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The 2nd Law of Thermodynamics: Reversibility & Irreversibility

Explore the concept of reversibility and irreversibility in the 2nd Law of Thermodynamics and its applications in entropy and energy quality. Learn why so much energy is wasted in power plants and understand the efficiency of heat engines and refrigerators.

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The 2nd Law of Thermodynamics: Reversibility & Irreversibility

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  1. 19. 2nd Law of Thermodynamics Reversibility & Irreversibility The 2nd Law of Thermodynamics Applications of the 2nd Law Entropy & Energy Quality

  2. Most energy extracted from the fuel in power plants is dumped to the environment as waste heat, here using a large cooling tower. Why is so much energy wasted? 2nd law: no Q  W with 100% efficiency

  3. Efficiencies

  4. 19.1. Reversibility & Irreversibility Block slowed down by friction: irreversible Bouncing ball: reversible • Examples of irreversible processes: • Beating an egg, blending yolk & white • Cups of cold & hot water in contact Spontaneous process: order  disorder ( statistically more probable )

  5. GOT IT? 19.1. • Which of these processes is irreversible: • stirring sugar into coffee. • building a house. • demolishing a house with a wrecking ball, • demolishing a house by taking it apart piece by piece, • harnessing the energy of falling water to drive machinery, • harnessing the energy of falling water to heat a house?

  6. 19.2. The 2nd Law of Thermodynamics • Heat engine extracts work from heat reservoirs. • gasoline & diesel engines • fossil-fueled & nuclear power plants • jet engines Perfect heat engine: coverts heat to work directly. 2nd law of thermodynamics ( Kelvin-Planck version ): There is no perfect heat engine. Heat dumped

  7. cylinder in contact with Th • ( T rises to Th adiabatically) • gas expands isothermally to do work • Wh = Qh absorbed • cylinder in contact with Tc • ( T drops to Tc adiabatically) • gas compressed isothermally • Wc = Qc dumped simple heat engine (any engine) Efficiency (any cycle) (Simple engine)

  8. Carnot Engine Ideal gas: AB: Heat abs. B  C: Work done C  D: Heat rejected: D  A: Work done Adiabatic processes: • isothermal expansion: T = Th , W1 = Qh • Adiabatic expansion: Th Tc • isothermal compression: T = Tc , W3 = Qc • Adiabatic compression : Tc Th 

  9. Example 19.1. Carnot Engine A Carnot engine extracts 240 J from its high T reservoir during each cycle, & rejects 100 J to the environment at 15C. How much work does the engine do in each cycle? What’s its efficiency? What’s the T of the hot reservoir? work done efficiency 

  10. Engines, Refrigerators, & the 2nd Law • Carnot’s theorem: • All Carnot engines operating between temperatures Th & Tc have the same efficiency. • No other engine operating between Th & Tc can have a greater efficiency. Refrigerator: extracts heat from cool reservoir into a hot one. work required

  11. perfect refrigerator: moves heat from cool to hot reservoir without work being done on it. 2nd law of thermodynamics ( Clausius version ): There is no perfect refrigerator.

  12. Perfect refrigerator  Perfect heat engine Clausius  Kelvin-Planck

  13. Hypothetical engine, e = 70% Carnot refrigerator, e = 60%  Carnot engine is most efficient eCarnot = thermodynamic efficiency eCarnot erev > eirrev

  14. 19.3. Applications of the 2nd Law Power plant Turbine Generator Electricity Boiler fossil-fuel : Th = 650 K Nuclear : Th = 570 K Tc = 310 K Condenser Heat source Cooling water Waste water Actual values: efossil ~ 40 % enuclear~ 34 % ecar ~ 20 % Prob 54 & 55

  15. Application: Combined-Cycle Power Plant Turbine engines: high Th ( 1000K  2000K ) & Tc ( 800 K ) … not efficient. Steam engines : Tc ~ ambient 300K. Combined-cycle : Th ( 1000K  2000K ) & Tc ( 300 K ) … e ~ 60%

  16. Example 19.2. Combined-Cycle Power Plant The gas turbine in a combined-cycle power plant operates at 1450 C. Its waste heat at 500 C is the input for a conventional steam cycle, with its condenser at 8 C. Find e of the combined-cycle, & compare it with those of the individual components.

  17. Refrigerators Coefficient of performance (COP) for refrigerators : 1st law Max. theoretical value (Carnot) COP is high if Th Tc . W = 0 ( COP =  ) for moving Q when Th= Tc .

  18. Example 19.3. Home Freezer A typical home freezer operates between Tc =  18C to Th = 30 C. What’s its maximum possible COP? With this COP, how much electrical energy would it take to freeze 500 g of water initially at 0 C? Table 17.1 2nd law: only a fraction of Q can become W in heat engines. a little W can move a lot of Q in refrigerators.

  19. Heat Pumps Heat pump: moves heat from Tc to Th . Heat pump as AC : Heat pump as heater : Ground temp ~ 10C year round

  20. GOT IT? 19.2. • A clever engineer decides to increase the efficiency of a Carnot engine by cooling the low-T reservoir using a refrigerator with the maximum possible COP. • Will the overall efficiency of this system • exceed. • be less than. • equal that of • the original engine alone? see Prob 32 for proof

  21. 19.4. Entropy & Energy Quality 2nd law: Energy of higher quality can be converted completely into lower quality form. But not vice versa. Energy quality Q measures the versatility of different energy forms.

  22. Application: Energy Quality, End Use, & Cogeneration Smart use: match quality (Q) to usage. e.g., fuel (low Q)  electricity (high Q) ( e ~ 40%)  heating by fuel ( good choice ) heating by electricity ( bad choice ) Cogeneration: Waste heat from electricity generation used for low Q needs. 600 kW cogenerating unit, Middlebury college.

  23. Entropy Carnot cycle (reversible processes): Qh = heat absorbed Qc = heat rejected lukewarm: can’t do W, Q  Qh , Qc = heat absorbed C = any closed path Irreversible processes can’t be represented by a path. S = entropy [ S ] = J / K

  24. S = 0 over any closed path  S12=S12 Entropy change is path-independent. ( S is a thermodynamic variable )

  25. Entropy in Carnot Cycle Ideal gas: Heat absorbed: Heat rejected: Adiabatic processes: 

  26. Irreversible Heat Transfer Cold & hot water can be mixed reversibly using extra heat baths. reversible processes irreversible processes

  27. Irreversible Heat Transfer Cold & hot water can be mixed reversibly using extra heat baths. T1 = some medium T. reversible processes T2 = some medium T. Actual mixing, irreversible processes

  28. Adiabatic Free Expansion Adiabatic  Qad.exp. = 0    S can be calculated by any reversible process between the same states. isothermal p = const. Can’t do work Q degraded.

  29. Entropy & Availability of Work Before adiabatic expansion, gas can do work isothermally After adiabatic expansion, gas cannot do work, while its entropy increases by  In a general irreversible process Coolest T in system

  30. Example 19.4. Loss of Q A 2.0 L cylinder contains 5.0 mol of compressed gas at 300 K. If the cylinder is discharged into a 150 L vacuum chamber & its temperature remains at 300 K, how much energy becomes unavailable to do work?

  31. Entropy & the 2nd Law of Thermodynamics 2nd Law of Thermodynamics : in any closed system S can decrease in an open system by outside work on it. However, S  0 for combined system.  S  0 in the universe Universe tends to disorder Life ?

  32. GOT IT? 19.3. • In each of the following processes, does the entropy of the named system increase, decrease, or stay the same? • a balloon inflates • cells differentiate in a growing embryo, forming different organs • an animal dies, its remains gradually decays • an earthquake demolishes a building • a plant utilize sunlight, CO2 , & water to manufacture sugar • a power plant burns coal & produces electrical energy • a car’s friction based brakes stop the car.       

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