The 2nd Law of Thermodynamics: Reversibility & Irreversibility

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.       