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EGR 334 Thermodynamics Chapter 5: Sections 10-11

EGR 334 Thermodynamics Chapter 5: Sections 10-11. Lecture 22: Carnot Cycle. Quiz Today?. Today’s main concepts:. State what processes make up a Carnot Cycle. Be able to calculate the efficiency of a Carnot Cycle Be able to give the Classius Inequality

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EGR 334 Thermodynamics Chapter 5: Sections 10-11

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  1. EGR 334 ThermodynamicsChapter 5: Sections 10-11 Lecture 22: Carnot Cycle Quiz Today?

  2. Today’s main concepts: • State what processes make up a Carnot Cycle. • Be able to calculate the efficiency of a Carnot Cycle • Be able to give the Classius Inequality • Be able to apply the Classisus Inequality to determine if a cycle is reversible, irreversible, or impossible as predicted by the 2nd Law. Reading Assignment: Read Chapter 6, Sections 1-5 Homework Assignment: Problems from Chap 5: 64, 79, 81,86

  3. Recall from last time: Energy Balance: Energy Rate Balance: Entropy Balance: Entropy Rate Balance:

  4. Carnot Cycle 2 p 2 3 • The Carnot cycle provides a specific example of a reversible cycle that operates between two thermal reservoirs. Other examples covered in Chapter 9 are the Ericsson and Stirling cycles. • In a Carnot cycle, the system executing the cycle undergoes a series of four internally reversible processes: two adiabatic processes (Q = 0) alternated with two isothermal processes ( T = constant) T 3 1 4 1 4 v v

  5. The p-v diagram and schematic of a gas in a piston-cylinder assembly executing a Carnot cycle are shown below: Carnot Power Cycles

  6. In each of these cases the thermal efficiency is given by The p-v diagram and schematic of water executing a Carnot cycle through four interconnected components are shown below: Carnot Power Cycles

  7. Sec 5.10 : The Carnot Cycle • The Carnot cycle: • QH • Gas only cycle T 2 3 • Area = Work • QH • QC • Process 1-2 : Adiabatic Compression. • Process 2 -3 : Isothermal Expansion receiving QH. • Process 3 – 4 : Adiabatic Expansion. • Process 4 – 1 : Isothermal Compression, rejecting QC. 1 4 v • QC

  8. Sec 5.10 : The Carnot Cycle • Analyzing the Carnot cycle: • Energy Balance: • First look at the two isothermal processes • Process 2 -3 : Isothermal Expansion receiving QH. • Process 4 – 1 : Isothermal Compression, rejecting QC. • and

  9. Sec 5.10 : The Carnot Cycle • Analyzing the Carnot cycle: • Energy Balance: • Then look at the two adiabatic processes (Q = 0) • Process 1-2 : Adiabatic Compression. • The term • Thus, • Process 3 – 4 :Adiabatic Expansion.

  10. Sec 5.10 : The Carnot Cycle • Analyzing the Carnot cycle: • With • and • Therefore, • We have now proven

  11. The Carnot Model of a Hurricane • Added heat causes further rising. • As T to dew point, vapor condenses, releasing hfg and warming air • Cools & Expands as P • Adiabatic cooling • Warm air rises • Warm moist air

  12. The Carnot Model of a Hurricane Isothermal Compression Adiabatic vcore>>v outer A Adiabatic B D Isothermal Expansion C

  13. Sec 5.10 : The Carnot Cycle • Example: (5.76) One-half pound of water executes a Carnot power cycle. During the isothermal expansion, the water is heated at 600°F from a saturated liquid to a saturated vapor. The vapor then expands adiabatically to a temperature of 90°F and a quality of 64.3% • Sketch the cycle on a P-v diagram. • Evaluate the heat and work for each process in BTU • Evaluate the thermal efficiency. p v

  14. Sec 5.10 : The Carnot Cycle • Example: (5.76) One-half pound of water executes a Carnot power cycle. During the isothermal expansion, the water is heated at 600°F from a saturated liquid to a saturated vapor. The vapor then expands adiabatically to a temperature of 90°F and a quality of 64.3% • Sketch the cycle on a P-v diagram. • Evaluate the heat and work for each process in BTU • Evaluate the thermal efficiency.

  15. Sec 5.10 : The Carnot Cycle • Example: (5.76) One-half pound of water executes a Carnot power cycle. During the isothermal expansion, the water is heated at 600°F from a saturated liquid to a saturated vapor. The vapor then expands adiabatically to a temperature of 90°F and a quality of 64.3% • Sketch the cycle on a P-v diagram. • Evaluate the heat and work for each process in BTU • Evaluate the thermal efficiency. • Isothermal • Adiabatic • Isothermal • Adiabatic • Using Table A-2

  16. Sec 5.10 : The Carnot Cycle • Example: (5.76) • Evaluate the heat and work for each process in BTU • Adiabatic • Isothermal • Adiabatic • Isothermal Process 1-2

  17. Sec 5.10 : The Carnot Cycle • Example: (5.76) • Evaluate the heat and work for each process in BTU • Adiabatic • Isothermal • Adiabatic • Isothermal Process 2-3 (adiabatic process)

  18. Sec 5.10 : The Carnot Cycle • Example: (5.76) • Evaluate the heat and work for each process in Btu • Isothermal • Adiabatic • Isothermal • Adiabatic • For Process 3 – 4: for the Carnot cycle: where QH = Q12 = 274.8 Btu

  19. Sec 5.10 : The Carnot Cycle • Example: (5.76) • Evaluate the heat and work for each process in Btu • Isothermal • Adiabatic • Isothermal • Adiabatic • continuing for Process 3 – 4: recalling h = u + pv

  20. Sec 5.10 : The Carnot Cycle • Example: (5.76) • Evaluate the heat and work for each process in Btu • Isothermal • Adiabatic • Isothermal • Adiabatic • continuing for Process 3 – 4: • Then using Table A2 at h4 = 441.5 Btu/lbm and T4 = 90 deg. • which let the state 4 intensive properties be found:

  21. Sec 5.10 : The Carnot Cycle • Example: (5.76) • Evaluate the heat and work for each process in BTU • Isothermal • Adiabatic • Isothermal • Adiabatic and for Process 4-1:

  22. Sec 5.10 : The Carnot Cycle • Example: (5.76) • Evaluate the heat and work for each process in BTU • Isothermal • Adiabatic • Isothermal • Adiabatic • Finally, process 4 – 1:

  23. Sec 5.10 : The Carnot Cycle • Example: (5.76) • (c) Evaluate the thermal efficiency. • Isothermal • Adiabatic • Isothermal • Adiabatic • Thermal efficiency of the cycle:

  24. Clausius Inequality • The Clausius inequality is developed from the Kelvin-Planck statement of the second law and can be expressed as: The nature of the cycle executed is indicated by the value of scycle: scycle = 0 no irreversibilities present within the system scycle > 0 irreversibilities present within the system scycle < 0 impossible

  25. Sec 5.11 : The Clausius Inequality • We have shown that: • and thus • Therefore: • For an ideal/reversible process • Now consider a general process • Each part of the cycle is divided into an infinitesimally small process p • Then sum (integrate) the entire process • dQ = heat transfer at boundary • T = absolute T at that part of the cycle. v

  26. Sec 5.11 : The Clausius Inequality • For a real process, Qreal > Qreversible • Therefore: • We can then define σ, where • and σcycle = 0  reversible process • σcycle > 0  irreversible process • σcycle < 0  impossible process P • We now also have the mathematical definition of enthalpy. • More on this in Chapter 6 v

  27. Sec 5.11 : The Clausius Inequality • Example: (5.81) A system executes a power cycle while receiving heat transfer at a temperature of 500 K and discharging 1000 kJ by heat transfer at a temperature of 300 K. There are no other heat transfers. Appling the Clausius Inequality, determine σcycle if the thermal efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle reversible, or impossible? TH= 500 K W=? TH= 300 K Q = 50 kJ

  28. Sec 5.11 : The Clausius Inequality • Example: (5.81) A system executes a power cycle while receiving heat transfer at a temperature of 1000 K and discharging 1000 kJ by heat transfer at a temperature of 300 K. There are no other heat transfers. Appling the Clausius Inequality, determine σcycle if the thermal efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle reversible, or impossible? TH= 500 K W=? • and TH= 300 K • Therefore: Q = 50 kJ • Since σcycle is negative, the cycle is impossible.

  29. Sec 5.11 : The Clausius Inequality • Example: (5.81) A system executes a power cycle while receiving heat transfer at a temperature of 1000 K and discharging 1000 kJ by heat transfer at a temperature of 300 K. There are no other heat transfers. Appling the Clausius Inequality, determine σcycle if the thermal efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle reversible, or impossible? TH= 500 K W=? TH= 300 K Q = 50 kJ • Since σcycle is zero, the cycle is internally reversible.

  30. Sec 5.11 : The Clausius Inequality • Example: (5.81) A system executes a power cycle while receiving heat transfer at a temperature of 1000 K and discharging 1000 kJ by heat transfer at a temperature of 300 K. There are no other heat transfers. Appling the Clausius Inequality, determine σcycle if the thermal efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle reversible, or impossible? TH= 500 K W=? TH= 300 K Q = 50 kJ • Since σcycle is positive, the cycle is irreversible.

  31. End of Lecture 21 Slides

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