Modern Engineering Thermodynamics

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14.10 Absorption Refrigeration 561 cycles, since the compressor unit in the standard system is simply replaced by the absorption unit in the absorption system. The absorption cycle is a special type of vapor-compression refrigeration cycle, since it is driven by heat as opposed to work. Consequently, the coefficient of performance of an absorption cycle is not computed in the same way as a standard vapor-compression cycle, and the two COPs should not be directly compared. The heat energy necessary to drive an absorption cycle is at a much lower availability than the electrical energy necessary to power a work-driven cycle. There is a need for some work in absorption systems with a generator feed pump, but the amount of pump work required is negligible in comparison to the compressor work needed in a standard vapor-compression cycle. The coefficient of performance of an ideal absorption refrigerator is ðCOPÞ absorption refrigerator = Refrigeration ðevaporatorÞ cooling Generator heat + Pump work = _Q evaporator _Q generator + _W pump (14.19) A Carnot absorption cycle can be constructed by driving a Carnot refrigerator with a Carnot engine, as shown in Figure 14.20. In this system, both the engine and the refrigerator exhaust heat to the local environment at temperature T a . The thermal efficiency of the Carnot engine used here is ðη T Þ Carnot engine = _ W engine _Q generator and the thermal efficiency (COP) of the Carnot refrigerator used here is ðη T Þ Carnot refrigerator = ðCOPÞ Carnot refrigerator = _Q evaporator _W refrigerator Since the overall efficiency of a combined system is equal to the product of the efficiencies of its components, the COP of this combined system is equal to the product of the thermal efficiency of the Carnot engine multiplied by the thermal efficiency of the Carnot refrigerator, or ðCOPÞ Carnot absorption refrigerator = ! _W engine _Q generator ! Q _ evaporator = _W refrigerator _Q evaporator _Q generator (14.20) as all the work produced by the engine is used to drive the refrigerator, or _W engine = _W refrigerator = _W. For a Carnot cycle, _Q generator = _W /ð1 − T a /T g Þ and _Q evaporator = _WT e /ðT a − T e Þ. Consequently, ðCOPÞ Carnot absorption refrigerator = T e T g − T a T g T a − T e (14.21) Gas generator at temperature T g Condenser at ambient temperature T a Carnot engine Q generator Q ambient T a Ambient at temperature W engine = W ref. Carnot refrigerator Q evaporator Evaporator at temperature T e Q condenser FIGURE 14.20 A Carnot absorption refrigeration cycle.
562 CHAPTER 14: Vapor and Gas Refrigeration Cycles Since no real system can be more efficient than a Carnot system, Eq. (14.21) represents the maximum possible thermal efficiency of an absorption refrigeration system operating with a generator temperature T g , an evaporator temperature T e , and an ambient temperature T a . Because absorption systems are heat rather than work driven, their practical COP values tend to be around 1.0 or less. Common absorption refrigeration fluid systems are: ammonia (refrigerant)-water (carrier), water (refrigerant)-lithium bromide (carrier), and water (refrigerant)- lithium chloride (carrier). The ammonia-water system was widely used in domestic refrigerators until about 1950. The lithium salt–water systems that use water as the refrigerant cannot go below 32°F (0°C) and consequently are mainly used in air conditioning applications. Absorption refrigeration dominated the refrigeration market before 1875, which is remarkable considering its inherent complexity and the fact that its design was empirical. A suitable theory for the operation of absorption refrigeration did not appear until 1913, and today multistage regenerative absorption refrigerators can produce temperatures as low as 65 K. EXAMPLE 14.8 A new absorption refrigeration system with a generator temperature of 100.°C and an evaporator temperature of 5.00°C is being designed to operate in an environment at a temperature of 20.0°C. To provide an upper limit for the operating efficiency, determine the Carnot absorption refrigeration coefficient of performance of this system. Solution Use Figure 14.20 as the equipment schematic for this example. Equation (14.21) gives the COP for this system as ðCOPÞ Carnot = T e T g − T a 5:00 + 273:15 100: − 20:0 = = 3:98 T g T a − T e 100: + 273:15 20:0 − 5:00 absorption refrigerator Since no system can be more efficient than a Carnot system, this represents the maximum COP of any absorption system operating under these conditions. Exercises 22. The environmental temperature of the absorption refrigerator being designed in Example 14.8 is suddenly increased from 20°C to30°C. Determine the new Carnot absorption refrigeration COP, assuming all the other variables remain unchanged. Answer: (COP) Carnot absorption ref = 2.09. 23. Suppose now we want to convert the absorption refrigeration system design discussed in Example 14.8 into a cryogenic unit with an evaporator temperature of only 65 K. What would be the maximum possible coefficient of performance of this system assuming all the other variables remain unchanged? Answer: (COP) Carnot absorption ref = 0.06. 24. Explain why the COP given by Eq. (14.21) goes to zero as the evaporator temperature approaches absolute zero. Answer: In this instance, as T e → 0, the evaporator cooling load _Q e also goes to zero. Since the COP is defined as the ratio of the system cooling to the energy input, the COP must vanish as the cooling vanishes. 14.11 COMMERCIAL AND HOUSEHOLD REFRIGERATORS Commercial and household refrigeration technology essentially developed together, because commercial refrigeration in shops and supermarkets requires the same basic technological advances as household refrigerators. Also, once frozen or chilled food products were purchased by the consumer, similar refrigeration needs were created in the home. Thus. the parallel development of household and commercial refrigeration was advantageous, if the market for chilled and frozen foods was to expand beyond the needs of a single day’s foodsupply. Throughout the 19th century, mechanical vapor-compression refrigeration systems had been limited to largescale industrial units powered by steam engines or internal combustion engines. Several major technical bottlenecks prevented small commercial and household vapor-compression refrigerators from being successfully developed. The first problem was the development of a power source suitable for use in a household. The traditional commercial power sources (steam and internal combustion engines) were not suitable for household use. The second problem was the enormous friction in the mechanical seals on the shaft between the power source and the compressor. Without a complex and tight sealing system, refrigerant leaked out, causing serious environmental and maintenance problems. The third problem was the development of an automatic
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Modern Engineering Thermodynamics
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Modern Engineering Thermodynamics R
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Dedication WHAT IS AN ENGINEER AND
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Contents PREFACE . . . . . . . . .
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Contents ix 7.6 Heat Engines Runnin
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Contents xi 14.13 Air Standard Gas
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Preface TEXT OBJECTIVES This textbo
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Preface xv Step 5. Write down the b
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Acknowledgments I wish to acknowled
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Resources That Accompany This Book
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List of Symbols A a B COP CR c c p
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Prologue PARIS FRANCE, 10:35 AM, AU
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CHAPTER 1 The Beginning CONTENTS 1.
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1.2 Why Is Thermodynamics Important
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1.3 Getting Answers: A Basic Proble
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1.5 How Do We Measure Things? 7 bei
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1.6 Temperature Units 9 THE DEVELOP
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1.7 Classical Mechanical and Electr
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1.7 Classical Mechanical and Electr
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1.9 Modern Units Systems 15 where M
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1.10 Significant Figures 17 CRITICA
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1.10 Significant Figures 19 WHAT AB
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1.11 Potential and Kinetic Energies
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1.11 Potential and Kinetic Energies
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Summary 25 FIGURE 1.19 Case study 1
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Problems 27 Problems (* indicates p
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Problems 29 14. Determine the mass
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Problems 31 49. Using the CGS units
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CHAPTER 2 Thermodynamic Concepts CO
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2.3 Phases of Matter 35 System boun
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2.4 System States and Thermodynamic
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2.6 Thermodynamic Processes 39 WHAT
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2.7 Pressure and Temperature Scales
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2.9 The Continuum Hypothesis 43 Sur
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2.10 The Balance Concept 45 Solutio
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2.11 The Conservation Concept 47 or
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2.11 The Conservation Concept 49 Th
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Summary 51 change in the mass of X
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Problems 53 Problems (* indicates p
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Problems 55 and Death rate = α 2 +
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CHAPTER 3 Thermodynamic Properties
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3.3 Fun with Mathematics 59 CRITICA
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3.4 Some Exciting New Thermodynamic
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3.6 Enthalpy 63 WHO WAS AMALIE EMMY
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3.7 Phase Diagrams 65 WHO WAS EMMY
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Pressure Vapor 3.7 Phase Diagrams 6
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3.7 Phase Diagrams 69 WHAT IS A “
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3.7 Phase Diagrams 71 considerable
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3.8 Quality 73 10 5 300 C Critical
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3.8 Quality 75 Although Eq. (3.26)
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3.9 Thermodynamic Equations of Stat
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3.10 Thermodynamic Tables 85 Critic
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3.12 Thermodynamic Charts 87 vð100
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3.13 Thermodynamic Property Softwar
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Summary 91 and e = E/m = u + V2 2g
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Problems 93 c. For a saturated mixt
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Problems 95 specific enthalpy of sa
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Problems 97 Table 3.23 Problem 65 M
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CHAPTER 4 The First Law of Thermody
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4.3 The First Law of Thermodynamics
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4.3 The First Law of Thermodynamics
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4.4 Energy Transport Mechanisms 105
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4.5 Point and Path Functions 107 4.
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4.6 Mechanical Work Modes of Energy
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4.7 Nonmechanical Work Modes of Ene
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4.9 Work Efficiency 125 In the case
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4.12 Heat Modes of Energy Transport
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4.13 Heat Transfer Modes 129 Table
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4.14 A Thermodynamic Problem Solvin
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4.14 A Thermodynamic Problem Solvin
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4.15 How to Write a Thermodynamics
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4.15 How to Write a Thermodynamics
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Summary 139 The general open system
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Problems 141 11.* Determine the hea
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Problems 143 where K = 0.810 lbf. D
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Problems 145 Table 4.12 Problem 67
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CHAPTER 5 First Law Closed System A
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5.2 Sealed, Rigid Containers 149 In
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5.4 Power Plants 151 Step 7. Calcul
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5.5 Incompressible Liquids 153 The
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1Q 2 − 1 W 2 = mðu 2 − u 1 Þ
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5.8 Closed System Unsteady State Pr
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5.9 The Explosive Energy of Pressur
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Problems 161 Problems (* indicates
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Problems 163 31. A thermoelectric g
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Problems 165 where v, T, and r are
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CHAPTER 6 First Law Open System App
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6.2 Mass Flow Energy Transport 169
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6.3 Conservation of Energy and Cons
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6.4 Flow Stream Specific Kinetic an
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6.5 Nozzles and Diffusers 175 m (a)
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6.5 Nozzles and Diffusers 177 We ar
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6.6 Throttling Devices 179 These as
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6.6 Throttling Devices 181 For an i
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6.7 Throttling Calorimeter 183 The
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6.8 Heat Exchangers 185 Convective
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6.9 Shaft Work Machines 187 6.9 SHA
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6.9 Shaft Work Machines 189 1 Basem
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6.10 Open System Unsteady State Pro
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Problems 197 we have _m R / _m D =
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Problems 201 32.* A commercial slid
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Summary 203 59. Incompressible liqu
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CHAPTER 7 Second Law of Thermodynam
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7.3 The Second Law of Thermodynamic
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7.4 Carnot’s Heat Engine and the
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7.4 Carnot’s Heat Engine and the
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7.5 The Absolute Temperature Scale
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7.5 The Absolute Temperature Scale
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7.6 Heat Engines Running Backward 2
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7.7 Clausius’s Definition of Entr
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7.8 Numerical Values for Entropy 22
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7.10 Differential Entropy Balance 2
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7.11 Heat Transport of Entropy 229
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7.13 Entropy Production Mechanisms
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7.14 Heat Transfer Production of En
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7.15 Work Mode Production of Entrop
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7.15 Work Mode Production of Entrop
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7.16 Phase Change Entropy Productio
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Summary 241 ■ Indirect method inv
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Problems 243 amount of work W irr i
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Problems 245 a. If the heat pump is
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Problems 247 51. Develop a program
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CHAPTER 8 Second Law Closed System
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8.2 Systems Undergoing Reversible P
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8.3 Systems Undergoing Irreversible
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8.4 Diffusional Mixing 271 Exercise
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Summary 273 Note that this example
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Problems 275 15. A 20.0 ft 3 tank c
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Problems 277 46. a. Determine a for
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CHAPTER 9 Second Law Open System Ap
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9.4 Open System Entropy Balance Equ
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9.4 Open System Entropy Balance Equ
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9.5 Nozzles, Diffusers, and Throttl
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9.5 Nozzles, Diffusers, and Throttl
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9.6 Heat Exchangers 289 Exercises 1
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9.6 Heat Exchangers 291 (T H ) (T H
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9.7 Mixing 293 Now, _m air is given
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9.7 Mixing 295 Critical value of y,
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9.9 Unsteady State Processes in Ope
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Problems 309 Multiplying this equat
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Problems 311 entropy production rat
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Problems 313 heat is added to the b
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Problems 315 55.* Determine the max
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Problems 317 The cavitation process
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CHAPTER 10 Availability Analysis CO
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10.3 What Are Conservative Forces?
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10.6 Availability 323 WHAT IS A SYS
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10.6 Availability 325 and the total
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10.7 Closed System Availability Bal
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10.7 Closed System Availability Bal
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10.8 Flow Availability 331 EXAMPLE
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s − s 0 = c ln T T 0 10.8 Flow Av
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10.10 Modified Availability Rate Ba
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10.10 Modified Availability Rate Ba
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10.11 Energy Efficiency Based on th
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Summary 351 In this problem, we hav
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Summary 353 7. The general open sys
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Problems 355 12.5 Btu/hr·ft ·R. I
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Problems 357 flow rate of 15.0 lbm/
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Problems 359 85. Create a specific
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CHAPTER 11 More Thermodynamic Relat
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11.2 Two New Properties: Helmholtz
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11.2 Two New Properties: Helmholtz
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11.4 Maxwell Equations 367 11.4 MAX
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11.4 Maxwell Equations 369 then,
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11.5 The Clapeyron Equation 371 and
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11.6 Determining u, h, and s from p
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11.7 Constructing Tables and Charts
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11.8 Thermodynamic Charts 381 so th
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11.9 Gas Tables 383 where p r is th
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11.10 Compressibility Factor and Ge
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11.11 Is Steam Ever an Ideal Gas? 3
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Summary 399 equation, and a series
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Problems 401 20.* Estimate h fg for
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Problems 403 Design Problems The fo
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CHAPTER 12 Mixtures of Gases and Va
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12.2 Thermodynamic Properties of Ga
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12.3 Mixtures of Ideal Gases 413 Th
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12.3 Mixtures of Ideal Gases 415 Co
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12.4 Psychrometrics 417 Finally, th
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12.4 Psychrometrics 419 EXAMPLE 12.
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12.6 The Sling Psychrometer 421 The
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12.6 The Sling Psychrometer 423 p w
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12.7 Air Conditioning 425 WHAT ENVI
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12.8 Psychrometric Enthalpies 427 E
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12.8 Psychrometric Enthalpies 429 o
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12.9 Mixtures of Real Gases 431 Whe
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12.9 Mixtures of Real Gases 433 and
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12.9 Mixtures of Real Gases 435 The
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12.9 Mixtures of Real Gases 437 Sol
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Summary 439 Last, we combine Dalton
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Problems 443 exhaust gas is an idea
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Problems 445 57.* Cooling towers ar
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CHAPTER 13 Vapor and Gas Power Cycl
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13.2 Part I. Engines and Vapor Powe
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13.4 Rankine Cycle 457 A B Reversib
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13.5 Operating Efficiencies 459 For
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13.5 Operating Efficiencies 461 13.
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13.5 Operating Efficiencies 463 b.
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13.5 Operating Efficiencies 465 Sta
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13.6 Rankine Cycle with Superheat 4
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13.7 Rankine Cycle with Regeneratio
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13.8 The Development of the Steam T
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13.9 Rankine Cycle with Reheat 477
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13.9 Rankine Cycle with Reheat 479
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13.10 Modern Steam Power Plants 481
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13.12 Air Standard Power Cycles 487
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13.13 Stirling Cycle 489 Q H 4 1 4
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13.14 Ericsson Cycle 491 Q H 4 1 3
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13.15 Lenoir Cycle 493 Exercises 28
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13.16 Brayton Cycle 495 Solution Us
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13.16 Brayton Cycle 497 Equation (7
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13.17 Aircraft Gas Turbine Engines
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13.17 Aircraft Gas Turbine Engines
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13.18 Otto Cycle 503 T Q H 1 1 1 v
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13.18 Otto Cycle 505 Exercises 40.
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13.18 Otto Cycle 507 and, since the
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13.20 Miller Cycle 509 13.19.1 Mode
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Page 552 and 553: Problems 527 7. The thermal efficie
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Page 556 and 557: Problems 531 horsepower hour. Assum
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Page 564 and 565: 14.4 In the Beginning There Was Ice
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Page 604 and 605: 14.18 Second Law Analysis of Refrig
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Page 610 and 611: Problems 585 13.* A refrigeration u
Page 612 and 613: Problems 587 Loop B Station 1B Stat
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Page 616 and 617: CHAPTER 15 Chemical Thermodynamics
Page 618 and 619: 15.2 Stoichiometric Equations 593 1
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Page 624 and 625: 15.4 Fuel Modeling 599 Hydrocarbon
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Page 630 and 631: 15.6 Heat of Formation 605 and H 2
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15.7 Heat of Reaction 611 Then, h P
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15.8 Adiabatic Flame Temperature 61
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15.9 Maximum Explosion Pressure 619
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15.10 Entropy Production in Chemica
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15.10 Entropy Production in Chemica
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15.11 Entropy of Formation and Gibb
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15.12 Chemical Equilibrium and Diss
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H 2 O !ð1 − yÞH 2 O + yðv H2
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15.14 The van’t Hoff Equation 635
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15.15 Fuel Cells 637 Anode Electrol
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15.15 Fuel Cells 639 The maximum po
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15.16 Chemical Availability 641 and
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ðn i /n fuel Þðc pi Þ E system
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Problems 645 where j = 4n + m kgmol
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Problems 647 c. The percent excess
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Problems 649 77. Determine the mola
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CHAPTER 16 Compressible Fluid Flow
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16.3 Isentropic Stagnation Properti
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16.4 The Mach Number 655 Note that
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16.4 The Mach Number 657 Table 16.1
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16.4 The Mach Number 659 Since for
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16.5 Converging-Diverging Flows 661
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16.5 Converging-Diverging Flows 663
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16.6 Choked Flow 665 T a a p a = co
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16.6 Choked Flow 667 Exercises 16.
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16.7 Reynolds Transport Theorem 669
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16.7 Reynolds Transport Theorem 671
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16.8 Linear Momentum Rate Balance 6
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16.9 Shock Waves 675 16.9 SHOCK WAV
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16.9 Shock Waves 677 and, for an id
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16.9 Shock Waves 679 6 5 4 S P /(m
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16.10 Nozzle and Diffuser Efficienc
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16.10 Nozzle and Diffuser Efficienc
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Summary 685 Since for a diffuser, M
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43. 0.800 lbm/s of air passes throu
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Problems 691 A plot of this functio
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CHAPTER 17 Thermodynamics of Biolog
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17.3 Thermodynamics of Biological C
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17.3 Thermodynamics of Biological C
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17.4 Energy Conversion Efficiency o
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17.4 Energy Conversion Efficiency o
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17.5 Metabolism 703 Table 17.3 Brea
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17.6 Thermodynamics of Nutrition an
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17.7 Limits to Biological Growth 71
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17.7 Limits to Biological Growth 71
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17.8 Locomotion Transport Number 71
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17.9 Thermodynamics of Aging and De
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17.9 Thermodynamics of Aging and De
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1/3 V most efficient = P o ρAC D S
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Problems 723 a. If the monster cons
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Problems 725 the officer asks Paul
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CHAPTER 18 Introduction to Statisti
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18.3 Kinetic Theory of Gases 729 3.
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U trans = 3 2 NkT (Continued ) 18.3
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18.4 Intermolecular Collisions 733
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18.5 Molecular Velocity Distributio
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18.5 Molecular Velocity Distributio
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18.6 Equipartition of Energy 739 We
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18.7 Introduction to Mathematical P
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18.8 Quantum Statistical Thermodyna
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18.9 Three Classical Quantum Statis
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18.11 Monatomic Maxwell-Boltzmann G
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18.12 Diatomic Maxwell-Boltzmann Ga
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18.12 Diatomic Maxwell-Boltzmann Ga
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18.13 Polyatomic Maxwell-Boltzmann
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Summary 759 In this chapter, we sum
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Problems 761 4. Find the temperatur
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CHAPTER 19 Introduction to Coupled
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19.3 Linear Phenomenological Equati
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19.4 Thermoelectric Coupling 767 Eq
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19.4 Thermoelectric Coupling 769 Th
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19.4 Thermoelectric Coupling 771 wh
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19.4 Thermoelectric Coupling 773 Th
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19.4 Thermoelectric Coupling 775 b.
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19.5 Thermomechanical Coupling 777
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water), it seems reasonable that li
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Problems 785 b. For a Knudson gas,
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Appendix A: Physical Constants and
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Appendix B: Greek and Latin Origins
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Appendix B 791 Table B.4 Plural End
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Index Page numbers followed by f in
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Index 795 E e (specific energy), 10
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Index 797 Isobaric coefficient of v
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Index 799 Ranque, Georges Joseph, 3
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Index 801 William III, King of Engl
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Modern Engineering Thermodynamics

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