1、Chapter 2 Energy and the First Law of Thermodynamics深刻認識熱力學(xué)第一定律的實質(zhì)能量守恒了解熱和功是系統(tǒng)與外界交換能量的兩種方式，定義、特性及計算方法熱力學(xué)第一定律能量方程的基本表達式運用熱力學(xué)第一定律進行工程實踐分析本章要求Work is done by a system on its surroundings if the sole effect on everything external to the system could have been the raising of a weight. 功是系統(tǒng)間相互作用而傳遞的能量當(dāng)系統(tǒng)完

2、成功時，其對外界的作用可以用在外界舉起重物的單一效果來代替。注意：“舉起重物”在熱力學(xué)定義中是“過程產(chǎn)生的效果相當(dāng)于重物舉起，而不一定是真實舉起。WorkSign convention and NotationW0: work done by the systemW 0Perpetual Motion Machine of the First Kind (PMM1)Proof: For a cycle consists of two processes A and B, one can writecycdWadi = 1-A-2dWadi + 2-B-1dWadisince cyc dWadi

3、 01-A-2dWadi -2-B-1dWadi 1-B-2dWadi which violates the First Law. Therefore PMM1 is impossible.PMM1Consider two cycles linking two states 1 and 2 of a system that involves both work and heat. Since for a cycle cycQ - cycW = 0Therefore, 1-A-2Q + 2-B-1 Q = 1-A-2W + 2-B-1W 1-A-2Q + 2-C-1 Q = 1-A-2W + 2

4、-C-1WFirst Law for a Process involving both Work and Heatand,2-B-1Q - 2-C-1Q = 2-B-1W - 2-C-1W or,2-B-1 (Q W) = 2-C-1 (Q W) = E1 E2Or, dE = Q W; de = q - wThe energy defined from the First Law consists of all different forms of energy a substance may have. They include:E = U (internal energy) + KE (

5、kinetic energy) + PE (potential energy) + ChE (chemical energy) + NuE (nuclear energy) + .KE = 1/2 (mV2)PE = mghEnergyTherefore,Q - W = dE = dU + dKE + dPE + dChE + dNuE+ For a system contains only internal energy, (舉例參見P44.)Q - W = dU; 熱力學(xué)第一定律的表達式Or, q - w = du per unit massThe first law of thermod

6、ynamicsEnergyEnergy can be transformed from one form to another and transferred between systems.For closed systems, energy can be transferred by work and heat transfer In all transformations and transfers, the total amount of energy is conservedEnergyEnergy is the ability to do work.It has units of

7、Joules.It is a “Unit of Exchange”.Example1 car = $20k1 house = $100k5 cars = 1 house=Energy EquivalentsWhat is the case for nuclear power?1 kg coal 42,000,000 joules1 kg uranium 82,000,000,000,000 joules1 kg uranium 2,000,000 kg coal!Kinetic EnergyKinetic Energy is the energy of motion.Kinetic Energ

8、y = mass speed2The work of the force Fs as the body moves from S1 to S2 along the path is expressed by followingKinetic Energy根據(jù)牛頓第二定律的推導(dǎo)，式2.6 說明系統(tǒng)動能的變化等于合力F對系統(tǒng)做的功。Potential EnergyAccordingly，the total work equals the change in kinetic energy.式2.9說明，施加在物體上的所有力除了重力外，對物體所做的功，等于物體動能變化與勢能變化的和。Potential

9、Energy從一個特殊的例子來說明能量守恒定律：當(dāng)物體僅僅受到重力時，那么2.9式的右邊等于零，則有：通過2.11式也表達了能量可以從一種形式轉(zhuǎn)換成另外一種形式。100 kg100 kg100 kg1 meternailFor a system involves moving boundary, the work involved is given by:W = - FextdX work done on the systemFext = -pressure x AreaW = pA dX = pdV“+” dV work done by the system; “-” dV work don

10、e on the systemExpansion and Compression WorkFor a system surrounded by an atmospheric air, the useful work isW = (p-p0)dV = pdV - p0dV W1-2 = pdV - p0dVOr, Example 2.1Problem: A gas in a piston-cylinder assembly undergoes an expansion process for which the relationship between pressure and volume i

11、s given by What we have known:Initial pressure is 3 bars, Initial volume is 0.1m3 Final volume is 0.2m3 Determine: Work for the process, in KJ.a) n=1.5, b) n=1.0, c) n=0.Schematic and Given DataThe gas is a closed systemThe moving boundary is the only work modeThe expansion is a polytropic process A

12、ssumptionsAnalysis在上述分析中，不同過程所做的功都可用p-V圖上過程曲線下面的面積進行說明，面積大小與做功數(shù)值相符合。所計算出的功值取決于經(jīng)歷的具體過程和最終狀態(tài)。關(guān)于多變過程的假設(shè)是非常重要的，如果給定的壓力體積關(guān)系是從實驗數(shù)據(jù)擬合獲得的，那么僅當(dāng)測量所得的壓力等于施加在活塞截面上的壓力時，pdV的數(shù)值才會提供一個比較可靠的做功估計值。CommentsQuasistatic work has the following features:1. The force F depends only on the state of the system and is depende

13、nt of the direction of the displacement x.2. The process is reversible. That is the initial state of the system can be restored by reversing the process.3. The value of force F remains finite as dx approaches zero.Quasi-static and Non-quasistatic WorkFor non-quasistatic work:1. F depends on the rate

14、 of change of state. 2. The work interaction is unidirectional. 3. F approaches zero as dx goes to zero.系統(tǒng)在準(zhǔn)平衡過程中完成的功量稱為準(zhǔn)靜功準(zhǔn)靜功可以僅通過系統(tǒng)內(nèi)部的參數(shù)來描述，而無須考慮外界的情況對于非平衡過程，系統(tǒng)完成的功量需要利用對系統(tǒng)進行實際測量來確定；而準(zhǔn)平衡過程的概念對工程上獲得可以接受的近似結(jié)果，帶來很大的方便。準(zhǔn)靜功The number of independent properties required to define a state of a system is e

15、qual to one plus the number of possible quasistatic work modes.對于組成一定封閉系的給定平衡狀態(tài)而言，可用N1個獨立的狀態(tài)參數(shù)來限定它。這里N是系統(tǒng)可能出現(xiàn)的準(zhǔn)靜功形式的數(shù)量，1則是考慮了系統(tǒng)與外界的熱交換。The State PrincipleA simple compressible system is defined as one for which the only relevant quasistatic work interaction is boundary pdV work. For such a system th

16、e number of independent properties required to define a state is two.對于簡單可壓縮系而言，熱力系與外界交換的準(zhǔn)靜功只有氣體的體積變化功（膨脹或壓縮）一種形式，根據(jù)狀態(tài)公理，決定該系統(tǒng)的平衡狀態(tài)的獨立狀態(tài)參數(shù)只有2個Simple Compressible SystemdU = Q W; or, du=q wper unit mass general differential formw = -pdv; only true for a simple systemu = q w; general integrated form =

17、 q pv true for a simple systemSome textbooks will usedu = q + w; work done by system in “-”u = q + wFirst Law for a Simple Compressible SystemConsider a gas confined in a cylinder-piston arrangement, the piston is loaded in such a way that the pressure of the gas is constant. From the First Law,dE =

18、 Q - WFor a simple system, dE = dU and W = pdVdU = Q pdV; orQ = dU + pdVEnthalpy and Specific HeatsBut the pressure is maintained constant,dp = 0.Therefore,Q = d(U + pV)DefineU + pV H EnthalpydQ = dHFor a simple system,u = u(T, v) and h = h(T,p)du = (u/T)v dT + (u/v)T dv dh = (h/T)p dT + (h/p)T dpDe

19、fine,Constant volume specific heat cv by cv = (u/T)v ; du = cvdT + (u/v)T dv Constant pressure specific heat cp bycp = (h/T)p ; dh = cp dT + (h/p)T dp Heat Capacity for Constant Volume Processes (Cv)Heat is added to a substance of mass m in a fixed volume enclosure, which causes a change in internal

20、 energy, U. Thus,Q = U2 - U1 = DU = m Cv DTThe v subscript implies constant volumeHeat, QaddedmmDTinsulationHeat Capacity for Constant Pressure Processes (Cp)Heat is added to a substance of mass m held at a fixed pressure, which causes a change in internal energy, U, AND some PV work.Heat, QaddedDTm

21、mDx Cp DefinedThus,Q = DU + PDV = DH = m Cp DTThe p subscript implies constant pressureH, enthalpy. is defined as U + PV, so DH = D(U+PV) = DU + VDP + PDV = DU + PDVExperimentally, it is easier to add heat at constant pressure than constant volume, thus you will typically see tables reporting Cp for

22、 various materials.Joules ExperimentJoule showed that mechanical energy could beconverted into heat energy.FMDxH2ODTW = FDxIntroduction1843-1848年， 英國釀酒商 James Prescott Joule (1818 - 1889) 以確鑿無疑的定量實驗結(jié)果為基礎(chǔ)，論述了能量守恒和轉(zhuǎn)化定律。焦耳的熱功當(dāng)量實驗是熱力學(xué)第一定律的實驗基礎(chǔ)。 Joule (1818 - 1889) The First law of a closed system in a c

23、ycle容器、攪拌器和水組成一個熱力系，這是一個閉口系統(tǒng)。讓熱力系從初始態(tài)經(jīng)歷一個循環(huán)過程而回到原態(tài)。例如：使容器絕熱，讓重物落下使攪拌器回轉(zhuǎn)。此時有功加入到熱力系中，依靠摩擦功轉(zhuǎn)變成熱，使水溫升高。然后水對環(huán)境放熱，溫度下降而回到原態(tài)。利用不同重物并多次測量后焦耳首先發(fā)現(xiàn)，加入的功量總是與放出的熱量成比例：即： 1 calorie = 4.184 JoulesWhere did the energy go?By the First Law of Thermodynamics, the energy we put into the water (either work or heat) can

24、not be destroyed.The heat or work added increased the internal energy of the water.This is the energy stored in the atoms and molecules that make up the water; they move faster.Example 2.2A closed system initially at rest on the surface of the earth undergoes a process for which there is a net energ

25、y transfer to the system by work of magnitude 200 Btu. During the process there is a net heat transfer of energy from the system to its surroundings of 30 Btu. At the end of the process, the system has a velocity of 200 ft/s at an elevation of 200 ft. The mass of the system is 50 lb, and the local a

26、cceleration of gravity is g=32.0 ft/s2. Determine the change of internal energy of the system for the process, in Btu.SolutionKnown: ms=50lb, g=32.0 ft/s2During the process,Work done on the system, W=-200BtuHeat transfer from the system, Q=-30Btuthe initial state, system is at rest, therefore, V1 =0

27、the final state, V2=200ft/s, h=200ftTo find: the change of internal energy of the system for the process, in Btu.Schematic diagramAssumptionsA closed system is under consideration.At the end of the process, the system is moving with a uniform velocity.The local acceleration of gravity is constant at

28、 g=32.0ft/s2.AnalysisThe Equation of an energy balance,Substituting the value of the parameter in above, we can evaluate: CommentsThe positive sign for U indicates that the internal energy of the system increase during the process.Note the unit conversions when using English system: the energy “bala

29、nce sheet” Comments3. the energy “balance sheet” Input: 200Btu(work) Change of system energy +39.9Btu(KE) +12.8Btu(PE) Output: 30Btu(heat transfer) +117.3Btu(internal energy) +170.0BtuThe net input exceeds the net output by 170Btu, and the system energy increases by this amount.Example 2.3Consider 5

30、 kg of steam (water vapor) contained within a piston-cylinder assembly. The steam undergoes an expansion from state 1, where the specific internal energy (the internal energy per unit mass) is u1=2709.9kJ/kg, to state 2, where u2=2659.6kJ/kg. During the process, there is a heat transfer of energy to

31、 the steam with magnitude of 80 kJ. Also, a paddle wheel transfer energy to the steam by work in the amount of 18.5kJ. There is no significant change in the kinetic or potential energy of the steam. Determine the amount of energy transfer by work from the steam to the piston during the process, in k

32、J.SolutionKnown: ms=5kg, u1=2709.9kJ/kg u2=2659.6kJ/kg Q=+80 kJ Wpw=-18.5 kJTo find: WpistonSchematic diagramAssumptionsThe steam is a closed system.There is no change in the kinetic or potential energy of the steam.AnalysisAn energy balance for the closed system illustrated by :KE+ PE+ U=Q-WThen, t

33、he energy equation es,The net work from the system consists of two modes: from a paddle wheel and by the piston. AnalysisSo, Incorporating the above expressions, the results is,Substituting values into above, Wpiston=+350 kJCommentsNotice the positive sign for Wpiston, which implies the energy transfer is from the steam to the piston as the system expands during this process.The energy “balance sheet” is as following, Input Output 18.5(work, paddle wheel) 350(work,piston) 80.0(heat transfer)Total:98.5 350Example 2.4Example 2.5The rate of heat transfer between a certain eletric