Modern Engineering Thermodynamics

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15.12 Chemical Equilibrium and Dissociation 631 which is also subject to an overall irreversible reaction equation in which y% oftheA present dissociates into B and C as A !ð1 − yÞA + y½A dissociated Š Then, the overall irreversible reaction equation can be written as A !ð1 − yÞA + y ðv B /v A ÞB + ðv C /v A ÞC and the equilibrium constant for this reaction is given by Eq. (15.37) as K e = ðχ BÞ vB ðχ C Þ vC ð vB+v p C−v A Þ m ðχ A Þ vA p° where χ A , χ B , and χ C are determined from the products of the overall irreversible reaction equation as and χ A = χ B = 1 − y 1 − y + yðv B /v A + v C /v A Þ yðv B /v A Þ 1 − y + yðv B /v A + v C /v A Þ yv ð C /v A Þ χ C = 1 − y + yv ð B /v A + v C /v A Þ (15.38) Table C.17 lists values of the base-10 logarithm of K e for a variety of simple dissociation reactions of this form at various temperatures. EXAMPLE 15.15 The overall irreversible carbon dioxide dissociation reaction equation wherein y% of the CO 2 dissociates into CO and O 2 is CO 2 !ð1 − yÞðCO 2 Þ + yðCO 2 Þ dissociated subject to the reversible equilibrium dissociation reaction CO 2 ⇆ CO + 1 2 O 2 Then the overall irreversible dissociation reaction equation is For this reaction, determine CO 2 !ð1 − yÞðCO 2 Þ + yðCOÞ + ðy/2ÞðO 2 Þ a. The variation in the degree of dissociation (y) with temperature at a total pressure of 0.100 MPa. b. The influence of total pressure on the degree of dissociation (y) at 3000. K. Solution The total number of moles of product in the overall irreversible reaction equation is ð1 − yÞ + y + y/2 = ð2 + yÞ/2. Then the mole fractions of the products are χ CO2 = 2ð1 − yÞ/ð2 + yÞ χ CO = 2y/ð2 + yÞ χ O2 = y/ð2 + yÞ The reversible equilibrium dissociation reaction equation gives the stoichiometric coefficients, v i ,asv CO2 = 1:0, v CO = 1:0, and v O2 . Then, Eq. (15.38) gives the equilibrium constant as = 1 2 " K e = ðχ # COÞðχ O2 Þ 1/2 ðχ CO2 Þ p m p° 3 7 5 ð1+1/2−1Þ 2 1/2 2y y 2 + y 2 + y 1/2 p m = 6 4 2ð1 − yÞ p° 2 + y " # 1/2 1/2 y y p m = 1 − y 2 + y p° (Continued )
632 CHAPTER 15: Chemical <strong>Thermodynamics</strong> EXAMPLE 15.15 (Continued) a. To determine the required results for part a of this problem, we set p m = p° =0:1 MPa, and then K e can be found in Table C.17 for various temperatures. Our task is now to pick several reaction temperatures, look up their corresponding K e values, and solve the previous equation for y. Squaring both sides of the preceding reversible equilibrium dissociation reaction equation and rearranging gives a cubic equation in y: where y 3 + 3α 1 − α y − 2α 1 − α = 0 α = Ke 2 p° p m A cubic equation in y has a simple algebraic solution. If we let a = 3α 1 − α and b = − 2α 1 − α then the cubic equation becomes y 3 + ay + b = 0, and the three roots of this equation are where y 1 y 2 y 3 = A + B = − A + B + A − B pffiffiffiffiffiffiffiffi − 3 2 2 = − A + B − A − B pffiffiffiffiffiffiffiffi − 3 2 2 r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! 1/3 r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! 1/3 A = − b 2 + b 2 4 + a3 and B = − b 27 2 − b 2 4 + a3 27 Note that, in evaluating these expressions, the cube root of a negative number is computed as the negative of the cube root of the positive value of the number; that is, (−8) 1/3 = −(8 1/3 ) = −2. If α is less than 1, then b 2 /4 + a 3 /27 > 0 and there is just one real root, given by y 1 .Butifα is greater than 1, then b 2 /4 + a 3 /27 < 0 and there are three real roots. The following procedure is a simple way of calculating these roots. Let J = [(− (a/3) 3 ] 1/2 and K = J 1/3 . Then calculate L = arccos[−b/(2J)], M = cos(L/3), and N = (3 1/2 )sin(L/3). The three roots are then given by y 1 = 2K cosðL/3Þ y 2 = − KðM + NÞ y 3 = − KðM − NÞ Note that the only valid root here is the one between 0 and 1. Table 15.8 lists several typical results using these solution equations with p°=0.1 MPa. These results can then be plotted either by hand or with the use of computer software. The final plot showing the variation in the degree of dissociation (y) versus reaction equilibrium temperature is shown in Figure 15.10. b. With the solution from part a, it is a simple matter to generate reaction equations for various total pressures at T = 3000. K. The curve in Figure 15.11 shows the results of this effort and provides a graphical representation of the effect of total pressure on the degree of dissociation (y) at 3000. K. Table 15.8 Typical Values for the Degree of Dissociation of Carbon Dioxide at Various Temperatures and Pressures with p°=0.100 MPa Reaction Temperature (K) Reaction Pressure p m (MPa) log 10 (K e ) K e y 2000. K 0.100 MPa 2.863 0.0014 0.0154 3000. K 0.100 MPa 0.469 0.3396 0.4436 5000. K 0.100 MPa +1.387 24.378 0.9770 3000. K 1.00 MPa 0.469 0.3396 0.2453 3000. K 10.0 MPa 0.469 0.3396 0.1235
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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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13.20 Miller Cycle 511 Then, from F
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13.21 Diesel Cycle 513 to stroll on
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13.21 Diesel Cycle 515 Solution a.
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13.22 Modern Prime Mover Developmen
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13.23 Second Law Analysis of Vapor
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Summary 525 Fill port 160 mm End of
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Problems 527 7. The thermal efficie
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efficiency increase if the condense
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Problems 531 horsepower hour. Assum
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Problems 533 72. Determine the valu
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CHAPTER 14 Vapor and Gas Refrigerat
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14.3 Carnot Refrigeration Cycle 537
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14.4 In the Beginning There Was Ice
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14.4 In the Beginning There Was Ice
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14.5 Vapor-Compression Refrigeratio
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14.5 Vapor-Compression Refrigeratio
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14.6 Refrigerants 547 T 3 2s 2 p 3
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14.7 Refrigerant Numbers 549 14.7 R
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14.7 Refrigerant Numbers 551 R-110
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14.8 CFCs and the Ozone Layer 553 H
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14.9 Cascade and Multistage Vapor-C
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14.10 Absorption Refrigeration 561
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14.11 Commercial and Household Refr
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14.14 Reversed Brayton Cycle Refrig
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14.14 Reversed Brayton Cycle Refrig
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14.15 Reversed Stirling Cycle Refri
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14.16 Miscellaneous Refrigeration T
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14.16 Miscellaneous Refrigeration T
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14.18 Second Law Analysis of Refrig
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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
Page 620 and 621: 15.2 Stoichiometric Equations 595 C
Page 622 and 623: 15.3 Organic Fuels 597 ANSWERS SOME
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
Page 632 and 633: 15.7 Heat of Reaction 607 EXAMPLE 1
Page 634 and 635: 15.7 Heat of Reaction 609 CRITICAL
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Page 652 and 653: 15.12 Chemical Equilibrium and Diss
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Page 664 and 665: 15.15 Fuel Cells 639 The maximum po
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Page 684 and 685: 16.4 The Mach Number 659 Since for
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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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