课件工程热力学yu chapter.2energy and the first law of thermodynamics

1、Chapter 2 Energy and the First Law of Thermodynamics深刻认识热力学第一定律的实质能量守恒了解热和功是系统与外界交换能量的两种方式，定义、特性及计算方法热力学第一定律能量方程的基本表达式运用热力学第一定律进行工程实践分析本章要求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. 功是系统间相互作用而传递的能量当系统完

2、成功时，其对外界的作用可以用在外界举起重物的单一效果来代替。注意：“举起重物”在热力学定义中是“过程产生的效果相当于重物举起，而不一定是真实举起。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, (举例参见P44.)Q - W = dU; 热力学第一定律的表达式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根据牛顿第二定律的推导，式2.6 说明系统动能的变化等于合力F对系统做的功。Potential EnergyAccordingly，the total work equals the change in kinetic energy.式2.9说明，施加在物体上的所有力除了重力外，对物体所做的功，等于物体动能变化与势能变化的和。Potential

9、Energy从一个特殊的例子来说明能量守恒定律：当物体仅仅受到重力时，那么2.9式的右边等于零，则有：通过2.11式也表达了能量可以从一种形式转换成另外一种形式。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，不同过程所做的功都可用p-V图上过程曲线下面的面积进行说明，面积大小与做功数值相符合。所计算出的功值取决于经历的具体过程和最终状态。关于多变过程的假设是非常重要的，如果给定的压力体积关系是从实验数据拟合获得的，那么仅当测量所得的压力等于施加在活塞截面上的压力时，pdV的数值才会提供一个比较可靠的做功估计值。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.系统在准平衡过程中完成的功量称为准静功准静功可以仅通过系统内部的参数来描述，而无须考虑外界的情况对于非平衡过程，系统完成的功量需要利用对系统进行实际测量来确定；而准平衡过程的概念对工程上获得可以接受的近似结果，带来很大的方便。准静功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.对于组成一定封闭系的给定平衡状态而言，可用N1个独立的状态参数来限定它。这里N是系统可能出现的准静功形式的数量，1则是考虑了系统与外界的热交换。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.对于简单可压缩系而言，热力系与外界交换的准静功只有气体的体积变化功（膨胀或压缩）一种形式，根据状态公理，决定该系统的平衡状态的独立状态参数只有2个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， 英国酿酒商 James Prescott Joule (1818 - 1889) 以确凿无疑的定量实验结果为基础，论述了能量守恒和转化定律。焦耳的热功当量实验是热力学第一定律的实验基础。 Joule (1818 - 1889) The First law of a closed system in a c

23、ycle容器、搅拌器和水组成一个热力系，这是一个闭口系统。让热力系从初始态经历一个循环过程而回到原态。例如：使容器绝热，让重物落下使搅拌器回转。此时有功加入到热力系中，依靠摩擦功转变成热，使水温升高。然后水对环境放热，温度下降而回到原态。利用不同重物并多次测量后焦耳首先发现，加入的功量总是与放出的热量成比例：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