubled, Equation (11.34) shows that the poweroutput is increased by a factor of four, assuming there is no change in the flow orload conditions.____________________________________________________________________

485

Other MHD Configurations

The preceding discussion of the segmented electrode MHD channel provides abrief insight into the fundamentals of MHD power. Other electrode configurationsinclude the Faraday continuous electrode generator and the so-called Hall generator.In the former case, with continuous electrodes, the electric field in the y-directionleads to a streamwise electrical current in a circuit that is completed in the continuousconducting electrodes. The resulting current along the channel axis, called the Hallcurrent, contributes nothing to external power delivered in the Faraday generator. Itis shown in references 1 and 44 that the power output for the continuous electrodegenerator is influenced by the Hall parameter and is given by Equation (11.34) withthe right-hand side divided by the factor 1 + (�B)2. Thus high values of the Hallparameter drastically reduce the channel power output relative to the segmentedelectrode generator.

Rather than attempting to circumvent the Hall current, the Hall generator seeks toemploy it by using the electrode configuration shown in Figure 11.22. In the Hallconfiguration, the electrodes are designed to short circuit the cross-channel electriccurrent and pass the axial Hall current through the load. The Hall generator poweroutput also depends on the Hall parameter, but in a more complex way than in thecontinuous electrode Faraday configuration. The reader is referred to references 1and 44 for further detail.

MHD channels, like gas turbines, may be operated on open or closed cycles.Normally, combustion-driven MHD systems use an open-cycle, such as that shown inFigure 11.23, where pressurized combustion takes place upstream of the MHDchannel. As with the gas turbine, the temperature of the channel exhaust is high,making it advantageous to employ regenerative heat exchange and/or a bottomingcycle such as the steam turbine in Figure 11.23. For MHD channels with steambottoming cycles, overall efficiencies in the range of 50�60% are predicted.

486

If a DC source, rather that a load, is applied to the electrodes of a properlycontoured Faraday channel, the energy in the flow may be increased and the flowaccelerated. The electric field in this case exceeds the uB term in Equation (11.32),reversing the direction of the current density and thus the direction of the J × B force.The MHD channel may then be used for propulsion through expulsion of mass in thesame way as in a jet engine or rocket.

487

11.5 Solar Energy

The preceding pages have dealt with the conversion of energy from fossil and nuclearfuels, all finite resources. While the need may not be immediate, the day isapproaching when the Earth�s fossil and nuclear resources will no longer satisfyman�s need to grow and prosper. We must therefore consider more seriously theeffective utilization of the readily available and, for practical purposes, eternal energysource provided by the sun.

Ancient peoples learned to use the sun for survival and comfort. Not very longago a clothesline and wind formed the mechanism of a solar clothes dryer. Poweredby the sun, the atmospheric engine that produces weather change provides the windto fill sails and turn windmills to produce small amounts of power at a few selectlocations. Passive solar architecture has recently attracted widespread interest, andsolar collectors mounted on roofs for water heating and space heating have becomecommonplace in many areas. Figure 11.24 shows fixed and tracking solar arrays thatproduce electricity to supplement or replace grid power at a remote location.

A major problem connected with using the sun for power generation is the same characteristic that allows the survival of human life on earth: the low intensity levelsof solar radiation. It is well known that the two planets closest to the sun haveaverage temperatures higher than the Earth�s and that the outer planets are colderbecause of the inverse square law of radiative transfer of the sun�s radiation field.Higher solar fluxes at Earth�s surface would clearly make the sun a more easilyengineered radiation source. Let us estimate the intensity of the solar radiationarriving at Earth.

488

Consider the geometry of the Earth�sun system and the spherically symmetricnature of the sun's radiation field presented in Figure 11.25. Assuming that the sunradiates energy uniformly in all directions at a rate of E kW, the radiant flux density� crossing any sphere concentric with the sun is given by dividing E by the area ofthe sphere. Thus the radiant flux density, or irradiance, at a distance r from the suncenter is

� = E/(4� r 2) [kW/m2] (11.41)

In order to evaluate �, it is necessary to calculate the rate of energy emitted by thesun per unit area of its surface, �s. While the temperature at the center of the sun ismuch hotter, solar radiation approximates blackbody radiation at a temperature of5762 K. Using the Stefan-Boltzmann constant, � = 5.66961×10�8 W/m2-°K4, theblackbody radiation law (discussed later in more detail) gives the rate of energyemission per unit area of the sun�s surface as

�s = �T4 = 5.66961 × 10-8 (5762)4 (10�3)= 62,495 kW/m2 Applying conservation of solar energy to concentric spheres makes it clear thatenergy emitted uniformly from a sphere at the sun�s radius rs will be distributed over awider area when passing through a sphere of larger radius. Thus, eliminating E fortwo concentric spheres from Equation (11.41) shows that radiant flux density scalesinversely as the area ratio or square of the radius ratio of the spheres:

� = �s( rs /r ) 2 [kW/m2] (11.42)

This and Equation (11.41) are both mathematical expressions of the inverse squarelaw for solar radiation.

489

As in Figure 11.25, taking the sun�s diameter to be 865,400 miles and theEarth�sun distance as 93,000,000 miles, the solar radiant flux density, or solarconstant of radiation, reaching Earth�s orbit is

�e = �s( rs /re ) 2 = 62,495[(865,400/2)/93,000,000]2 = 1.353 kW/m2

Thus the value of the solar constant is the rate at which solar radiation crosses a unitarea normal to the sun�s rays at the distance of Earth from the sun. A more precisevalue for the solar constant is �e = 1.357 kW/m2, or 430.2 Btu/hr-ft2 (ref. 39).

While solar radiation may be viewed as a divergent spherical radiation field on thescale of the solar system, a calculation of the angle of divergence at the earth'sdistance indicates that its rays are essentially parallel on the scale of almost all humanactivities. Thus it is common to assume a constant radiation flux density with parallelrays in analyzing terrestrial solar radiation problems.

Some of the solar radiation arriving at the earth�s surface is scattered andreflected in the atmosphere. Thus solar radiation consists of direct (parallel) anddiffuse components. Moreover, absorption, reflection, and scattering in theatmosphere reduces the maximum direct radiation flux density arriving at sea level toabout �dir = 1 kW/m2. When the sky is clear, the direct component dominates. Butthe direct component may essentially vanish on overcast days, leaving only a diffusecomponent. The calculation of the solar performance of space vehicles is usuallysimplified by the absence of a diffuse component. However, for near-Earth satellites,radiation emitted from Earth and solar radiation reflected from Earth could besignificant.

Matter interacts with solar radiation in three basic ways: it can absorb theradiation, transmit it, or reflect it. These actions are represented by characteristicscalled, respectively, absorptance A, transmittance T, and reflectance R. Each isexpressed as a fraction of the total incident radiation. Thus the radiation energyreflected from a surface with reflectance R and a given area S is RS�.

Consider a layer of material with surface area S normal to a radiation field withradiant flux �. Conservation of energy requires that the rate of energy incident on thesurface equal the sum of the rates of energy reflected, transmitted, and absorbed.Thus �S = RS� + TS� + AS�. Thus the sum of the reflectance, the transmittance,and the absorptance must equal 1:

R + T + A = 1 [dl] (11.43)

Let us apply this statement of energy conservation to the Earth�s atmosphere: On aclear day, the atmospheric reflectance is small, hence any radiation not absorbed istransmitted to the surface (T + A � 1). On a cloudy day, reflection from andabsorption by clouds are significant, and the transmitted radiation becomes a smallfraction of the incoming solar irradiance (R + A � 1); only the diffuse componentscattered by the clouds remains.

490

Spectral Characteristics

Though it is usually convenient to deal with total radiation quantities, sometimes it isnecessary to consider wavelength-dependent, or spectral, effects. The materialradiative properties just introduced above are sometimes frequency or wavelengthdependent. Thus, while solar radiation in space approximates a blackbody spectraldistribution, radiation penetrating to Earth�s surface deviates significantly from the blackbody distribution because of scattering and strong absorption bands due toatmospheric water vapor, carbon dioxide, dust, and other substances that removesignificant amounts of energy from the incident radiation at certain wave lengths.

The eye itself is a highly wavelength-selective radiation sensor, being limited toobserving radiation only in the range of about 0.4 to 0.7 microns. A micron is 10� 6

meters.Consider the Planck equation for the blackbody spectral distribution of radiation

Fb(�), where wavelength � is in microns [� ] (refs. 37, 43 and 55):

Fb(�) = 2�hc2�--5/(ehc/(k�T) � 1) [Btu/(ft2-hr-�) | W/(m2-�)]

The Planck equation may be written in terms of a single parameter �T, wherewavelength is in microns and temperature is in appropriate absolute units:

Fb(�T)/T5 = 2�hc2 (�T) --5/(ehc/(k�T) � 1)

= 3.742×108(�T) �5/(e14381/(�T) � 1) [W/(m2-�-°K5)] (11.44a)

= 1.187x108( �T)--5/(e25896/(�T) � 1) [Btu/(ft2-hr-�-°R5)] (11.44b)

Figure 11.26 shows the Planck blackbody radiation distributions for two different temperatures as a function of wave length, one approximating solar radiation at10,000 °R (5555.5°K) and the other much closer to, but above, the normal terrestrialtemperature of 1000°R (555.5°K). At first glance the radiation levels appearcomparable; but they actually they are vastly different from each other (by a factor of105), because the Planck function is divided by �T5, to allow display on the samescale.

The figure makes clear that in the spectral range to which the eye is sensitive, thevisible range, little radiation emitted by relatively cool terrestrial objects is visible; andalmost all of what we see is reflected solar radiation or other high-temperature-sourceradiation (such as from high-temperature lamp filaments or from flames). The vastdifference in the wave lengths of the peaks shows why it is important to considerspectral effects in dealing with solar radiation.

491

The Wien displacement law, a simple relation derivable from the Planck equation,given in both English and SI units,

�maxT = 5215.6 [�-°R] (11.45a)

= 2897.6 [�-°K] (11.45b)

shows that the wavelength of the peak of the blackbody radiation distribution, �max ,is inversely proportional to the temperature, as may be verified numerically for thetwo peaks in Figure 11.26 ( �max = 5.215� and 0.5215�).

Integration of Fb over all wavelengths from zero to infinity, using Equation(11.44) yields the Stefan-Boltzmann law for the blackbody flux �b:

�

�

�b = � Fb d� = T4� [Fb(�T)/T5]d(�T) = � T 4 [Btu/hr-ft2|W/m2] (11.46) 0 0

which was used earlier to estimate the radiative flux from the surface of the sun andthe terrestrial solar constant.

It is useful to sum the contributions to blackbody emission from zero to a givenwavelength by an integration of Fb, as was done for the entire spectral range inEquation (11.46). Letting � = �T, we define a function �(�T) as a dimensionlessfraction of the blackbody irradiance from � = 0 to � = �T:

492

�T

�(�T) = T4/�b � [Fb(�)/T5]d� [dl] 0

Figure 11.27 shows a log-log graph of the Planck distribution and �(�T). A table ofthe latter function, calculated by numerical integration using a spreadsheet, isprovided in Appendix I. The dimensional blackbody flux between any two wavelengths for a given temperature may be determined from a difference of sigmafunctions as:

�(�T)2 � �(�T)1 =

(�T)2

= T 4�[Fb(�)/T5]d� = [�(�T)2 - �(�T)1]�T4 [Btu/hr-ft2|W/m2] (�T)1

EXAMPLE 11.7

Consider a window, normal to the sun�s rays, covering a solar collector mounted on asatellite in near-Earth orbit. The window has a transmittance of 0.9 in the range ofwavelengths from 0.2�2.0 � and 0 at other wavelengths. What is the fraction ofincident solar radiation transmitted by the window, and what is the rate of usefulenergy transfer to a circulating fluid in a 4 × 4-m flat plate collector if the collectorabsorbs 85% of the transmitted radiation?

493

SolutionThe only radiation transmitted is between 0.2 and 2.0 �. Because the effective

temperature of the solar radiation is 10,000° R, the values of �T and � for these twowavelengths are

�T = (0.2)(10000) = 2000 �-°R and �(2000) = 0.0013and

�T = (2.0)(10000) = 20,000 �-°R and �(20,000) = 0.934

using the � table in Appendix I.The transmitted fraction of the blackbody incident radiant energy is the product of

the fraction of radiation in the transmitted spectral range and the windowtransmittance:

(0.9)(0.934 � 0.0013) = 0.839

The radiation that heats the circulating fluid is then the product of the transmittedspectral fraction, the absorptance, the area, and the near-earth solar constant:

(0.839)(0.85)(42)(1.357) = 15.5 kW.____________________________________________________________________

Example 11.7 dealt with a constant transmittance over a single spectral interval.More complex radiation characteristics may be treated in a similar way by summingthe contributions of more than one spectral band.

Earth-Sun Geometry and Solar Collectors

Radiative transfer from the sun to a plane surface depends on the orientation of thesurface to the sun�s rays. If parallel rays irradiate a surface, the total energy rate perunit area is given by the product of the irradiation � and the cosine of the anglebetween the surface normal and the sun�s rays: �cos�. The following discusses waysin which the cosine function can be evaluated.

Consider the component of radiation falling on a terrestrial horizontal surfacewith unit normal n, when the sun�s rays are in the direction of a second unit vector s.Following the notation of Figure 11.28, the two unit vectors n and s can be expressedin terms of the orthogonal unit vectors i, j, and k:

n = nxi + nyj + nzk and s = sxi + syj + szk

where nx, ny, nz, and sx, sy and sz are the direction cosines of the two vectors. Notethat the direction cosines must satisfy the following conditions:

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n � n = nx2 + ny

2 + nz2 = 1 and s � s = sx

2 + sy2 + sz

2 = 1

for n and s each to be of unit magnitude. The fraction of the maximum solar radiationfalling on the unit area, then, is given by the scalar product:

cos � = n � s = nxsx + nysy + nzsz (11.47a)

Figure 11.29 shows these unit vectors set at the center of the earth. The unitvector n points to the zenith of the observer and is normal to the horizontal surface atP. Here the x-axis is taken through the intersection of the meridional plane containingn and the equatorial plane. The z-axis points north and the y axis is selected tocomplete a right-handed coordinate system. The angle L is the latitude of n. In thissystem the position of the sun is determined by two convenient angles: H and .Examination of Figure 11.29 shows that the components of s on the coordinate axesare

sx = cos cos H sy = cos sin H and sz = sin

Similarly, nx = cos L, ny = 0, and nz = sin L. Thus the cosine of the sun-surface anglegiven by n � s using Equation (11.47a) is

cos � = cos cos H cos L + sin sin L (11.47)

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Thus the cosine of the angle of the sun�s rays falling on a horizontal surface dependson the latitude of the surface and the angles H and of the sun.

Now consider the determination of H and . First we define the ecliptic as theplane of earth�s annual motion around the sun, as shown in Figure 11.30. Because theearth�s axis always points in approximately the same direction in space and is inclinedat 23.5° to the normal to the ecliptic, the apparent position of the sun as viewed fromEarth ranges above and below the equator by 23.5° annually. This movement isdesignated by the solar declination angle in Figure 11.29, measured from theequatorial plane along a meridian of longitude. It can be seen from Figure 11.30 that varies from + 23.5° degrees on June 21 to � 23.5° on December 21.

The hour angle H of the sun in Figure 11.29 is the angle measured in the plane ofthe equator between the observer�s meridional plane and the meridional plane of thesun. Earth rotates once or 360° about its axis in 24 hours; hence, the hour angledepends on time, as its name implies. If solar time is measured from solar noon(when the sun is at its highest point in the sky at the observer�s meridian), the hourangle may be computed from the solar time using the factor 360/24 = 15° per hour.The hour angle is positive when the sun is east of the observer and negative when it iswest. Thus at 3:00 P.M. solar time, the sun is 3 hours past solar noon and, therefore,has an hour angle of 3×(�15) = � 45° (west of the observer).

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Care should be taken to distinguish between solar time and time defined by lawcalled civil time. In the continental United States, for instance, there are four timezones defined by law in which standard time decreases by one hour for each 15° oflongitude from the east to the west coast. This allows a uniform time within a zonewhile approximating solar time. Exceptions have been made to accomodate irregularstate boundaries. Thus standard time deviates from solar time by several minutes,depending on the longitude of the observer. Moreover, by law, daylight saving timediffers from standard time. In the spring and fall, clocks are advanced and set back anhour, respectively. Under daylight saving time, the civil time in the summer would be1 pm at any location where it is noon standard time. Thus solar time can deviate fromdaylight saving time by more than an hour at some locations.

EXAMPLE 11.8

What is the incident radiative energy rate falling on a 7 × 10-m horizontal roof on aclear day at 40° north latitude, at a solar time of 10 A.M. on December 21?

SolutionFigure 11.30 shows that on December 21 the solar declination angle = � 23.5°.

At 10 A.M. sun time, the hour angle is 2 hr × 15°/hr = 30°. By Equation (11.47), the angle of the sun relative to the roof vertical is

cos � = cos cos H cos L + sin sin L

= cos(�23.5°) cos(30°) cos(40°) + sin(�23.5°) sin(40°) = 0.352.

Taking the solar constant at Earth�s surface as 1.0 kW / m2, the incident energy onthe roof is

(0.352)(70)(1.0) = 24.64 kW

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Note that the solar angle � is independent of the sign of the hour angle. Thuscos� is a symmetric function of H, and the same irradiance occurs at 10 A.M. and2 P.M. solar time.____________________________________________________________________

EXAMPLE 11.9

What is the irradiance on an east-facing vertical wall in kW / m2, under the conditionsof Example 11.8?

SolutionA unit vector on an east-facing wall at the observer location in Figure 11.29 is

directed along the y-axis and is therefore simply j. The cosine of the angle betweenthe east-facing-wall normal and the solar direction vector s is

j � s = sy = cos sin H = cos(�23.5°) sin(30°) = 0.4585

The solar irradiance is then (1.0)(0.4585) = 0.4585 kW/m2. Here, the hour angledependency is antisymmetric. It would give a negative result for 2 P.M. solar time (H = � 30°). Thus, after solar noon, when j � s < 0, the negative sign indicates thatthe sun is behind the east-facing wall and that the wall of the building is in the shade.____________________________________________________________________

Solar collectors can be fixed in position or they can track the sun. Fixed collectorsfor year-round use in the northern hemisphere are normally oriented facing the southwith a tilt angle ß that is approximately equal to the collector latitude. For a south-facing collector tilted at an angle ß with respect to the horizontal, the collectornormal makes an angle of L � ß with the equatorial plane in Figure 11.29. The angleof the collector normal, n, with the sun direction may then be determined using

n = i cos (L � ß) + k sin (L � ß)

Then cos � is

cos � = n � s = cos(L � ß) cos cosH + sin(L � ß)sin

Note that this equation may also be obtained by replacing the latitude in Equation(11.47) with L � ß.

For sun-tracking solar collectors, two degrees of angular freedom are required forperfect tracking. Typically, such collectors would pivot about horizontal and verticalaxes dictated by the local solar azimuth and elevation. Their angular motion may bepreprogrammed using astronomical data such as those from reference 56, or, it couldbe controlled by a sun-seeking control system.

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Solar Thermal Electric Generating Stations

Large thermal applications of solar energy to produce electrical power are located attwo sites in the Mohave desert of California (refs. 65 and 66). These facilities,(Figure 11.31) called solar electric generating stations�-SEGS, consist of 14-MW,30-MW, and 80-MW units employing line-focusing parallel-trough solar collectors toprovide heat for reheat steam turbine systems. The SEGS plants in the Mojave Desertmake up the world's largest parabolic trough facility. In the year 2000, there werenine plants, which provide a combined capacity of 354 MW (ref. 89). Deployment ofadditional plants is not expected to occur until at least 2002. When completed, the 12SEGS units will have a total electrical generating capacity of about 600 megawatts.

Reflected solar radiation from the mirrored-glass sun-tracking horizontal-axisparabolic collectors is focused on an evacuated tubular-heat-collection elementthrough which a heat transfer fluid flows. After leaving the collector, the hot fluidheats water in a steam generator before returning to loop through the collectors, asseen in Figure 11.32. The resulting super heated steam is used in Rankine-cyclereheat steam turbine generators. Cooling towers reject heat from the condensers tothe surrounding desert air.

Supplemental heat is provided, when required, by boilers burning natural gas, as shown in Figure 11.32, or, in the newer designs, by gas-fired heaters that heat thecollector heat transfer fluid. This supplemental heat is required to ensure that fullSEGS plant design output is obtained during the periods of peak demand forelectricity in Southern California�from noon to 6 pm between June and September.

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The large land area required by SEGS plants is evident from Figure 11.31. References 65 and 66 give insight into some of the technical issues faced in thedevelopment of SEGS systems; issues that include periodic mirror washing, breakageof the heat collection elements, wind loading, and other operational considerations. Updated information on the SEGS system and other solar-thermal programs is givenin references 89 to 92.

Solar Photovoltaics

Photovoltaic cells have gained wide use in recent years as small scale power sourcesfor watches, calculators and other such devices. Less evident to the public is thewidespread research and development effort on electric power production using solarphotovoltaics. Over two billion dollars in corporate funds have been invested inphotovoltaics (ref. 58). The major goal is to provide cheap and efficient directconversion of solar photons into electricity, using light-sensitive devices with few or

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no moving parts. Success could provide a sorely-needed long-term energy source thatcould benefit most of the peoples of the world.

Photovoltaics rely on specially prepared materials called semiconductors, whichproduce useful current flows when illuminated with solar or artificial radiation. Theyare typically thin layers of materials such as crystalline silicon, with electrical leadsmounted on opposite faces, as shown in Figure 11.33. The silicon is usually mountedon a substrate that provides both structural support and cooling to conduct away heatcreated by nonproductive radiation absorption. The exposed silicon is usually coveredwith a transparent, antireflecting protective coating to reduce energy loss via surfacereflection.

Crystalline semiconductors are near perfect geometric lattices of atoms that wereusually produced by a crystal growth method known as the Czochralski process (CZ),which involves slowly turning a seed crystal of pure silicon as it is withdrawn from abath of high-purity molten silicon. The CZ process forms an ingot that may be 15 cmin diameter and over one meter long. Thin wafers of silicon are then sliced from theingot in a slow, delicate process. Other methods, such as casting of square ingots, arein use also. The research and development of more cost-efficient techniques for thelarge-scale production of semiconductors has a continuing high priority. Theextensive use of semiconductors in the computer and electronics industries providessubstantial motivation for continued research in semiconductor production methods.

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After slicing, silicon crystals are "doped" with occasional special impurity atoms,perhaps a few foreign atoms per million silicon atoms in the lattice. This is typicallyachieved by heating the wafers and exposing their surfaces to vapors of the dopingmaterial, called a dopant. This process allows the dopant to diffuse to a controlleddepth in the crystals. Semiconducting crystals are electrical insulators at lowtemperature and weak conductors of electricity at room temperature. The presence ofimpurities provides charge carriers that substantially enhance the material electrical conductivity when it is exposed to radiation. The following considers how photonabsorption in the crystals produces an electrical potential and direct current.

Electrons in semiconductors may exist in a band of bound energy states called avalence band. This is analogous to the bound energy levels of individual atoms. Aswith individual atoms, the absorption of a photon in a semiconductor may change theenergy state of an electron. If the photon energy equals or exceeds a threshold level,the electron may be excited into a conduction band, where it is free to movethroughout the material, obviously enhancing the electrical conductivity of thematerial.

Consider silicon again as an example of semiconductor behavior. Silicon atomshave four electrons in their outer shell. If impurity atoms in a silicon crystalline latticehave five electrons in their outer shell (atoms such as arsenic or phosphorus) thesilicon crystal has an overabundance of electrons (and positive charge) and is there-fore known as an n-type (negative) semiconductor. The extra electrons associatedwith impurity atoms are loosely bound just below the conduction band. Smallamounts of energy from the thermal motions in the lattice or from absorption ofphotons allow these electrons from the impurity atoms (called donor atoms) to movefreely in the crystal. The result is a neutral material containing mobile negative-chargecarriers.

Now consider another type of crystalline silicon semiconductor, the p-type, whichhas positive charge carriers. If, in the silicon crystal,, the impurity atoms have threeelectrons instead of five in their outer shells (as in boron or gallium), the impurityatoms can easily snatch electrons from nearby silicon atoms leaving positively chargedsilicon sites called holes and fixed negative dopant atoms. These impurity sites arecalled acceptors because they draw electrons from silicon atoms, thus creatingpositive holes in the lattice. This type of semiconductor is called a p-type, because theholes are positive charge carriers. When an electron abandons one silicon site foranother, a hole disappears at its destination and new hole is created at its point oforigin. Thus a p-type semiconductor has holes that may move through the material aspositive-charge carriers.

When an external electric field is applied to an n-type semiconductor, electronswill move toward the region of high potential. However, the electrons are restrainedby the positive fixed donor sites, resulting in a charge distribution in balance with theapplied field. In a p-type semiconductor, its positive holes would move toward lowpotential until a stable charge distribution with the negative acceptor sites isestablished. Thus, because of the fixed charges at donor and acceptor sites, the

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charge carriers will move just far enough to set up charge distributions that are inbalance with the external field.

Consider layers of p-type and n-type semiconductors brought together to form anelectrically neutral slab with a common interface called a p-n junction. Because ofconcentration gradients of free electrons and holes across the junction, someelectrons from the n-type will cross into the p-type, while holes move a short distanceinto the n-type. The diffusion of holes and electrons rapidly creates and is balanced byan electric field in a thin space-charge region less than a micron thick, as indicated inFigure 11.33. As a result, the p-type material is at a higher potential than the n-type.Thus the two materials on either side of the junction are each at different potentialswithout the benefit of an external electric field, thanks to the electric field at the p-njunction.

Now let�s consider radiation incident on the p-n semiconductor. Photons have anenergy h, where h is Planck�s constant and is the frequency of the radiation.When a p-type material near a p-n junction is irradiated by an energetic photon, thephoton may be absorbed, raising an electron into the conduction band and leaving ahole behind. Thus a single photon creates an electron-hole pair. Conduction bandelectrons readily move across the junction into the n-type material to a region oflower potential energy, while the holes tend to move in the opposite direction andthus remain in the p-material. Likewise, electron-hole pairs created by irradiation ofn-type material cause holes to move into the p-material, leaving the electrons behind.Thus the photons provide the energy for a pumping process that separates andselectively drives charges across the interface. These actions increase the positivecharge in the p-type and the negative charge in the n-type materials, opposing theelectric field produced by diffusion due to charge concentration gradients. Thisprocess continues until the electric fields balance each other.

When an external load is connected to the irradiated cell, electrons flow from then-type material through the external circuit while holes move in the oppositedirection. The net current flow through the load is the sum of both the electron flowand the hole flow through the semiconductor layers. Since the number of chargecarriers driven through the load depends on the number of photon excitations, thetotal cell current and power output depend on the junction surface area and on theintensity of solar irradiance.

A silicon solar cell may be 10 cm by 10 cm, and have an open circuit voltage ofabout 0.5 V. At 10% efficiency it would deliver a power output of 1 W whenreceiving peak solar irradiance on Earth of 1.0 kW/m2.

Power = (1.0)(10/100)2(0.1)(1000) = 1.0 W

The band gap, or energy separation, between conduction and valence bands is 1.1 eVfor silicon p-n junctions. This is the energy necessary to move an electron into theconduction band of the semiconductor.

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Thus, to excite electrons into the conduction band, photons must have afrequency equal to or greater than E/h, the photon energy (equal to band gap energy)divided by Planck�s constant

= (1.1 eV)(1.602×10�19 J/eV)/(6.626×10�34 J-s) = 2.66×1014 s �1

and a wave length no greater than c/, the speed of light divided by the photonfrequency:

� = (2.9979×108 m/s)(10 6 �/m)/(2.66×1014 s�1) = 1.127 �

Photons with lower frequencies and longer wave lengths, and therefore energiesless than the band-gap energy, cannot raise electrons to the conduction band and maybe transmitted through the material without effect. On the other hand, one photon canexcite only one electron into the conduction band, and any excess photon energybeyond the band gap value only heats the semiconductor. Thus photon wave lengthsgreater and less than the bandgap value result in inefficiency in converting incidentradiation to electricity.

The p-n junction acts like a combination of a current source and a diode inparallel as shown in Figure 11.34. For a given level of cell irradiation, a cell electricalcharacteristic may be represented by

I = Isc � Io(e qV/kT � 1) [A] (11.48)

where k is the Boltzmann constant, q is the electronic charge, and Io is called the darkcurrent. The short-circuit current Isc (corresponding to a cell potential difference =0)is proportional to the rate of incident irradiance. The dark current Io is related to thecell open-circuit potential difference Voc by setting I = 0 in Equation (11.48). Thus:

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Thus

Io /Isc = 1 /(e qVoc/kT � 1).

By combining these two equations, we can write the cell characteristic innondimensional form as

I /Isc = 1 � (Io /Isc)(e qV/kT � 1) = 1 � (e qV /kT � 1)/(e qVoc /kT � 1) [dl] (11.49)

The cell power output is the product of the current and cell potential difference:

P = IV = [Isc � Io (eqV/kT � 1)]V [W]

Nondimensionalizing the power by dividing it by IscVoc gives

P/IscVoc = [1 � (e qV /kT � 1)/(e qVoc /kT � 1)](V/Voc) [dl] (11.50)

The nondimensional current and power characteristics are shown in Figure (11.35). Itmay be seen from the figure that peak cell power increases linearly with outputvoltage for small values of voltage, reaches a maximum at between 80% and 90% ofthe open circuit voltage, and drops rapidly thereafter. The maximum power is about80% of IscVoc. For given values of Isc and Voc, the maximum power occurs at particularvalues of current and voltage (indicated by the circle on the current characteristic),and thus at a single load resistance. For other resistive loads the power output isreduced.

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EXAMPLE 11.10

For a photocell with an open-circuit voltage of 0.6 V at 350K, evaluate qVoc/kT andthe nondimensional current and power at V/ Voc = 0.8. Propose a simplification of themodel given earlier for large values of the voltage parameter.

SolutionFor T = 350K and Voc = 0.6 V, the dimensionless voltage parameter is

qVoc/kT = 1.602×10�19×0.6 / (1.38×10�23×350) = 19.9

Using Equation (11.49), the current ratio is then

I /Isc = 1 � (e 0.8×19.9 � 1)/(e 19.9 � 1) = 0.9813

and the power ratio is

P/IscVoc = (I/Isc)(V/Voc) = 0.9813×0.8 = 0.785

It is clear that for large values of the voltage parameter, 1 may be neglected withrespect to the exponentials; hence we may write the following excellentapproximations:

I /Isc = 1 � e (qVoc/kT)(V/Voc � 1)

and

P/IscVoc = [1 � e (qVoc/kT)(V/Voc � 1)](V/Voc)

The reader may verify numerically the accuracy of these approximations.____________________________________________________________________

The cell efficiency is the ratio of the maximum cell power output to the rate ofincident radiant energy normal to the cell, Pm /Einc, where Pm is the product of the cellcurrent and the voltage at maximum power point. For a given irradiance, doublingthe cell efficiency implies doubling the cell current and power output. Let us considera few of the types of energy losses that contribute to low cell efficiency in convertingenergy from the radiation field into electricity.

It has been seen that in silicon, the minimum photon energy required to create anelectron-hole pair is 1.1 ev and that (1) photons with energies less than 1.1 ev are notabsorbed by the semiconductor; and (2) the excess energy of photons with greaterthan the required energy is converted to heat, since a photon can create only oneelectron-hole pair. For silicon cells, these losses exceed 50% of the incident energy(ref. 11.59). Thus the nature of the solar spectrum and the excitation energy of

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particular junction materials influences the cells� efficiency. Analyses of these lossesindicate that the silicon band-gap energy is lower than optimum for the solar specraldistribution. References 39 and 42 indicate that cadmium telluride, CdTe, and galliumarsenide, GaAs, have near-optimum band-gap energies around 1.4 eV for the solarapplication.

Another loss that contributes to inefficiency is reflection at the cell surface.Antireflection coatings are currently used to reduce this loss to a few percent of theincident energy. When these and other losses are accounted for, a maximumconversion efficiency of single-crystal silicon cells in sunlight is about 25%. Values of23.2% and 22.3% have been obtained in two laboratories when high efficiency wasthe objective (ref. 69). The best efficiency of commercial silicon cells is about 15%.

Solar Cell Design

Other semiconducting materials that have received less attention than silicon havebeen found to use the solar spectrum more efficiently, as indicated above, and thusoffer potential for higher conversion efficiency. Research is also in progress on cellsmade of thin layers of several junction materials that each work best in different partsof the spectrum. Thus, if a top layer converts high-frequency photons efficiently andallows lower frequency photons to pass to a lower layer to be converted by asemiconductor with a lower band-gap energy, a larger part of the spectrum may beused efficiently. For instance, as Figure 11.36 shows for a two junction cell,

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ultraviolet and visible wavelength photons may be absorbed near the top junction andinfrared photons near the bottom.

Regardless of the individual cell characteristics, in order to achieve power outputsof significance for utility and industrial applications it is necessary to string the cellstogether in arrays. Identical cells connected in series all have the same current; andtheir combined voltage is the sum of the individual cell voltages at the given currentflow, as shown in Figure 11.37. For example, using a typical value for the open-circuit voltage and the voltage condition for maximum cell power, in order to get avoltage of about 14 V to charge a 12-V battery it is necessary to bring together about

14 V/[(0.5Voc/cell) (.8V/Voc)] = 35 cells

in series. For 10-cm by 10-cm cells, such a string would have a maximum poweroutput of about

(35 cells)(1 watt /cell) = 35 W

and an area of 35×10×10 = 3500 cm2 or 0.35 m2.In order to achieve high current flow, it is necessary to connect cells or strings of

cells in parallel. In this case the currents are additive for a given cell potential, as seenin Figure 11.38. While high voltages are desirable to reduce ohmic losses and reducewire size, according to reference 38 limitations on insulations and safety concerns willlimit the voltage levels achieved by solar arrays. Since the National Electric Codedoes not provide for equipment ratings above 250 V DC, higher voltages may implyincreased costs for solar power systems.

In the event of cell damage, loss of one cell in a parallel string of cells means theloss of only the power of that cell, whereas loss of a single cell in a series string

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causes a large loss in power because of the large increase in series resistance of thestring. Shading from the sun of a single cell in series has a similar effect, because thereduction in incident radiation cuts the current carrying capability of the cell andhence the entire string. Circuitry is usually provided to allow current to bypass adefective cell.

EXAMPLE 11.11

A near-Earth satellite requires a maximum of 10 kW of electrical power. Estimate therequired collector area and the number of 10-cm by 10-cm cells, each having aconversion efficiency of 14%.

SolutionUsing a solar constant of 1.357 kW/m2, the required area is determined by

dividing the design power requirement by the amount of the incident radiation that isconverted to electricity by the cells:

10/[(0.14)(1.357)] = 52.64 m2

The number of 10 by 10 cells is then

52.64/(10/100)2 = 5264 cells____________________________________________________________________

The direct conversion of solar radiation to electricity is very attractive from thepoint of view of the long-term reliability of the source and environmental

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acceptability. There are, however, several fairly obvious adverse characteristics ofsolar conversion. First, terrestrial applications suffer from the limited daily availabilityof sunshine. The frequent coincidence of the daily peak irradiance and utility demandpeaks is advantageous but does not compensate for the wide daily variability of thesolar source. The occurrence of cloudy days, sometimes several successive ones, andthe inevitable daily sunset indicate the need for energy storage if solar energy is toprovide a reliable primary source of electricity. For remote, stand-alone locations thisimplies the use of batteries or other storage devices to satisfy 24 hr/day demand.Figure 11.39 shows a remote photovoltaic-powered telecommunication site locatedon a mountain top. The system comprises the cell array, the storage batteries, thetelecommunications load, and the electronics needed to control battery charging andload matching. The system shown is said to have recovered its cost in a year and toproduce annual savings of $60,000.

In industrialized countries, such as the United States, where utility grids penetrateto most parts of the country, linkages with utilities seem to offer the best hope forwidespread photovoltaic use, because the existing utility grids can provide the neededbackup. Both decentralized and central-station solar conversion appear to bepossibilities for utility involvement. Moreover, the Public Utilities Regulatory PolicyAct (PURPA) offers other possibilities by encouraging third-party as well as utilityand customer ownership arrangements.

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The high cost and low efficiency of solar cells are major deterrents to theirwidespread application for power generation. Although the costs of solar cells havebeen declining and cell efficiencies increasing over the years, solar cells still areeconomically suited for only a limited number of applications such as those wherecost is not a major factor (e.g., military and aerospace applications) or where thealternatives are expensive (e.g., energy sources for remote installations). Neverthelessthe annual production of photovoltaics has been increasing and the resulting product-ion experience will help develop production techniques that, together with increasedvolume, should further drive down their prices. It is likely that the demand for photo-voltaic power generation would increase dramatically if solar cells become availablein a price range below $1/W, depending on the price of natural gas, oil and coal.

Concentrating Solar Cell Systems

One approach to reducing the cost and increasing the efficiency of photovoltaicsis the development of a concentrating solar cell system that focuses radiation on thecell. The use of a Fresnel lens or other type of concentrator increases the incidentenergy on a given cell area and, consequently, the cell power output. A photograph ofsuch a concentrating array is shown in Figure 11.40. Each of the 14 elements in each

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parquet concentrates energy from a 5-inch-square Fresnel lens onto a 1/4-inch siliconcell. Unlike nonconcentrating cells, the concentrating units only function under clearsky conditions and derive no benefit from diffuse radiation.

Cells with high concentration ratios must be cooled because of high solarintensities on the cells. Some of the details of the EPRI-Stanford cell system adaptedfrom reference 63 are given in Figure 11.41. According to that reference andreference 49, efficiencies as high as 28% have been reached in experimentalcrystalline silicon concentrator cells. Reference 59 suggests that the required highefficiencies for commercialization of concentrating cells are being achieved today andthat attention can be focussed on production and cost reduction efforts for this type.

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Other Approaches

Major reasons for the high cost of solar arrays is the complexity and laborintensiveness of production of high-quality crystalline silicon. Noncrystalline silicon,or amorphous silicon, may also be used for photovoltaic cells, but with a practicalconversion efficiency limit of about 14%. The best commercially available amorphouscells have an efficiency of about 6%. Crystalline silicon currently has most of themarket but amorphous silicon and other non-crystalline semiconductors are receivingconsiderable research attention. Whereas crystalline cells are cut from carefully growncrystals, amorphous cells may be created by vacuum deposition of a thin film ofsilicon on a substrate such as glass. While not without problems, manufacturingtechniques for amorphous materials appear to offer greater potential for low-costmass production than those for crystalline cells. One of the current research problems,however, is that amorphous silicon loses some of its effectiveness rapidly uponexposure to light. This has not stopped its use in some applications but is a seriousconcern.

An exciting approach to photovoltaic power technology was announced in 1991by Texas Instruments and Southern California Edison. The concept, referred to asSpheral SolarTM* technology, involves the use of 17,000 tiny silicon balls per 100 cm2

supported on flexible aluminum foil sheets, as seen in Figure 11.42. The spheres aremetallurgical grade p-type silicon with surfaces doped with an n-type material so thateach sphere is an individual solar cell. The spheres are set in perforations in a foil

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layer that forms the contact with the n-layer. Both sides of the spheres are coatedwith transparent ethylene vinyl acetate. The rear tips of the spheres are etched,exposing p-type material where another foil layer forms the contact with the p-layer.

Several features claimed for the spheral technology offer hope for a breakthroughin the production of solar electricity:

� The silicon used is inexpensive, low purity, metallurgical grade.

� The technology uses low-cost production methods that provide an almost 100%yield of manufactured laminates.

� Since each sphere is a solar cell, the failure of an individual sphere has anegligible influence on system performance.

� The flexible nature of the laminate makes it adaptable to a wide range ofapplications (Figure 11.43).

Module efficiencies of about 8�10% are expected (ref. 76). Reference 80indicates that the highest efficiency attained during R&D testing was 11.5%. A linefor pilot production prototype modules was established in 1991, with productioncommercial planned for 1994. More recent information is available in reference 96.

The application and sales of photovoltaics have been growing continually in thelast 20 years. Though photovoltaics continue to offer great hope for large-scalepower production, there remains a diversity of opinion regarding the avenues andrates of commercialization of photovoltaics in power production applications (ref.49).

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11.6 A Hydrogen Economy

Coal, natural gas, petroleum, and uranium are all primary energy sources, mainlyfor heat. On the other hand, electricity is a convenient form of energy, or work,derived from a primary source and may therefore may be called a secondary energysource. Electricity plays an important role in the world energy system because it is aconvenient and inexpensive source of both heat and work. It is used as readily forlighting and cooling as for heating and mechanical power delivery. It is readilytransmitted over moderate distances and is relatively safe and non-polluting. Itspervasive influences on society within the world�s energy system are such that wehave what is sometimes referred to as �an electrical economy.�

However, electricity has failed to gain a foothold in most modes of transportation,primarily because of the lack of a compact and flexible means of storing it. Someprogress in this area may be anticipated because of renewed interest in electric cars,trucks and other road vehicles. A primary dissatisfaction with electricity as a means ofenergy application lies in the facts that an electrical economy requires theconsumption of about three units of primary energy for each unit of secondary energyproduced and that most of the primary energy conversion processes have adverseenvironmental consequences.

Much interest has been expressed in supplementing or replacing electricity withanother secondary energy source, with hydrogen being the leading contender.Methanol, ethanol, liquefied natural gas (LNG), and compressed natural gas (CNG)are other possibilities. A number of advantages may be attributed to employinghydrogen as a secondary energy source:

� Hydrogen may be produced by chemical processes from fossil fuels, byhydrolysis of water with oxygen as a by-product, or by thermal decomposition.According to reference 16, electrolysis cells produce hydrogen at efficienciesbetween 60 and 69%; 75 to 80% was feasible with 1973 technology, and 85% isexpected in a well-developed hydrogen economy.

� Hydrogen can be used as a fuel to produce power with almost no directenvironmental pollution, because the principal product is water vapor.Combustion of hydrogen produces no carbon products and therefore no�greenhouse gases.� Oxides of nitrogen would remain as pollutants but should bemore easily controllable because of the absence of carbon and sulfur-containingpollutants.

� Internal combustion engines can use hydrogen with minor modifications, some-times with improved efficiency.

� Fuel cells employing hydrogen can be used to produce electricity at highefficiencies not bounded by the Carnot limit and with little environmental impact.

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� Hydrogen could be produced by electrolysis at large base-loaded electricalpower plants during periods of low demand for electricity, thus providing anopportunity for electrical energy storage when used with fuel cells or otherefficient energy converters.

� Hydrogen could be transported in pipelines similar to natural gas pipelines. Insome cases, existing pipelines could be used with relatively little modification.Liquid hydrogen has been routinely shipped in special rail cars and trucks for manyyears.

� Hydrogen can be stored in the same ways that natural gas and helium are storednow, underground or in liquefied form in special storage tanks.

� Hydrogen has a very high heating value on a mass basis but a low volumetricvalue because of its high specific volume at standard pressure and temperature.

The numerous attractive features of hydrogen as a fuel are to a certain extent offsetby some serious concerns, which are the objects of extensive engineering research:

First, the ignition energy of hydrogen is low and the flammability limits are wide, sohydrogen is readily ignited. While this is advantageous in combustion systems, greatcare must be taken in storing and handling hydrogen. Though an atypical event, theHindenburg disaster perpetually reminds us of these concerns. On the other hand,because hydrogen is lighter than air and diffuses more rapidly than other gases, it ismore readily diluted and disperses quickly, characteristics that tend to reduce fire andexplosion hazards.

The hazards associated with using hydrogen gas should not be minimized, but toput things in perspective, we should remember that society has enthusiastically adoptedan automobile that commonly carries ten to twenty gallons of gasoline, a veryhazardous fuel. Large storage tanks of gasoline are maintained and freely accessed byuntrained users in the midst of residential areas. Trucks loaded with gasoline routinelyservice and pass through these same areas.

Second, hydrogen�s high specific volume makes its storage a problem, especially intransportation applications, where space for fuel storage is costly. In addition to storageas a compressed gas, it may be liquefied and stored at cryogenic temperatures, or itmay be stored in metal hydrides. In the latter cases, the additional equipment space,expense and complexity must be considered in evaluating feasibility and design.

Finally, while gasoline and natural gas remain readily available and at a modest price, hydrogen will have little opportunity to become established as an importantsecondary fuel. For ground transportation, as a replacement fuel for gasoline, hydrogenmust compete with relatively inexpensive methanol and ethanol (ref. 9). Heavilysubsidized ethanol is currently available in the United States in a 10% blend withgasoline and there is considerable interest in compressed natural gas and richermethanol mixtures for ground vehicles (ref. 74)

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EXAMPLE 11.12

Using reasonable estimates of unit efficiencies for generation, transport, and utilization, compare the overall energy efficiencies of nuclear-electric and nuclear-hydrogeneconomies in providing mechanical work.

SolutionFor the nuclear-electric economy, mechanical energy is produced in the sequence

Fission �Electricity �Transmission �Electric motor

which, using unit efficiencies in the order given, would have an efficiency of about

(0.32)(0.9)(0.9) = 0.259

For the nuclear-powered hydrogen economy the appropriate chain is

Fission �Electricity �Electrolysis �Hydrogen �Transmission �Fuel cell �Motor

Here the overall efficiency is estimated to be

(0.32)(0.85)(0.9)(0.6)(0.9) = 0.132

Note that neither case includes the energy costs of mining and processing nuclear fuel.Nevertheless, this example suggests that, despite hydrogen�s many advantages, its useas a secondary fuel may lead to inefficient resource utilization. This conclusion isdiscussed more fully in reference 16._____________________________________________________________________

One can readily appreciate the beauty of the concept of a solar-energy-drivenhydrogen-powered cycle in which hydrogen produced by electrolysis of seawater usingsolar photovoltaic energy, reacts and delivers power and heat in a high-efficiency and non-polluting fuel cell at the point of use, with product water ascending into theatmosphere, where it eventually returns to the sea as rain. The problems in developingand adopting such a cycle for commercial use are great, but the future needs of Earthare so serious that such concepts at least provide directions in which research shouldproceed. While the hydrogen economy may not displace the electric economy, when thepolitics and economics are right, society will find ways to use hydrogen as a supplementto electricity as a secondary fuel.

11.7 Concluding Remarks

It is clear that the seriousness of Earth's energy and environmental problems and thegrowing diversity of propulsion and power production options ensures stimulating

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careers for engineers far into the future. Attractive, mid-term alternatives exist to theconventional, state-of-the-art, coal-burning power plant with advanced scrubbertechnology. Despite the risks involved, it seems clear that numerous other poweroptions will find commercial use as dictated by local and regional conditions andavailable technologies and resources.

A recurring question in all of the developments we have discussed exists: Once thefeasibility of a new technology has been established, how can it be rapidly brought to acommercial stage in the absence of massive government support for development anddemonstration plants? Under existing economic and regulatory structures in the UnitedStates, power utilities cannot make massive investments in technologies that involvesignificant risks or costs that are not competitive with existing methods. The prices andrisks of new technologies will remain high, however, until a significant number of unitshave been ordered that can absorb research, development, and production costs. Inmany cases it is unlikely that timely U.S. investment capital will be available fordemonstration plants to provide visible confidence-building utility experience. WhileU.S. governmental support exists, it appears that much of the costs and risks, togetherwith the successes, will be borne internationally.

An important long-term question remains: How can humankind make the transitionfrom a resource-depleting, highly polluting society to some semblance of a steady-state,environmentally acceptable, energy economy based on renewable energy sources? Thisis not a question for the energy conversion engineer alone. Indeed, it involves political,moral, and social issues that transcend the boundaries of science and engineering. Can asmooth transition be achieved that avoids cataclysmic change and minimizes painfulsocietal dislocations? Perhaps the internationalization of the leadership in technologywill help bring suitable global responses to these formidable challenges. Regardless ofthe future course, it must be a responsibility of engineers to keep the long-term needsof Earth and its inhabitants in mind as they make technical decisions to satisfy presentneeds and desires and continue the quest for technical progress.

Bibliography and References

1. Angrist, Stanley W., Direct Energy Conversion, 3rd ed. Newton, Mass.: Allyn andBacon, 1976.

2. Spitzer, Lyman, Physics of Fully Ionized Gases. New York: Wiley-Interscience,1956.

3. Appleby, A. J., "Advanced Fuel Cells and Their Future Market," Ann. Rev. Energy,13 (1988): 267�316.

4. McLarnon, Frank R., and Cairns, Elton J., "Energy Storage," Ann. Rev. Energy, 14 (1989): 241�271.

518

5. Mellde, Rolf W., "Advanced Automobile Engines For Fuel Economy, LowEmissions, and Multifuel Capability," Ann. Rev. Energy, 14 (1989): 425�444.

6. Jackson, John David, Classical Electrodynamics. New York: Wiley, 1962.

7. Fraas, Arthur P., Engineering Evaluation of Energy Systems. New York:McGraw-Hill, 1982.

8. Vincent, Colin A., et al., Modern Batteries. Baltimore: Edward Arnold, 1984.

9. Sperling, Daniel, and DeLuchi, Mark A., �Transportation Energy Futures,� Ann.Rev. Energy, 14 ( 1989): 375�424.

10. Fickett, A. P., "Fuel Cell Power Plants," Scientific American, December 1978: 70.

11. Society of Automotive Engineers, Automotive Handbook, 4th ed., Robert BoschGmbH, 1996.

12. Mayfield, Manville J., Beyma, Edmund F., and Nelkin, Gary A., "Update on U.S.Department of Energy�s Phosphoric Acid Fuel Cell Program." Proceedings of theSixteenth Energy Technology Conference, February 28�March 2, 1989, GovernmentInstitutes, Inc., pp. 184�196.

13. Myles, K. M., and Krumpelt, M., "Status of Molten Carbonate Fuel Cell Technol-ogy." Proceedings of the Sixteenth Energy Technology Conference, February 28�March 2, 1989, Government Institutes, Inc., pp. 197-204.

14. Bates, J. Lambert, "Solid Oxide Fuel Cells: A Materials Challenge." Proceedings ofthe Sixteenth Energy Technology Conference, February 28�March 2, 1989, Govern-ment Institutes, Inc., pp. 205�219.

15. Gillis, Edward, "Fuel Cells," EPRI Journal, September 1989: 34�36.

16. Harder, Edwin L., Fundamentals of Energy Production. New York: Wiley, 1982.

17. Sissine, F., "Fuel Cells for Electric Power Production: Future Potential, FederalRole and Policy Options," in Fuel Cells:Trends In Research and Applications, ed. A. J.Appleby. Washington: Hemisphere, 1987.

18. Appleby, A. J., "Phosphoric Acid Fuel Cells," in Fuel Cells:Trends In Researchand Applications, ed. A. J. Appleby. Washington: Hemisphere, 1987.

519

19. Appleby, A. J. (Ed.), Fuel Cells:Trends In Research and Applications,Washington: Hemisphere. 1987.

20. Ketelaar, J. A. A., "Molten Carbonate Fuel Cells," Fuel Cells:Trends In Researchand Applications, Ed. A. J. Appleby. Washington: Hemisphere, 1987.

21. Kinoshita, K., McLarnon, R. R., and Cairns, E.J., Fuel Cells: A Handbook. U.S.Department of Energy METC-88/6096, 1988.

22. Berry, D.A. and Mayfield, M.J., Fuel Cells, Technology Status Report, U.S.Department of Energy METC-89/0266, 1988.

23. Anon., Fuel Cell Systems Program Plan, FY 1990, U.S.Department of Energy,FE-0106P, October 1989.

24. Huber, W. J., Proceedings of the First Annual Fuel Cells Contractors ReviewMeeting, U.S.Department of Energy METC-89/6105, May 1989.

25. Anon., 1988 Fuel Cell Seminar. Washington, D.C.: Courtesy Associates, 1988.

26. Feynman, Richard P., Leighton, Robert B., and Sands, Matthew, The FeynmanLectures on Physics, Vol. 2. Reading, Mass.: Addison-Wesley, 1964.

27. Anon., Battery Service Manual. Chicago: Battery Council International, 1987.

28. Kotz, John C., and Purcell, Keith F., Chemistry and Chemical Reactivity.Philadelphia: Saunders, 1987.

29. Goodman, Frank, "Power Electronics for Renewables," EPRI Journal,January/February 1988: 44�47.

30. Schaefer, John, "Photovoltaic Operating Experience," EPRI Journal, March 1988:40�42.

31. Peterson, Terry, "Amorphous Silicon Thin Film Solar Cells," EPRI Journal, June1988: 41�44.

32. Dostalek, Frank, "High Concentration Photovoltaics," EPRI Journal, March 1989: 46�49.

33. Morris, Douglas, "The Chino Battery Facility," EPRI Journal, March 1988: 46�50.

520

34. Purcell, Gary, and Driggans, Rick, "Electric Vehicle Testing at TVA," EPRIJournal, March 1988: 44�46.

35. Anon., The Storage Battery. Horsham, Pa.: Exide Corp., 1980.

36. Makanski, Jason, "Fuel Cells Extend Boundaries of Process/Power Integration,"Power, May 1990: 82�86.

37. Chapman, Alan J., Heat Transfer. New York: Macmillan, 1968.

38. Swanson, Theodore D., "Large-Scale Photovoltaic System Design Considerations,"The Handbook of Photovoltaic Applications. Atlanta, Ga.: Fairmont Press, 1986, pp.19�32.

39. Van Overstraeten, R.J., and Mertens, R.P., Physics, Technology, and Use ofPhotovoltaics. Boston: Adam Hilger, 1986.

40. McDaniels, David K., The Sun: Our Future Energy Source, 2nd ed. New York:Wiley, 1984.

41. Feynman, Richard P., Leighton, Robert B., and Sands, Matthew, The FeynmanLectures on Physics, Vol. 3. Reading, Mass.: Addison-Wesley, 1965.

42. Takahashi, K., and Konagai, M., Amorphous Silicon Solar Cells. New York:Wiley, 1986.

43. Duffie, John A., and Beckman, William A., Solar Engineering of ThermalProcesses. New York: Wiley, 1980.

44. Rosa, Richard J., Magnetohydrodynamic Energy Conversion, rev. ed. Washington:Hemisphere, 1987.

45. Gregory, Derek P., �The Hydrogen Economy,� Scientific American, Vol. 228,No.1. (January 1973) 13�21.

46. Gregory, D. P., and Pangborn, J.B., "Hydrogen Energy," Ann. Rev. Energy, 1 (1976): 279�310.

47. Cogineni, M. Rao, Andrus, H.E. Jr., and Jones, T.J., "Advanced Energy Systems."Proceedings of the American Power Conference, Vol. 50, 1988, pp. 282�287.

521

48. Bajura, R. A., and Halow, J.S., " Looking Beyond the Demonstration Plants:Longer-Term, Coal-Based Technology Options." Proceedings of the American PowerConference, Vol. 50, 1988, pp. 49�56.

49. DeMeo, Edgar, et al., "Thin Films: Expanding the Solar Marketplace," EPRIJournal, March 1989: pp. 4�15.

50. Wood, Bernard W., Applications of Thermodynamics, 2nd ed. Reading, Mass.:Addison-Wesley, 1982.

51. Womack, G. J., MHD Power Generation: Engineering Aspects. London: Chapmanand Hall, 1969.

52. Soo, S. L., Direct Energy Conversion. Engelwood Cliffs, N.J.: Prentice Hall, 1968.

53. Kettani, M. Ali, Direct Energy Conversion. Reading, Mass.: Addison Wesley,1970.

54. Sutton, George W., Direct Energy Conversion. New York: McGraw-Hill, 1966.

55. Pitts, Donald R., and Sissom, Leighton E., Heat Transfer, Shaum's Outline Series.New York: McGraw-Hill, 1977.

56. Anon., The Astronomical Almanac. Washington, D.C.: U. S. Government PrintingOffice, 1990.

57. Parsons, Robert A. (Ed.), The ASHRAE Handbook, HVAC Systems and Appli-cations Volume. Atlanta, Ga.: American Society of Heating, Refrigerating, and AirConditioning Engineers, 1987.

58. DeMeo, Edgar, "Getting Down to Business with Thin Films," EPRI Journal,March 1989: 4�15.

59. Taylor, Roger W., Cummings, John E., and Swanson, Richard M., "High EfficiencyPhotovoltaic Device Development: An Example of the R&D Process," Proceedings ofthe American Power Conference, Vol. 47, 1985, pp. 255�259.

60. Bockris, J. O. M., Energy: The Solar-Hydrogen Alternative. New York: Wiley,1975.

61. Williams, L. O., Hydrogen Power. Elmsford, N. Y.: Pergamon Press, 1980.

522

62. Sze, S. M., Semiconductor Devices Physics and Technology. New York: Wiley,1985.

63. Moore, Taylor, et al., "Opening the Door for Utility Photovoltaics," EPRI Journal,January�February 1987: 5�15.

64. Griffiths, David, J., Introduction to Electrodynamics, 2nd ed. Engelwood Cliffs, N. J.: Prentice-Hall, 1989.

65. Anon., �Solar Electric Generating Stations (SEGS),� IEEE Power EngineeringReview, August, 1989: 4�8.

66. Anon., �Promise of Solar Energy Being Fulfilled in California,� Power, October1989: s32�s36.

67. Romano, Samuel, �Fuel Cells for Transportation,� Mechanical Engineering,August 1989: 74�77.

68. Hirschenhofer, J. H., �International Developments in Fuel Cells,� MechanicalEngineering, August 1989: 78�83.

69. Carlson, D. E., �Photovoltaic Techniques for Commercial Power Generation,�Annual Review of Energy, Vol. 15, 1990, pp. 85�98.

70. MacDonald, Gordon J., �The Future of Methane as an Energy Resource,� AnnualReview of Energy, Vol. 15, 1990, pp. 53�83.

71. Moore, Taylor, et al., �On-Site Utility Applications for Photovoltaics,� EPRIJournal, March 1991: 26�37.

72. Moore, Taylor, et al., �Thin Films: Expanding the Solar Marketplace,� EPRIJournal, March 1989: 4�15.

73. Smock, Robert W., �Second Generation Fuel Cell Technology Moves TowardDemos,� Power Engineering, June 1990: 10.

74. Sapre, Alex R., �Properties, Performance and Emissions of Medium-ConcentrationMethanol-Gasoline Blends in a Single-Cylinder, Spark-Ignition Engine,� SAE Paper 881679, October 1988.

75. Douglas, John, �Beyond Steam: Breaking Through Performance Limits,� EPRIJournal, December 1990: 5�11.

523

76. Graff, Eric, Texas Instruments Co. Personal Communication, May 24, 1991.

77. Chase, M. W. Jr., et al., JANAF Thermochemical Tables, 3rd ed., J. Phys. Chem.Ref. Data 14, Supplement No. 1, 1985.

78. Douglas, John, et al., �Fuel Cells for Urban Power,� EPRI Journal, September1991: 5�11.

79. Howell, John R., Bannerot, Richard B., and Vliet, Gary C., �Solar-Thermal EnergySystems.� New York: McGraw-Hill, 1982.

80. Levine, Jules D., et al., �Basic Properties of the Spheral Solar Cell.� 22nd

Photovoltaic Specialists Conference, Las Vegas, Nev., 1991.

81. �International Fuel Cells: Clean, Reliable Fuel Cell Energy,�www.internationalfuelcells.com/index_fl1.shtml (November 14, 2000).

82. Rulseh, Ted, �Fuel Cells: From Promise to Performance,� Grid, Spring/Summer2000: 15.

83. �Santa Clara Demonstration Project,� www.ttcorp.com/fccg/scdpnew1.htm(November 14, 2000).

84. �Fuel Cell Energy, Carbonate Fuel Cell Manufacturer,�www.ttcorp.com/fccg/erc_abt.htm (November 15, 2000).

85. �Welcome to Fuel Cell Energy, Inc.,� www.fuelcellenergy.com/homeframe.html(December 9, 2000).

86. �Library of Fuel Cell Related Publications,� http://216.51.18.233/biblio.html(November 14, 2000)

87. Archer, David H. and Wimer, John G., � A Phosphoric Acid Fuel Cell CogenerationSystem Retrofit to a Large Office Building,� Department of Energy FETC-97/1044,April 1997., www.fetc.doe.gov/netltv/index.html (November 15, 2000).

88. �DaimlerChrysler Offers First Commercial Fuel Cell Buses to Transit Agencies,�www.hfcletter.com/letter/may00/feature.html (November 15, 2000).

89. �TroughNet Projects - Projects Deployed,� www.eren.doe.gov/troughnet/deployed.html (November 23, 2000).

524

90. �Solar Fact Sheets: Solar Thermal Electricity, Solar Energy IndustriesAssociation,� www.seia.org/sf/sfsolthe.htm (November 23, 2000).

91. �Solar Thermal Electric Power Plant,�www.magnet.consortia.org.il/ConSolar/stepp.html (November 23, 2000).

92. �About FPL Energy,� www.fplenergy.com/aboutfpl/solar-1.htm (November 23,2000).

93. �US Department of Energy PV Home Page,�www.eren.doe.gov/pv/pvmenu.cgi?site=pv&idx=0&body=video.html (November 24,2000).

94. �Solar Information Center - Photovoltaic generation systems - Kyocera Solar,Inc.,� www.kyocerasolar.com/info/solarenergy.html (November 23, 2000).

95. �Photovoltaics Program,� http://www.sandia.gov/pv/ ( November 24, 2000).

96. �TI Seeks Buyer For Spheral Solar Technology,�www.ti.com/corp/docs/press/company/1995/510no.shtml (November 25, 2000).

EXERCISES

11.1* Derive an expression for the battery power output, in terms of E and Ri /Ro, forthe linear model discussed in connection with Example 11.2. Nondimensionalizethe power by dividing by E2/Ro. Use a spreadsheet to tabulate and plot thedimensionless power as a function of Ri /Ro.

11.2* Derive an expression for the linear-battery-model power output nondimension-alized by E2/Ri in terms of the internal-to-external resistance ratio. Use aspreadsheet to tabulate the dimensionless power function, and plot it. Is there acondition that produces an extreme value of the dimensionless power? If so,use calculus methods to derive the condition.

11.3 An automobile storage battery with and open-circuit voltage of 12.8V is ratedat 260 A-hr. The internal resistance of the battery is 0.2 �. Estimate themaximum duration of current flow and its value through an external resistanceof 1.8 �.

_________________________*Exercise numbers with an asterisk involve computer usage.

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11.4 A battery electrical storage plant is to be designed for 20-MW peak powerdelivery for a duration of four hours. The plant uses 600 A-hr batteriesoperating at 400 volts DC. Estimate the minimum number of batteries and thecurrent in each during peak operation.

11.5 A hydrogen-oxygen fuel cell operates with a voltage of 0.7 volts with water-vapor product. Calculate the work per kg-mole of hydrogen, in kJ and in kW-hr, and determine the cell efficiency.

11.6 A neighborhood fuel cell power plant is to be designed for an electrical poweroutput of 2000 kW with liquid-water product. Estimate the flow rates ofhydrogen and oxygen during peak power production, assuming that an 80%efficient power conditioner is used to convert DC to AC power and the fuel cellefficiency is 55%. What is the plant heat rate?

11.7 A hydrogen-oxygen fuel cell has liquid water as product when it operates at0.82 volts. What is the electrical energy output in kJ / kg-mole of hydrogen,and the cell efficiency?

11.8 Prepare a typed three-page, double-spaced memorandum outlining the design ofthe power train of a hydrogen-oxygen fuel-cell-powered automobile, giving thedesign criteria, system description, and quantitative preliminary design data onthe power and fuel supply system.

11.9 A hydrogen-oxygen fuel cell stack produces 50 kW of DC power at anefficiency of 60%, with water vapor as product. What is the hydrogen massflow rate, in g/s, and the cell voltage?

11.10 A fuel cell power plant is to be designed for an electrical power output of 200MW with liquid-water product. Estimate the flow rates of hydrogen andoxygen during peak power production with an 85% efficient power conditionerand fuel cell efficiency of 55%. What is the plant heat rate?

11.11 Determine the fraction of the solar spectrum that lies in the visible wavelengthsbetween 0.4 and 0.7 �. What is the ratio of the energy in this visible range tothat in a range of the same width between 2.0 and 2.3 �?

11.12 It is desired to have a window in a south-facing wall shaded by a 1-m overhangat noon on June 21 and fully exposed to the sun at noon on December 21.Determine the maximum vertical size of the window for a house at latitude 35°north.

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11.13 A skylight in a horizontal roof is 10ft. above the floor of the room below.Determine the distance that the image of the skylight on the floor at noonmoves between December 21 and June 21 at latitude 30° north. Sketch andlabel a plan view of the two positions.

11.14 Derive an equation for the duration of daylight as a function of latitude forDecember 21 and June 21. Use a spreadsheet to create a table and a plot of thehours of daylight as a function of latitude for December 21 and June 21.

11.15 Perform an analysis analogous to Example 11.12, comparing the energy

efficiency of using utility central solar photovoltaic electricity with (1) anelectric heat pump for space heating and (2) with solar hydrogen production forspace heating using a furnace.

11.16 An MHD generator uses helium seeded with cesium at 2200K to give anelectrical conductivity of 10 (�-m)�1. The gas travels at 1000 m/s in a magneticfield of 3 Webers/m2. What MHD generator volume is needed to produce50MW of output power with a load factor is 0.6?

11.17 The electrical conductivity of C2H4 burned in oxygen at 3000K with a smallamount of potassium seed is 60 (�-m)�1. An MHD generator operates with agas velocity of 1000 m/s, the magnetic field intensity is 5 tesla, the channel is 1meter square in cross-section, and the load factor is 0.5. What is the open-circuit voltage, the load potential difference, and the short-circuit and operatingcurrent density? If the electrode area is 50% of the wall area and the channel is10 m long, what is the current and the power output?

11.18 Prepare a preliminary design of a solar photovoltaic system to provide 1.0 kWof stand-alone, 24 hour / day power to a travel trailer. Write a report givingdetails, including a schematic of the overall system, a cost estimate, the cellarray design, and a discussion of the assumptions on which the design is based.

11.19 Estimate the electrical power requirement in kW, of a 1400-ft2 floor area (threebedroom home) with three occupants. Using your home power estimate, predictthe power requirement for a city of 300,000 people. Use these results toestimate the area of silicon solar cells required to satisfy the community powerrequirements. Write a short narrative discussing your assumptions and analysis.

11.20 Estimate the efficiency of a parquet of an EPRI-Stanford concentrating solarcell.

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11.21 According to reference 65, a 400-MW expansion of the SEGS plants insouthern California is expected to cost $1.4 billion. What is the expectedcapital cost of the generation facilities, in dollars per kW of capacity? Estimatethe minimum cost of electricity per kW-h generated by the new facilities if theyare operated at full capacity for six hours per day for 30 years.

11.22 Evaluate the efficiency of a solar-photovoltaic-powered hydrogen economy forcomparison with the nuclear-powered version given in Example 11.12. Consider the cases where the energy source is (a) photovoltaic electricity, and(b) the sun.

11.23 Assuming 8% efficiency for a spheral solar laminate, what is the maximumelectrical output of a one square meter sheet? Estimate the peak and the annualelectrical loads for three-person family home. Determine the required area for afixed spheral solar installation. Indicate clearly the assumptions made.

11.24 Assuming 10% efficiency for a spheral solar laminate, what is the maximumelectrical output of a one square foot sheet? Estimate the peak and the annualelectrical loads for three-person family home. Determine the required area for afixed spheral solar installation. Indicate clearly the assumptions made.

Table A.1 Physical Constants__________________________________________________Name Value Units of measure

Speed of light in a vacuum 2.9979x108 m/sGravitational constant 6.673x10 �11 N-m2 / kg2

Avogadro's number 6.022x1023 atoms/g-moleBoltzmann constant 1.38x10 �23 J / °KUniversal gas constant 0.082057 L-atm/g-mole -°K

8.314 kJ /kg-mole - °K1545.3 ft-lbf /lb-mole - °R1.986 Btu /lb-mole - °R1.986 cal / g-mole - °K0.730 atm-ft3 /lb-mole - °R10.73 psia - ft3 /lb-mole - °R

Volume of ideal gas, at STP 22.41 m3/kg-moleUnified atomic mass unit, amu 1.660531x10�27 kgPlanck's constant 6.626196x10�34 J-sElectron charge 1.602192×10�19 CElectron rest mass 9.109558x10�31 kg

5.485930x10�4 amuProton rest mass 1.672614x10�27 kg

1.00727661 amuNeutron rest mass 1.674920x10�27 kg

1.00866520 amuElectron-charge-to-mass ratio 1.758803x1011 C / kgStefan-Boltzmann constant 5.66961x10�8 W / m2 - °K4

Faraday's constant 9.6487x107 C / kg-mole________________________________________________________

Table A.2 Conversion Factors________________________________________________________________________________

Length Volume1 ft = 0.3084 m I liter = 10 �3 m3

1 in = 2.540 cm = 1000 cm3

1 mile = 5280 ft = 0.035313 ft3

= 1.6093 km = 0.26417 gal1 km = 1000 m = 61.025 in3

1 m = 100 cm 1 gal = 3.7854 liter= 1.0936 yd = 0.13368 ft3

= 3.2808 ft I ft3 = 28.317 liter1 cm = 0.3937 in. = 7.4805 gal

1 in3 = 16.387 cm3

Mass1 Ibm = 0.45359237 kg

= 7000 grain Force1 slug = 32.174 Ibm I Ibf = 4.4482 N

= 14.5939 kg = 444,800 dyne1 kg = 1000 g = 32.174 Ibm-ft/s2

= 2.2046 Ibm 1 N = 1 kg-m/s2

1 ton: = 0.22481 Ibf

metric = 1000 kg 1 dyne= 1 g-cm/s2

short = 2000 Ibm = 10 �5 N

Pressure Time1 kPa = I000 N / m2 1 h = 3600 s

= 20.886 Ibf /ft2 I min = 60 s1 atm = 760 torr 1 ms = 10 � 3s

= 1.01325 x 105 N/m2 1 �s = 10 �6 s= 14.696 Ibf /in2

= 29.92 in. Hg Energy1 torr = 1 mm Hg 1 J = 1 kg-m2/s2

= 1.933 × 10�2 psi = 9.478×10 � 4 Btu1 mm Hg = 0.01934 Ibf /in2 = 107 erg1 bar = 0.9869 atm 1 erg = 1 dyne-cm

= 105 N /m2 1cal = 4.186 J= 106 dyne/cm2 1 Btu = 252.16 cal

1 in. Hg = 0.491 Ibf /in2 = 1.05504 kJ= 0.0334 atm = 778.16 ft-lbf

= 33,864 dyne/cm2 1 ft-lbf = 1.3558 J1 dyne/cm2 = 10 �1 N/m2 1 ev = 1.602 x 10 �19 J

Table A.2 (continued)________________________________________________________________________________

Temperature Power1 °K = 1.8 °R 1 W = 1 J/s°K = °C + 273.15 = 1 kg-m2/s3

°C = (°F � 32)/1.8 = 860.42 cal /hr°R = °F + 459.67 1 kW = 1000 W

= 3412.2 Btu/hElectromagnetic units = 1.341 hp1 ampere = 1 C/s = 737.56 Ibf -ft/s

= 1 W/v 1 hp = 2544.5 Btu/h1 volt = 1 J/C = 745.7 W1 ohm = 1 v/A = 550 ft-lbf /s1 farad = 1 A-s/v = 33,000 ft-lbf /min1 henry = 1 v-s/A1 weber = 1 v-s Specific Energy1 tesla = 1 Wb/m2 1 Btu/Ibm = 2.3259 kJ/kg

= 1 N/A-m= 104 gauss Energy Flux

1 Btu/ft2-s = 11.356 kW/m2

Density1 Ibm/ft3 = 0.01602 g/cm3

= 16.02 kg / m3

APPENDIX B

Properties of Steam (English units)

Table B.1 Saturated Steam: Temperature TableTable B.2 Saturated Steam: Pressure TableFigure B.1 Mollier Diagram for Steam (two parts)Figure B.2 Temperature�Entropy Diagram for SteamFigure B.3 Pressure�Enthalpy Diagram for Steam

This page is intentionally blank.

APPENDIX C

Properties of Steam (SI units)

Table C.1 Saturated Steam (Temperature)

From NBS/NRC Steam Tables by L. Haar, et.al., New York: Hemisphere Publishing,1984; with permission

APPENDIX C

Properties of Steam (SI units)

Figure C.1 Isentropic Exponent for SteamFigure C.2 Mollier Diagram for SteamFigure C.3 Pressure�Enthalpy Diagram for SteamFigure C.4 Temperature�Entropy Diagram for Steam

From NBS/NRC Steam Tables by L. Haar, et.al., New York: Hemisphere Publishing,1984; with permission

AP P E N D I X D

Enthalpy of Selected Substances

Table D.1 Enthalpy (Btu/lb-mole)Table D.2 Enthalpy (Btu/Ibm)Table D.3 Enthalpy (kJ/kg)Table D.4 Enthalpy (kJ/g-mole)

Reproduced with Permission from: Chase, M.W. Jr., Davies, C.A., Downey, J.R., FruripD.J., McDonald, R.A., JANAF Thermochernical Tables, Third Edition, Part1 (Al-Co) and Part H (Cr-Zr), J. Phys. Chem. Ref. Data 14, Supplement No. 1, 1985. The primary data in the JANAF Tables is presented on a gram-mole basis in S1 units. Hethe primary enthalpy data is presented in units of kJ/gram-mole. Tables in other sets of uniare derived from this primary table. For instance enthalpy in kJ/kg is obtained by dividinthe primary values by the appropriate molecular masses and multiplying by 1000. Mass-basevalues in the English system are obtained from the SI mass-based values by multiplying by thconversion factor 0.43021 Btu-kg/kJ-lbm. Mole-based values in the English system are obtainedirectly from the primary values by multiplying by (0.43021)(1000 JANAF Table temperatures are given in degrees Kelvin. Rankine temperatures listed in tpresent tables are a factor of 1.8 times the Kelvin values

Table D.1 Enthalpy (Btu/lb-mole TEMPERATURE ENTHALPY (Btu/lb-mole)

R K O2 N2 C CO CO2 H2 H2O SO2_______________ _____________________________________________________________________

0 0 -3736 -3730 -452 -3730 -4028 -3643 -4261 -4540

180 100 -2486 -2481 -426 -2482 -2777 -2352 -2846 -3105360 200 -1234 -1229 -286 -1230 -1469 -1193 -1412 -1607

536.7 298.15 0 0 0 0 0 0 0 0

540 300 23 23 7 23 30 23 27 32720 400 1301 1278 447 1280 1722 1273 1485 1828900 500 2617 2543 1017 2552 3573 2530 2979 3768

1080 600 3977 3826 1696 3847 5553 3791 4518 58271260 700 5377 5135 2459 5172 7638 5055 6106 79801440 800 6812 6473 3286 6529 9811 6325 7745 102051620 900 8278 7840 4161 7916 12059 7604 9438 124861800 1000 9767 9234 5074 9331 14368 8897 11185 148111980 1100 11277 10652 6018 10770 16728 10204 12988 171712160 1200 12803 12093 6987 12231 19133 11528 14845 195592340 1300 14345 13553 7976 13710 21574 12871 16753 219702520 1400 15899 15030 8982 15205 24047 14232 18711 244012700 1500 17466 16522 10004 16714 26546 15612 20715 268472880 1600 19044 18028 11038 18234 29069 17011 22762 293073060 1700 20632 19544 12085 19766 31612 18428 24848 317803240 1800 22230 21071 13142 21307 34172 19862 26971 342633420 1900 23839 22607 14208 22860 36748 21313 29128 367553600 2000 25458 24151 15283 24412 39338 22780 31315 392573780 2100 27086 25702 16367 25974 41940 24263 33531 417653960 2200 28725 27259 17458 27542 44553 25759 35773 442814140 2300 30373 28822 18557 29118 47177 27270 38040 468034320 2400 32030 30390 19662 30684 49809 28793 40328 493324500 2500 33698 31963 20774 32259 52450 30329 42637 518664680 2600 35374 33540 21893 33846 55098 31877 44966 544054860 2700 37059 35122 23018 35436 57754 33436 47312 569505040 2800 38753 36707 24148 37030 60416 35006 49674 594995220 2900 40455 38295 25285 38627 63084 36586 52051 620535400 3000 42166 39887 26427 40226 65759 38177 54443 64611

Table D.1 continued. Enthalpy (Btu/lb-mole TEMPERATURE ENTHALPY (Btu/lb-mole)

R K O2 N2 C CO CO2 H2 H2O SO2_______________ _____________________________________________________________________

5580 3100 43885 41481 27574 41829 68438 39777 56848 671735760 3200 45612 43079 28727 43434 71123 41387 59264 697405940 3300 47346 44678 29885 45041 73812 43006 61693 723106120 3400 49088 46281 31049 46651 76507 44634 64133 748856300 3500 50836 47885 32218 48263 79206 46271 66583 774626480 3600 52591 49492 33392 49877 81909 47917 69042 800446660 3700 54352 51101 34570 51493 84616 49571 71510 826306840 3800 56120 52712 35754 53111 87328 51233 73988 852197020 3900 57893 54324 36943 54731 90043 52904 76473 878117200 4000 59672 55939 38136 56353 92763 54583 78966 904077380 4100 61457 57555 39335 57976 95486 56269 81466 930057560 4200 63247 59173 40538 59601 98213 57964 83973 956087740 4300 65043 60793 41746 61228 100943 59667 86487 982147920 4400 66843 62414 42959 62856 103677 61377 89007 1008228100 4500 68650 64037 44165 64486 106414 63095 91533 1034348280 4600 70461 65661 45398 66118 109155 64820 94065 1060498460 4700 72277 67287 46625 67751 111899 66553 96603 1086688640 4800 74099 68914 47857 69385 114646 68292 99145 1112898820 4900 75927 70543 49093 71021 117397 70039 101693 1139139000 5000 77760 72173 50333 72658 120150 71792 104246 1165419180 5100 79599 73805 51579 74297 122907 73552 106803 1191719360 5200 81444 75438 52829 75937 125668 75317 109366 1218059540 5300 83295 77073 54083 77578 128433 77089 111934 1244419720 5400 85153 78709 55342 79222 131202 78867 114507 1270819900 5500 87017 80347 56606 80861 133974 80649 117085 129723

10080 5600 88889 81986 57874 82504 136751 82437 119668 13236810260 5700 90769 83626 59146 84150 139532 84230 122256 13501610440 5800 92657 85268 60423 85799 142316 86026 124849 13766710620 5900 94552 86912 61705 87448 145105 87827 127447 14032110800 6000 96457 88558 62991 89099 147897 89630 130050 142979

Table D.2. Enthalpy (Btu/lbm) TEMPERATURE ENTHALPY (Btu/lbm)

R K O2 N2 C CO CO2 H2 H2O SO2_______________ ______________________________________________________________________

0 0 -116.7 -133.2 -37.6 -133.2 -91.5 -1806.8 -236.5 -252.0180 100 -77.7 -88.6 -35.5 -88.6 -63.1 -1166.9 -158.0 -172.3360 200 -38.6 -43.9 -23.8 -43.9 -33.4 -592.0 -78.4 -89.2

536.7 298.15 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

540 300 0.7 0.8 0.6 0.8 0.7 11.3 1.5 1.8720 400 40.7 45.6 37.2 45.7 39.1 631.4 82.4 101.5900 500 81.8 90.8 84.7 91.1 81.2 1255.2 165.4 209.1

1080 600 124.3 136.6 141.2 137.3 126.2 1880.2 250.8 323.41260 700 168.0 183.3 204.7 184.7 173.6 2507.2 338.9 442.91440 800 212.9 231.1 273.5 233.1 222.9 3137.4 429.9 566.41620 900 258.7 279.9 346.4 282.6 274.0 3772.0 523.9 693.01800 1000 305.2 329.7 422.5 333.1 326.5 4413.1 620.9 822.11980 1100 352.4 380.3 501.1 384.5 380.1 5061.6 720.9 953.12160 1200 400.1 431.7 581.7 436.7 434.7 5718.4 824.0 1085.62340 1300 448.3 483.9 664.0 489.5 490.2 6384.4 929.9 1219.52520 1400 496.9 536.6 747.8 542.8 546.4 7059.6 1038.6 1354.42700 1500 545.8 589.9 832.9 596.7 603.2 7744.2 1149.8 1490.22880 1600 595.1 643.6 919.0 651.0 660.5 8438.0 1263.4 1626.73060 1700 644.8 697.8 1006.1 705.7 718.3 9140.9 1379.2 1764.03240 1800 694.7 752.3 1094.1 760.7 776.5 9852.4 1497.1 1901.83420 1900 745.0 807.1 1182.9 816.1 835.0 10571.9 1616.8 2040.23600 2000 795.6 862.2 1272.4 871.5 893.8 11299.6 1738.2 2179.03780 2100 846.5 917.6 1362.7 927.3 953.0 12035.0 1861.2 2318.23960 2200 897.6 973.2 1453.5 983.3 1012.3 12777.4 1985.6 2457.94140 2300 949.2 1029.0 1545.0 1039.5 1072.0 13526.6 2111.4 2597.94320 2400 1001.0 1085.0 1637.0 1095.5 1131.8 14282.3 2238.5 2738.24500 2500 1053.0 1141.1 1729.6 1151.7 1191.8 15044.1 2366.6 2878.94680 2600 1105.4 1197.4 1822.7 1208.3 1251.9 15811.9 2495.9 3019.84860 2700 1158.1 1253.9 1916.4 1265.1 1312.3 16585.3 2626.1 3161.15040 2800 1211.0 1310.5 2010.5 1322.0 1372.8 17364.0 2757.2 3302.65220 2900 1264.2 1367.2 2105.1 1379.0 1433.4 18148.0 2889.2 3444.35400 3000 1317.7 1424.0 2200.2 1436.1 1494.2 18936.9 3021.9 3586.3

Table D.2 continued. Enthalpy (Btu/lbm TEMPERATURE ENTHALPY (Btu/lbm)

R K O2 N2 C CO CO2 H2 H2O SO2_______________ ______________________________________________________________________

5580 3100 1371.4 1480.9 2295.8 1493.3 1555.1 19730.8 3155.4 3728.55760 3200 1425.4 1538.0 2391.7 1550.6 1616.1 20529.3 3289.5 3871.05940 3300 1479.6 1595.1 2488.2 1608.0 1677.2 21332.5 3424.4 4013.76120 3400 1534.0 1652.3 2585.1 1665.5 1738.4 22140.0 3559.8 4156.66300 3500 1588.6 1709.6 2682.4 1723.0 1799.7 22952.0 3695.8 4299.66480 3600 1643.5 1766.9 2780.1 1780.7 1861.1 23768.2 3832.3 4442.96660 3700 1698.5 1824.4 2878.2 1838.4 1922.7 24588.6 3969.3 4586.56840 3800 1753.7 1881.9 2976.8 1896.1 1984.3 25413.3 4106.8 4730.27020 3900 1809.2 1939.5 3075.8 1953.9 2046.0 26242.0 4244.7 4874.07200 4000 1864.8 1997.1 3175.1 2011.9 2107.8 27074.6 4383.1 5018.17380 4100 1920.5 2054.8 3274.9 2069.8 2169.6 27911.4 4521.9 5162.47560 4200 1976.5 2112.6 3375.1 2127.8 2231.6 28751.9 4661.0 5306.87740 4300 2032.6 2170.4 3475.7 2185.9 2293.6 29596.6 4800.6 5451.57920 4400 2088.9 2228.3 3576.6 2244.0 2355.8 30444.8 4940.4 5596.38100 4500 2145.3 2286.2 3677.0 2302.2 2418.0 31296.9 5080.7 5741.28280 4600 2201.9 2344.2 3779.7 2360.5 2480.2 32152.6 5221.2 5886.48460 4700 2258.7 2402.2 3881.9 2418.8 2542.6 33012.2 5362.1 6031.78640 4800 2315.6 2460.3 3984.4 2477.1 2605.0 33875.0 5503.2 6177.28820 4900 2372.7 2518.5 4087.3 2535.5 2667.5 34741.4 5644.6 6322.99000 5000 2430.0 2576.7 4190.6 2594.0 2730.1 35611.0 5786.3 6468.79180 5100 2487.5 2635.0 4294.3 2652.5 2792.7 36484.0 5928.2 6614.79360 5200 2545.1 2693.3 4398.3 2711.0 2855.4 37359.8 6070.5 6760.99540 5300 2603.0 2751.6 4502.8 2769.6 2918.3 38238.8 6213.0 6907.39720 5400 2661.0 2810.0 4607.6 2828.3 2981.2 39120.5 6355.8 7053.89900 5500 2719.3 2868.5 4712.8 2886.8 3044.2 40004.6 6498.9 7200.4

10080 5600 2777.8 2927.0 4818.4 2945.5 3107.3 40891.5 6642.3 7347.310260 5700 2836.5 2985.6 4924.3 3004.3 3170.5 41780.5 6786.0 7494.210440 5800 2895.5 3044.2 5030.7 3063.1 3233.7 42671.7 6929.9 7641.410620 5900 2954.8 3102.9 5137.3 3122.0 3297.1 43564.7 7074.1 7788.710800 6000 3014.3 3161.7 5244.4 3180.9 3360.5 44459.5 7218.6 7936.2

Table D.3. Enthalpy (kJ/kg) TEMPERATURE ENTHALPY (kJ/kg)

R K O2 N2 C CO CO2 H2 H2O SO2_______________ ______________________________________________________________________

0 0 -271.3 -309.5 -87.5 -309.6 -212.8 -4199.9 -549.7 -585.7180 100 -180.6 -205.9 -82.5 -206.0 -146.7 -2712.3 -367.2 -400.6360 200 -89.6 -102.0 -55.4 -102.0 -77.6 -1376.0 -182.2 -207.4

536.7 298.15 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

540 300 1.7 1.9 1.3 1.9 1.6 26.3 3.4 4.1720 400 94.5 106.1 86.5 106.2 91.0 1467.8 191.6 235.9900 500 190.1 211.0 196.9 211.7 188.7 2917.7 384.4 486.1

1080 600 288.9 317.5 328.3 319.2 293.3 4370.5 582.9 751.81260 700 390.6 426.2 475.9 429.2 403.4 5827.9 787.7 1029.51440 800 494.8 537.2 635.8 541.8 518.2 7292.7 999.2 1316.71620 900 601.3 650.6 805.3 656.9 636.9 8767.9 1217.7 1611.01800 1000 709.5 766.3 982.0 774.4 758.9 10257.9 1443.2 1911.01980 1100 819.1 884.0 1164.7 893.8 883.5 11765.4 1675.8 2215.52160 1200 930.0 1003.5 1352.1 1015.0 1010.5 13292.2 1915.3 2523.52340 1300 1042.0 1124.7 1543.5 1137.7 1139.5 14840.3 2161.5 2834.62520 1400 1154.9 1247.3 1738.3 1261.8 1270.1 16409.7 2414.1 3148.22700 1500 1268.7 1371.1 1936.0 1387.0 1402.1 18001.0 2672.7 3463.82880 1600 1383.3 1496.0 2136.2 1513.2 1535.3 19613.6 2936.7 3781.33060 1700 1498.7 1621.9 2338.7 1640.3 1669.6 21247.5 3205.9 4100.23240 1800 1614.8 1748.6 2543.3 1768.1 1804.8 22901.3 3479.9 4420.63420 1900 1731.7 1876.0 2749.6 1897.0 1940.9 24573.9 3758.1 4742.23600 2000 1849.2 2004.2 2957.7 2025.8 2077.7 26265.4 4040.3 5064.93780 2100 1967.5 2132.9 3167.4 2155.5 2215.1 27974.7 4326.2 5388.63960 2200 2086.5 2262.1 3378.7 2285.6 2353.1 29700.4 4615.5 5713.24140 2300 2206.3 2391.8 3591.2 2416.4 2491.7 31442.0 4907.9 6038.64320 2400 2326.7 2522.0 3805.2 2546.3 2630.7 33198.4 5203.2 6364.84500 2500 2447.8 2652.5 4020.4 2677.0 2770.2 34969.2 5501.1 6691.84680 2600 2569.5 2783.4 4236.9 2808.7 2910.1 36754.0 5801.5 7019.44860 2700 2691.9 2914.6 4454.5 2940.7 3050.4 38551.6 6104.2 7347.75040 2800 2815.0 3046.2 4673.3 3072.9 3190.9 40361.6 6409.0 7676.65220 2900 2938.6 3178.0 4893.3 3205.5 3331.9 42184.0 6715.7 8006.15400 3000 3062.9 3310.1 5114.2 3338.2 3473.1 44017.9 7024.3 8336.1

Table D.3 continued. Enthalpy (kJ/kg) TEMPERATURE ENTHALPY (kJ/kg)

R K O2 N2 C CO CO2 H2 H2O SO2_______________ ______________________________________________________________________

5580 3100 3187.8 3442.4 5336.4 3471.2 3614.7 45863.1 7334.5 8666.75760 3200 3313.2 3574.9 5559.5 3604.4 3756.4 47719.2 7646.4 8997.95940 3300 3439.2 3707.7 5783.6 3737.8 3898.5 49586.3 7959.8 9329.56120 3400 3565.7 3840.7 6008.8 3871.3 4040.8 51463.3 8274.5 9661.76300 3500 3692.7 3973.8 6235.0 4005.1 4183.3 53350.7 8590.6 9994.36480 3600 3820.2 4107.1 6462.2 4139.1 4326.1 55248.0 8907.9 10327.46660 3700 3948.1 4240.7 6690.3 4273.2 4469.1 57154.8 9226.4 10661.06840 3800 4076.5 4374.3 6919.4 4407.4 4612.3 59071.9 9546.0 10995.07020 3900 4205.3 4508.2 7149.4 4541.8 4755.8 60998.0 9866.6 11329.47200 4000 4334.5 4642.2 7380.4 4676.4 4899.4 62933.5 10188.3 11664.47380 4100 4464.2 4776.3 7612.4 4811.1 5043.2 64878.5 10510.8 11999.77560 4200 4594.2 4910.6 7845.2 4946.0 5187.2 66832.3 10834.3 12335.47740 4300 4724.6 5044.9 8079.0 5081.0 5331.4 68795.6 11158.6 12671.67920 4400 4855.4 5179.5 8313.7 5216.1 5475.8 70767.4 11483.8 13008.28100 4500 4986.6 5314.2 8547.0 5351.4 5620.4 72748.0 11809.7 13345.28280 4600 5118.2 5448.9 8785.8 5486.8 5765.2 74737.1 12136.4 13682.68460 4700 5250.2 5583.9 9023.2 5622.3 5910.1 76735.1 12463.8 14020.48640 4800 5382.5 5718.9 9261.5 5757.9 6055.2 78740.6 12791.9 14358.68820 4900 5515.3 5854.1 9500.8 5893.7 6200.5 80754.5 13120.6 14697.29000 5000 5648.4 5989.4 9740.8 6029.5 6345.9 82775.8 13449.9 15036.29180 5100 5782.0 6124.8 9981.8 6165.5 6491.5 84805.1 13779.9 15375.69360 5200 5916.0 6260.3 10223.7 6301.6 6637.3 86840.8 14110.5 15715.49540 5300 6050.4 6396.0 10466.5 6437.8 6783.3 88883.9 14441.8 16055.69720 5400 6185.4 6531.8 10710.2 6574.2 6929.6 90933.5 14773.8 16396.19900 5500 6320.8 6667.7 10954.7 6710.3 7076.0 92988.6 15106.4 16737.0

10080 5600 6456.8 6803.7 11200.1 6846.6 7222.7 95050.1 15439.7 17078.310260 5700 6593.3 6939.8 11446.3 6983.2 7369.6 97116.6 15773.6 17420.010440 5800 6730.5 7076.1 11693.5 7120.0 7516.6 99188.0 16108.1 17762.010620 5900 6868.2 7212.5 11941.5 7256.9 7663.9 101264 16443.4 18104.410800 6000 7006.6 7349.1 12190.4 7393.9 7811.4 103344 16779.3 18447.3

Table D.4. Enthalpy (kJ/g-mole) TEMPERATURE ENTHALPY (kJ/g-mole)

R K O2 N2 C CO CO2 H2 H2O SO2_______________ ______________________________________________________________________

0 0 -8.683 -8.670 -1.051 -8.671 -9.364 -8.467 -9.904 -10.552180 100 -5.779 -5.768 -0.991 -5.769 -6.456 -5.468 -6.615 -7.217360 200 -2.868 -2.857 -0.665 -2.858 -3.414 -2.774 -3.282 -3.736

536.7 298.15 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

540 300 0.054 0.054 0.016 0.054 0.069 0.053 0.062 0.074720 400 3.025 2.971 1.039 2.976 4.003 2.959 3.452 4.250900 500 6.084 5.911 2.365 5.931 8.305 5.882 6.925 8.758

1080 600 9.244 8.894 3.943 8.942 12.907 8.811 10.501 13.5441260 700 12.499 11.937 5.716 12.023 17.754 11.749 14.192 18.5481440 800 15.835 15.046 7.637 15.177 22.806 14.702 18.002 23.7211620 900 19.241 18.223 9.672 18.401 28.030 17.676 21.938 29.0231800 1000 22.703 21.463 11.795 21.690 33.397 20.680 26.000 34.4281980 1100 26.212 24.760 13.989 25.035 38.884 23.719 30.191 39.9142160 1200 29.761 28.109 16.240 28.430 44.473 26.797 34.506 45.4642340 1300 33.344 31.503 18.539 31.868 50.148 29.918 38.942 51.0692520 1400 36.957 34.936 20.879 35.343 55.896 33.082 43.493 56.7182700 1500 40.599 38.405 23.253 38.850 61.705 36.290 48.151 62.4042880 1600 44.266 41.904 25.658 42.385 67.569 39.541 52.908 68.1233060 1700 47.958 45.429 28.090 45.945 73.480 42.835 57.758 73.8703240 1800 51.673 48.978 30.547 49.526 79.431 46.169 62.693 79.6423420 1900 55.413 52.548 33.026 53.136 85.419 49.541 67.706 85.4363600 2000 59.175 56.137 35.525 56.744 91.439 52.951 72.790 91.2503780 2100 62.961 59.742 38.044 60.376 97.488 56.397 77.941 97.0813960 2200 66.769 63.361 40.581 64.021 103.562 59.876 83.153 102.9294140 2300 70.600 66.995 43.134 67.683 109.660 63.387 88.421 108.7924320 2400 74.453 70.640 45.704 71.324 115.779 66.928 93.741 114.6694500 2500 78.328 74.296 48.289 74.985 121.917 70.498 99.108 120.5594680 2600 82.224 77.963 50.889 78.673 128.073 74.096 104.520 126.4624860 2700 86.141 81.639 53.503 82.369 134.246 77.720 109.973 132.3765040 2800 90.079 85.323 56.131 86.074 140.433 81.369 115.464 138.3025220 2900 94.036 89.015 58.773 89.786 146.636 85.043 120.990 144.2385400 3000 98.013 92.715 61.427 93.504 152.852 88.740 126.549 150.184

Table D.4 continued. Enthalpy (kJ/g-mole TEMPERATURE ENTHALPY (kJ/g-mole)

R K O2 N2 C CO CO2 H2 H2O SO2_______________ ______________________________________________________________________

5580 3100 102.009 96.421 64.095 97.229 159.081 92.460 132.139 156.1405760 3200 106.023 100.134 66.775 100.960 165.321 96.202 137.757 162.1065940 3300 110.054 103.852 69.467 104.696 171.573 99.966 143.403 168.0816120 3400 114.102 107.577 72.172 108.438 177.836 103.750 149.073 174.0656300 3500 118.165 111.306 74.889 112.185 184.109 107.555 154.768 180.0576480 3600 122.245 115.041 77.617 115.937 190.393 111.380 160.485 186.0586660 3700 126.339 118.781 80.357 119.693 196.686 115.224 166.222 192.0686840 3800 130.447 122.525 83.109 123.454 202.989 119.089 171.980 198.0867020 3900 134.569 126.274 85.872 127.219 209.301 122.972 177.757 204.1117200 4000 138.705 130.027 88.646 130.989 215.622 126.874 183.552 210.1457380 4100 142.854 133.784 91.432 134.762 221.951 130.795 189.363 216.1867560 4200 147.015 137.545 94.229 138.540 228.290 134.734 195.191 222.2357740 4300 151.188 141.309 97.037 142.321 234.637 138.692 201.034 228.2927920 4400 155.374 145.078 99.856 146.106 240.991 142.667 206.892 234.3568100 4500 159.572 148.850 102.658 149.895 247.354 146.660 212.764 240.4278280 4600 163.783 152.625 105.526 153.687 253.725 150.670 218.650 246.5068460 4700 168.005 156.405 108.378 157.483 260.103 154.698 224.548 252.5928640 4800 172.240 160.187 111.240 161.282 266.489 158.741 230.458 258.6858820 4900 176.488 163.973 114.114 165.084 272.882 162.801 236.380 264.7859000 5000 180.749 167.763 116.997 168.890 279.283 166.876 242.313 270.8939180 5100 185.023 171.556 119.892 172.699 285.691 170.967 248.258 277.0079360 5200 189.311 175.352 122.797 176.511 292.109 175.071 254.215 283.1289540 5300 193.614 179.152 125.713 180.326 298.535 179.190 260.184 289.2579720 5400 197.933 182.955 128.640 184.146 304.971 183.322 266.164 295.3929900 5500 202.267 186.761 131.577 187.957 311.416 187.465 272.157 301.534

10080 5600 206.618 190.571 134.524 191.775 317.870 191.621 278.161 307.68310260 5700 210.987 194.384 137.482 195.603 324.334 195.787 284.177 313.83810440 5800 215.375 198.201 140.451 199.434 330.806 199.963 290.204 320.00010620 5900 219.782 202.023 143.429 203.268 337.288 204.148 296.244 326.16910800 6000 224.210 205.848 146.419 207.106 343.779 208.341 302.295 332.346

APPENDIX E

Properties of Selected Coals

Table E.1 Analyses of Selected U.S. Coals, as minedTable E.2 Typical analyses of Coals of the World

APPENDIX F

Thermodynamic properties of Refrigerants

Figure F.1 Thermodynamic Properties of R-12 (English units)Figure F.2 Thermodynamic Properties of R-22 (English units)Figure F.3 Thermodynamic Properties of R-123 (English units)Figure F.4 Thermodynamic Properties of R-123a (English units)Figure F.5 Thermodynamic Properties of R-134a (SI units)

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APPENDIX H

Properties of the 1976 U.S. Standard Atmosphere

Table H.1 Properties of the 1976 U.S. Standard Atmosphere (SI units)Table H.2 Properties of the 1976 U.S. Standard Atmoshere (English units)

APPENDIX IBlackbody Spectral Distribution Function

The Planck function divided by the fifth power of the temperature and the Stefan-Boltzmann constant is multiplied by 100,000 so that it plots nicelyversus the wavelength-temperature product, Lambda-T, as in Figure 11.27.SIGMA was calculated by summing the product of the tabulated Planckfunction and the abscissa interval (100) to the desired value of lambda-T.The sum must then be divided by 1000 to compensate for the 100,000 factor.Note the distinction between the Stefan-Boltzmann constant (Sigma) andthe cumulative radiation function (SIGMA).

c1 = 118700000 Btu-micron^4/ft^2-hrc2 = 25896 micron-degree R

Lambda-T Fb(Lambda-T)/Sigma-T^5 SIGMA micron-R 1/(micron-R) dimensionless

x100,000

1000 3.926E-05 3.926E-081100 0.000256682 2.95942E-071200 0.001181568 1.47751E-061300 0.004164636 5.64215E-061400 0.011928687 1.75708E-051500 0.028995038 4.65659E-051600 0.061770887 0.0001083371700 0.11819881 0.0002265361800 0.207027327 0.0004335631900 0.336873778 0.0007704372000 0.515283226 0.001285722100 0.747952401 0.0020336722200 1.038219036 0.0030718912300 1.386845559 0.0044587372400 1.792070875 0.0062508082500 2.249871296 0.0085006792600 2.754359627 0.0112550392700 3.298254226 0.0145532932800 3.87336131 0.0184266542900 4.471028633 0.0228976833000 5.082543507 0.0279802263100 5.699460885 0.0336796873200 6.313857168 0.0399935443300 6.918512359 0.0469120573400 7.507027618 0.0544190843500 8.073887592 0.0624929723600 8.614477758 0.071107453700 9.12506675 0.0802325163800 9.602762863 0.0898352793900 10.04545268 0.099880732

Lambda-T Fb(Lambda-T)/Sigma-T^5 SIGMA micron-R 1/(micron-R) dimensionless

x100,000

4000 10.45172852 0.1103324614100 10.82081001 0.1211532714200 11.152464 0.1323057354300 11.44692597 0.1437526614400 11.70482499 0.1554574864500 11.9271139 0.1673845994600 12.11500549 0.1794996054700 12.26991521 0.191769524800 12.39341038 0.2041629314900 12.48716597 0.2166500965000 12.55292639 0.2292030235100 12.59247315 0.2417954965200 12.6075976 0.2544030945300 12.60007853 0.2670031725400 12.5716639 0.2795748365500 12.52405626 0.2920988925600 12.45890147 0.3045577945700 12.37778011 0.3169355745800 12.28220131 0.3292177755900 12.17359857 0.3413913746000 12.05332729 0.3534447016100 11.92266363 0.3653673656200 11.78280457 0.3771501696300 11.6348688 0.3887850386400 11.47989844 0.4002649376500 11.31886126 0.4115837986600 11.15265333 0.4227364516700 10.98210197 0.4337185536800 10.80796898 0.4445265226900 10.63095392 0.4551574767000 10.4516975 0.4656091737100 10.27078501 0.4758799587200 10.08874964 0.4859687087300 9.906075829 0.4958747847400 9.723202485 0.5055979867500 9.540526075 0.5151385137600 9.358403632 0.5244969167700 9.177155612 0.5336740727800 8.997068608 0.542671147900 8.818397933 0.5514895388000 8.641370049 0.5601309088100 8.466184858 0.5685970938200 8.293017853 0.5768901118300 8.12202213 0.5850121338400 7.95333027 0.5929654638500 7.787056092 0.600752528600 7.623296286 0.6083758168700 7.462131926 0.6158379488800 7.303629879 0.6231415788900 7.147844105 0.630289422

Lambda-T Fb(Lambda-T)/Sigma-T^5 SIGMA micron-R 1/(micron-R) dimensionless

x100,000

9000 6.994816861 0.6372842399100 6.844579811 0.6441288189200 6.697155054 0.6508259739300 6.552556067 0.657378539400 6.410788573 0.6637893189500 6.271851341 0.6700611699600 6.135736916 0.6761969069700 6.002432297 0.6821993399800 5.871919548 0.6880712589900 5.744176362 0.69381543510000 5.619176579 0.69943461110100 5.496890652 0.70493150210200 5.377286082 0.71030878810300 5.260327805 0.71556911610400 5.145978545 0.72071509410500 5.034199146 0.72574929310600 4.924948859 0.73067424210700 4.818185608 0.73549242810800 4.713866235 0.74020629410900 4.611946716 0.74481824111000 4.512382358 0.74933062311100 4.415127973 0.75374575111200 4.320138045 0.75806588911300 4.227366866 0.76229325611400 4.136768666 0.76643002511500 4.048297729 0.77047832211600 3.961908497 0.77444023111700 3.877555654 0.77831778711800 3.795194213 0.78211298111900 3.714779583 0.7858277612000 3.636267632 0.78946402812100 3.559614744 0.79302364312200 3.484777862 0.79650842112300 3.41171453 0.79992013512400 3.340382931 0.80326051812500 3.270741911 0.8065312612600 3.202751009 0.80973401112700 3.136370475 0.81287038112800 3.071561285 0.81594194312900 3.008285156 0.81895022813000 2.946504554 0.82189673213100 2.886182704 0.82478291513200 2.82728359 0.82761019913300 2.769771958 0.83037997113400 2.713613318 0.83309358413500 2.658773939 0.83575235813600 2.60522085 0.83835757913700 2.552921826 0.84091050113800 2.501845393 0.84341234613900 2.451960812 0.845864307

Lambda-T Fb(Lambda-T)/Sigma-T^5 SIGMA micron-R 1/(micron-R) dimensionless

x100,000

14000 2.403238073 0.84826754514100 2.355647887 0.85062319314200 2.309161674 0.85293235414300 2.263751554 0.85519610614400 2.219390335 0.85741549614500 2.1760515 0.85959154814600 2.133709199 0.86172525714700 2.09233823 0.86381759514800 2.051914034 0.86586950914900 2.012412676 0.86788192215000 1.973810837 0.86985573315100 1.936085795 0.87179181915200 1.89921542 0.87369103415300 1.863178154 0.87555421215400 1.827953001 0.87738216515500 1.793519515 0.87917568515600 1.759857785 0.88093554315700 1.726948422 0.88266249115800 1.694772551 0.88435726315900 1.663311791 0.88602057516000 1.632548251 0.88765312416100 1.60246451 0.88925558816200 1.573043612 0.89082863216300 1.544269048 0.89237290116400 1.51612475 0.89388902516500 1.488595076 0.89537762116600 1.4616648 0.89683928516700 1.435319102 0.89827460416800 1.409543555 0.89968414816900 1.384324117 0.90106847217000 1.359647121 0.90242811917100 1.335499261 0.90376361817200 1.311867588 0.90507548617300 1.288739494 0.90636422617400 1.266102709 0.90763032817500 1.243945289 0.90887427417600 1.222255604 0.91009652917700 1.201022337 0.91129755217800 1.180234467 0.91247778617900 1.159881267 0.91363766718000 1.139952293 0.9147776218100 1.120437377 0.91589805718200 1.10132662 0.91699938418300 1.082610382 0.91808199418400 1.06427928 0.91914627318500 1.046324175 0.92019259718600 1.028736168 0.92122133418700 1.011506593 0.9222328418800 0.994627013 0.92322746718900 0.978089209 0.924205556

Lambda-T Fb(Lambda-T)/Sigma-T^5 SIGMA micron-R 1/(micron-R) dimensionless

x100,000

19000 0.961885177 0.92516744219100 0.946007123 0.92611344919200 0.930447451 0.92704389619300 0.915198768 0.92795909519400 0.900253868 0.92885934919500 0.885605733 0.92974495419600 0.871247525 0.93061620219700 0.857172583 0.93147337519800 0.843374417 0.93231674919900 0.829846703 0.93314659620000 0.816583279 0.93396317920100 0.80357814 0.93476675720200 0.790825435 0.93555758320300 0.778319459 0.93633590220400 0.766054655 0.93710195720500 0.754025605 0.93785598220600 0.742227027 0.93859820920700 0.730653775 0.93932886320800 0.719300828 0.94004816420900 0.708163296 0.94075632721000 0.697236407 0.94145356421100 0.686515509 0.94214007921200 0.675996069 0.94281607521300 0.665673662 0.94348174921400 0.655543976 0.94413729321500 0.645602804 0.94478289621600 0.635846044 0.94541874221700 0.626269693 0.94604501121800 0.616869849 0.94666188121900 0.607642702 0.94726952422000 0.598584538 0.94786810822100 0.589691733 0.948457822200 0.580960749 0.94903876122300 0.572388137 0.94961114922400 0.563970528 0.9501751222500 0.555704637 0.95073082422600 0.547587256 0.95127841122700 0.539615256 0.95181802722800 0.531785581 0.95234981222900 0.524095249 0.95287390823000 0.516541348 0.95339044923100 0.509121036 0.9538995723200 0.501831539 0.95440140223300 0.494670147 0.95489607223400 0.487634214 0.95538370623500 0.480721158 0.95586442723600 0.473928455 0.95633835523700 0.467253643 0.95680560923800 0.460694314 0.95726630323900 0.454248117 0.957720552

Lambda-T Fb(Lambda-T)/Sigma-T^5 SIGMA micron-R 1/(micron-R) dimensionless

x100,000

24000 0.447912759 0.95816846424100 0.441685995 0.9586101524200 0.435565634 0.95904571624300 0.429549536 0.95947526524400 0.42363561 0.95989890124500 0.417821812 0.96031672324600 0.412106146 0.96072882924700 0.406486658 0.96113531624800 0.400961444 0.96153627724900 0.395528638 0.96193180625000 0.390186419 0.96232199225100 0.384933006 0.96270692525200 0.379766658 0.96308669225300 0.374685673 0.96346137825400 0.369688389 0.96383106625500 0.364773178 0.96419583925600 0.359938449 0.96455577825700 0.355182648 0.9649109625800 0.350504254 0.96526146425900 0.345901779 0.96560736626000 0.341373769 0.9659487426100 0.336918802 0.96628565926200 0.332535485 0.96661819426300 0.328222458 0.96694641726400 0.323978389 0.96727039526500 0.319801976 0.96759019726600 0.315691944 0.96790588926700 0.311647046 0.96821753626800 0.307666063 0.96852520226900 0.3037478 0.9688289527000 0.299891089 0.96912884127100 0.296094787 0.96942493627200 0.292357775 0.96971729427300 0.288678958 0.97000597327400 0.285057264 0.9702910327500 0.281491643 0.97057252127600 0.277981069 0.97085050327700 0.274524537 0.97112502727800 0.27112106 0.97139614827900 0.267769677 0.97166391828000 0.264469442 0.97192838728100 0.261219432 0.97218960728200 0.258018743 0.97244762528300 0.254866487 0.97270249228400 0.251761797 0.97295425428500 0.248703823 0.97320295828600 0.245691732 0.97344864928700 0.242724709 0.97369137428800 0.239801956 0.97393117628900 0.236922689 0.974168099

Lambda-T Fb(Lambda-T)/Sigma-T^5 SIGMA micron-R 1/(micron-R) dimensionless

x100,000

29000 0.234086142 0.97440218529100 0.231291563 0.97463347629200 0.228538218 0.97486201529300 0.225825385 0.9750878429400 0.223152357 0.97531099229500 0.220518442 0.97553151129600 0.217922962 0.97574943429700 0.215365251 0.97596479929800 0.212844659 0.97617764429900 0.210360545 0.97638800430000 0.207912285 0.97659591630100 0.205499264 0.97680141630200 0.20312088 0.97700453730300 0.200776545 0.97720531330400 0.198465679 0.97740377930500 0.196187716 0.97759996730600 0.1939421 0.97779390930700 0.191728285 0.97798563730800 0.189545739 0.97817518330900 0.187393935 0.97836257731000 0.185272361 0.97854784931100 0.183180512 0.97873102931200 0.181117894 0.97891214731300 0.179084023 0.97909123131400 0.177078421 0.9792683131500 0.175100623 0.9794434131600 0.173150172 0.97961656131700 0.171226618 0.97978778731800 0.16932952 0.97995711731900 0.167458447 0.98012457532000 0.165612974 0.98029018832100 0.163792686 0.98045398132200 0.161997174 0.98061597832300 0.160226038 0.98077620432400 0.158478884 0.98093468332500 0.156755326 0.98109143832600 0.155054987 0.98124649332700 0.153377494 0.98139987132800 0.151722482 0.98155159332900 0.150089594 0.98170168333000 0.148478478 0.98185016133100 0.146888789 0.9819970533200 0.145320188 0.9821423733300 0.143772343 0.98228614333400 0.142244927 0.98242838833500 0.14073762 0.98256912533600 0.139250107 0.98270837533700 0.137782079 0.98284615733800 0.136333232 0.98298249133900 0.134903269 0.983117394

Lambda-T Fb(Lambda-T)/Sigma-T^5 SIGMA micron-R 1/(micron-R) dimensionless

x100,000

34000 0.133491896 0.98325088634100 0.132098827 0.98338298534200 0.130723778 0.98351370834300 0.129366474 0.98364307534400 0.128026641 0.98377110134500 0.126704012 0.98389780534600 0.125398324 0.98402320434700 0.124109319 0.98414731334800 0.122836744 0.9842701534900 0.121580348 0.9843917335000 0.120339889 0.9845120735100 0.119115124 0.98463118535200 0.117905819 0.98474909135300 0.11671174 0.98486580335400 0.115532659 0.98498133535500 0.114368353 0.98509570435600 0.1132186 0.98520892235700 0.112083185 0.98532100635800 0.110961895 0.98543196735900 0.10985452 0.98554182236000 0.108760855 0.98565058336100 0.107680698 0.98575826436200 0.10661385 0.98586487736300 0.105560116 0.98597043836400 0.104519305 0.98607495736500 0.103491227 0.98617844836600 0.102475696 0.98628092436700 0.101472532 0.98638239636800 0.100481554 0.98648287836900 0.099502586 0.9865823837000 0.098535454 0.98668091637100 0.097579989 0.98677849637200 0.096636023 0.98687513237300 0.095703391 0.98697083537400 0.09478193 0.98706561737500 0.093871482 0.98715948937600 0.092971889 0.98725246137700 0.092082998 0.98734454437800 0.091204657 0.98743574837900 0.090336717 0.98752608538000 0.08947903 0.98761556438100 0.088631454 0.98770419538200 0.087793845 0.98779198938300 0.086966064 0.98787895538400 0.086147973 0.98796510338500 0.085339438 0.98805044338600 0.084540326 0.98813498338700 0.083750505 0.98821873438800 0.082969847 0.98830170338900 0.082198226 0.988383902

Lambda-T Fb(Lambda-T)/Sigma-T^5 SIGMA micron-R 1/(micron-R) dimensionless

x100,000

39000 0.081435515 0.98846533739100 0.080681594 0.98854601939200 0.079936341 0.98862595539300 0.079199638 0.98870515539400 0.078471367 0.98878362639500 0.077751414 0.98886137839600 0.077039664 0.98893841739700 0.076336008 0.98901475339800 0.075640335 0.98909039439900 0.074952537 0.98916534640000 0.074272508 0.98923961940100 0.073600143 0.98931321940200 0.07293534 0.98938615440300 0.072277996 0.98945843240400 0.071628013 0.9895300640500 0.070985292 0.98960104540600 0.070349736 0.98967139540700 0.069721249 0.98974111640800 0.069099739 0.98981021640900 0.068485113 0.98987870141000 0.067877279 0.98994657841100 0.067276149 0.99001385541200 0.066681634 0.99008053641300 0.066093647 0.9901466341400 0.065512104 0.99021214241500 0.064936919 0.99027707941600 0.06436801 0.99034144741700 0.063805295 0.99040525241800 0.063248693 0.99046850141900 0.062698126 0.99053119942000 0.062153516 0.99059335342100 0.061614785 0.99065496742200 0.061081858 0.99071604942300 0.06055466 0.99077660442400 0.060033118 0.99083663742500 0.059517159 0.99089615442600 0.059006713 0.99095516142700 0.058501708 0.99101366342800 0.058002076 0.99107166542900 0.057507748 0.99112917243000 0.057018658 0.99118619143100 0.056534738 0.99124272643200 0.056055925 0.99129878243300 0.055582153 0.99135436443400 0.055113359 0.99140947743500 0.054649481 0.99146412743600 0.054190457 0.99151831743700 0.053736227 0.99157205343800 0.053286731 0.9916253443900 0.05284191 0.99167818244000 0.052401706 0.991730584