19. 2 nd Law of Thermodynamics 1. Reversibility & Irreversibility 2. The 2 nd Law of Thermodynamics 3. Applications of the 2 nd Law 4. Entropy & Energy Quality

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? 2 nd law: no Q  W with 100% efficiency

Efficiencies EngineEfficiency Gasoline18~20% Dieselup to 40% Steam~8% Gas Turbine~ Diesel Power PlantEfficiency Coal36% Nuclear30%

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 )

GOT IT? Which of these processes is irreversible: (a) stirring sugar into coffee. (b) building a house. (c) demolishing a house with a wrecking ball, (d) demolishing a house by taking it apart piece by piece, (e) harnessing the energy of falling water to drive machinery, (f) harnessing the energy of falling water to heat a house?

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

simple heat engine cylinder in contact with T h ( T rises to T h adiabatically) gas expands isothermally to do work W h = Q h absorbed cylinder in contact with T c ( T drops to T c adiabatically) gas compressed isothermally W c = Q c dumped Efficiency (any engine) (Simple engine) (any cycle)

Carnot Engine 1.isothermal expansion: T = T h, W 1 = Q h 2.Adiabatic expansion: T h  T c 3.isothermal compression: T = T c, W 3 = Q c 4.Adiabatic compression : T c  T h Ideal gas: Adiabatic processes:  A  B: Heat abs. C  D: Heat rejected: B  C: Work done D  A:  Work done

Example 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 

Engines, Refrigerators, & the 2 nd Law Carnot’s theorem: 1.All Carnot engines operating between temperatures T h & T c have the same efficiency. 2.No other engine operating between T h & T c can have a greater efficiency. Refrigerator: extracts heat from cool reservoir into a hot one. work required

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

Perfect refrigerator  Perfect heat engine Clausius  Kelvin-Planck

 Carnot engine is most efficient e Carnot = thermodynamic efficiency e Carnot  e rev > e irrev Carnot refrigerator, e = 60% Hypothetical engine, e = 70%

19.3. Applications of the 2 nd Law Power plant fossil-fuel : T h = 650 K Nuclear : T h = 570 K T c = 310 K Actual values: e fossil ~ 40 %e nuclear ~ 34 %e car ~ 20 % Prob 54 & 55 Heat source Boiler Turbine Generator Electricity Condenser Waste water Cooling water

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

Example 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.

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

Example Home Freezer A typical home freezer operates between T c =  18  C to T h = 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 nd law: only a fraction of Q can become W in heat engines. a little W can move a lot of Q in refrigerators.

Heat Pumps Heat pump as AC : Heat pump as heater : Ground temp ~ 10  C year round Heat pump: moves heat from T c to T h.

GOT IT? 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 (a) exceed. (b) be less than. (c) equal that of the original engine alone? see Prob 32 for proof

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

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.

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

Entropy change is path-independent. ( S is a thermodynamic variable )  S = 0 over any closed path   S 12 =  S 12

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

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

Irreversible Heat Transfer Cold & hot water can be mixed reversibly using extra heat baths. Actual mixing, irreversible processes reversible processes T 1 = some medium T. T 2 = some medium T.

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

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

Example 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?

Entropy & the 2 nd Law of Thermodynamics 2 nd 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 ?

GOT IT? In each of the following processes, does the entropy of the named system increase, decrease, or stay the same? (a) a balloon inflates (b) cells differentiate in a growing embryo, forming different organs (c) an animal dies, its remains gradually decays (d) an earthquake demolishes a building (e) a plant utilize sunlight, CO 2, & water to manufacture sugar (f) a power plant burns coal & produces electrical energy (g) a car’s friction based brakes stop the car.       