Presentation on theme: "Short Version : 19. 2 nd Law of Thermodynamics. 19.1. Reversibility & Irreversibility Block slowed down by friction: irreversible Bouncing ball: reversible."— Presentation transcript:
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 )
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 Perfect heat engine: coverts heat to work directly. Heat dumped 2 nd 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.
Carnot Engine 1.isothermal expansion: T = T h, W 1 = Q h > 0 2.Adiabatic expansion: T h T c, W 2 > 0 3.isothermal compression: T = T c, W 3 = Q c < 0 Adiabatic compression : T c T h, W 4 = W 2 < 0 Ideal gas: Adiabatic processes: A B: Heat abs. C D: Heat rejected: B C: Work done D A: Work done
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. No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature.
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 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 : 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 19.3. 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 17.1 2 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 (US) Heat pump: moves heat from T c to T h.
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.
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 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. C = Carnot cycle Contour = sum of Carnot cycles.
Entropy change is path-independent. ( S is a thermodynamic variable ) S = 0 over any closed path S 21 + S 12 = 0 S 21 = S 21
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. 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 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 MicrostatesMacrostatesprobability of macrostate 1/4 2 ¼ = ½ 1/4
Gas of 4 distinguishable molecules occupying 2 sides of a box MicrostatesMacrostatesprobability of macrostate 1/16 = 0.06 4 1/16 = ¼ =0.25 1/16 = 0.06 6 1/16 = 3/8 = 0.38
Gas of 100 molecules Gas of 10 23 molecules Equal distribution of molecules Statistical definition of entropy : # of micro states
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 ?