19.2ndLawofThermodynamics熱力學(xué)第二定律課件

1、19. 2nd Law of Thermodynamics 熱力學(xué)第二定律 Reversibility & Irreversibility1. 可逆性和不可逆性 The 2nd Law of Thermodynamics2. 熱力學(xué)第二定律 Applications of the 2nd Law3. 第二定律的應(yīng)用 Entropy & Energy Quality4. 熵和能量品質(zhì)Most energy extracted from the fuel in power plants is dumped to the environment as waste heat, here using a

2、 large cooling tower. 發(fā)電廠內(nèi)從燃料抽取的能量，大部份都被當(dāng)成廢熱而丟到環(huán)境裏；圖中用的是一個(gè)大型冷卻塔。Why is so much energy wasted?為甚麼浪費(fèi)這麼多能量？ 2nd law: no Q W with 100% efficiency第二定律：沒(méi)有 Q W可以達(dá)到 100% 効率。Efficiencies効率Engine 引擎Efficiency 効率Gasoline 汽油1820%Diesel 柴油up to 可達(dá) 40%Steam 蒸氣8%Gas Turbine 氣渦輪up to 可達(dá) 40%Power Plant 發(fā)電廠Efficiency

3、 効率Coal 燃煤36%Nuclear 核子30%19.1. Reversibility & Irreversibility 可逆性和不可逆性Block slowed down 質(zhì)塊因摩擦力by friction: 而慢下來(lái):Irreversible 不可逆Bouncing ball: 彈跳的球: reversible 可逆Examples of irreversible processes:不可逆程序範(fàn)例: Beating an egg, blending yolk & white打一個(gè)蛋，蛋黃蛋白混成一團(tuán) Cups of cold & hot water in contact幾杯冷和熱水

4、貼在一起Spontaneous process: 自發(fā)性程序: order disorder 有序 無(wú)序 ( statistically more probable ) (統(tǒng)計(jì)上較有可能)時(shí)間GOT IT? 19.1.Which of these processes are irreversible:以下那些程序是不可逆的： stirring sugar into coffee,(a) 把糖攪進(jìn)咖啡中， building a house,(b) 建一幢房子， demolishing a house with a wrecking ball, (c) 用鐵球撞毀一幢房子， demolishing

5、 a house by taking it apart piece by piece,(d) 把房子逐塊拆下， harnessing the energy of falling water to drive machinery, (e) 利用水下墮的能量驅(qū)動(dòng)機(jī)器， harnessing the energy of falling water to heat a house? (f) 利用水下墮的能量加熱房子？19.2. The 2nd Law of Thermodynamics 熱力學(xué)第二定律Heat engine extracts work from heat reservoirs. 熱引擎

6、從熱庫(kù)取熱。 gasoline & diesel engines汽油和柴油引擎 fossil-fueled & nuclear power plants化石燃料和核子發(fā)電廠 jet engines噴射引擎2nd law of thermodynamics ( Kelvin-Planck version ):There is no perfect heat engine.熱力學(xué)第二定律 ( 凱爾文- 普朗克 版 ) ： 沒(méi)有完美熱引擎。 Perfect heat engine: coverts heat to work directly.完美熱引擎: 把熱直接變成功。熱庫(kù)Heat dumped倒

7、掉的廢熱No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.沒(méi)有任何程序的淨(jìng)効果可以是從單一個(gè)熱庫(kù)吸收熱量，然後把它全部變成功。simple heat engine簡(jiǎn)單熱引擎 cylinder compressed adiabatically 氣缸絕熱地壓縮( T rises to Th adiabatically T 絕熱地升至Th ) gas in contact with Th , e

8、xpands isothermally to do work; heat Qh = Wh absorbed氣缸與Th 接觸，氣體等溫膨脹作工；吸進(jìn) Qh=Wh的熱 cylinder expands adiabatically 氣缸絕熱地膨脹 ( T drops to Tc adiabatically T 絕熱地降至Tc ) cylinder in contact with Tc , gas compressed isothermally heat Qc = Wc dumped氣缸與Tc 接觸，氣體等溫壓縮；倒出 Qc = Wc 的熱Efficiency効率(any engine) 任何引擎(S

9、imple engine) 簡(jiǎn)單引擎(any cycle) 任何循環(huán)Carnot Engine (Cycle)卡諾引擎 (循環(huán))Isothermal expansion 等溫膨脹 :T = Th , W1 = Qh 0Adiabatic expansion 絕熱膨脹 :Th Tc, W2 0Isothermal compression 等溫壓縮 :T = Tc , W3 = Qc 04. Adiabatic compression 絕熱壓縮 :Tc Th , W4 = W2 eirrevCarnot refrigerator卡諾冷凍機(jī), e = 60%Hypothetical engine 假

10、想引擎, e = 70%完美熱引擎19.3. Applications of the 2nd Law 第二定律的應(yīng)用Power plant發(fā)電廠fossil-fuel 化石燃料 : Th = 650 K Nuclear 核能 : Th = 570 K Tc = 310 KActual values 實(shí)在數(shù)值 :efossil 40 %enuclear 34 %ecar 20 %Prob 54 & 55Heat source熱源Boiler鍋爐Turbine渦輪機(jī)Generator發(fā)電機(jī)Electricity電Condenser冷凝器Waste water 廢水Cooling water冷卻用水

11、溫度溫度溫度水蒸氣水入出Application: Combined-Cycle Power Plant應(yīng)用: 聯(lián)合-循環(huán)發(fā)電廠Turbine engines: high Th ( 1000K 2000K ) & Tc ( 800 K ) not efficient.渦輪機(jī): 高 Th ( 1000K 2000K ) & Tc ( 800 K ) 無(wú)効率。Steam engines : Tc ambient 300K.蒸氣機(jī) : Tc 周圍的 300K.Combined-cycle 聯(lián)合-循環(huán) : Th ( 1000K 2000K ) & Tc ( 300 K ) e 60%Example 19

12、.2. Combined-Cycle Power Plant聯(lián)合-循環(huán)發(fā)電廠The gas turbine in a combined-cycle power plant operates at 1450 C.一個(gè)聯(lián)合-循環(huán)發(fā)電廠內(nèi)的氣渦輪機(jī)在1450 C下運(yùn)作。Its waste heat at 500 C is the input for a conventional steam cycle, with its condenser at 8 C.它的 500 C 廢熱輸入另一個(gè)冷凝器在 8 C 的傳統(tǒng)蒸氣循環(huán)。Find e of the combined-cycle, & compare

13、it with those of the individual components.求此聯(lián)合-循環(huán)的 e ，並與其個(gè)別組件的值比較。氣渦輪機(jī)燃燒溫度氣渦輪機(jī)蒸氣循環(huán)中介溫度冷卻水溫度Refrigerators 冷凍機(jī)Coefficient of performance (COP) for refrigerators :冷凍機(jī)的績(jī)效系數(shù)COP is high if Th Tc .Th Tc 時(shí) COP 高M(jìn)ax. theoretical value (Carnot cycle)理論最大值(卡諾循環(huán))1st lawW = 0 ( COP = ) for moving Q when Th = Tc

14、 .若 Th = Tc ，移動(dòng) Q 時(shí) W = 0 ( COP = ) 。Example 19.3. Home Freezer 家用冰箱A typical home freezer operates between Tc = 18C to Th = 30 C.家用冰箱通常都在 Tc = 18C 到 Th = 30 C 之間運(yùn)作。Whats its maximum possible COP? 它可能的最大 COP 值為何？With this COP, how much electrical energy would it take to freeze 500 g of water initial

15、ly at 0 C?在這COP 值之下，需要多少電能才可以把 500 g 在 0 C 的水凝固？Table 17.12nd law: only a fraction of Q can become W in heat engines.第二定律：熱引擎中祇有一部份 Q 能變成 W 。 a little W can move a lot of Q in refrigerators.冷凍機(jī)內(nèi)一點(diǎn) W 可以產(chǎn)生很多 Q 。Heat Pumps 熱泵Heat pump as AC :以熱泵作冷氣機(jī)Heat pump as heater :以熱泵作暖爐Ground temp 10C year round土

16、地溫度整年都 10CHeat pump: moves heat from Tc to Th .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.一個(gè)聰明的工程師決定要提高卡諾引擎的効率，方法是用一部具有最大 COP可能值的冷凍機(jī)來(lái)冷卻 低溫?zé)釒?kù)。Will the overall efficiency of this

17、 system這系統(tǒng)的整體效率會(huì) exceed超過(guò)， be less than低於， equal that of等於the original engine alone原來(lái)引擎本身的值？see Prob 32 for proof證明可參考 Prob 3219.4. Entropy & Energy Quality熵和能量品質(zhì)Energy quality Q measures the versatility of different energy forms.能量品質(zhì) Q 衡量各種能量的可用性。2nd law:Energy of higher quality can be converted co

18、mpletely into lower quality form.第二定律：高質(zhì)能量可以完全變成較低質(zhì)能量But not vice versa. 反之卻不成。最高質(zhì)最低質(zhì)機(jī)械，電能高溫低溫Conceptual Example 19.1. Energy Quality, End Use, & Cogeneration能量品質(zhì)，終端使用，和共發(fā)電You need a new water heater, & youre trying to decide between gas & electric.你需要一個(gè)新的熱水器；現(xiàn)正在瓦斯和電熱兩者之間作一選擇。The gas heater is 85% e

19、fficient, meaning 85% of the fuel energy goes into heating water.瓦斯熱水器的効率是 85% ，亦即 85% 的燃料能可以用來(lái)熱水。The electric heater is essentially 100% efficient.電熱水器的効率可說(shuō)是 100% 。Thermodynamically, which heater makes the most sense?從熱力學(xué)來(lái)看，那一個(gè)熱水器比較值得用?Cogeneration 共發(fā)電 :Waste heat from electricity generation used f

20、or low Q needs.發(fā)電時(shí)產(chǎn)生的廢熱用在低 Q 的需求上。Ans.答Only 1/3 of fuel energy is converted to electricity at a power plant.發(fā)電廠祗把燃料能的 1/3 轉(zhuǎn)成電能。With this in mind, the gas heater is a better choice. 在這考量下，瓦斯熱水器是較佳選擇。Making the Connection連起來(lái)If the electricity comes from a more efficient gas-fired power plant with e =

21、48%,假設(shè)電力來(lái)自効率較佳，e = 48%，的天然氣發(fā)電廠，compare the gas consumption of your two heater choices.比較上述兩種熱水器的瓦斯消耗。Gas heater:1 unit of fuel energy becomes 0.85 unit of heat.瓦斯熱水器:1單位的燃料能可變成 0.85單位的熱。Electric heater: 1 unit of fuel energy becomes 0.48 unit of electric energy,電熱水器: 1單位的燃料能可變成 0.48單位的電能，then become

22、s 0.48 unit of heat.再變成 0.48單位的熱。Electric heater consumes 0.85/0.48 = 1.8 times the fuel consumed by gas heater. 電熱水器消耗 0.85/0.48 = 1.8 倍瓦斯熱水器所需的燃料。Entropy熵lukewarm: cant do W, Q 微溫：不能作功， Q Carnot cycle (reversible processes):卡諾循環(huán) (可逆程序)Qh = heat absorbed 所吸熱Qc = heat rejected 所排熱Qh , Qc = heat abso

23、rbed所吸熱C = any closed path 任何閉口路徑S = entropy 熵 S = J / KIrreversible processes cant be represented by a path.不可逆程序不能以路徑表達(dá)。等溫線體積壓力絕熱線C = Carnot cycle 卡諾循環(huán)Contour = sum of Carnot cycles.路徑 = 所有卡諾循環(huán)之和Entropy熵lukewarm: cant do W, Q 微溫：不能作功， Q Carnot cycle (reversible processes):卡諾循環(huán) (可逆程序)Qh = heat abso

24、rbed 所吸熱Qc = heat rejected 所排熱Qh , Qc = heat absorbed所吸熱C = any closed path 任何閉口路徑S = entropy 熵 S = J / KIrreversible processes cant be represented by a path.不可逆程序不能以路徑表達(dá)。等溫線體積壓力絕熱線C = Carnot cycle 卡諾循環(huán)Entropy change is path-independent.熵變量與路徑無(wú)關(guān)。( S is a thermodynamic variable )( S 是一個(gè)熱力變數(shù))S = 0 ove

25、r any closed path任何閉口路徑上， S = 0 S21 + S12= 0 S21 = S21 Entropy in Carnot Cycle卡諾循環(huán)的熵Ideal gas 理想氣體：Adiabatic processes 絕熱程序：Heat absorbed:吸收的熱Heat rejected:排出的熱Irreversible Heat Transfer不可逆熱傳遞Cold & hot water can be mixed reversibly using extra heat baths.冷和熱水可以用額外的熱庫(kù)來(lái)達(dá)成可逆性混合。Actual mixing, irrevers

26、ible processes原來(lái)的混合，不可逆程序reversible processes可逆程序T1 = some medium T. 某個(gè)中介T 。T2 = some medium T. 某個(gè)中介T 。Adiabatic Free Expansion絕熱性自由膨脹Adiabatic 絕熱 Qad.exp. = 0S can be calculated by any reversible process between the same states.p = const. p = 定值Cant do work不能作功Q degraded. Q 被貶。 S 可從兩態(tài)之間任何可逆程序計(jì)算真空提取

27、功拿走隔板氣隔板isothermal等溫 Entropy & Availability of Work 熵與可用的功Before adiabatic expansion, gas can do work isothermally絕熱膨脹之前，氣體可以在等溫下作功After adiabatic expansion, gas cannot do work, while its entropy increases by絕熱膨脹之後，氣體不能作功，它的熵則增加了In a general irreversible process在一個(gè)廣泛的不可逆程序中Coolest T in systemExample

28、 19.4. Loss of Q Q的損耗A 2.0 L cylinder contains 5.0 mol of compressed gas at 300 K.一個(gè) 2.0 L 氣筒灌了 5.0 mol 在 300 K 的壓縮氣體。If the cylinder is discharged into a 150 L vacuum chamber & its temperature remains at 300 K,如果氣筒往一個(gè) 150 L真空倉(cāng)放氣，而且它的溫度保持在 300 K ，how much energy becomes unavailable to do work?有多少能量變

29、成不能作功？閥打開(kāi)之前氣閥關(guān)上真空之後A Statistical Interpretation of Entropy 熵的統(tǒng)計(jì)詮釋Gas of 2 distinguishable molecules occupying 2 sides of a box由可存在盒子兩邊的兩個(gè)可分辨分子所組成的氣體Microstates微觀態(tài)Macrostates巨觀態(tài)probability of macrostate巨觀態(tài)的或然率 1/42 = 1/4Gas of 4 distinguishable molecules occupying 2 sides of a box由可存在盒子兩邊的 4 個(gè)可分辨分子所組成的氣體Microstates微觀態(tài)Macrostates巨觀態(tài)probability of macrostate巨觀態(tài)的或然率 1/16 = 0.064 1/16 = =0.254 1/16 = =0.251/16 = 0.066 1/16 = 3/8 = 0.38Gas of 100 molecules100個(gè)分子的氣體Gas of 1023 molecules1023 個(gè)分子的氣體Equal distri