SSL9 The Second Law of Thermodynamics

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Introduction to the second law
Thermal energy reservoirs
Heat engines
Thermal efficiency
The 2nd law: Kelvin-Planck statement
Refrigerators and heat pumps
Coefficient of performance (COP)
The 2nd law: Clasius statement
Perpetual motion machines
Reversible and irreversible processes
Irreversibility's, Internal and externally reversible processes
The Carnot cycle
The reversed Carnot cycle
The Carnot principles
The thermodynamic temperature scale
The Carnot heat engine
The quality of energy
The Carnot refrigerator and heat pump

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  • A process must satisfy both the first and second laws of thermodynamics to proceed.\n
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  • Qin = amount of heat supplied to steam in boiler from a high temperature source (furnace)\nQout = amount of heat rejected from steam in condenser to a low-temperature sink (the atmosphere, a river, etc...)\nWin = amount of work required to compress water to boiler pressure\nWout = amount of work delivered by steam as it expands in turbine\n\n
  • Qin = amount of heat supplied to steam in boiler from a high temperature source (furnace)\nQout = amount of heat rejected from steam in condenser to a low-temperature sink (the atmosphere, a river, etc...)\nWin = amount of work required to compress water to boiler pressure\nWout = amount of work delivered by steam as it expands in turbine\n\n
  • Qin = amount of heat supplied to steam in boiler from a high temperature source (furnace)\nQout = amount of heat rejected from steam in condenser to a low-temperature sink (the atmosphere, a river, etc...)\nWin = amount of work required to compress water to boiler pressure\nWout = amount of work delivered by steam as it expands in turbine\n\n
  • Qin = amount of heat supplied to steam in boiler from a high temperature source (furnace)\nQout = amount of heat rejected from steam in condenser to a low-temperature sink (the atmosphere, a river, etc...)\nWin = amount of work required to compress water to boiler pressure\nWout = amount of work delivered by steam as it expands in turbine\n\n
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  • In a household refrigerator, the freezer compartment where heat is absorbed by the refrigerant serves as the evaporator, and the coils usually behind the refrigerator where heat is dissipated to the kitchen air serve as the condenser.\n
  • Can the value of COPR be greater than unity?\n
  • Can the value of COPR be greater than unity?\n
  • Can the value of COPR be greater than unity?\n
  • Most heat pumps in operation today have a seasonally averaged COP of 2 to 3.\nMost existing heat pumps use the cold outside air as the heat source in winter (air-source HP).\nIn cold climates their efficiency drops considerably when temperatures are below the freezing point. \nIn such cases, geothermal (ground-source) HP that use the ground as the heat source can be used.\nSuch heat pumps are more expensive to install, but they are also more efficient.\nAir conditioners are basically refrigerators whose refrigerated space is a room or a building instead of the food compartment.\nThe COP of a refrigerator decreases with decreasing refrigeration temperature. \nTherefore, it is not economical to refrigerate to a lower temperature than needed.\n
  • Most heat pumps in operation today have a seasonally averaged COP of 2 to 3.\nMost existing heat pumps use the cold outside air as the heat source in winter (air-source HP).\nIn cold climates their efficiency drops considerably when temperatures are below the freezing point. \nIn such cases, geothermal (ground-source) HP that use the ground as the heat source can be used.\nSuch heat pumps are more expensive to install, but they are also more efficient.\nAir conditioners are basically refrigerators whose refrigerated space is a room or a building instead of the food compartment.\nThe COP of a refrigerator decreases with decreasing refrigeration temperature. \nTherefore, it is not economical to refrigerate to a lower temperature than needed.\n
  • Most heat pumps in operation today have a seasonally averaged COP of 2 to 3.\nMost existing heat pumps use the cold outside air as the heat source in winter (air-source HP).\nIn cold climates their efficiency drops considerably when temperatures are below the freezing point. \nIn such cases, geothermal (ground-source) HP that use the ground as the heat source can be used.\nSuch heat pumps are more expensive to install, but they are also more efficient.\nAir conditioners are basically refrigerators whose refrigerated space is a room or a building instead of the food compartment.\nThe COP of a refrigerator decreases with decreasing refrigeration temperature. \nTherefore, it is not economical to refrigerate to a lower temperature than needed.\n
  • Most heat pumps in operation today have a seasonally averaged COP of 2 to 3.\nMost existing heat pumps use the cold outside air as the heat source in winter (air-source HP).\nIn cold climates their efficiency drops considerably when temperatures are below the freezing point. \nIn such cases, geothermal (ground-source) HP that use the ground as the heat source can be used.\nSuch heat pumps are more expensive to install, but they are also more efficient.\nAir conditioners are basically refrigerators whose refrigerated space is a room or a building instead of the food compartment.\nThe COP of a refrigerator decreases with decreasing refrigeration temperature. \nTherefore, it is not economical to refrigerate to a lower temperature than needed.\n
  • Most heat pumps in operation today have a seasonally averaged COP of 2 to 3.\nMost existing heat pumps use the cold outside air as the heat source in winter (air-source HP).\nIn cold climates their efficiency drops considerably when temperatures are below the freezing point. \nIn such cases, geothermal (ground-source) HP that use the ground as the heat source can be used.\nSuch heat pumps are more expensive to install, but they are also more efficient.\nAir conditioners are basically refrigerators whose refrigerated space is a room or a building instead of the food compartment.\nThe COP of a refrigerator decreases with decreasing refrigeration temperature. \nTherefore, it is not economical to refrigerate to a lower temperature than needed.\n
  • Most heat pumps in operation today have a seasonally averaged COP of 2 to 3.\nMost existing heat pumps use the cold outside air as the heat source in winter (air-source HP).\nIn cold climates their efficiency drops considerably when temperatures are below the freezing point. \nIn such cases, geothermal (ground-source) HP that use the ground as the heat source can be used.\nSuch heat pumps are more expensive to install, but they are also more efficient.\nAir conditioners are basically refrigerators whose refrigerated space is a room or a building instead of the food compartment.\nThe COP of a refrigerator decreases with decreasing refrigeration temperature. \nTherefore, it is not economical to refrigerate to a lower temperature than needed.\n
  • Most heat pumps in operation today have a seasonally averaged COP of 2 to 3.\nMost existing heat pumps use the cold outside air as the heat source in winter (air-source HP).\nIn cold climates their efficiency drops considerably when temperatures are below the freezing point. \nIn such cases, geothermal (ground-source) HP that use the ground as the heat source can be used.\nSuch heat pumps are more expensive to install, but they are also more efficient.\nAir conditioners are basically refrigerators whose refrigerated space is a room or a building instead of the food compartment.\nThe COP of a refrigerator decreases with decreasing refrigeration temperature. \nTherefore, it is not economical to refrigerate to a lower temperature than needed.\n
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  • The Kelvin–Planck and the Clausius statements are equivalent in their consequences, and either statement can be used as the expression of the second law of thermodynamics. \nAny device that violates the Kelvin–Planck statement also violates the Clausius statement, and vice versa.\n
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  • Some processes are more irreversible than others.\n\nWe try to approximate reversible processes. Why?\n
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  • http://en.wikipedia.org/wiki/Carnot_cycle\n
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  • How do you increase the thermal efficiency of a Carnot heat engine? How about for actual heat engines? \n
  • How do you increase the COP of a Carnot refrigerator or heat pump? How about for actual ones? \n
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  • SSL9 The Second Law of Thermodynamics

    1. 1. SECOND LAW OFTHERMODYNAMICS Lecture 9 Keith Vaugh BEng (AERO) MEngReference text: Chapter 7 - Fundamentals of Thermal-Fluid Sciences, 3rd Edition Yunus A. Cengel, Robert H. Turner, John M. Cimbala McGraw-Hill, 2008 KEITH VAUGH
    2. 2. OBJECTIVES } KEITH VAUGH
    3. 3. OBJECTIVESIntroduce the second law of thermodynamics. } KEITH VAUGH
    4. 4. OBJECTIVESIntroduce the second law of thermodynamics.Identify valid processes as those that satisfy boththe first and second laws of thermodynamics. } KEITH VAUGH
    5. 5. OBJECTIVESIntroduce the second law of thermodynamics.Identify valid processes as those that satisfy boththe first and second laws of thermodynamics. }Discuss thermal energy reservoirs, reversible andirreversible processes, heat engines, refrigerators,and heat pumps. KEITH VAUGH
    6. 6. OBJECTIVESIntroduce the second law of thermodynamics.Identify valid processes as those that satisfy boththe first and second laws of thermodynamics. }Discuss thermal energy reservoirs, reversible andirreversible processes, heat engines, refrigerators,and heat pumps.Describe the Kelvin–Planck and Clausiusstatements of the second law of thermodynamics. KEITH VAUGH
    7. 7. OBJECTIVESIntroduce the second law of thermodynamics.Identify valid processes as those that satisfy boththe first and second laws of thermodynamics. }Discuss thermal energy reservoirs, reversible andirreversible processes, heat engines, refrigerators,and heat pumps.Describe the Kelvin–Planck and Clausiusstatements of the second law of thermodynamics.Discuss the concepts of perpetual-motionmachines. KEITH VAUGH
    8. 8. }KEITH VAUGH
    9. 9. Apply the second law of thermodynamics to cyclesand cyclic devices. } KEITH VAUGH
    10. 10. Apply the second law of thermodynamics to cyclesand cyclic devices.Apply the second law to develop the absolute }thermodynamic temperature scale. KEITH VAUGH
    11. 11. Apply the second law of thermodynamics to cyclesand cyclic devices.Apply the second law to develop the absolute }thermodynamic temperature scale.Describe the Carnot cycle. KEITH VAUGH
    12. 12. Apply the second law of thermodynamics to cyclesand cyclic devices.Apply the second law to develop the absolute }thermodynamic temperature scale.Describe the Carnot cycle.Examine the Carnot principles, idealised Carnotheat engines, refrigerators, and heat pumps. KEITH VAUGH
    13. 13. Apply the second law of thermodynamics to cyclesand cyclic devices.Apply the second law to develop the absolute }thermodynamic temperature scale.Describe the Carnot cycle.Examine the Carnot principles, idealised Carnotheat engines, refrigerators, and heat pumps.Determine the expressions for the thermalefficiencies and coefficients of performance forreversible heat engines, heat pumps, andrefrigerators. KEITH VAUGH
    14. 14. INTRODUCTION TO THE SECOND LAW These processes cannot occur even though they are not in violation of the first law. }A cup of hot coffee does Transferring heat to a paddle Transferring heat to a wire willnot get hotter in a wheel will not cause it to not generate electricity.cooler room. rotate. KEITH VAUGH
    15. 15. occur in a certain directionPROCESSES & not in the reverse direction KEITH VAUGH
    16. 16. Major uses of the second lawThe second law may be used to identify thedirection of processes.The second law also asserts that energy has qualityas well as quantity. The first law is concerned withthe quantity of energy and the transformations ofenergy from one form to another with no regard toits quality. The second law provides the necessarymeans to determine the quality as well as thedegree of degradation of energy during a process.The second law of thermodynamics is also used indetermining the theoretical limits for theperformance of commonly used engineeringsystems, such as heat engines and refrigerators, aswell as predicting the degree of completion ofchemical reactions. KEITH VAUGH
    17. 17. THERMAL ENERGY RESERVOIRS A hypothetical body with a relatively large thermal energy capacity (mass x specific heat) that can supply or absorb } finite amounts of heat without undergoing any change in temperature isBodies with relatively large called a thermal energy reservoir, or justthermal masses can be modelled a reservoir.as thermal energy reservoirs. In practice, large bodies of water such as oceans, lakes, and rivers as well as the atmospheric air can be modelled accurately as thermal energy reservoirs because of their large thermal energy storage capabilities or thermal masses. KEITH VAUGH
    18. 18. HEAT ENGINESThe devices that convert heat to work ✓ They receive heat from a high-temperature source (solar } energy, oil furnace, nuclear reactor, etc.) ✓ They convert part of this heat to work (usually in the form of a rotating shaft.) ✓ They reject the remaining waste heat to a low-temperature sink (the atmosphere, rivers, etc.) ✓ They operate on a cycleHeat engines and other cyclic devices usually involve afluid to and from which heat is transferred whileundergoing a cycle. This fluid is called the working fluid KEITH VAUGH
    19. 19. Work can always beconverted to heat directly and completely, but the reverse is not true KEITH VAUGH
    20. 20. A source supplies energy in theform of heat and a sinkabsorbs it KEITH VAUGH
    21. 21. A source supplies energy in theform of heat and a sinkabsorbs it Part of the heat received by a heat engine is converted to work, while the rest is rejected to a sink KEITH VAUGH
    22. 22. Steam power plant KEITH VAUGH
    23. 23. Steam power plant A portion of the work output of a heat engine is consumed internally to maintain continuous operation. KEITH VAUGH
    24. 24. Steam power plant A portion of the work output of a heat engine is consumed internally to maintain continuous operation. Wnet , out = Wout − Win ( kJ ) Wnet , out = Qin − Qout ( kJ ) KEITH VAUGH
    25. 25. Thermal efficiencySome heat engines perform better than others(convert more of the heat they receive to work). KEITH VAUGH
    26. 26. Thermal efficiency Net work output Thermal efficiency = Total heat inputSome heat engines perform better than others(convert more of the heat they receive to work). KEITH VAUGH
    27. 27. Thermal efficiency Net work output Thermal efficiency = Total heat input Wnet , out ηth = QinSome heat engines perform better than others(convert more of the heat they receive to work). KEITH VAUGH
    28. 28. Thermal efficiency Net work output Thermal efficiency = Total heat input Wnet , out ηth = Qin Qout ηth = 1− QinSome heat engines perform better than others(convert more of the heat they receive to work). KEITH VAUGH
    29. 29. Thermal efficiency Net work output Thermal efficiency = Total heat input Wnet , out ηth = Qin Qout ηth = 1− Qin Wnet , out = Qin − QoutSome heat engines perform better than others(convert more of the heat they receive to work). KEITH VAUGH
    30. 30. Schematic ofa heat engine KEITH VAUGH
    31. 31. Schematic ofa heat engine KEITH VAUGH
    32. 32. Schematic of Wnet , out = QH − QLa heat engine Wnet , out ηth = QH QL ηth = 1− QH KEITH VAUGH
    33. 33. Schematic of Even the most efficient heat Wnet , out = QH − QLa heat engine engines reject almost one-half of Wnet , out the energy they receive as waste ηth = heat QH QL ηth = 1− QH KEITH VAUGH
    34. 34. Can we save Qout?In a steam power plant thecondenser is the device wherelarge quantities of waste heatis rejected to rivers, lakes, orthe atmosphere.Can we not just take thecondenser out of the plant andsave all that waste energy? A heat-engine cycle cannot be completed without rejecting someNo - without a heat rejection heat to a low-temperature sink.process in a condenser thecycle cannot be completed. KEITH VAUGH
    35. 35. Can we save Qout?In a steam power plant thecondenser is the device wherelarge quantities of waste heatis rejected to rivers, lakes, orthe atmosphere.Can we not just take thecondenser out of the plant andsave all that waste energy? A heat-engine cycle cannot be completed without rejecting someNo - without a heat rejection heat to a low-temperature sink.process in a condenser thecycle cannot be completed. Every heat engine must waste some energy by transferring it to a low-temperature reservoir in order to complete the cycle, even under idealised conditions. KEITH VAUGH
    36. 36. The second law of thermodynamics:Kelvin - Planck statement It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work. No heat engine can have a thermal efficiency of 100 percent, or as for a power plant to operate, the working fluid must exchange heat with the environment as well as the furnace. The impossibility of having a 100% efficient heat engine is not due to friction or other dissipative effects. It is a limitation that applies to both the idealised and the actual heat engines. KEITH VAUGH
    37. 37. A heat engine that violates theKelvin–Planck statement ofthe second law. KEITH VAUGH
    38. 38. REFRIGERATORS & HEAT PUMPS The transfer of heat from a low-temperature medium to a high-temperature one requires special devices } called refrigerators. Refrigerators, like heat engines, are cyclic devices. The working fluid used in the refrigeration cycle is called a refrigerant. The most frequently used refrigeration cycle is the vapour-compression refrigeration cycle. KEITH VAUGH
    39. 39. Basic components of arefrigeration system andtypical operating conditions. KEITH VAUGH
    40. 40. Coefficient of Performance The efficiency of a refrigerator is expressed in terms of the coefficient of performance (COP). The objective of a refrigerator is to remove heat (QL) from the refrigerated space.The objective of a refrigerator is toremove QL from the cooled space. KEITH VAUGH
    41. 41. Coefficient of Performance The efficiency of a refrigerator is expressed in terms of the coefficient of performance (COP). The objective of a refrigerator is to remove heat (QL) from the refrigerated space. Desired output QL COPR = = Required input Wnet , inThe objective of a refrigerator is toremove QL from the cooled space. KEITH VAUGH
    42. 42. Coefficient of Performance The efficiency of a refrigerator is expressed in terms of the coefficient of performance (COP). The objective of a refrigerator is to remove heat (QL) from the refrigerated space. Desired output QL COPR = = Required input Wnet , in Wnet , out = Qin − QoutThe objective of a refrigerator is toremove QL from the cooled space. KEITH VAUGH
    43. 43. Coefficient of Performance The efficiency of a refrigerator is expressed in terms of the coefficient of performance (COP). The objective of a refrigerator is to remove heat (QL) from the refrigerated space. Desired output QL COPR = = Required input Wnet , in Wnet , out = Qin − Qout QL 1 COPR = = á QH − QL QH −1 QLThe objective of a refrigerator is toremove QL from the cooled space. KEITH VAUGH
    44. 44. Heat Pumps The objective of a heat pump is to supply heat QH into the warmer space. KEITH VAUGH
    45. 45. Heat Pumps Desired output QH COPHP = = Required input Wnet , in The objective of a heat pump is to supply heat QH into the warmer space. KEITH VAUGH
    46. 46. Heat Pumps Desired output QH COPHP = = Required input Wnet , in QH 1 COPHP = = QH − QL 1− QL Q H The objective of a heat pump is to supply heat QH into the warmer space. KEITH VAUGH
    47. 47. Heat Pumps Desired output QH COPHP = = Required input Wnet , in QH 1 COPHP = = QH − QL 1− QL Q H COPHP = COPR + 1 for fixed values of QL and QH The objective of a heat pump is to supply heat QH into the warmer space. KEITH VAUGH
    48. 48. Heat Pumps Desired output QH COPHP = = Required input Wnet , in QH 1 COPHP = = QH − QL 1− QL Q H COPHP = COPR + 1 for fixed values of QL and QH The work supplied to a The objective of a heat pump is to heat pump is used to supply heat QH into the warmer extract energy from the space. cold outdoors and carry it into the warm indoors. KEITH VAUGH
    49. 49. The second law of thermodynamics:Clasius statement It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body. It states that a refrigerator cannot operate unless its compressor is driven by an external power source, such as an electric motor. This way, the net effect on the surroundings involves the consumption of some energy in the form of work, in addition to the transfer of heat from a colder body to a warmer one. To date, no experiment has been conducted that contradicts the second law, and this should be taken as sufficient proof of its validity. KEITH VAUGH
    50. 50. A refrigerator that violates theClausius statement of the second law. KEITH VAUGH
    51. 51. Equivalence of the two statements Proof that the violation of the Kelvin–Planck statement leads to the violation of the Clausius statement. KEITH VAUGH
    52. 52. PERPETUAL-MOTION MACHINESPerpetual-motion machine - any device that violates the first or thesecond law of Thermodynamics }A device that violates the first law (by creating energy) is called aPMM1.A device that violates the second law is called a PMM2.Despite numerous attempts, no perpetual-motion machine isknown to have worked.If something sounds too good to be true, it probably is. KEITH VAUGH
    53. 53. A perpetual-motions machine thatviolates the first law (PMM1) KEITH VAUGH
    54. 54. A perpetual-motions machine thatviolates the first law (PMM2) KEITH VAUGH
    55. 55. REVERSIBLE &IRREVERSIBLE PROCESSES Reversible process - A process that can be reversed without leaving any trace on the surroundings. } Irreversible process - A process that is not reversible. All the processes occurring in nature are irreversible. Why are we interested in reversible processes? ✓ they are easy to analyse and ✓ they serve as idealised models (theoretical limits) to which actual processes can be compared. KEITH VAUGH
    56. 56. Reversible processes deliver the most and consume the least work. KEITH VAUGH
    57. 57. Irreversibilitys The factors that cause a process to be irreversible are called irreversibilitys. ✓ They include friction, unrestrained expansion, mixing of two fluids, heat transfer across a finite temperature difference, electric resistance, inelastic deformation of solids, and chemical reactions. ✓ The presence of any of these effects renders a process irreversible. (a) Heat transfer through a temperature difference is irreversible, and (b) the reverse process is impossible. KEITH VAUGH
    58. 58. Irreversibilitys The factors that cause a process to be irreversible are called irreversibilitys. ✓ They include friction, unrestrained expansion, mixing of two fluids, heat transfer across a finite temperature difference, electric resistance, inelastic deformation of solids, and chemical reactions. ✓ The presence of any of these effects renders a process irreversible. (a) Heat transfer through a temperature difference is irreversible, and (b) the reverse process is impossible. Friction renders a process Irreversible compression irreversible. and expansion processes. KEITH VAUGH
    59. 59. Internally & externally reversible processesInternally reversible processIf no irreversibilitys occur within theboundaries of the system during the process.Externally reversibleIf no irreversibilitys occur outside the systemboundaries.Totally reversible processIt involves no irreversibilitys within thesystem or its surroundings.A totally reversible process involves no heattransfer through a finite temperaturedifference, no nonquasi-equilibrium changes,and no friction or other dissipative effects. A reversible process involves no internal and external irreversibilitys. KEITH VAUGH
    60. 60. Totally and internally reversible heat transfer processes. KEITH VAUGH
    61. 61. THE CARNOT CYCLEIn an ideal heat engine, all processes are reversible andthere are no losses; such an engine is the most efficientpossible between the chosen reservoirs. This is known }as Carnots principle and the efficiency of an ideal heatengine is called the Carnot efficiencyCarnots principle may be demonstrated by consideringthe existence of a reversed heat engine, which is adevice, which accepts work transfer in order to transferheat from a cold reservoir to a hot reservoir.An ideal reversed heat engine will perform reversibleprocesses and will be the exact reverse to the ideal heatengine. KEITH VAUGH
    62. 62. Execution of Carnot cycle in a closed system KEITH VAUGH
    63. 63. Reversible Isothermal Expansion (process 1-2, TH = constant)Execution of Carnot cycle in a closed system KEITH VAUGH
    64. 64. Reversible Isothermal Expansion (process 1-2, TH = constant) Reversible Adiabatic Expansion (process 2-3, temperature drops from TH to TL)Execution of Carnot cycle in a closed system KEITH VAUGH
    65. 65. Execution of Carnot cycle in a closed system KEITH VAUGH
    66. 66. Reversible Isothermal Compression (process 3-4, TL = constant)Execution of Carnot cycle in a closed system KEITH VAUGH
    67. 67. Reversible Isothermal Compression (process 3-4, TL = constant) Reversible Adiabatic Compression (process 4-1, temperature rises from TL to TH)Execution of Carnot cycle in a closed system KEITH VAUGH
    68. 68. Internally & externally reversible processesThe Carnot heat-engine cycle is a totally reversible cycle.Therefore, all the processes that comprise it can be reversed,in which case it becomes the Carnot refrigeration cycle. P-V diagram of the Carnot cycle KEITH VAUGH
    69. 69. Internally & externally reversible processesThe Carnot heat-engine cycle is a totally reversible cycle.Therefore, all the processes that comprise it can be reversed,in which case it becomes the Carnot refrigeration cycle. P-V diagram of the Carnot cycle KEITH VAUGH
    70. 70. Internally & externally reversible processesThe Carnot heat-engine cycle is a totally reversible cycle.Therefore, all the processes that comprise it can be reversed,in which case it becomes the Carnot refrigeration cycle. P-V diagram of the Carnot cycle P-V diagram of the reversed Carnot cycle KEITH VAUGH
    71. 71. The Carnot principlesThe efficiency of an irreversible heatengine is always less than theefficiency of a reversible one operatingbetween the same two reservoirs.The efficiencies of all reversible heatengines operating between the sametwo reservoirs are the same. The Carnot principles KEITH VAUGH
    72. 72. Proof of the first Carnot principle KEITH VAUGH
    73. 73. The thermodynamic temperature scaleA temperature scale that is independent ofthe properties of the substances that are usedto measure temperature is called athermodynamic temperature scale.Such a temperature scale offers greatconveniences in thermodynamic calculations The arrangement of heat engines used to develop the thermodynamic temperature scale. KEITH VAUGH
    74. 74. The thermodynamic temperature scale A temperature scale that is independent of the properties of the substances that are used to measure temperature is called a thermodynamic temperature scale. Such a temperature scale offers great conveniences in thermodynamic calculations All reversible heat enginesoperating between the same The arrangement of heat engines two reservoirs have the used to develop the thermodynamic same efficiency. temperature scale. KEITH VAUGH
    75. 75. For reversible cycles, theheat transfer ratio QH /QLcan be replaced by theabsolute temperature ratioTH /TL. KEITH VAUGH
    76. 76. For reversible cycles, theheat transfer ratio QH /QLcan be replaced by theabsolute temperature ratioTH /TL. KEITH VAUGH
    77. 77. For reversible cycles, the A conceptualheat transfer ratio QH /QL experimental setup tocan be replaced by the determineabsolute temperature ratio thermodynamicTH /TL. temperatures on the Kelvin scale by measuring heat transfers QH and QL. KEITH VAUGH
    78. 78. ⎛ QH ⎞ TH ⎜ Q ⎟ = T ⎝ L ⎠ rev LFor reversible cycles, the A conceptualheat transfer ratio QH /QL experimental setup tocan be replaced by the determineabsolute temperature ratio thermodynamicTH /TL. temperatures on the Kelvin scale by measuring heat transfers QH and QL. KEITH VAUGH
    79. 79. ⎛ QH ⎞ TH ⎜ Q ⎟ = T ⎝ L ⎠ rev L This temperature scale is called the Kelvin scale, and the temperatures on this scale are called absolute temperaturesFor reversible cycles, the A conceptualheat transfer ratio QH /QL experimental setup tocan be replaced by the determineabsolute temperature ratio thermodynamicTH /TL. temperatures on the Kelvin scale by measuring heat transfers QH and QL. KEITH VAUGH
    80. 80. ⎛ QH ⎞ TH ⎜ Q ⎟ = T ⎝ L ⎠ rev L This temperature scale is called the Kelvin scale, and the temperatures on this scale are called absolute temperatures T ( °C ) = T ( K ) − 273.15For reversible cycles, the A conceptualheat transfer ratio QH /QL experimental setup tocan be replaced by the determineabsolute temperature ratio thermodynamicTH /TL. temperatures on the Kelvin scale by measuring heat transfers QH and QL. KEITH VAUGH
    81. 81. The Carnot heat engine Any heat engine QL ηth = 1− QH Carnot heat engine TL ηth ,rev = 1− TH ⎧< ηth, rev irreversible heat engine ⎪ ηth ⎨= ηth, rev reversible heat engine ⎪> η impossible heat engine ⎩ th, rev The Carnot heat engine is the most efficient of all heat engines operating between the same high and low temperature reservoirs. KEITH VAUGH
    82. 82. No heat engine can have a higher efficiency than a reversible heatengine operating between the same high- and low-temperaturereservoirs. KEITH VAUGH
    83. 83. The quality of energy TL Can we use °C unit for ηth ,rev = 1− TH temperature here? The higher the temperature of the thermal energy,The fraction of heat that can be converted to the higher itswork as a function of source temperature. quality. KEITH VAUGH
    84. 84. The Carnot refrigerator and heat pump Any refrigerator or heat pump 1 COPR = QH −1 QL 1 COPHP = QL 1− QH Carnot refrigerator or heat pump 1 COPHP, rev = TL 1−No refrigerator can have a higher COP than a THreversible refrigerator operating between the sametemperature limits. 1 COPR, rev = TH −1 TL KEITH VAUGH
    85. 85. Introduction to the second lawThermal energy reservoirsHeat engines  Thermal efficiency  The 2nd law: Kelvin-Planck statementRefrigerators and heat pumps  Coefficient of performance (COP)  The 2nd law: Clasius statementPerpetual motion machinesReversible and irreversible processes  Irreversibilitys, Internal and externally reversible processesThe Carnot cycle  The reversed Carnot cycleThe Carnot principlesThe thermodynamic temperature scaleThe Carnot heat engine  The quality of energyThe Carnot refrigerator and heat pump KEITH VAUGH

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