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19. 2 nd Law of Thermodynamics. Reversibility & Irreversibility The 2 nd Law of Thermodynamics Applications of the 2 nd Law 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.
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19. 2nd Law of Thermodynamics Reversibility & Irreversibility The 2nd Law of Thermodynamics Applications of the 2nd Law 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? 2nd law: no Q W with 100% efficiency
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? 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?
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. No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. Heat dumped
cylinder compressed adiabatically ( T rises to Th adiabatically) • gas in contact with Th , expands isothermally to do work; heat Qh = Wh absorbed • cylinder expands adiabatically ( T drops to Tc ) • cylinder in contact with Tc , gas compressed isothermally heatQc = Wc dumped simple heat engine (any engine) Efficiency (any cycle) (Simple engine)
Carnot Engine Ideal gas: AB: Heat abs. B C: Work done C D: Heat rejected: D A: Work done Adiabatic processes: • isothermal expansion: T = Th , W1 = Qh > 0 • Adiabatic expansion: Th Tc, W2 > 0 • isothermal compression: T = Tc , W3 = Qc < 0 Adiabatic compression : Tc Th , W4 = W2 < 0
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 15C. 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 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
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. No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature.
Perfect refrigerator Perfect heat engine Clausius Kelvin-Planck
Hypothetical engine, e = 70% Carnot refrigerator, e = 60% Carnot engine is most efficient eCarnot = thermodynamic efficiency eCarnot erev > eirrev
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
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%
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.
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 .
Example 19.3. Home Freezer A typical home freezer operates between Tc = 18C 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.
Heat Pumps Heat pump: moves heat from Tc to Th . Heat pump as AC : Heat pump as heater : Ground temp ~ 10C year round (US)
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
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.
Conceptual Example 19.1. Energy Quality, End Use, & Cogeneration You need a new water heater, & you’re trying to decide between gas & electric. The gas heater is 85% efficient, meaning 85% of the fuel energy goes into heating water. The electric heater is essentially 100% efficient. Thermodynamically, which heater makes the most sense? Ans. Only 1/3 of fuel energy is converted to electricity at a power plant. With this in mind, the gas heater is a better choice. Cogeneration: Waste heat from electricity generation used for low Q needs.
Making the Connection If the electricity comes from a more efficient gas-fired power plant with e = 48%, compare the gas consumption of your two heater choices. Gas heater: 1 unit of fuel energy becomes 0.85 unit of heat. Electric heater: 1 unit of fuel energy becomes 0.48 unit of electric energy, then becomes 0.48 unit of heat. Electric heater consumes 0.85/0.48 = 1.8 times the fuel consumed by gas heater.
Entropy Carnot cycle (reversible processes): Qh = heat absorbed Qc = heat rejected Qh , Qc = heat absorbed lukewarm: can’t do W, Q C = Carnot cycle C = any closed path Irreversible processes can’t be represented by a path. S = entropy [ S ] = J / K Contour = sum of Carnot cycles.
S = 0 over any closed path S21+S12= 0 S21=S21 Entropy change is path-independent. ( S is a thermodynamic variable )
Entropy in Carnot Cycle Ideal gas: Heat absorbed: Heat rejected: Adiabatic processes:
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
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.
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 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?
A Statistical Interpretation of Entropy Gas of 2 distinguishable molecules occupying 2 sides of a box 1/4 2 ¼ = ½ 1/4
Gas of 4 distinguishable molecules occupying 2 sides of a box 1/16 = 0.06 4 1/16 = ¼ =0.25 6 1/16 = 3/8 = 0.38 4 1/16 = ¼ =0.25 1/16 = 0.06
Gas of 1023 molecules Gas of 100 molecules Equal distribution of molecules Statistical definition of entropy : • # of micro states
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 ?
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.