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Lecture on Entropy

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- 1. ENTROPY Lecture 10 Keith Vaugh BEng (AERO) MEngReference text: Chapter 8 - Fundamentals of Thermal-Fluid Sciences, 3rd Edition Yunus A. Cengel, Robert H. Turner, John M. Cimbala McGraw-Hill, 2008
- 2. OBJECTIVES }
- 3. OBJECTIVESApply the second law of thermodynamics toprocesses. }
- 4. OBJECTIVESApply the second law of thermodynamics toprocesses. }Deﬁne a new property called entropy to quantifythe second-law effects.
- 5. OBJECTIVESApply the second law of thermodynamics toprocesses. }Deﬁne a new property called entropy to quantifythe second-law effects.Establish the increase of entropy principle.
- 6. OBJECTIVESApply the second law of thermodynamics toprocesses. }Deﬁne a new property called entropy to quantifythe second-law effects.Establish the increase of entropy principle.Calculate the entropy changes that take placeduring processes for pure substances*,incompressible substances, and ideal gases.* we’re limiting our studies to pure substances (2010)
- 7. ENTROPY }
- 8. ENTROPYEntropy is a somewhat abstract property and it isdifﬁcult to give a physical description of withoutconsidering the microscopic state of the system }
- 9. ENTROPYEntropy is a somewhat abstract property and it isdifﬁcult to give a physical description of withoutconsidering the microscopic state of the system }The second law of thermodynamics often leads toexpressions that involve inequalities.
- 10. ENTROPYEntropy is a somewhat abstract property and it isdifﬁcult to give a physical description of withoutconsidering the microscopic state of the system }The second law of thermodynamics often leads toexpressions that involve inequalities.An irreversible heat engine is less efﬁcient than areversible one operating between the same twothermal reservoirs
- 11. ENTROPYEntropy is a somewhat abstract property and it isdifﬁcult to give a physical description of withoutconsidering the microscopic state of the system }The second law of thermodynamics often leads toexpressions that involve inequalities.An irreversible heat engine is less efﬁcient than areversible one operating between the same twothermal reservoirsAnother important inequality that has majorconsequences in thermodynamics is Clausiusinequality
- 12. The system considered in thedevelopment of the Clausius inequality.
- 13. Clausius inequality is expressed as; δQ — ≤0 ∫T the validity of the Clausius inequality is demonstrated on pages 298/9 of the reference text.The system considered in thedevelopment of the Clausius inequality.
- 14. Clausius inequality is expressed as; δQ — ≤0 ∫T the validity of the Clausius inequality is demonstrated on pages 298/9 of the reference text. Clausius realized he had discovered a new thermodynamic property and named it entropy with designation of S. It is deﬁned as; δQ dS = T int, rev ( kJ K )The system considered in thedevelopment of the Clausius inequality.
- 15. Clausius inequality is expressed as; δQ — ≤0 ∫T the validity of the Clausius inequality is demonstrated on pages 298/9 of the reference text. Clausius realized he had discovered a new thermodynamic property and named it entropy with designation of S. It is deﬁned as; δQ dS = T int, rev ( kJ K ) The entropy change of a system during a process can be determined by integratingThe system considered in the between the initial and ﬁnal statesdevelopment of the Clausius inequality. 2 δQ ΔS = S2 − S1 = ∫ 1 T int, rev ( kJ K )
- 16. The entropy change between two speciﬁedstates is the same whether the process isreversible or irreversible
- 17. A quantity whose cyclic integral is zero (i.e. a property like volume) δQ —T int, rev = 0 ∫ The entropy change between two speciﬁedstates is the same whether the process isreversible or irreversible
- 18. A quantity whose cyclic integral is zero (i.e. a property like volume) δQ —T int, rev = 0 ∫ The entropy change between two speciﬁedstates is the same whether the process isreversible or irreversible The net change in volume (a property) during a cycle is always zero
- 19. A quantity whose cyclic integral is zero (i.e. a property like volume) δQ —T int, rev = 0 ∫ The entropy change between two speciﬁedstates is the same whether the process isreversible or irreversible The net change in volume (a property) during a cycle is always zero Entropy is an extensive property of a system
- 20. A special case: Internally reversibleisothermal heat transfer processes
- 21. A special case: Internally reversibleisothermal heat transfer processes 2 δQ ΔS = ∫ 1 T int, rev
- 22. A special case: Internally reversibleisothermal heat transfer processes 2 δQ ΔS = ∫ 1 T int, rev 2 δQ = ∫ 1 T 0 int, rev
- 23. A special case: Internally reversibleisothermal heat transfer processes 2 δQ ΔS = ∫ 1 T int, rev 2 δQ = ∫ 1 T 0 int, rev 1 2 = T0 ∫ (δ Q ) 1 int, rev
- 24. A special case: Internally reversibleisothermal heat transfer processes 2 δQ ΔS = ∫ 1 T int, rev 2 δQ = ∫ 1 T 0 int, rev 1 2 = T0 ∫ (δ Q ) 1 int, rev ΔS = Q T0 } This equation is particularly useful for determining the entropy changes of thermal energy reservoirs
- 25. EXAMPLE 8 - 1A piston cylinder device contains liquid vapormixture at 300 K. During a constant pressure process, }750 kJ of heat is transferred to the water. As a result,part of the liquid in the cylinder vaporizes. Determinethe entropy change of the water during this process.
- 26. SOLUTIONHeat is transferred to a liquid vapor mixtureof water in a piston cylinder device atconstant pressure. The entropy change ofwater is to be determined
- 27. SOLUTIONHeat is transferred to a liquid vapor mixtureof water in a piston cylinder device atconstant pressure. The entropy change ofwater is to be determinedASSUMPTIONSNo irreversibilitys occur within the systemboundaries during the process.
- 28. SOLUTION ANALYSISHeat is transferred to a liquid vapor mixture We take the entire water (liquid + vapor) in theof water in a piston cylinder device at cylinder as the system. This is a closed systemconstant pressure. The entropy change of since no mass crosses the system boundary duringwater is to be determined the process. We note that the temperature of the system remains constant at 300 K since theASSUMPTIONS temperature of a pure substance remains constant at the saturation value during a phase changeNo irreversibilitys occur within the system process at constant pressure.boundaries during the process.
- 29. SOLUTION ANALYSISHeat is transferred to a liquid vapor mixture We take the entire water (liquid + vapor) in theof water in a piston cylinder device at cylinder as the system. This is a closed systemconstant pressure. The entropy change of since no mass crosses the system boundary duringwater is to be determined the process. We note that the temperature of the system remains constant at 300 K since theASSUMPTIONS temperature of a pure substance remains constant at the saturation value during a phase changeNo irreversibilitys occur within the system process at constant pressure.boundaries during the process. The system undergoes an internally reversible, isothermal process and thus its entropy change can be determined directly Q 750kJ ΔS = = = 2.5 kJ K Tsys 300K NOTE - Entropy change of the system is positive as expected since heat transfer is to the system
- 30. The increase of entropy principleA cycle composed of a reversible and anirreversible process
- 31. The increase of entropy principle δQ 2 δQ 1 δQ — ≤ 0 → ∫1 T + ∫2 T int, rev ≤ 0 ∫T A cycle composed of a reversible and anirreversible process
- 32. The increase of entropy principle δQ 2 δQ 1 δQ — ≤ 0 → ∫1 T + ∫2 T int, rev ≤ 0 ∫T A cycle composed of a reversible and anirreversible process
- 33. The increase of entropy principle δQ 2 δQ 1 δQ — ≤ 0 → ∫1 T + ∫2 T int, rev ≤ 0 ∫T 2 δQ 2 δQ ∫ + S1 − S0 ≤ 0 → S2 − S1 ≥ ∫ 1 T 1 TA cycle composed of a reversible and anirreversible process
- 34. The increase of entropy principle δQ 2 δQ 1 δQ — ≤ 0 → ∫1 T + ∫2 T int, rev ≤ 0 ∫T 2 δQ 2 δQ ∫ + S1 − S0 ≤ 0 → S2 − S1 ≥ ∫ 1 T 1 TA cycle composed of a reversible and anirreversible process
- 35. The increase of entropy principle δQ 2 δQ 1 δQ — ≤ 0 → ∫1 T + ∫2 T int, rev ≤ 0 ∫T 2 δQ 2 δQ ∫ + S1 − S0 ≤ 0 → S2 − S1 ≥ ∫ 1 T 1 T δQ The equality holds for an internally dS ≥ reversible process and the inequality T for an irreversible processA cycle composed of a reversible and anirreversible process
- 36. The increase of entropy principle δQ 2 δQ 1 δQ — ≤ 0 → ∫1 T + ∫2 T int, rev ≤ 0 ∫T 2 δQ 2 δQ ∫ + S1 − S0 ≤ 0 → S2 − S1 ≥ ∫ 1 T 1 T δQ The equality holds for an internally dS ≥ reversible process and the inequality T for an irreversible process 2 δQ ΔSsys = S2 − S1 = ∫ + Sgen 1 TA cycle composed of a reversible and anirreversible process
- 37. The increase of entropy principle δQ 2 δQ 1 δQ — ≤ 0 → ∫1 T + ∫2 T int, rev ≤ 0 ∫T 2 δQ 2 δQ ∫ + S1 − S0 ≤ 0 → S2 − S1 ≥ ∫ 1 T 1 T δQ The equality holds for an internally dS ≥ reversible process and the inequality T for an irreversible process 2 δQ ΔSsys = S2 − S1 = ∫ + Sgen 1 TA cycle composed of a reversible and anirreversible process Sgen = ΔStotal = ΔSsys + ΔSsurr ≥ 0 Some entropy is generated or created during an irreversible process and this generation is due entirely to the presence of irreversibilitys.
- 38. A system and its surrounding form anisolated system
- 39. The entropy change of an isolated system is the sum of the entropy changes of its components, and is never less than zero.A system and its surrounding form anisolated system
- 40. The entropy change of an isolated system is the sum of the entropy changes of its components, and is never less than zero.A system and its surrounding form an ΔSisolated ≥ 0isolated system
- 41. The entropy change of an isolated system is the sum of the entropy changes of its components, and is never less than zero.A system and its surrounding form an ΔSisolated ≥ 0isolated system Sgen = ΔStotal = ΔSsys + ΔSsurr ≥ 0
- 42. The entropy change of an isolated system is the sum of the entropy changes of its components, and is never less than zero.A system and its surrounding form an ΔSisolated ≥ 0isolated system Sgen = ΔStotal = ΔSsys + ΔSsurr ≥ 0 > Irreversible process The increase of Sgen = Reversible process entropy principle < Impossible process
- 43. A comment on entropyThe entropy change of a system canbe negative, but the entropygeneration cannot.
- 44. A comment on entropy Processes can occur in a certain direction only, not in any direction. A process must proceed in the direction that complies with the increase of entropy principle, that is, Sgen ≥ 0. A process that violates this principle is impossible.The entropy change of a system canbe negative, but the entropygeneration cannot.
- 45. A comment on entropy Processes can occur in a certain direction only, not in any direction. A process must proceed in the direction that complies with the increase of entropy principle, that is, Sgen ≥ 0. A process that violates this principle is impossible. Entropy is a non-conserved property, and there is no such thing as the conservation of entropy principle. Entropy is conserved during the idealized reversible processes only and increases during all actual processes.The entropy change of a system canbe negative, but the entropygeneration cannot.
- 46. A comment on entropy Processes can occur in a certain direction only, not in any direction. A process must proceed in the direction that complies with the increase of entropy principle, that is, Sgen ≥ 0. A process that violates this principle is impossible. Entropy is a non-conserved property, and there is no such thing as the conservation of entropy principle. Entropy is conserved during the idealized reversible processes only and increases during all actual processes. The performance of engineering systems isThe entropy change of a system can degraded by the presence of irreversibilitys, andbe negative, but the entropygeneration cannot. entropy generation is a measure of the magnitudes of the irreversibilitys during that process. It is also used to establish criteria for the performance of engineering devices.
- 47. Entropy change of pure substances Entropy is a property and therefore its value for a system is ﬁxed provided that state of that system is also ﬁxed. Schematic of the T-s diagram for water.
- 48. Entropy change ΔS = mΔS = m ( s2 − s1 ) ( kJ K )The entropy of a pure substance isdetermined from the tables (like otherproperties).
- 49. QUESTION 8 - 39A well insulated rigid tank contains 2kg of a saturatedliquid vapor mixture of water at 100 kPa. Initiallythree quarters of the mass is in the liquid phase. An }electric resistance heater placed in the tank is nowturned on and kept on until all the liquid in the tank isvaporized. Determine the entropy change of thesteam during this process.
- 50. SOLUTIONAn insulated rigid tank contains asaturated liquid-vapor mixture ofwater at a speciﬁed pressure. Anelectric heater inside is turned onand kept on until all the liquidvaporized. The entropy change ofthe water during this process is tobe determined.
- 51. SOLUTIONAn insulated rigid tank contains asaturated liquid-vapor mixture ofwater at a speciﬁed pressure. Anelectric heater inside is turned onand kept on until all the liquidvaporized. The entropy change ofthe water during this process is tobe determined. H2O 2 kg 100 kPa We
- 52. SOLUTION ANALYSISAn insulated rigid tank contains a From the steam tables (A-4 through A-6)saturated liquid-vapor mixture ofwater at a speciﬁed pressure. Anelectric heater inside is turned onand kept on until all the liquidvaporized. The entropy change ofthe water during this process is tobe determined. H2O 2 kg 100 kPa We
- 53. SOLUTION ANALYSISAn insulated rigid tank contains a From the steam tables (A-4 through A-6)saturated liquid-vapor mixture of P1 v1 = v f + x1v fgwater at a speciﬁed pressure. An electric heater inside is turned on x1 s1 = s f + x1s fgand kept on until all the liquidvaporized. The entropy change ofthe water during this process is tobe determined. H2O 2 kg 100 kPa We
- 54. SOLUTION ANALYSISAn insulated rigid tank contains a From the steam tables (A-4 through A-6)saturated liquid-vapor mixture of P1 v1 = v f + x1v fgwater at a speciﬁed pressure. An electric heater inside is turned on x1 s1 = s f + x1s fgand kept on until all the liquidvaporized. The entropy change of v1 = v f + x1v fgthe water during this process is to v1 = 0.001 + ( 0.25 ) (1.6941 − 0.001)be determined. 3 v1 = 0.4243 m kg H2O 2 kg 100 kPa We
- 55. SOLUTION ANALYSISAn insulated rigid tank contains a From the steam tables (A-4 through A-6)saturated liquid-vapor mixture of P1 v1 = v f + x1v fgwater at a speciﬁed pressure. An electric heater inside is turned on x1 s1 = s f + x1s fgand kept on until all the liquidvaporized. The entropy change of v1 = v f + x1v fgthe water during this process is to v1 = 0.001 + ( 0.25 ) (1.6941 − 0.001)be determined. 3 v1 = 0.4243 m kg s1 = s f + x1s fg H2O s1 = 1.3028 + ( 0.25 ) ( 6.0562 ) 2 kg 100 kPa s1 = 2.8168 kJ kg ⋅ K We
- 56. SOLUTION ANALYSISAn insulated rigid tank contains a From the steam tables (A-4 through A-6)saturated liquid-vapor mixture of P1 v1 = v f + x1v fgwater at a speciﬁed pressure. An electric heater inside is turned on x1 s1 = s f + x1s fgand kept on until all the liquidvaporized. The entropy change of v1 = v f + x1v fgthe water during this process is to v1 = 0.001 + ( 0.25 ) (1.6941 − 0.001)be determined. 3 v1 = 0.4243 m kg s1 = s f + x1s fg H2O s1 = 1.3028 + ( 0.25 ) ( 6.0562 ) 2 kg 100 kPa s1 = 2.8168 kJ kg ⋅ K v2 = v1 We s2 = 6.8649 kJ kg ⋅ K sat. vapor Then the entropy change of the steam becomes
- 57. SOLUTION ANALYSISAn insulated rigid tank contains a From the steam tables (A-4 through A-6)saturated liquid-vapor mixture of P1 v1 = v f + x1v fgwater at a speciﬁed pressure. An electric heater inside is turned on x1 s1 = s f + x1s fgand kept on until all the liquidvaporized. The entropy change of v1 = v f + x1v fgthe water during this process is to v1 = 0.001 + ( 0.25 ) (1.6941 − 0.001)be determined. 3 v1 = 0.4243 m kg s1 = s f + x1s fg H2O s1 = 1.3028 + ( 0.25 ) ( 6.0562 ) 2 kg 100 kPa s1 = 2.8168 kJ kg ⋅ K v2 = v1 We s2 = 6.8649 kJ kg ⋅ K sat. vapor Then the entropy change of the steam becomes ΔS = m ( s2 − s1 ) = ( 2kg ) ( 6.8649 − 2.8168 ) kJ kg ⋅ K ΔS = 8.10 kJ K
- 58. Isentropic processesA process during which the entropyremains constant is called an isentropicprocessDuring an internally reversible, adiabatic(isentropic) process, the entropy remainsconstant.
- 59. Isentropic processesA process during which the entropyremains constant is called an isentropicprocess The isentropic process appears as a vertical line segment on a T-s diagram.During an internally reversible, adiabatic(isentropic) process, the entropy remainsconstant.
- 60. Isentropic processesA process during which the entropyremains constant is called an isentropicprocess The isentropic process appears as a vertical line segment on a T-s diagram.During an internally reversible, adiabatic(isentropic) process, the entropy remains Δs = 0 or s 2 − s1 ( kJ kg ⋅K)constant.
- 61. QUESTION 8 - 44An insulated piston-cylinder device contains 0.05 m3of saturated refrigerant 134a vapor at 0.8 MPapressure. The refrigerant is now allowed to expand in }a reversible manner until the pressure drops to 0.4MPa. Determine;a. The ﬁnal temperature in the cylinderb. The work done by the refrigerant
- 62. SOLUTIONAn insulated cylinder is initiallyﬁlled with saturated R-134a vaporat a speciﬁed pressure. Therefrigerant expands in a reversiblemanner until the pressure drops toa speciﬁed value. The ﬁnaltemperature in the cylinder and thework done by the refrigerant are tobe determined.
- 63. SOLUTIONAn insulated cylinder is initiallyﬁlled with saturated R-134a vaporat a speciﬁed pressure. Therefrigerant expands in a reversiblemanner until the pressure drops toa speciﬁed value. The ﬁnaltemperature in the cylinder and thework done by the refrigerant are tobe determined. R-134a 0.8 MPa 0.05 m3
- 64. SOLUTION ASSUMPTIONSAn insulated cylinder is initially 1. The kinetic and potential energyﬁlled with saturated R-134a vapor changes are negligible.at a speciﬁed pressure. Therefrigerant expands in a reversiblemanner until the pressure drops toa speciﬁed value. The ﬁnaltemperature in the cylinder and thework done by the refrigerant are tobe determined. R-134a 0.8 MPa 0.05 m3
- 65. SOLUTION ASSUMPTIONSAn insulated cylinder is initially 1. The kinetic and potential energyﬁlled with saturated R-134a vapor changes are negligible.at a speciﬁed pressure. The 2. The cylinder is well-insulated and thusrefrigerant expands in a reversible heat transfer is negligible.manner until the pressure drops toa speciﬁed value. The ﬁnaltemperature in the cylinder and thework done by the refrigerant are tobe determined. R-134a 0.8 MPa 0.05 m3
- 66. SOLUTION ASSUMPTIONSAn insulated cylinder is initially 1. The kinetic and potential energyﬁlled with saturated R-134a vapor changes are negligible.at a speciﬁed pressure. The 2. The cylinder is well-insulated and thusrefrigerant expands in a reversible heat transfer is negligible.manner until the pressure drops to 3. The thermal energy stored in thea speciﬁed value. The ﬁnal cylinder itself is negligible.temperature in the cylinder and thework done by the refrigerant are tobe determined. R-134a 0.8 MPa 0.05 m3
- 67. SOLUTION ASSUMPTIONSAn insulated cylinder is initially 1. The kinetic and potential energyﬁlled with saturated R-134a vapor changes are negligible.at a speciﬁed pressure. The 2. The cylinder is well-insulated and thusrefrigerant expands in a reversible heat transfer is negligible.manner until the pressure drops to 3. The thermal energy stored in thea speciﬁed value. The ﬁnal cylinder itself is negligible.temperature in the cylinder and the 4. The process is stated to be reversible.work done by the refrigerant are tobe determined. R-134a 0.8 MPa 0.05 m3
- 68. SOLUTION ASSUMPTIONSAn insulated cylinder is initially 1. The kinetic and potential energyﬁlled with saturated R-134a vapor changes are negligible.at a speciﬁed pressure. The 2. The cylinder is well-insulated and thusrefrigerant expands in a reversible heat transfer is negligible.manner until the pressure drops to 3. The thermal energy stored in thea speciﬁed value. The ﬁnal cylinder itself is negligible.temperature in the cylinder and the 4. The process is stated to be reversible.work done by the refrigerant are tobe determined. ANALYSIS a. This is a reversible adiabatic (i.e., isentropic) process, and thus s2 = s1. From the refrigerant tables (Tables A-11 through A-13), R-134a 0.8 MPa 0.05 m3
- 69. SOLUTION ASSUMPTIONSAn insulated cylinder is initially 1. The kinetic and potential energyﬁlled with saturated R-134a vapor changes are negligible.at a speciﬁed pressure. The 2. The cylinder is well-insulated and thusrefrigerant expands in a reversible heat transfer is negligible.manner until the pressure drops to 3. The thermal energy stored in thea speciﬁed value. The ﬁnal cylinder itself is negligible.temperature in the cylinder and the 4. The process is stated to be reversible.work done by the refrigerant are tobe determined. ANALYSIS a. This is a reversible adiabatic (i.e., isentropic) process, and thus s2 = s1. From the refrigerant tables (Tables A-11 through A-13), v1 = vg@0.8 MPa = 0.02561 m3 kg P1 = 0.8MPa u1 = u g@0.8 MPa = 246.79 kJ kg R-134a sat. vapor 0.8 MPa s1 = sg@0.8 MPa = 0.91835 kJ kg ⋅ K 0.05 m3
- 70. V 0.05m 3m= = m3 = 1.952kg v1 0.025621 kg
- 71. V 0.05m 3m= = m3 = 1.952kg v1 0.025621 kg s2 − s f 0.91835 − 024761P2 = 0.4MPa x2 = = = 0.9874 s fg 0.67929s2 = s1 u2 = u f + x2 u fg = 63.62 + ( 0.9874 ) (171.45 ) = 232.91 kJ kg
- 72. V 0.05m 3m= = m3 = 1.952kg v1 0.025621 kg s2 − s f 0.91835 − 024761P2 = 0.4MPa x2 = = = 0.9874 s fg 0.67929s2 = s1 u2 = u f + x2 u fg = 63.62 + ( 0.9874 ) (171.45 ) = 232.91 kJ kgT2 = Tsat @0.4 MPa = 8.91°C
- 73. V 0.05m 3m= = m3 = 1.952kg v1 0.025621 kg s2 − s f 0.91835 − 024761P2 = 0.4MPa x2 = = = 0.9874 s fg 0.67929s2 = s1 u2 = u f + x2 u fg = 63.62 + ( 0.9874 ) (171.45 ) = 232.91 kJ kgT2 = Tsat @0.4 MPa = 8.91°Cb. We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this adiabatic closed system can be expressed as
- 74. V 0.05m 3m= = m3 = 1.952kg v1 0.025621 kg s2 − s f 0.91835 − 024761P2 = 0.4MPa x2 = = = 0.9874 s fg 0.67929s2 = s1 u2 = u f + x2 u fg = 63.62 + ( 0.9874 ) (171.45 ) = 232.91 kJ kgT2 = Tsat @0.4 MPa = 8.91°Cb. We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this adiabatic closed system can be expressed as Ein − Eout = ΔEsystem 14 2 4 3 123 Net energy transfer Change in internal, kinetic,by heat, work and mass potential, etc... energies − Wb, out = ΔU Wb, out = m ( u1 − u2 )
- 75. V 0.05m 3m= = m3 = 1.952kg v1 0.025621 kg s2 − s f 0.91835 − 024761P2 = 0.4MPa x2 = = = 0.9874 s fg 0.67929s2 = s1 u2 = u f + x2 u fg = 63.62 + ( 0.9874 ) (171.45 ) = 232.91 kJ kgT2 = Tsat @0.4 MPa = 8.91°Cb. We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this adiabatic closed system can be expressed as Ein − Eout = ΔEsystem Substituting, the work done during this isentropic 14 2 4 3 123 process is determined to be Net energy transfer Change in internal, kinetic,by heat, work and mass potential, etc... energies − Wb, out = ΔU Wb, out = m ( u1 − u2 ) Wb, out = m ( u1 − u2 ) = (1.952kg ) ( 246.79 − 232.91) kJ kg = 27.09kJ
- 76. Property diagrams involving entropyOn a T-S diagram, the area under the process curverepresents the heat transfer for internally reversibleprocesses.
- 77. Property diagrams involving entropy δ Qint, rev = TdSOn a T-S diagram, the area under the process curverepresents the heat transfer for internally reversibleprocesses.
- 78. Property diagrams involving entropy δ Qint, rev = TdS 2 Qint, rev = ∫ T dS 1On a T-S diagram, the area under the process curverepresents the heat transfer for internally reversibleprocesses.
- 79. Property diagrams involving entropy δ Qint, rev = TdS 2 Qint, rev = ∫ T dS 1 δ qint, rev = TdsOn a T-S diagram, the area under the process curverepresents the heat transfer for internally reversibleprocesses.
- 80. Property diagrams involving entropy δ Qint, rev = TdS 2 Qint, rev = ∫ T dS 1 δ qint, rev = Tds 2 qint, rev = ∫ T ds 1On a T-S diagram, the area under the process curverepresents the heat transfer for internally reversibleprocesses.
- 81. Property diagrams involving entropy δ Qint, rev = TdS 2 Qint, rev = ∫ T dS 1 δ qint, rev = Tds 2 qint, rev = ∫ T ds 1 Qint, rev = T0 ΔSOn a T-S diagram, the area under the process curverepresents the heat transfer for internally reversibleprocesses.
- 82. Property diagrams involving entropy δ Qint, rev = TdS 2 Qint, rev = ∫ T dS 1 δ qint, rev = Tds 2 qint, rev = ∫ T ds 1 Qint, rev = T0 ΔS qint, rev = T0 ΔsOn a T-S diagram, the area under the process curverepresents the heat transfer for internally reversibleprocesses.
- 83. Property diagrams involving entropy For adiabatic steady-ﬂow devices, the vertical distance ∆h on an h-s diagram is a measure of work, and the horizontal distance ∆s is a measure of irreversibilitys.Mollier diagram: The h-s diagram
- 84. EntropyClausius inequalityThe Increase of entropy principle Example 8-1A comment on entropyEntropy change of pure substances Question 8-39Isentropic processes Question 8-44Property diagrams involving entropyWhat is entropy?The T ds relationsEntropy change of liquids and solidsThe entropy change of ideal gasesReversible steady-ﬂow workMinimizing the compressor workIsentropic efﬁciencies of steady-ﬂow devicesEntropy balance

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