ME Thermodynamics I. Lecture Notes and Example Problems

Transcription

1 ME Thermodynamics I Lecture Notes and Example Problems James D. Bugg September 2018 Department of Mechanical Engineering

2 Introduction Part I: Lecture Notes This part contains handout versions of the lecture notes that will be presented during the lectures of ME Thermodynamics I. Students should be aware of the following points: 1. These notes are not stand-alone. Students must attend lectures and pay close attention to the material presented. 2. Not all information that will be presented in the lectures is contained in these handouts. Some slides have missing information that will have to be noted by the students during the lectures. 3. Some additional slides may be added to the lecture notes subsequent to when this booklet was produced. 4. Please make sure you have the version of this booklet from the current year. The aim in providing these notes is for students to spend less time copying down information and more time thinking about the material being presented, listening to the instructor, and participating in the lecture. Please feel free to provide the instructor with feedback at any time during or after the course. Part II: Solved Examples This section contains fully solved example problems. These examples will not be reviewed in class but are for you to use as model solutions. Part III: Extra Problems This section contains extra practice questions without solutions. These are all drawn from old examinations in this course. Many of these will be done in class as examples. Having them available in this book will avoid the need to copy them down in class.

3 ME Thermodynamics I Department of Mechanical Engineering Fall 2018 Definition of Thermodynamics The study of energy and the relationships among the properties of matter. 1.0 Introduction ME Thermodynamics I 2 1

4 Thermodynamics How is heat converted into work? 1.0 Introduction ME Thermodynamics I 4 Examples of Thermodynamic Systems 1.0 Introduction ME Thermodynamics I 5 2

5 Power Generation 1.0 Introduction ME Thermodynamics I 6 Jet Engine High Kinetic Energy 1.0 Introduction ME Thermodynamics I 8 3

6 Gas Turbine Compressor Combustion Chamber Turbine 1.0 Introduction ME Thermodynamics I 10 Refrigeration Condenser Throttling Valve Compressor Evaporator 1.0 Introduction ME Thermodynamics I 13 4

7 Dimensions and Units 1.0 Introduction ME Thermodynamics I 14 SI Units 1.0 Introduction ME Thermodynamics I 15 5

8 English Engineering Units 1.0 Introduction ME Thermodynamics I 16 A Thermodynamic Glossary 1.0 Introduction ME Thermodynamics I 17 6

9 Thermodynamic System a portion of mass or space which we want to study Surroundings everything in the universe EXCEPT the system 1.0 Introduction ME Thermodynamics I 18 System Boundary (control surface) the surface separating the system from its surroundings two types of systems 1.0 Introduction ME Thermodynamics I 19 7

10 Closed System no mass flow across system boundary 1.0 Introduction ME Thermodynamics I 20 Open System mass flow occurs across the system boundary System Boundary (control surface) 1.0 Introduction ME Thermodynamics I 21 8

11 Thermodynamic State condition of a system as described by its thermodynamic properties 1.0 Introduction ME Thermodynamics I 22 Thermodynamic Properties 1.0 Introduction ME Thermodynamics I 23 9

12 Thermodynamic Properties 1.0 Introduction ME Thermodynamics I 24 Thermodynamic Properties 1.0 Introduction ME Thermodynamics I 25 10

13 Process a change from one state to another 1.0 Introduction ME Thermodynamics I 26 Cycle a series of processes which return the system to its original state 1.0 Introduction ME Thermodynamics I 27 11

14 Energy Total Energy = Internal Energy + Kinetic Energy + Potential Energy 1.0 Introduction ME Thermodynamics I Internal Energy energy associated with the molecular state of the system 1.0 Introduction ME Thermodynamics I 29 12

15 Kinetic Energy System 1.0 Introduction ME Thermodynamics I 30 Potential Energy 1.0 Introduction ME Thermodynamics I 31 13

16 Thermal Equilibrium 1.0 Introduction ME Thermodynamics I 32 Temperature The property which defines the state of systems in thermal equilibrium. Proportional to the average kinetic energy of the random molecular motion. 1.0 Introduction ME Thermodynamics I 33 14

17 Temperature Scales S.I. Units Kelvin (absolute scale) Celsius English Engineering Units Rankine (absolute scale) Fahrenheit 1.0 Introduction ME Thermodynamics I 34 Heat When a system is not in thermal equilibrium with its surroundings, heat can flow across the system boundary. Heat is a means of energy transfer. It is not energy. 1.0 Introduction ME Thermodynamics I 35 15

18 Heat (continued) Amount of heat transferred depends upon path of process. SIGN CONVENTION: heat is positive when it is transferred into the system 1.0 Introduction ME Thermodynamics I 36 Heat transfer during process 1.0 Introduction ME Thermodynamics I 37 16

19 For properties This has no meaning 1.0 Introduction ME Thermodynamics I 38 Motion on a Plane State function: Position Path function: Distance travelled 1.0 Introduction ME Thermodynamics I 39 17

20 Work Thermodynamic Definition Work is done by a system on its surroundings if the sole effect on the surroundings could have been the raising of a weight. If there is energy crossing the system boundary and it is not due to a temperature difference, it is work. Work is not a property. 1.0 Introduction ME Thermodynamics I 40 Mechanical Work 1.0 Introduction ME Thermodynamics I 41 18

21 Work 1.0 Introduction ME Thermodynamics I 42 Other Forms of Work (Electrical) 1.0 Introduction ME Thermodynamics I 43 19

22 Sign Convention for Work 1.0 Introduction ME Thermodynamics I 44 Quasiequilibrium Process As each mass is removed, system is momentarily disturbed from equilibrium. A new equilibrium is established. If masses are infinitesimally small, we have a quasiequilibrium process. 1.0 Introduction ME Thermodynamics I 45 20

23 Quasiequilibrium Process Equilibrium States 1.0 Introduction ME Thermodynamics I 46 Isometric (isochoric)process 1.0 Introduction ME Thermodynamics I 47 21

24 Isobaric Process 1.0 Introduction ME Thermodynamics I 48 Isothermal Process 1.0 Introduction ME Thermodynamics I 49 22

25 Isothermal Process 1.0 Introduction ME Thermodynamics I 50 Polytropic Process (isobaric) (isometric) 1.0 Introduction ME Thermodynamics I 51 23

26 Polytropic Process 1.0 Introduction ME Thermodynamics I Introduction ME Thermodynamics I 53 24

27 Properties of Working Substances Pure Substances uniform & invariable chemical composition e.g. H 2 O, N 2, O 2 Simple Compressible Systems 2.0 Properties ME Thermodynamics I 1 State Principle For simple compressible systems, two independent properties are required to fix the equilibrium state. 2.0 Properties ME Thermodynamics I 2 J.D. Bugg 25

28 A Constant Pressure Experiment add heat to a substance which is initially a solid keep pressure constant 2.1 Properties - PvT Surface ME Thermodynamics I 3 W W W 1 W W W W 2 W Properties - PvT Surface 4 Solid Liquid Gas ME Thermodynamics I 4 J.D. Bugg 26

29 Diagram 2.1 Properties - PvT Surface ME Thermodynamics I 5 Saturated Liquid Line Saturated Vapour Line 2.1 Properties - PvT Surface ME Thermodynamics I 6 J.D. Bugg 27

30 2.1 Properties - PvT Surface ME Thermodynamics I 7 Recall the State Principle For simple compressible systems, two independent properties are required to fix the equilibrium state. Inside the liquid-vapour region P and T are NOT independent. 2.0 Properties ME Thermodynamics I 8 J.D. Bugg 28

31 2.1 Properties - PvT Surface ME Thermodynamics I 9 W - specific volume of saturated vapour - specific volume of saturated liquid 2.1 Properties - PvT Surface ME Thermodynamics I 10 J.D. Bugg 29

32 2.1 Properties - PvT Surface ME Thermodynamics I 11 Note: 2.1 Properties - PvT Surface ME Thermodynamics I 12 J.D. Bugg 30

33 Critical Point 2.1 Properties - PvT Surface ME Thermodynamics I 13 Critical Point 2.1 Properties - PvT Surface ME Thermodynamics I 14 J.D. Bugg 31

34 Critical Point 2.1 Properties - PvT Surface ME Thermodynamics I 15 A Constant Temperature Experiment Bath keeps temperature constant Measure pressure and volume during process Begin with liquid 2.1 Properties - PvT Surface ME Thermodynamics I 16 J.D. Bugg 32

35 Diagram 2.1 Properties - PvT Surface ME Thermodynamics I 17 Diagram 2.1 Properties - PvT Surface ME Thermodynamics I 18 J.D. Bugg 33

36 2.1 Properties - PvT Surface ME Thermodynamics I 19 Diagram 2.1 Properties - PvT Surface ME Thermodynamics I 21 J.D. Bugg 34

37 2.1 Properties - PvT Surface ME Thermodynamics I 22 Other Properties 2.1 Properties - PvT Surface ME Thermodynamics I 23 J.D. Bugg 35

38 Finding Properties What do we need to know? What is the substance? Do we have enough properties to fix the state (2 usually)? What phases are present? (liquid, liquid-vapour mixture, gas) 2.1 Properties - PvT Surface ME Thermodynamics I 24 Finding Properties 2.1 Properties - PvT Surface ME Thermodynamics I 25 J.D. Bugg 36

39 Superheat and Subcooling Degrees of Superheat Amount by which the temperature of a superheated vapour is above the saturation temperature Degrees of Subcooling Amount by which the temperature of a subcooled (compressed) liquid is below the saturation temperature 2.3 Properties - Liquids ME Thermodynamics I 26 Liquid-vapour Mixtures Most complicated region Tabular data (tutorial #1) Computer programs e.g. NIST 2.2 Properties - Mixtures ME Thermodynamics I 27 J.D. Bugg 37

40 Liquid Properties compressed liquid 2.3 Properties - Liquids ME Thermodynamics I 28 Liquid Properties subcooled liquid 2.1 Properties - PvT Surface ME Thermodynamics I 29 J.D. Bugg 38

41 Option 1: Special liquid tables e.g. Table A-5 Option 2: use saturated liquid properties at the same temperature 2.3 Properties - Liquids ME Thermodynamics I 30 Example: Note that 2.3 Properties - Liquids ME Thermodynamics I 31 J.D. Bugg 39

42 Ideal Gas Model 2.4 Properties - Ideal Gas Model ME Thermodynamics I 32 Ideal Gas Model 2.4 Properties - Ideal Gas Model ME Thermodynamics I 33 J.D. Bugg 40

43 Ideal Gas Model 2.4 Properties - Ideal Gas Model ME Thermodynamics I 34 Deviations from the Ideal Gas Law Define compressibility factor for ideal gases 2.4 Properties - Ideal Gas Model ME Thermodynamics I 35 J.D. Bugg 41

44 Critical Point 2.4 Properties - Ideal Gas Model ME Thermodynamics I 36 Generalised Compressibility Charts 1.0 many gases lie on the same lines when plotted this way provides a check as to whether gas can be considered ideal 2.4 Properties - Ideal Gas Model ME Thermodynamics I 37 J.D. Bugg 42

45 Example: Table A Properties - Ideal Gas Model ME Thermodynamics I 38 Specific Heats specific heat at constant volume specific heat at constant pressure ratio of specific heats 2.1 Properties - PvT Surface ME Thermodynamics I 40 J.D. Bugg 43

46 Recall that, for ideal gases. Also, the definitions 2.4 Properties - Ideal Gas Model ME Thermodynamics I Properties - Ideal Gas Model ME Thermodynamics I 42 J.D. Bugg 44

47 - tedious but quite accurate - e.g. Air (A-22) and other gases (A-23) 2.4 Properties - Ideal Gas Model ME Thermodynamics I Assume constant specific heats - OK for limited temperature changes - gives very convenient analytic expressions 2.4 Properties - Ideal Gas Model ME Thermodynamics I 44 J.D. Bugg 45

48 Example: We will calculate this two ways. 1. Use Ideal Gas Tables (A-22) 2.4 Properties - Ideal Gas Model ME Thermodynamics I 45 Example: We will calculate this two ways. 1. Use Ideal Gas Tables (A-22) 2.4 Properties - Ideal Gas Model ME Thermodynamics I 47 J.D. Bugg 46

49 2. Assume constant specific heats PROBLEM: 2.4 Properties - Ideal Gas Model ME Thermodynamics I 48 Ideal Gas Tables 2.4 Properties - Ideal Gas Model ME Thermodynamics I 49 J.D. Bugg 47

50 Ideal Gas Tables 2.4 Properties - Ideal Gas Model ME Thermodynamics I Properties - Ideal Gas Model ME Thermodynamics I 51 J.D. Bugg 48

51 Polytropic Processes & Ideal Gases Polytropic Process: Ideal Gas Law: 2.4 Properties - Ideal Gas Model ME Thermodynamics I 52 So, 2.4 Properties - Ideal Gas Model ME Thermodynamics I 53 J.D. Bugg 49

52 only for ideal gases 2.4 Properties - Ideal Gas Model ME Thermodynamics I 54 J.D. Bugg 50

53 First Law of Thermodynamics Recall 3.0 First Law ME Thermodynamics I 1 Conservation of Energy 3.1 First Law Closed Systems ME Thermodynamics I 2 51

54 3.1 First Law Closed Systems ME Thermodynamics I 3 Other Forms of the 1 st Law 3.1 First Law Closed Systems ME Thermodynamics I 4 52

55 Evaluating U Water, refrigerants tables, Appendix A ideal gases 3.1 First Law Closed Systems ME Thermodynamics I 5 Constant Volume Process 3.1 First Law Closed Systems ME Thermodynamics I 6 53

56 Constant Pressure Process 3.1 First Law Closed Systems ME Thermodynamics I First Law Closed Systems ME Thermodynamics I 8 54

57 Summary 3.1 First Law Closed Systems ME Thermodynamics I 9 Control Volume Energy Analysis Closed System Open System Control Mass Control Volume 3.2 First Law Open Systems ME Thermodynamics I 10 55

58 Conservation of Energy 3.2 First Law Open Systems ME Thermodynamics I 11 Develop an equation for conservation of energy for a control volume by applying the first law for a control mass as follows: 3.2 First Law Open Systems ME Thermodynamics I 12 56

59 for a closed system Now, 3.2 First Law Open Systems ME Thermodynamics I 13 Now, 3.2 First Law Open Systems ME Thermodynamics I 14 57

60 3.2 First Law Open Systems ME Thermodynamics I 15 Flow Work Consider the inlet: Control volume boundary 3.2 First Law Open Systems ME Thermodynamics I 16 58

61 Similarly, at the exit: OR 3.2 First Law Open Systems ME Thermodynamics I 17 Energy balance for a control volume. 3.2 First Law Open Systems ME Thermodynamics I 18 59

62 3.2 First Law Open Systems ME Thermodynamics I 19 Conservation of Mass 3.2 First Law Open Systems ME Thermodynamics I 20 60

63 Open System Components Governing equations for all components are 3.2 First Law Open Systems ME Thermodynamics I 21 Turbine: mass: Common Assumptions: Steady Potential energy negligible Kinetic energy negligible Heat transfer negligible Single inlet not always Single outlet energy: 3.2 First Law Open Systems ME Thermodynamics I 22 61

64 3.2 First Law Open Systems ME Thermodynamics I 23 Compressor: Common Assumptions: Steady Potential energy negligible Kinetic energy negligible Heat transfer negligible Single inlet Single outlet 3.2 First Law Open Systems ME Thermodynamics I 24 62

65 Nozzle: Common Assumptions: Steady Potential energy negligible Work negligible Heat transfer negligible Single inlet Single outlet 3.2 First Law Open Systems ME Thermodynamics I First Law Open Systems ME Thermodynamics I 26 63

66 Diffuser: Common Assumptions: Steady Potential energy negligible Work negligible Heat transfer negligible Single inlet Single outlet 3.2 First Law Open Systems ME Thermodynamics I 27 Heat Exchangers: Common Assumptions: Steady Potential energy negligible Kinetic energy negligible Work negligible 3.2 First Law Open Systems ME Thermodynamics I 28 64

67 Heat Exchangers: The many names for heat exchangers Heat Exchanger Boiler Condenser Superheater Economiser Evaporator Feedwater Heater Regenerator Intercooler Aftercooler Radiator Reheater 3.2 First Law Open Systems ME Thermodynamics I Separate Fluid Streams Two fluid streams do not mix but do exchange energy Control volume 3.2 First Law Open Systems ME Thermodynamics I 30 65

68 mass: and energy: The enthalpy change of the two streams must balance. 3.2 First Law Open Systems ME Thermodynamics I Fluid Streams Mix Control volume mass: energy: 3.2 First Law Open Systems ME Thermodynamics I 32 66

69 3. External Heat Source/Sink Control volume mass: energy: 3.2 First Law Open Systems ME Thermodynamics I First Law Open Systems ME Thermodynamics I 34 67

70 Throttling Valve: Common Assumptions: Steady Potential energy negligible Kinetic energy negligible Work negligible Heat transfer negligible 3.2 First Law Open Systems ME Thermodynamics I 35 68

71 Second Law of Thermodynamics First Law: Energy must be conserved. 4.0 Second Law ME Thermodynamics I 1 two identical masses in thermal contact Insulated 4.0 Second Law ME Thermodynamics I 2 J.D. Bugg 69

72 4.0 Second Law ME Thermodynamics I 3 Clausius Statement of the 2 nd Law It is impossible for any system to operate in such a way that the sole result would be an energy transfer by heat from a cooler to a hotter body. 4.0 Second Law ME Thermodynamics I 4 J.D. Bugg 70

73 Hot Reservoir Cycle Cold Reservoir 4.0 Second Law ME Thermodynamics I 5 Reversible and Irreversible Processes Reversible Process: A process where both the system and surroundings can be returned to their initial states. 4.2 Second Law Irreversibilities ME Thermodynamics I 6 J.D. Bugg 71

74 vacuum Ideal gas 1 unrestrained expansion 2 Is this process reversible? 4.2 Second Law Irreversibilities ME Thermodynamics I 7 Replace the divider with a piston 1 unrestrained expansion Second Law Irreversibilities ME Thermodynamics I 8 J.D. Bugg 72

75 Now try to reverse the process 4.2 Second Law Irreversibilities ME Thermodynamics I Second Law Irreversibilities ME Thermodynamics I 10 J.D. Bugg 73

76 What if the original process had taken place SLOWLY 4.2 Second Law Irreversibilities ME Thermodynamics I 11 Therefore, the process was reversible. What makes processes irreversible? 1. Unrestrained expansion 2. Friction 3. Inelastic deformation 4. Heat transfer through a non-zero temperature difference 5. Spontaneous chemical reaction 6. Electric current through a resistance 4.2 Second Law Irreversibilities ME Thermodynamics I 12 J.D. Bugg 74

77 Consider heat transfer through a non-zero temperature difference 4.2 Second Law Irreversibilities ME Thermodynamics I 13 Converting Work to Heat Consider Cycle Is this possible? 4.0 Second Law ME Thermodynamics I 14 J.D. Bugg 75

78 4.0 Second Law ME Thermodynamics I 15 Cycle 4.0 Second Law ME Thermodynamics I 16 J.D. Bugg 76

79 What about the opposite? Cycle 4.0 Second Law ME Thermodynamics I Second Law Heat Engines ME Thermodynamics I 18 J.D. Bugg 77

80 PROBLEM: 4.1 Second Law Heat Engines ME Thermodynamics I Second Law Heat Engines ME Thermodynamics I 20 J.D. Bugg 78

81 4.1 Second Law Heat Engines ME Thermodynamics I Second Law Heat Engines ME Thermodynamics I 22 J.D. Bugg 79

82 4.1 Second Law Heat Engines ME Thermodynamics I 23 Cycle 4.1 Second Law Heat Engines ME Thermodynamics I 24 J.D. Bugg 80

83 Kelvin-Planck Statement of the 2 nd Law It is impossible for any system to operate in a thermodynamic cycle and deliver a net amount of work to its surroundings while receiving energy by heat transfer from a single reservoir. 4.1 Second Law Heat Engines ME Thermodynamics I 25 Heat Engine A device that operates in a cycle and produces net positive work while receiving heat from a hot reservoir and rejecting heat to a cold reservoir. 4.1 Second Law Heat Engines ME Thermodynamics I 26 J.D. Bugg 81

84 Thermal Efficiency 4.1 Second Law Heat Engines ME Thermodynamics I 27 Carnot Cycle The cycle described above was a Carnot cycle. It consisted of 4 processes: Constant temperature heat addition Adiabatic expansion Constant temperature heat rejection Adiabatic compression All four processes must be reversible 4.1 Second Law Heat Engines ME Thermodynamics I 28 J.D. Bugg 82

85 We will show that, Carnot Efficiency e.g. 4.1 Second Law Heat Engines ME Thermodynamics I 29 Development of Carnot Efficiency Consider an adiabatic process involving an ideal gas with constant specific heats It can be shown that where, 4.2 Second Law Irreversibilities ME Thermodynamics I 30 J.D. Bugg 83

86 describes a reversible, adiabatic process for an ideal gas with constant specific heats Recall, the Carnot cycle with an ideal gas 4.2 Second Law Irreversibilities ME Thermodynamics I 31 We will now show that 4.2 Second Law Irreversibilities ME Thermodynamics I 32 J.D. Bugg 84

87 4.2 Second Law Irreversibilities ME Thermodynamics I 33 and 4.2 Second Law Irreversibilities ME Thermodynamics I 34 J.D. Bugg 85

88 4.2 Second Law Irreversibilities ME Thermodynamics I 35 Recall that, Therefore, 4.2 Second Law Irreversibilities ME Thermodynamics I 36 J.D. Bugg 86

89 The Clausius Inequality Carnot A 4.3 Second Law Entropy ME Thermodynamics I Second Law Entropy ME Thermodynamics I 38 J.D. Bugg 87

90 Carnot Cycle: but, so, 4.3 Second Law Entropy ME Thermodynamics I 39 Irreversible Cycle: since, C A 4.3 Second Law Entropy ME Thermodynamics I 40 J.D. Bugg 88

91 and, C A 4.3 Second Law Entropy ME Thermodynamics I 41 Entropy The integral of a system property over a cycle is zero. Therefore, is a small change in a system property. 4.3 Second Law Entropy ME Thermodynamics I 42 J.D. Bugg 89

92 for this cycle since at least part of it is irreversible 4.3 Second Law Entropy ME Thermodynamics I 43 but since, So, for an irreversible process 4.3 Second Law Entropy ME Thermodynamics I 44 J.D. Bugg 90

93 or 4.3 Second Law Entropy ME Thermodynamics I Second Law Entropy ME Thermodynamics I 46 J.D. Bugg 91

94 If entropy of system INCREASES then 4.3 Second Law Entropy ME Thermodynamics I 47 Entropy is a very useful coordinate for property diagrams. Consider the Carnot cycle. 4.3 Second Law Entropy ME Thermodynamics I 48 J.D. Bugg 92

95 also, since it is reversible Similarly, = area under the curve 4.3 Second Law Entropy ME Thermodynamics I 49 Evaluating Changes in Entropy Entropy is a property of state. Therefore, given any two independent properties it can be found. 4.3 Second Law Entropy ME Thermodynamics I 50 J.D. Bugg 93

96 First law: For a reversible process: 4.3 Second Law Entropy ME Thermodynamics I 51 Recall that, 4.3 Second Law Entropy ME Thermodynamics I 52 J.D. Bugg 94

97 1. Water, refrigerants, etc.: 2. Ideal gases: 4.3 Second Law Entropy ME Thermodynamics I Second Law Entropy ME Thermodynamics I 54 J.D. Bugg 95

98 3. Ideal gases with constant specific heats 4.3 Second Law Entropy ME Thermodynamics I 56 Isentropic Processes Ideal gas: given any 3, the 4 th can be found 4.3 Second Law Entropy ME Thermodynamics I 57 J.D. Bugg 96

99 In table A-22 (air) Only has meaning when used as a ratio Only for isentropic processes 4.3 Second Law Entropy ME Thermodynamics I 58 Ideal gas with constant specific heats: Combined with 4.3 Second Law Entropy ME Thermodynamics I 60 J.D. Bugg 97

100 These in turn give, 4.3 Second Law Entropy ME Thermodynamics I 61 Examples of Entropy Calculation Example 1: Consider an expansion of air where Calculate model for air. assuming the ideal gas 4.3 Second Law Entropy ME Thermodynamics I 62 J.D. Bugg 98

101 4.3 Second Law Entropy ME Thermodynamics I 63 Example 2: Consider an isentropic expansion of air where 4.3 Second Law Entropy ME Thermodynamics I 64 J.D. Bugg 99

102 OR 4.3 Second Law Entropy ME Thermodynamics I Second Law Entropy ME Thermodynamics I 66 J.D. Bugg 100

103 Redo Example 1 assuming that From Example 1, 4.3 Second Law Entropy ME Thermodynamics I 67 Example 3: Consider an isentropic expansion of air where 4.3 Second Law Entropy ME Thermodynamics I 68 J.D. Bugg 101

104 Table A-22 gives 4.3 Second Law Entropy ME Thermodynamics I Second Law Entropy ME Thermodynamics I 70 J.D. Bugg 102

105 Closed Systems: Entropy Balances 4.3 Second Law Entropy ME Thermodynamics I 71 Control Volume Entropy Balance: 4.3 Second Law Entropy ME Thermodynamics I 72 J.D. Bugg 103

106 Isentropic Efficiency of a Turbine First law for turbine: Second Law Entropy ME Thermodynamics I 73 Isentropic Efficiency of a Turbine 4.3 Second Law Entropy ME Thermodynamics I 74 J.D. Bugg 104

107 Isentropic Efficiency of a Compressor 4.3 Second Law Entropy ME Thermodynamics I 75 Isentropic Nozzle Efficiency 4.3 Second Law Entropy ME Thermodynamics I 76 J.D. Bugg 105

108 Work Calculations Open (Control volume) Closed (Control mass) Direct (path required)? Indirect 4.3 Second Law CV Work ME Thermodynamics I 77 Energy equation Control Volume Work Assume steady state, single inlet, single outlet, no KE or PE 4.3 Second Law CV Work ME Thermodynamics I 78 J.D. Bugg 106

109 4.3 Second Law CV Work ME Thermodynamics I Second Law CV Work ME Thermodynamics I 80 J.D. Bugg 107

110 Recall that for a closed system or But 4.3: Second Law CV Work ME Thermodynamics I Second Law CV Work ME Thermodynamics I 82 J.D. Bugg 108

111 4.3 Second Law CV Work ME Thermodynamics I 83 Work Calculations Open (Control volume) Closed (Control mass) Direct (path required) Indirect 4.3 Second Law CV Work ME Thermodynamics I 84 J.D. Bugg 109

112 Polytropic Process Consider a reversible process described by 4.3 Second Law Isentropic Proc. ME Thermodynamics I Second Law Isentropic Proc. ME Thermodynamics I 86 J.D. Bugg 110

113 4.3 Second Law Isentropic Proc. ME Thermodynamics I 87 J.D. Bugg 111

114 Vapour Compression Refrigeration Cycle 5.0 Refrigeration ME Thermodynamics I 1 Consider a Carnot cycle operating in reverse 1 2: reversible, adiabatic compression 2 3: constant temperature heat rejection 3 4: reversible, adiabatic expansion 4 1: constant temperature heat addition 5.0 Refrigeration ME Thermodynamics I 2 J.D. Bugg 112

115 5.0 Refrigeration ME Thermodynamics I 3 T Problem: 5.0 Refrigeration ME Thermodynamics I 4 J.D. Bugg 113

116 Solution: Problem: 5.0 Refrigeration ME Thermodynamics I 5 Solution: 5.0 Refrigeration ME Thermodynamics I 6 J.D. Bugg 114

117 5.0 Refrigeration ME Thermodynamics I Refrigeration ME Thermodynamics I 8 J.D. Bugg 115

118 cp 5.0 Refrigeration ME Thermodynamics I Refrigeration ME Thermodynamics I 11 J.D. Bugg 116

119 5.0 Refrigeration ME Thermodynamics I Refrigeration ME Thermodynamics I 13 J.D. Bugg 117

120 5.0 Refrigeration ME Thermodynamics I Refrigeration ME Thermodynamics I 15 J.D. Bugg 118

121 Heat Engine Refrigerator Heat Pump 5.0 Refrigeration ME Thermodynamics I 16 J.D. Bugg 119

122 Basic Rankine Cycle Reheat Vapour Power Cycles (Rankine Cycle) Regeneration Open Feedwater Heater Closed Feedwater Heater 6.0 Rankine Cycle ME Thermodynamics I 1 Carnot Cycle Boiler C T Condenser 6.0 Rankine Cycle ME Thermodynamics I 2 120

123 PROBLEM: SOLUTION: 6.0 Rankine Cycle ME Thermodynamics I 3 Basic Rankine Cycle 6.0 Rankine Cycle ME Thermodynamics I 4 121

124 PROBLEM: SOLUTION: 6.0 Rankine Cycle ME Thermodynamics I 5 Rankine Cycle - Superheat 6.0 Rankine Cycle ME Thermodynamics I 6 122

125 PROBLEM: SOLUTION: 6.0 Rankine Cycle ME Thermodynamics I 7 Rankine Cycle - Reheat 6.0 Rankine Cycle ME Thermodynamics I 8 123

126 Rankine Cycle - Reheaters 6.0 Rankine Cycle ME Thermodynamics I 9 Rankine Cycle - Reheaters 6.0 Rankine Cycle ME Thermodynamics I

127 PROBLEM: SOLUTION: 6.0 Rankine Cycle ME Thermodynamics I 11 Regeneration Feedwater 6.0 Rankine Cycle ME Thermodynamics I

128 Open Feedwater Heater 6.0 Rankine Cycle ME Thermodynamics I 13 Closed Feedwater Heater 6.0 Rankine Cycle ME Thermodynamics I

129 Rankine Cycle - Feedwater Heaters Open FWH Closed FWH 6.0 Rankine Cycle ME Thermodynamics I 15 Advanced Rankine Cycle 6.0 Rankine Cycle ME Thermodynamics I

130 Rankine Cycle - Turbine Efficiency 6.0 Rankine Cycle ME Thermodynamics I 17 Rankine Cycle Pressure Effects 6.0 Rankine Cycle ME Thermodynamics I

131 Effects of Thermal Efficiency Consider a 1 GWe power plant Increase thermal efficiency from 45% to 46%. Environmentally (assume same power output) Economically (assume same heat input) 6.0 Rankine Cycle ME Thermodynamics I 19 Combined Cycles Cycle 6.0 Rankine Cycle ME Thermodynamics I

132 Combined Cycles Topping Cycle Bottoming Cycle 6.0 Rankine Cycle ME Thermodynamics I 21 Combined Cycles 6.0 Rankine Cycle ME Thermodynamics I

133 ME Thermodynamics I Solved Examples c James D. Bugg September 2018 Department of Mechanical Engineering University of Saskatchewan

134 Problems 1. The piston-cylinder device shown contains a substance whose initial pressure is 200 kpa. Heat is added to the substance until the pressure is 400 kpa. What is the work done by the substance? (Solution: page 10) 1 cm A 1 = 0.10 m 2 Q 2. How much heat is required to evaporate 1 kg of saturated liquid water at a constant pressure of 100 kpa? (Solution: page 12) 3. Air is contained in an insulated piston-cylinder device as shown. The initial pressure and temperature is 100 kpa and 300 K respectively. The air is compressed until T = 1000 K. Find the work required to do this on a per kg basis. (Solution: page 14) 1 ME Solved Examples

135 4. Air in a piston-cylinder device has an initial pressure of 150 kpa and an initial temperature of 27 C. Atmospheric pressure is 100 kpa and the initial volume is 0.4 m 3. The piston weighs kn. Heat is added until the volume is doubled. Find the final temperature, the work done by the air, and the total heat transfer. (Solution: page 16) kn D = 0.4 m 5. Two rigid tanks containing water are connected by a valve. Tank A has an initial pressure of 400 kpa, a quality of 80%, and a volume of 0.2 m 3. Tank B has an initial pressure of 150 kpa, an initial temperature of 240 C, and a volume of 0.5 m 3. The surroundings are at 25 C. Find the final pressure and the heat transfer to the surroundings if the valve is opened. (Solution: page 19) A B ME Solved Examples 2

136 6. Air enters an afterburner at 1200 K, 12 bar, and 150 kg/s. It then passes through a nozzle where it discharges to atmosphere at 1 bar with a velocity of 1500 m/s and a temperature of 900 K. Find the amount of heat added in the afterburner (in kj/kg) and the temperature at the afterburner exit. Also find the exit diameter of the nozzle. Assume that air is an ideal gas and is the working fluid throughout. (Solution: page 22) q =? 7. A gas turbine draws air in at 100 kpa and 27 C and discharges to atmosphere at 100 kpa. The temperature at the inlet of the combustion chamber is 227 C. The combustion chamber adds 640 kj/kg to the air. The processes in the compressor and turbine are polytropic with the same polytropic index. Assume that air is the working fluid throughout and that it is an ideal gas. Find the air temperature at the turbine inlet, the net work output, and the mass flowrate required to develop 1 MW net power output. (Solution: page 25) 640 kj/kg 227 C C T 27 C 100 kpa 100 kpa 3 ME Solved Examples

137 8. An evacuated chamber slowly receives steam from a supply at 10 bar and 400 C. The chamber is insulated. Find the temperature of the water in the tank once it reaches 10 bar. (Solution: page 29) Steam Supply vacuum 10 bar 400 C 9. A 2 m 3, perfectly insulated tank initially contains air at 10 bar and 20 C. A valve is opened and the air slowly leaks out until the pressure is 1 bar. Find the final temperature of the air which remains in the tank. Assume that air is an ideal gas with c v constant at 720 J/(kg K). (Solution: page 31) 10 bar 20 C Air 1 bar ME Solved Examples 4

138 10. A machine is developed which accepts two streams of air as shown and produces work. If the mass flowrate of the high temperature supply is three times that of the low temperature supply, what is the maximum theoretical work. Express the answer in kj/kg based on the total mass flow through the device. Assume that air is an ideal gas and that the device is insulated. Also, potential and kinetic energy eÿects are negligible. (Solution: page 34) 800 K 2 bar 300 K 3 bar 1 bar Ẇ cv 11. An ideal Rankine cycle operates with a boiler pressure of 8 MPa and a condenser pressure of 50 kpa. The inlet of the turbine is a saturated vapour. The exit of the condenser is a saturated liquid. Find the turbine work, the heat added in the boiler, and the thermal e ciency of the cycle. What is the e ciency of a Carnot cycle operating between these temperature limits? What is the quality at the turbine exit? (Solution: page 37) Boiler Pump Turbine Condenser 5 ME Solved Examples

139 12. An ideal Rankine cycle operates with a boiler pressure of 8 MPa and a condenser pressure of 50 kpa. The exit of the condenser is a saturated liquid. A superheater provides 200 C of superheat after the boiler. Find the thermal e ciency of the cycle. What is the quality at the turbine exit? (Solution: page 41) Boiler Superheater Pump Turbine Condenser 13. An ideal Rankine cycle operates with a boiler pressure of 8 MPa and a condenser pressure of 50 kpa. The exit of the condenser is a saturated liquid. A superheater provides 200 C of superheat after the boiler. A reheater is placed at 800 kpa. What is the minimum amount of reheat required to eliminate liquid from the second turbine stage? Find the thermal e ciency of the cycle. (Solution: page 43) Boiler/Superheater Turbine #1 Pump Reheater Turbine #2 Condenser ME Solved Examples 6

140 14. An ideal Rankine cycle operates with a boiler pressure of 8 MPa and a condenser pressure of 50 kpa. The exit of the condenser is a saturated liquid. A superheater provides 200 C of superheat after the boiler. 5% of the flow through the first stage turbine is extracted and combined with the feedwater in an open feedwater heater. The extraction pressure is 800 kpa. Find the thermal e ciency of the cycle. (Solution: page 46) Boiler/Superheater Pump #2 Turbine #1 Feedwater Heater Pump #1 Turbine #2 Condenser 15. An ideal Rankine cycle operates with a boiler pressure of 8 MPa and a condenser pressure of 50 kpa. The exit of the condenser is a saturated liquid. A superheater provides 200 C of superheat after the boiler. 5% of the flow through the first stage turbine is extracted and passes through a closed feedwater heater into the condenser. The extraction pressure is 800 kpa. The feedwater side of the feedwater heater is at the boiler pressure. Find the thermal e ciency of the cycle. (Solution: page 50) Boiler/Superheater Turbine #1 Closed Feedwater Heater Turbine #2 Pump Condenser 7 ME Solved Examples

141 16. A vapour compression refrigeration cycle uses R134a as its working fluid. The pressure in the evaporator is 100 kpa and the pressure in the condenser is 700 kpa. The inlet and outlet temperatures for the compressor are 20 C and 50 C respectively. The temperature at the outlet of the condenser is 24 C. Calculate the coe cient of performance. (Solution: page 52) Condenser Compressor Evaporator ME Solved Examples 8

142 Solutions 9 ME Solved Examples

143 Problem 1 Given: A piston-cylinder device Initial pressure is 200 kpa Pressure increased to 400 kpa by heat addition P 1 = 200 kpa 1 cm A 1 = 0.10 m 2 system boundary Q Find: Find the work done by the substance. Assumptions: Frictionless piston Quasi-equilibrium process Analysis: Choose the substance as the system. The pressure-volume plot will be as follows: ME Solved Examples 10

144 P (kpa) V 1W 3 = 1 W W 3 2W 3 = 0 (constant volume process) 1W 2 = Z 2 1 P dv = P Z 2 1 dv 1W 2 = P (V 2 V 1 ) = P A(x 2 x 1 ) 1W 2 = 200[kPa]0.10[m 2 ]0.01[m] 1W 2 = 0.2 kj = 200 J The work done by the substance is 200 J. 11 ME Solved Examples

145 Problem 2 Given: A piston-cylinder device containing 1 kg of saturated liquid water. The water evaporates at a constant pressure of 100 kpa. P 1 = P 2 = 100 kpa W W 1 Saturated Liquid system boundary 2 Saturated Vapour Find: How much heat is required. Assumptions: Constant pressure process Initial and final states are at equilibrium Quasi-equilibrium process Analysis: METHOD I: 1Q 2 1 W 2 = U 1Q 2 = (U 2 U 1 ) + 1 W 2 ME Solved Examples 12

146 1Q 2 = m(u 2 u 1 ) + mp (v 2 v 1 ) 1Q 2 = m(u g u f ) + mp (v g v f ) Properties can be obtained from Table A-3. 1Q 2 = 1[kg]( )[kJ/kg] + 1[kg]10 5 [Pa]( )[m 3 /kg] = 2258 kj The heat required to evaporate the water is 2258 kj. METHOD II: Since this is a constant pressure process, 1Q 2 = (H 2 H 1 ) 1Q 2 = m(h 2 h 1 ) 1Q 2 = m(h g h f ) = mh fg 1Q 2 = 1[kg]2258[kJ/kg] = 2258 kj The heat required to evaporate the water is 2258 kj by this method also. 13 ME Solved Examples

147 Problem 3 Given: Air in an insulated piston-cylinder device Compressed until T 2 = 1000 K. P 1 = 100 kpa T 1 = 300 K system boundary Find: Work required. Assumptions: Air is an ideal gas Perfectly insulated Initial and final conditions are equilibrium states Analysis: Define the system boundary as the contents of the piston cylinder device (air). 1q 2 1 w 2 = u 2 u 1 1q 2 = 0 (insulated) Therefore 1 w 2 = u 2 u 1 ME Solved Examples 14

148 1w 2 = u 1 u 2 From Table A 22 T (K) u(kj/kg) w 2 = = kj/kg The work required is kj/kg What if we had used constant specific heats? 1 w 2 = u = Z 2 1 Z 2 c v dt = c v dt 1 1w 2 = c v (T 2 T 1 ) 1w 2 = c v (T 1 T 2 ) From Table A-20 we can try various values for the specific heat. T (K) c v (kj/(kg K)) w(kj/kg) ME Solved Examples

149 Problem 4 Given: Air in a piston-cylinder device Heat is added until the volume is doubled system boundary kn D = 0.4 m P atm = 100 kpa V 1 = 0.4 m 3 P 1 = 150 kpa T 1 = 27 C Find: (a) Find the final temperature, (b) the work done by the air, and (c) the total heat transfer. Assumptions: Air behaves as an ideal gas Quasi-equilibrium process Frictionless piston No change in K.E. or P.E. Analysis: Define the system as shown. (a) P V = mrt (ideal gas law) ME Solved Examples 16

150 Therefore, P 1 V 1 T 1 = P 2V 2 T 2 = P 3V 3 T 3 The pressure will increase until the piston begins to move. Therefore, the pressurevolume diagram will be as follows: P V Need to find the final pressure. Consider a free-body diagram of the piston. P atm A P 2 A W P atm A + W = P 2 A P 2 = P atm + W A = 100[kPa] [kN] = 350 kpa π(0.4) 2 /4[m 2 ] P 1 V 1 T 1 = P 3V 3 T 3 17 ME Solved Examples

151 P3 V3 T 3 = T 1 P 1 V 1 T 3 = 350 (2) 300[K] = 1400 K 150 The final temperature is 1400 K. (b) 1W 3 = 1 W W 3 1W 2 = 0 (constant volume process) 1W 3 = Z 3 2 P dv = P (V 3 V 2 ) = 350[kPa]0.4[m 3 ] = 140 kj The work required is 140 kj. (c) 1Q 3 1 W 3 = U 3 U 1 1Q 3 1 W 3 = m(u 3 u 1 ) From Table A-22, T (K) u(kj/kg) m = P 1V 1 = 150, 000[Pa]0.4[m3 ] = kg RT 1 287[J/(kg K)]300[K] 1Q 3 = [kg]( )[kJ/kg] + 140[kJ] = 767 kj The total heat transfer is 767 kj. ME Solved Examples 18

152 Problem 5 Given: Two rigid tanks connected by a valve Surroundings are at 25 C Tank A contains water, 0.2 m 3, 400 kpa, 80% quality. Tank B contains water, 0.5 m 3, 150 kpa, 240 C. A B system boundary Find: Find the final pressure and the heat transfer to surroundings if the valve is opened. Assumptions: Initial states in equilibrium Final state is in equilibrium No change in kinetic or potential energy Analysis: Define the system boundary as shown. This is a closed system. Therefore, 1Q 2 1 W 2 = U 2 U 1 There is no work transfer. Therefore, 1Q 2 = U 2 U 1 19 ME Solved Examples

153 Consider U 1 first. We cannot look up a single U because the properties are not uniform. Therefore, consider U 1 = U A1 + U B1 (extensive property) TANK A: At 400 kpa, v f = m 3 /kg, v g = m 3 /kg, u f = kj/kg, and u g = kj/kg. Therefore, for x = 0.8, v = ( ) = m 3 /kg u = ( ) = kj/kg Now, Therefore, TANK B: m A = V A v = 0.2[m 3 ] [m 3 /kg] = kg U A1 = [kg] [kJ/kg] = kj At 150 kpa and 240 o C, v = m 3 /kg and u = kj/kg. Therefore, Therefore, Therefore, m B = V B v = 0.5[m 3 ] = kg 1.570[m 3 /kg] U B1 = [kg]2717.2[kJ/kg] = kj U 1 = U A1 + U B1 = kj Now, what is U 2? ME Solved Examples 20

154 We know that the temperature is 25 C. Also, the final specific volume can be calculated. v 2 = ( )[m 3 ] ( )[kg] = m3 /kg At 25 C, v f = m 3 /kg and v g = m 3 /kg. Therefore, the final state is a mixture. x 2 = v 2 v f v g v f = At 25 C, u f = kj/kg and u g = kj/kg. u 2 = ( ) = kj/kg Therefore, U 2 = (m A + m B )u 2 = ( )[kg]148.16[kJ/kg] = kj 1Q 2 = U 2 U 1 = [kJ] [kJ] = 1907 kj The heat transfer to the surroundings is 1907 kj. Since the final state is a liquid-vapour mixture, P = P sat for 25 C. The final pressure is kpa. 21 ME Solved Examples

155 Problem 6 Given: Air is supplied to an afterburner at 1200 K and 12 bar at 150 kg/s Air leaves the afterburner and enters a nozzle The air leaves the nozzle at 1 bar, 900 K, and 1500 m/s q =? 150 kg/s 12 bar 1200 K K 1 bar 1500 m/s Find: (a) Heat added in the afterburner (b) Temperature between afterburner and nozzle (c) Nozzle exit diameter Assumptions: Steady state Air behaves as an ideal gas Negligible kinetic energy at 1 and 2 Potential energy is negligible everywhere No heat transfer in the nozzle Analysis: Consider the control volume shown. ME Solved Examples 22

156 q =? (a) de cv dt = Q cv Ẇcv + X inlets With the assumptions listed this becomes ṁ i h i + V i 2! 2 + gz i X exits ṁ e h e + V e 2! 2 + gz e 0 = Q ab + ṁh 1 ṁ(h 3 + V 2 3 /2) or q ab = h 3 + V 2 3 /2 h 1 The enthalpy can be found in Table A-22. q ab = " # J kg 2 " # m s 2 " # J = kj/kg kg The heat added in the afterburner is kj/kg. (b) Now put a control volume around only the afterburner. de cv dt = Q cv Ẇcv + X inlets With the assumptions listed this becomes ṁ i h i + V i 2! 2 + gz i X exits ṁ e h e + V e 2! 2 + gz e 0 = q ab + h 1 h 2 h 2 = = kj/kg T 2 = 1844 K (TableA 22 interpolation) The temperature at the afterburner exit is 1844 K. 23 ME Solved Examples

157 (c) ṁ = A 3V 3 v 3 = A 3V 3 P 3 RT 3 A 3 = ṁrt 3 V 3 P 3 = 150[kg/s]0.287[kJ/(kg K)]900[K] 1500[m/s]100[kPa] = m 2 The nozzle exit diameter is m. ME Solved Examples 24

158 Problem 7 Given: A gas turbine as shown 640 kj/kg 227 C 2 3 C T 27 C 100 kpa kpa Find: (a) Temperature at turbine inlet (b) Net work output (c) Flowrate to develop 1 MW Assumptions: Air behaves as an ideal gas Steady flow throughout Kinetic and potential energy changes negligible throughout No heat transfer in compressor or turbine No pressure change in combustion chamber 25 ME Solved Examples

159 Analysis: (a) Consider the first law applied to the combustion chamber. de cv dt = Q cv Ẇcv + X inlets ṁ i h i + V i 2! 2 + gz i X exits ṁ e h e + V e 2! 2 + gz e With the assumptions listed this becomes 0 = Q cv + ṁh 2 ṁh 3 q cc = h 3 h 2 From Table A-22, h 2 = kj/kg at T 2 = 500 K. 640[kJ/kg] = h [kJ/kg] h 3 = kj/kg From Table A 22, T 3 = K. The temperature at the turbine inlet is 1084 K. (b) Consider the first law applied to the turbine. de cv dt = Q cv Ẇcv + X inlets ṁ i h i + V i 2! 2 + gz i X exits ṁ e h e + V e 2! 2 + gz e With the assumptions listed this becomes 0 = Ẇt + ṁh 3 ṁh 4 w t = h 3 h 4 We do not know T 4 so h 4 is unknown. However, both the compressor and turbine are polytropic so P v n = constant ME Solved Examples 26

160 Also P v = RT or P v T = R and therefore v 4 v 3 = T 4 T 3 P 3 P 4 Across the turbine P 3 v n 3 = P 4v n 4 P 3 P 4 = v4 v 3 n = T4 P 3 T 3 P 4 n 1 n n P3 T4 = P 4 T 3 P 3 P 4 = T4 T 3 n 1 n Similarly, across the compressor P 2 P 1 = T1 T 2 Now, if P 3 = P 2 n 1 n P 2 P 1 = P 3 P 4 Therefore, T 4 T 3 = T 1 T 2 T1 T 4 = T 3 = [K] T 2 T 4 = K From Table A-22, h 4 = kj/kg by interpolation. w t = = kj/kg 27 ME Solved Examples

161 Likewise, for the compressor w c = h 1 h 2 w c = = kj/kg w net = w t + w c = = kj/kg The net work of the gas turbine is kj/kg. (c) Ẇ net = w net ṁ ṁ = 1000[kJ/s] [kJ/kg] = kg/s The mass flowrate through the gas turbine is 3.58 kg/s. ME Solved Examples 28

162 Problem 8 Given: An evacuated chamber. Steam Supply vacuum 10 bar 400 C Find: Final temperature in chamber. Assumptions: No heat transfer No work transfer No change in kinetic or potential energy Analysis: Consider the control volume shown. de cv dt = Q cv Ẇcv + X inlets ṁ i h i + V i 2! 2 + gz i X exits ṁ e h e + V e 2! 2 + gz e For the assumptions listed this becomes du cv dt = ṁ i h i Mass conservation for the same control volume gives dm cv dt = X ṁ i X ṁ e = ṁ i inlets exits 29 ME Solved Examples

163 U cv = m cv u cv d dt (m dm cv cvu cv ) = h i dt Z 2 1 d(m cv u cv ) = h i Z 2 1 dm cv Drop the cv subscript since it is on all variables. (mu) 2 (mu) 1 = h i (m 2 m 1 ) u 2 = h i At P = 10 bar and T = 400 o C, h = kj/kg. Therefore, u 2 = kj/kg and P 2 = 10 bar Therefore, T 2 = ( ) = 581 o C The final temperature is 581 o C. ME Solved Examples 30

164 Problem 9 Given: A perfectly insulated tank. 10 bar 20 o C Air 1 bar Find: Final air temperature Assumptions: No heat transfer No work transfer No change in kinetic or potential energy Constant specific heats Properties uniform in the tank Analysis: Consider the control volume shown. de cv dt = Q cv Ẇcv + X inlets ṁ i h i + V i 2! 2 + gz i X exits ṁ e h e + V e 2! 2 + gz e For the assumptions listed this becomes du cv dt = ṁ e h e 31 ME Solved Examples

165 Conservation of mass for the same control volume gives dm cv dt = X ṁ i X ṁ e = ṁ e inlets exits U cv = m cv u cv d(m cv u cv ) dt = h e dm cv dt Note that h e = h cv and drop the cv subscript. d(mu) dt = h dm dt m du dt + udm dt = (u + P v)dm dt Cancel u dm/dt from both sides and recall that P v = RT. m du dt = RT dm dt du = c v dt mc v dt dt = RT dm dt c v 1 dt R T dt = 1 dm m dt Z 2 1 Z c v dt 2 R T = dm 1 m c v R ln T2 = ln T 1 m2 m 1 cv/r T2 T 1 = m 2 = P 2V RT 1 m 1 RT 2 P 1 V ME Solved Examples 32

166 cv/r T2 T 1 = P 2 P 1 T 1 T 2 T2 T 1 cv R +1 = P 2 P 1 P2 T 2 = T 1 P 1 R cv+r T 2 = 293[K] = 152 K = C The final temperature of the air which remains in the tank is 121 C. 33 ME Solved Examples

167 Problem 10 Given: A chamber which mixes two air streams and produces work 800 K 2 bar 1 Ẇ cv ṁ 1 = 3ṁ K bar 1 bar Find: Maximum theoretical work (per kg total flow) Assumptions: Kinetic and potential energy changes negligible. No heat transfer. Reversible process. Air behaves as an ideal gas. Steady state Analysis: Consider the control volume shown. Mass: dm cv dt = X inlets ṁ i X exits ṁ e ṁ 1 + ṁ 2 = ṁ 3 ME Solved Examples 34

168 ṁ 1 = 3ṁ 2 3ṁ 2 + ṁ 2 = ṁ 3 4ṁ 2 = ṁ 3 ṁ 1 = 3 4ṁ3 Energy: de cv dt = Q cv Ẇcv + X inlets ṁ i h i + V i 2! 2 + gz i X exits ṁ e h e + V e 2! 2 + gz e With the assumptions listed this becomes Ẇ cv = ṁ 1 h 1 + ṁ 2 h 2 ṁ 3 h 3 Ẇ cv = 3 4ṁ3h ṁ3h 2 ṁ 3 h 3 Need to find h 3. Entropy balance: ds cv dt = X j Q j T j + X ṁ i s i X ṁ e s e + σ cv inlets exits For a steady, adiabatic, and reversible process this reduces to 0 = ṁ 1 s 1 + ṁ 2 s 2 ṁ 3 s 3. We have Since s B s A = s o B so A R ln PB P A ṁ 1 = 3ṁ 2 35 ME Solved Examples

169 and ṁ 3 = ṁ 1 + ṁ 2 0 = 3ṁ 2 s 1 + ṁ 2 s 2 (3ṁ 2 + ṁ 2 )s 3 0 = 3ṁ 2 (s 1 s 3 ) + ṁ 2 (s 2 s 3 ) 0 = 3 s o 1 so 3 R ln P1 P 3 + s o 2 so 3 R ln P2 P 3 From Table A-22 at 800 K, s o 1 = kj/(kg K), and h 1 = kj/kg. From Table A-22 at 300 K, s o 2 = kj/(kg K), and h 2 = kj/kg. " # " #! kj 0 = s o3 (kg K) kj ln(2) (kg K) " # " #! kj s o3 (kg K) kj ln(3) (kg K) s o 3 = kj/(kg K) which gives h 3 = kj/kg from Table A-22 (T = 508 K). Now return to the energy equation.! Ẇ cv = 3(821.95) = kj/kg ṁ The maximum theoretical work that can be produced by this machine is kj/kg. ME Solved Examples 36

170 Problem 11 Given: Ideal Rankine cycle 2 Boiler P 3 = 8 MPa, sat. vap. 3 Pump Turbine 1 sat. liq. Condenser 50 kpa 4 Find: (a) Turbine work (b) Heat added (c) Thermal e ciency (d) Carnot e ciency (e) Quality at turbine exit Assumptions: Steady state Negligible change in kinetic and potential energy 37 ME Solved Examples

171 Analysis: Energy equation: de cv dt = Q cv Ẇcv + X inlets ṁ i h i + V i 2! 2 + gz i X exits ṁ e h e + V e 2! 2 + gz e With the assumptions listed this becomes w = q + h i h e (e) Turbine: (q = 0, isentropic) w t = h 3 h 4 State 3 is saturated vapour at 80 bar. Therefore, h 3 = kj/kg and s 3 = kj/(kg K). s 4 = s 3 and P 4 = 50 kpa At 50 kpa, s f = kj/(kg K), s g = kj/(kg K), h f = kj/kg, and h fg = kj/kg. x 4 = s 4 s f = s g s f = The quality at the turbine exit is 71.54%. (a) h 4 = h f + x 4 h fg = (2305.4) = kj/kg w t = h 3 h 4 = = kj/kg The turbine work is kj/kg. (b) Boiler: (w = 0) q H = h 3 h 2 ME Solved Examples 38

172 Pump: (q = 0, isentropic) w p = h 1 h 2 h 1 = h f at 50 kpa h 1 = kj/kg Z 2 w p = vdp 1 Assume that v is constant across the pump. w p = v 1 (P 2 P 1 ) v 1 = v f at 50 kpa v 1 = m 3 /kg Therefore, w p = [m 3 /kg]( )[kPa] = kj/kg h 2 = h 1 w p h 2 = = kj/kg Therefore, q H = ( )[kJ/kg] = kj/kg The heat supplied by the boiler is kj/kg. (c) η = w net q H = w t + w p = = q H The thermal e ciency is 31.55%. (d) η c = 1 T C = 1 T H = The Carnot e ciency is 37.62%. 39 ME Solved Examples