工程热力学英语教材电子版
热能与动力工程专业英语第1,2,3,4,7,8,9章译文
第一章热科学基础1.1工程热力学基础热力学是一门研究能量储存、转换及传递的科学。
能量以内能（与温度有关）、动能（由物体运动引起）、势能（由高度引起）和化学能（与化学组成相关）的形式储存。
不同形式的能量可以相互转化，而且能量在边界上可以以热和功的形式进行传递。
在热力学中，我们将推导有关能量转化和传递与物性参数，如温度、压强及密度等关系间的方程。
因此，在热力学中，物质及其性质变得非常重要。
许多热力学方程都是建立在实验观察的基础之上，而且这些实验观察的结果已被整理成数学表达式或定律的形式。
其中，热力学第一定律和第二定律应用最为广泛。
1.1.1热力系统和控制体热力系统是一包围在某一封闭边界内的具有固定质量的物质。
系统边界通常是比较明显的（如气缸内气体的固定边界）。
然而，系统边界也可以是假想的（如一定质量的流体流经泵时不断变形的边界）。
系统之外的所有物质和空间统称外界或环境。
热力学主要研究系统与外界或系统与系统之间的相互作用。
系统通过在边界上进行能量传递，从而与外界进行相互作用，但在边界上没有质量交换。
当系统与外界间没有能量交换时，这样的系统称为孤立系统。
在许多情况下，当我们只关心空间中有物质流进或流出的某个特定体积时，分析可以得到简化。
这样的特定体积称为控制体。
例如泵、透平、充气或放气的气球都是控制体的例子。
包含控制体的表面称为控制表面。
因此，对于具体的问题，我们必须确定是选取系统作为研究对象有利还是选取控制体作为研究对象有利。
如果边界上有质量交换，则选取控制体有利；反之，则应选取系统作为研究对象。
1.1.2平衡、过程和循环对于某一参考系统，假设系统内各点温度完全相同。
当物质内部各点的特性参数均相同且不随时间变化时，则称系统处于热力学平衡状态。
当系统边界某部分的温度突然上升时，则系统内的温度将自发地重新分布，直至处处相同。
当系统从一个平衡状态转变为另一个平衡状态时，系统所经历的一系列由中间状态组成的变化历程称为过程。
工程热力学EngineeringThermodynamics
1. 2. 3.
.
工程热力学是一门研究热能有效利用及热能和其它形式能量 转换规律的科学 （能量转换）, , .
4.
热电厂(热能→机械能) ( →
汽车 (热能→机械能) ( → )
飞机(热能→机械能) ( →
冰箱 (机械能→热能) ( →
? ?
.自然界中的能源 （ )
自然能源的种类 ( .化学能（ ）: : 煤炭 （） 石油 （ ） 天然气（ ） 沼气 （, ）. )
() () () () () () () （问题陈述） （示意图） （假设） （物理定律） （物性参数） （计算） , , （推理、证明和讨论）
（对作业的要求）
（整洁） （有条理） （完整） （工整）
请在布置了作业以后的一周内交作业。 作业得分在及以下的同学可以重新做好之后 请老师另阅。
() 德国 发现能量子（量子理论） 热力学第二定律 年诺贝尔物理学奖
热力学方面获诺贝尔奖的科学家（）
.能斯特
() 德国 热化学，熵基准 年因发现热力学第三定律获诺贝尔化学奖。(年 补发)
热力学方面获诺贝尔奖的科学家（）
.昂萨格
() 美国 不可逆过程热力学理论 因创立多种热动力作用之间相互关系的理论获 诺贝尔化学奖。
（有效数字）
. （结果保留三位有效数字） .
考试成绩 、 (平时作业) 、 (课堂练习及讨论参与情况 ％）
、 (实验情况及报告)
、 （ 期末考试）
热力学方面获诺贝尔奖的科学家（）
.范.德瓦尔斯
() 荷兰 气体和液体状态方程 年诺贝尔物理学奖
热力学方面获诺贝尔奖的科学家（）
.普朗克
水蒸气 湿空气
工程热力学与传热学(英文) 绪论
3. Basic Principles
（1）The first law of thermodynamics （2）The second law of thermodynamics
4. Main Contents
（1）The thermal properties of substances （2）Processes and cycles （3）Ways and technical measures to enhance the energy conversion efficiency
0-1 Energy
0-1-1 Energy
1. Energy（能量）:
The measurement of the substance movement.
2. Various forms of energy
（1）Mechanical energy （2）Thermal energy
（3）Electrical energy
3. Research Approaches
Three basic approaches and their combination • Theory analysis • Numerical simulation • Experiment research
思考题
什么是能量？目前被人们认识和利用的能量形式有哪些？ 什么是能源？能源是如何分类的？ 热能利用的两种形式是什么？ 热工基础的研究是什么？ “热工基础”课与节能有怎样的关系？ 以任意一种热能动力装置为例，分析其在热功转换过程 中所经历的过程。 7. 你能够从各种热能动力装置的工作过程中，初步概括出 它们在实现热功转换时的某些共同特性吗？ 1. 2. 3. 4. 5. 6.
工程热力学英文版1
CHAPTER1INTRODUCTION1.1What is thermodynamics?Thermodynamics is the science which has evolved from the original investiga-tions in the19th century into the nature of“heat.”At the time,the leading theory of heat was that it was a type offluid,which couldflow from a hot body to a colder one when they were brought into contact.We now know that what was then called“heat”is not afluid,but is actually a form of energy–it is the energy associated with the continual,random motion of the atoms which compose macroscopic matter,which we can’t see directly.This type of energy,which we will call thermal energy,can be converted (at least in part)to other forms which we can perceive directly(for example, kinetic,gravitational,or electrical energy),and which can be used to do useful things such as propel an automobile or a747.The principles of thermodynamics govern the conversion of thermal energy to other,more useful forms.For example,an automobile engine can be though of as a device whichfirst converts chemical energy stored in fuel and oxygen molecules into thermal en-ergy by combustion,and then extracts part of that thermal energy to perform the work necessary to propel the car forward,overcoming friction.Thermody-namics is critical to all steps in this process(including determining the level of pollutants emitted),and a careful thermodynamic analysis is required for the design of fuel-efficient,low-polluting automobile engines.In general,thermody-namics plays a vital role in the design of any engine or power-generating plant, and therefore a good grounding in thermodynamics is required for much work in engineering.If thermodynamics only governed the behavior of engines,it would probably be the most economically important of all sciences,but it is much more than that.Since the chemical and physical state of matter depends strongly on how much thermal energy it contains,thermodynamic principles play a central role in any description of the properties of matter.For example,thermodynamics allows us to understand why matter appears in different phases(solid,liquid, or gaseous),and under what conditions one phase will transform to another.1CHAPTER1.INTRODUCTION2The composition of a chemically-reacting mixture which is given enough time to come to“equilibrium”is also fully determined by thermodynamic principles (even though thermodynamics alone can’t tell us how fast it will get there).For these reasons,thermodynamics lies at the heart of materials science,chemistry, and biology.Thermodynamics in its original form(now known as classical thermodynam-ics)is a theory which is based on a set of postulates about how macroscopic matter behaves.This theory was developed in the19th century,before the atomic nature of matter was accepted,and it makes no reference to atoms.The postulates(the most important of which are energy conservation and the impos-sibility of complete conversion of heat to useful work)can’t be derived within the context of classical,macroscopic physics,but if one accepts them,a very powerful theory results,with predictions fully in agreement with experiment.When at the end of the19th century itfinally became clear that matter was composed of atoms,the physicist Ludwig Boltzmann showed that the postu-lates of classical thermodynamics emerged naturally from consideration of the microscopic atomic motion.The key was to give up trying to track the atoms in-dividually and instead take a statistical,probabilistic approach,averaging over the behavior of a large number of atoms.Thus,the very successful postulates of classical thermodynamics were given afirm physical foundation.The science of statistical mechanics begun by Boltzmann encompasses everything in classical thermodynamics,but can do more also.When combined with quantum me-chanics in the20th century,it became possible to explain essentially all observed properties of macroscopic matter in terms of atomic-level physics,including es-oteric states of matter found in neutron stars,superfluids,superconductors,etc. Statistical physics is also currently making important contributions in biology, for example helping to unravel some of the complexities of how proteins fold.Even though statistical mechanics(or statistical thermodynamics)is in a sense“more fundamental”than classical thermodynamics,to analyze practical problems we usually take the macroscopic approach.For example,to carry out a thermodynamic analysis of an aircraft engine,its more convenient to think of the gas passing through the engine as a continuumfluid with some specified properties rather than to consider it to be a collection of molecules.But we do use statistical thermodynamics even here to calculate what the appropriate property values(such as the heat capacity)of the gas should be.CHAPTER1.INTRODUCTION3 1.2Energy and EntropyThe two central concepts of thermodynamics are energy and entropy.Most other concepts we use in thermodynamics,for example temperature and pres-sure,may actually be defined in terms of energy and entropy.Both energy and entropy are properties of physical systems,but they have very different characteristics.Energy is conserved:it can neither be produced nor destroyed, although it is possible to change its form or move it around.Entropy has a different character:it can’t be destroyed,but it’s easy to produce more entropy (and almost everything that happens actually does).Like energy,entropy too can appear in different forms and be moved around.A clear understanding of these two properties and the transformations they undergo in physical processes is the key to mastering thermodynamics and learn-ing to use it confidently to solve practical problems.Much of this book is focused on developing a clear picture of energy and entropy,explaining their origins in the microscopic behavior of matter,and developing effective methods to analyze complicated practical processes1by carefully tracking what happens to energy and entropy.1.3Some TerminologyMostfields have their own specialized terminology,and thermodynamics is cer-tainly no exception.A few important terms are introduced here,so we can begin using them in the next chapter.1.3.1System and EnvironmentIn thermodynamics,like in most other areas of physics,we focus attention on only a small part of the world at a time.We call whatever object(s)or region(s) of space we are studying the system.Everything else surrounding the system (in principle including the entire universe)is the environment.The boundary between the system and the environment is,logically,the system boundary. The starting point of any thermodynamic analysis is a careful definition of the system.EnvironmentSystemBoundarySystemCHAPTER 1.INTRODUCTION4Figure 1.1:Control masses and control volumes.1.3.2Open,closed,and isolated systemsAny system can be classified as one of three types:open,closed,or isolated.They are defined as follows:open system:Both energy and matter can be exchanged with the environ-ment.Example:an open cup of coffee.closed system:energy,but not matter,can be exchanged with the environ-ment.Examples:a tightly capped cup of coffee.isolated system:Neither energy nor matter can be exchanged with the envi-ronment –in fact,no interactions with the environment are possible at all.Example (approximate):coffee in a closed,well-insulated thermos bottle.Note that no system can truly be isolated from the environment,since no thermal insulation is perfect and there are always physical phenomena which can’t be perfectly excluded (gravitational fields,cosmic rays,neutrinos,etc.).But good approximations of isolated systems can be constructed.In any case,isolated systems are a useful conceptual device,since the energy and mass con-tained inside them stay constant.1.3.3Control masses and control volumesAnother way to classify systems is as either a control mass or a control volume .This terminology is particularly common in engineering thermodynamics.A control mass is a system which is defined to consist of a specified piece or pieces of matter.By definition,no matter can enter or leave a control mass.If the matter of the control mass is moving,then the system boundary moves with it to keep it inside (and matter in the environment outside).A control volume is a system which is defined to be a particular region of space.Matter and energy may freely enter or leave a control volume,and thus it is an open system.CHAPTER1.INTRODUCTION5 1.4A Note on UnitsIn this book,the SI system of units will be used exclusively.If you grew up anywhere but the United States,you are undoubtedly very familiar with this system.Even if you grew up in the US,you have undoubtedly used the SI system in your courses in physics and chemistry,and probably in many of your courses in engineering.One reason the SI system is convenient is its simplicity.Energy,no matter what its form,is measured in Joules(1J=1kg-m2/s2).In some other systems, different units are used for thermal and mechanical energy:in the English sys-tem a BTU(“British Thermal Unit”)is the unit of thermal energy and a ft-lbf is the unit of mechanical energy.In the cgs system,thermal energy is measured in calories,all other energy in ergs.The reason for this is that these units were chosen before it was understood that thermal energy was like mechanical energy, only on a much smaller scale.2Another advantage of SI is that the unit of force is indentical to the unit of(mass x acceleration).This is only an obvious choice if one knows about Newton’s second law,and allows it to be written asF=m a.(1.1)In the SI system,force is measured in kg-m/s2,a unit derived from the3primary SI quantities for mass,length,and time(kg,m,s),but given the shorthand name of a“Newton.”The name itself reveals the basis for this choice of force units.The units of the English system werefixed long before Newton appeared on the scene(and indeed were the units Newton himself would have used).The unit of force is the“pound force”(lbf),the unit of mass is the“pound mass”(lbm)and of course acceleration is measured in ft/s2.So Newton’s second law must include a dimensional constant which converts from Ma units(lbm ft/s2) to force units(lbf).It is usually written1F=2Mixed unit systems are sometimes used too.American power plant engineers speak of the “heat rate”of a power plant,which is defined as the thermal energy which must be absorbed from the furnace to produce a unit of electrical energy.The heat rate is usually expressed in BTU/kw-hr.CHAPTER1.INTRODUCTION6In practice,the units in the English system are now defined in terms of their SI equivalents(e.g.one foot is defined as a certain fraction of a meter,and one lbf is defined in terms of a Newton.)If given data in Engineering units,it is often easiest to simply convert to SI,solve the problem,and then if necessary convert the answer back at the end.For this reason,we will implicitly assume SI units in this book,and will not include the g c factor in Newton’s2nd law.。
工程热力学第三版答案【英文】第4章
4-4The work done during the isothermal process shown in the figure is to be determined. Assumptions The process is quasi-equilibrium. Analysis From the ideal gas equation,vRTP =For an isothermal process,/kgm 0.6kPa200kPa 600/kg)m (0.2331221===P P v vSubstituting ideal gas equation and this result into the boundary work integral produceskJ395.5-=⎪⎪⎭⎫⎝⎛⋅====⎰⎰333312112121out ,m kPa 1kJ 1m0.6m 0.2ln)m kPa)(0.6 kg)(200 (3lnv v v vvv mP d mRTd P W bThe negative sign shows that the work is done on the system.4-9Refrigerant-134a in a cylinder is heated at constant pressure until its temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium.Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-11 through A-13)/kgm 0.052427C 07kPa 005/kgm 0.0008059liquid Sat.kPa 00532223k Pa005@11=⎭⎬⎫︒====⎭⎬⎫=v vv T P P fAnalysiskg04.62/kgm 0.0008059m0.053311===v V mandvPkJ1600=⎪⎪⎭⎫⎝⎛⋅-=-=-==⎰33121221out ,m kPa 1kJ 1/kg m 0.0008059)427kPa)(0.052 kg)(500 (62.04)()( v v V V V mP P d P W bDiscussion The positive sign indicates that work is done by the system (work output).4-23A saturated water mixture contained in a spring-loaded piston-cylinder device is heated until the pressure and temperature rises to specified values. The work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis The initial state is saturated mixture at 90︒C. The pressure and the specific volume at this state are (Table A-4),/kgm 23686.0)001036.03593.2)(10.0(001036.0kPa183.70311=-+=+==fgfx P vvvThe final specific volume at 800 kPa and 250°C is (Table A-6)/kgm 29321.032=vSince this is a linear process, the work done is equal to the area under the process line 1-2:kJ24.52=⎪⎭⎫ ⎝⎛⋅-+=-+==331221out ,m kPa 1kJ 1)m 23686.01kg)(0.2932 (12)kPa 800(70.183)(2Area v v m P P W b4-29An insulated rigid tank is initially filled with a saturated liquid-vapor mixture of water. An electric heater in the tank is turned on, and the entire liquid in the tank is vaporized. The length of time the heater was kept on is to be determined, and the process is to be shown on a P-v diagram.Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The device is well-insulated and thus heat transfer is negligible. 3The energy stored in the resistance wires, and the heat transferred to the tank itself is negligible.Analysis We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as)(V 0)=PE =KE (since )(1212in ,energiesetc. potential, k inetic,internal,in Change system massand work ,heat,by nsfer energy tra Net out in u u m t I Q u u m U W E E E e -=∆=-=∆=∆=-The properties of water are (Tables A-4 through A-6)()[]()k J /k g2569.7v a p o rs a t ./k g m 0.29065k J /k g980.032052.30.25466.97/k g m 0.290650.0010531.15940.250.001053k J /k g3.2052,97.466/kg m 1.1594,001053.025.0kPa 150/k gm 0.29065@2312113113113==⎪⎭⎪⎬⎫===⨯+=+==-⨯+=+=====⎭⎬⎫==g fgffg f fgf g f u u ux uu x uu x P v v v vv vvSubstituting,min60.2==∆⎪⎪⎭⎫ ⎝⎛-=∆s 33613kJ/s1VA1000.03)kJ/kg 980kg)(2569.7(2)A8)(V 110(t t4-39A saturated water mixture contained in a spring-loaded piston-cylinder device is heated until the pressure and volume rise to specified values. The heat transfer and the work done are to be determined.Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed asv)(0)=PE =KE (since )(12ou ,in 12ou ,in energiesetc. potential, k inetic,internal,in Change system massand work ,heat,by nsfer energy tra Net out in u u m W Q u u m U W Q E E E t b t b -+=-=∆=-∆=-The initial state is saturated mixture at 75 kPa. The specific volume and internal energy at this state are (Table A-5),kJ/kg30.553)8.2111)(08.0(36.384/kg m 1783.0)001037.02172.2)(08.0(001037.0131=+=+==-+=+=fgffg f xuuu x v vvThe mass of water iskg22.11/kgm 1783.0m23311===v V mThe final specific volume is/kgm 4458.0kg22.11m53322===m V vThe final state is now fixed. The internal energy at this specific volume and 225 kPa pressure is (Table A-6) kJ/kg 4.16502=u Since this is a linear process, the work done is equal to the area under the process line 1-2:kJ450=⎪⎭⎫ ⎝⎛⋅-+=-+==331221out ,m kPa 1kJ 1)m 2(52)kPa225(75)(2Area V V P P W bSubstituting into energy balance equation giveskJ12,750=-+=-+=kJ/kg )30.553kg)(1650.422.11(kJ 450)(12out ,in u u m W Q b4-43Two tanks initially separated by a partition contain steam at different states. Now the partition is removed and they are allowed to mix until equilibrium is established. The temperature and quality of the steam at the final state and the amount of heat lost from the tanks are to be determined.Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions.Analysis (a ) We take the contents of both tanks as the system. This is a closed system since no massenters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as[][]0)=PE =KE (since )()(1212out energiesetc. potential, k inetic,internal,in Change systemmass and work ,heat,by nsfer energy tra Net out in =-+-=∆+∆=-∆=-W u u m u u m UUQ E E E B A BAThe properties of steam in both tanks at the initial state are (Tables A-4 through A-6)kJ/kg7.2793/kgm 25799.0C 300kPa 1000,13,1,1,1==⎪⎭⎪⎬⎫︒==A A A A u T P v()[]()kJ/kg4.15954.19270.50.66631/kgm 0.196790.0010910.392480.500.001091kJ/kg4.1927,66.631/kgm .392480,001091.050.0C 1501,131,131,1=⨯+=+==-⨯+=+=====⎭⎬⎫=︒=fgfB fg f B fgf gf B ux uu x u u x T v vv vvThe total volume and total mass of the system arekg523m106.1/kg)m 19679.0kg)( 3(/kg)m 25799.0kg)( 2(333,1,1=+=+==+=+=+=B A B B A A B A m m m m m v v V V VNow, the specific volume at the final state may be determined/kgm 22127.0kg5m 106.1332===m Vvwhich fixes the final state and we can determine other properties()kJ/kg8.12821.19820.3641.11561001073.060582.0001073.022127.0/kg m 22127.0kPa0032222k Pa 300 @sat 2322=⨯+=+==--=--=︒==⎪⎭⎪⎬⎫==fg f f g f u x u u x T T P 0.3641C133.5v v v v v(b ) Substituting,[][]kJ 3959kJ/kg )4.15958.1282(kg) 3(kJ/kg )7.27938.1282(kg) 2()()(1212out -=-+-=-+-=∆+∆=-BA BAu u m u u m UUQorkJ3959=out Q4-60The air in a rigid tank is heated until its pressure doubles. The volume of the tank and the amount of heat transfer are to be determined.Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -221︒F and 547 psia. 2 The kinetic and potential energy changes are negligible, ∆∆pe ke ≅≅0. 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. Properties The gas constant of air is R = 0.3704 psia.ft 3/lbm.R (Table A-1E).Analysis (a3ft80.0=⋅⋅==psia50R)R)(540/lbm ft psia 4lbm)(0.370 (20311P mRT V(b) We take the air in the tank as our system. The energy balance for this stationary closed system can be expressed as)()(1212in in energiesetc. potential, k inetic,internal,in Change system massand work ,heat,by nsfer energy tra Net out in T T mc u u m Q UQ E E E -≅-=∆=∆=-vThe final temperature of air isR1080R) (540211222211=⨯==−→−=T P P T T P T P V VThe internal energies are (Table A-17E)u u u u 12====@@540R 1080R 92.04Btu /lbm 186.93Btu /lbmSubstituting, Q in = (20 lbm)(186.93 - 92.04)Btu/lbm = 1898 BtuAlternative solutions The specific heat of air at the average temperature of T avg = (540+1080)/2= 810 R = 350︒F is, from Table A-2Eb, c v ,avg = 0.175 Btu/lbm.R. Substituting,Q in = (20 lbm)( 0.175 Btu/lbm.R)(1080 - 540) R = 1890 BtuDiscussion Both approaches resulted in almost the same solution in this case.4-64A student living in a room turns her 150-W fan on in the morning. The temperature inthe room when she comes back 10 h later is to be determined.Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -141︒C and 3.77 MPa. 2 The kinetic andQpotential energy changes are negligible,∆∆ke pe ≅≅0. 3 Constant specific heats atroom temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. 4 All the doors and windows are tightly closed, and heat transfer through the walls and the windows is disregarded.Properties The gas constant of air is R = 0.287 kPa.m 3/kg.K (Table A-1). Also, c v = 0.718 kJ/kg.K for air at room temperature (Table A-2).Analysis We take the room as the system. This is a closed system since the doors and the windows are said to be tightly closed, and thus no mass crosses the system boundary during the process. The energy balance for this system can be expressed as)()(1212,,energiesetc. potential, k inetic,internal,in Change system massand work ,heat,by nsfer energy tra Net T T mc u u m W UW E E E in e in e out in -≅-=∆=∆=-vThe mass of air iskg174.2K)K)(288/kg m kPa (0.287)m kPa)(144 (100m14466433113=⋅⋅===⨯⨯=RT P m V VThe electrical work done by the fan isW W t e e==⨯= ∆(0.15kJ /s)(103600s)5400kJ Substituting and using the c v value at room temperature, 5400 kJ = (174.2 kg)(0.718 kJ/kg ⋅︒C)(T 2 - 15)︒CT 2 = 58.2︒CDiscussion Note that a fan actually causes the internal temperature of a confinedspace to rise. In fact, a 100-W fan supplies a room with as much energy as a 100-W resistance heater.4-69Carbon dioxide contained in a spring-loaded piston-cylinder device is heated. The work done and the heat transfer are to be determined.Assumptions 1 CO 2 is an ideal gas since it is at a high temperature relative to its critical temperature of 304.2 K. 2 The kinetic and potential energy changes are negligible,0pe ke ≅∆≅∆.Properties The properties of CO 2 are R = 0.1889 kJ/kg ⋅K and c v = 0.657 kJ/kg ⋅K (Table A-2a ).PAnalysis We take CO 2 as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as)(12out ,in energiesetc. potential, k inetic,internal,in Change systemmass and work ,heat,by nsfer energy tra Net out in T T mc U W Q E E E b -=∆=-∆=-vThe initial and final specific volumes are 33111m5629.0k P a100K)K)(298/kg m kPa kg)(0.1889 (1=⋅⋅==P mRT V33222m1082.0kPa1000K)K)(573/kg m kPa kg)(0.1889 (1=⋅⋅==P mRT VPressure changes linearly with volume and the work done is equal to the area underthe process line 1-2:kJ1.250m kPa 1kJ 1)m5629.0(0.10822)kPa 1000(100)(2Area 331221out ,-=⎪⎪⎭⎫⎝⎛⋅-+=-+==V V P P W b Thus,kJ250.1=in ,b WUsing the energy balance equation,kJ4.69K )25K)(300kJ/kg 657.0(kg) 1(kJ 1.250)(12out ,in -=-⋅+-=-+=T T mc W Q b vThus, kJ 69.4=outQ4-76Air at a specified state contained in a piston-cylinder device with a set of stops is heated until a final temperature. The amount of heat transfer is to be determined.Assumptions 1 Air is an ideal gas since it is at a high temperature relative to its critical temperature of 304.2 K. 2 The kinetic and potential energy changes are negligible, 0pe ke ≅∆≅∆. Properties The properties of air are R = 0.287 kJ/kg ⋅K and c v = 0.718 kJ/kg ⋅K (Table A-2a ). Analysis We take air as the system. This is a closedsystem since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as)(12out ,in energiesetc. potential, k inetic,internal,in Change systemmassand work ,heat,by nsfer energy tra Net out in T T mc U W Q E E E b -=∆=-∆=-vThe volume will be constant until the pressure is 300 kPa:K900kPa100kPa 300K)(3001212===P P T TThe mass of the air iskg4646.0K)K)(300/kg m kPa (0.287)m kPa)(0.4 (10033111=⋅⋅==RT P m VThe boundary work done during process 2-3 iskJ04900)K -K)(1200/kg m kPa (0.287)kg 4646.0()()(323232out ,=⋅⋅=-=-=T T mR P W b V VSubstituting these values into energy balance equation,kJ340=-⋅+=-+=K )300K)(1200kJ/kg 718.0(kg) 4646.0(kJ 40)(13out ,in T T mc W Q b v4-85An egg is dropped into boiling water. The amount of heat transfer to the egg by the time it is cooked is to be determined.Assumptions 1 The egg is spherical in shape with a radius of r 0 = 2.75 cm. 2 The thermal properties of the egg are constant. 3 Energy absorption or release associated with any chemical and/or phase changes within the egg is negligible. 4 There are no changes in kinetic and potential energies.Properties The density and specific heat of the egg are given to be ρ = 1020 kg/m 3 and c p = 3.32 kJ/kg.︒C.Analysis We take the egg as the system. This is a closes system since no mass enters or leaves the egg. The energy balance for this closed system can be expressed as)()(1212egg in energiesetc. potential, k inetic,internal,in Change system massand work ,heat,by nsfer energy tra Net out in T T mc u u m U Q E E E -=-=∆=∆=-Then the mass of the egg and the amount of heat transfer becomeBoiling PV (m 3)kJ 21.2=︒-︒=-=====C )880)(C kJ/kg. 32.3)(kg 0889.0()(kg0889.06m)055.0()kg/m1020(612in 333T T mc Q Dm p ππρρV。
工程热力学与传热学(英文) 第8章 热量传递的基本方式
tf t
Φ
平壁上的对流换热
8-2-2 Newton’s law of cooling（牛顿冷却公式）
Ah(t w t f ) q h(t w t f ) ht
In 1701，by Newton
流动方向 u∞ u tw wall
t
f
t
Φ
平壁上的对流换热
A q A
dt dx
WW / m2源自Why the minus sign?
Where：
负号的含义
Heat is transferred in the direction of decreasing temp.
Φ — Heat- transfer rate（热流量） （单位时间通过给定面积传递的热量）W, kW
8-2 Convection Heat Transfer（热对流）
8-2-1 热对流和对流换热
1. Convection（对流） 是指流体各部分之间发生相对位移时，冷热流体相互 掺混所引起的热量传递现象。 2. Convection heat transfer（对流换热） 流体流过另一个物体表面时， 对流和导热联合起作用的热量 传递现象。 平壁表面的 传热机理
Heat Transfer
--- Chapter 8 Basic Modes of
Heat Transfer ---
Chapter 8 Basic Modes of Heat Transfer
Contents
Conduction heat transfer and Fourier’s law of onedimensional steady state heat-conduction Convection heat transfer and Newton’s law of cooling Radiation heat transfer and Stefan-Boltzmann Law Heat transfer processes
工程热力学与传热学(英文) 第6章 动力装置循环
t
h1 h1
h2 h3
h 1 f ( p1 , t1 ) h 2 f ( p1 , t1 , p2 ) h 3 f ( p2 )
t
h 1 h 2 h 1 h 3
1 T2 T1
T
1
T1
5 p1 6
4
Heat addition average temperature
吸热平均温度 T1 Thigh ,av
• Lowering the condenser pressure • Superheating the steam to high temperature • Increasing the boiler pressure
Improving Rankine cycle（改进朗肯循环） • Reheat Rankine cycle（再热循环） • Regenerative Rankine cycle（回热循环） • Combined thermal-electrical power cycles（热电联供循环等）
6-1-4 The ideal reheat Rankine cycle（再热循环）
Question
How can we take advantage of the increased efficiencies at higher boiler pressures without facing the problem of excessive moisture at the final stages of the turbine?
state 2 — state 3
1
saturated liquid-vapor mixture — saturated liquid
工程热力学与传热学(英文) 第1章 基本概念
Thermodynamic temperature scale in SI（热力学温标）
Thermodynamic temperature: symbol: T，unit: K. Datum point: the triple point of water.
The value on triple point is 273.16 K.
Ｑ＝０
adiabatic lysis
any non- isolated system
+
relative surrounding
=
Isolated system
环境
孤立系统
Most systems we discuss are:
The Universe
Simple compressible systems (简单可压缩系统).
（1）Definition: The force exerted by a fluid per unit area，p （2）Units（单位）
SI: Pa (pascal ), kPa Times of the atmospheric pressure: atm, at Height of liquid column: mH2O, mmHg bar, 1kgf/cm2 (in Europe): psi(lbf/in2) ( in English)
• •
Single-valued function of the system state. Related with States but have nothing to do with the path from one state to another.
1-2-2 Equilibrium State
工程热力学英语ppt
状态参数
( Q W ) ( Q W )
1a 2 1b 2
内能及闭口系热一律表达式
定义 dU = Q+W
闭口系热一律表达式
！！！两种特例 绝功系 Q = dU 绝热系 W = - dU
heat transfer to a system and work done by a system are positive; heat transfer from a system and work done on a system are negative. Heat is transferred by three mechanisms: conduction, convection, and radiation. Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interaction between particles. Convection is the transfer of energy between a solid surface and the adjacent fluid that is in motion, and it involves the combined effects of conduction and fluid motion. Radiation is the transfer of energy due to the emission of electromagnetic waves (or photons). An overview of the three mechanisms of heat transfer is given at the end of this chapter as a Topic of Special Interest.
工程热力学与传热学(英文) 第4章 热力学第二定律
1
T
q1
2
T1 q= 0 T2
3
1
2
q= 0
4
4
3
q2
0
v
0
Δs
s
3. The Efficiency of a Carnot Engine（卡诺热效率）
For any heat engine wnet q 1q 2 q2 t 1 q1 q1 q1 For a Carnot engine Method 1: From T-s diagram
A
the paddle wheel • The internal energy of the gas is increased
• The reverse process, raising the
B
mass by transferring heat from the fluid to the paddle wheel, does not occur in nature.
2. p-v and T-s Diagrams
p
1
T q1
2
T1
1
2
q= 0
4
q= 0
q2
3
T2 v 0
4
3
0
Δs
s
• 1-2 Reversible isothermal expansion
（T1=constant, heat absorbed q1） • 2-3 Reversible adiabatic expansion • 3-4 Reversible isothermal compression （T2=constant, heat rejected q2） • 4-1 Reversible adiabatic compression
高等工程热力学英文课件 Chapter the second law of thermodynamics
9
Байду номын сангаас
The importance of the Carnot cycle: • It is the theoretical foundation of the second law. • It directs how to increase the thermal efficiency • It is an ideal model of the real thermal engine. • It is not easy to achieve the Carnot cycle as (1) isothermal
1
3–1 Statements of the second law
1. The directionality of a spontaneous process (irreversibility)
a. From mechanical energy to thermal energy b. From electric energy to thermal energy c. Heat transfer from a hot object to a cold object
The overall effect of the Carnot cycle
8
• Carnot engine gives the upper limit of a heat engine. • No engine operating between the same two temperatures can be
Not in quasi-equilibrium
(process c, d and e)
工程热力学第三版答案【英文】第二章
2-8A person with his suitcase goes up to the 10th floor in an elevator. The part of the energy of the elevator stored in the suitcase is to be determined. Assumptions 1 The vibrational effects in the elevator are negligible.Analysis The energy stored in the suitcase is stored in the form of potential energy, which is mgz . Therefore,kJ 10.3=⎪⎭⎫⎝⎛=∆=∆=∆222suitcase /s m 1000 kJ/kg 1m) 35)(m/s(9.81) kg (30z mg PE ETherefore, the suitcase on 10th floor has 10.3 kJ more energy compared to an identical suitcase on the lobby level.Discussion Noting that 1 kWh = 3600 kJ, the energy transferred to the suitcase is 10.3/3600 = 0.0029 kWh, which is very small.2-14A river is flowing at a specified velocity, flow rate, and elevation. The total mechanical energy of the river water per unit mass, and the power generation potential of the entire river are to be determined.Assumptions 1 The elevation given is the elevation of the free surface of the river. 2 The velocity given is the average velocity. 3 The mechanical energy of water at the turbine exit is negligible. Properties We take the density of water to be ρ = 1000 kg/m 3. Analysis Noting that the sum of the flow energy and the potential energy is constant for a given fluid body, we can take theelevation of the entire river water to be the elevation of the free surface, and ignore the flowenergy. Then the total mechanical energy of the river water per unit mass becomesk J /k g 0.887=⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛+=+=+=22222me ch /s m 1000kJ/kg 12)m/s 3(m) 90)(m/s (9.812Vgh ke pe e The power generation potential of the river water is obtained by multiplying the totalmechanical energy by the mass flow rate,kg/s500,000/s)m 00)(5 kg/m 1000(33===V ρmMW444=====kW 000,444kJ/kg) 7kg/s)(0.88 000,500(mech mech max e m E WTherefore, 444 MW of power can be generated from this river as it discharges into thelake if its power potential can be recovered completely.Discussion Note that the kinetic energy of water is negligible compared to the potential energy, and it can be ignored in the analysis. Also, the power output of an actual turbine will be less than 444 MW because of losses and inefficiencies.2-21C(a ) From the perspective of the contents, heat must be removed in order to reduce and maintain the content's temperature. Heat is also being added to the contents from the room air since the room air is hotter than the contents.(b ) Considering the system formed by the refrigerator box when the doors are closed, there are three interactions, electrical work and two heat transfers. There is atransfer of heat from the room air to the refrigerator through its walls. There is also a transfer of heat from the hot portions of the refrigerator (i.e., back of the compressor where condenser is placed) system to the room air. Finally, electrical work is being added to the refrigerator through the refrigeration system.(c ) Heat is transferred through the walls of the room from the warm room air to the cold winter air. Electrical work is being done on the room through the electrical wiring leading into the room.2-27A man is pushing a cart with its contents up a ramp that is inclined at an angle of 10° from the horizontal. The work needed to move along this ramp is to be determined considering (a) the man and (b) the cart and its contents as the system.Analysis (a ) Considering the man as the system, letting l be the displacement along the ramp, and letting θ be the inclination angle of the ramp,Btu6.248ft lbf 4862=⎪⎭⎫⎝⎛⋅⋅=⋅=+==ft lbf 169.778Btu 1ft)lbf 4862(ft)sin(10) 100)(lbf 180100(sin θFl WThis is work that the man must do to raise the weight of the cart and contents, plus hisown weight, a distance of l sin θ.(b ) Applying the same logic to the cart and its contents givesBtu2.231ftlbf 1736=⎪⎭⎫⎝⎛⋅⋅=⋅===ft lbf 169.778Btu 1ft)lbf 1736(ft)sin(10) 100)(lbf 100(sin θFl W2-30The work required to compress a spring is to be determined.Analysis The force at any point during the deflection of the spring is given by F = F 0 + kx , where F 0 is the initial force and x is the deflection as measured from the point where the initial force occurred. From the perspective of the spring, this force acts in the direction opposite to that in which the spring is deflected. Then,[]Btu0.0214=⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛⋅⋅=⋅=-+-=-+-=+==⎰⎰in 12ft 1ft lbf 169.778Btu 1in)lbf 200(inlbf 200in)01(2lbf/in2000)in (1lbf) 100()(2)()(22221221202121x x k x x F dx kx FFds W2-43The classrooms and faculty offices of a university campus are not occupied anaverage of 4 hours a day, but the lights are kept on. The amounts of electricity and money the campus will save per year if the lights are turned off during unoccupied periods are to be determined.Analysis The total electric power consumed by the lights in the classrooms and faculty offices iskW528264264kW 264264,000= W)1106(400=lamps) of (mp)per consumed (Power kW264264,000= W)11012(200=lamps) of (mp)per consumed (Power officeslighting,classroomlighting,totallighting,offices lighting,classroom lighting,=+=+==⨯⨯⨯==⨯⨯⨯=E E E E ENoting that the campus is open 240 days a year, the total number of unoccupiedwork hours per year isUnoccupied hours = (4 hours/day)(240 days/year) = 960 h/yrThen the amount of electrical energy consumed per year during unoccupied work period and its cost are$41,564/yr======.082/kWh)kWh/yr)($0 506,880(energy) of cost nit savings)(U (Energy savings Cost kWh506,880h/yr) kW)(960 (528hours) Unoccupied )((savings Energy totallighting,EDiscussion Note that simple conservation measures can result in significant energyand cost savings.2-47A gasoline pump raises the pressure to a specified value while consuming electric power at a specified rate. The maximum volume flow rate of gasoline is to be determined.Assumptions 1 The gasoline pump operates steadily. 2 The changes in kinetic and potential energies across the pump are negligible.Analysis For a control volume that encloses the pump-motor unit, the energy balance can be written as/energiesetc. potential, k inetic, internal,in change of Rate (steady)0system massand work ,heat,by nsferenergy tra net of Rate out in ==-dtdE E E →outin E E =21in)()(v v P m P m W =+ →P P P m W ∆=-= V v )(12insince /v V =mand the changes in kinetic and potential energies of gasoline are negligible, Solving for volume flow rate and substituting, the maximum flow rate is determined to be/s m 0.5433=⎪⎪⎭⎫⎝⎛⋅=∆=kJ 1mkPa 1kPa 7kJ/s 8.3P 3inmaxWVDiscussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the volume flow rate will be less because of the losses associated with the conversion of electrical-to-mechanical shaft and mechanical shaft-to-flow energy.2-65Water is pumped from a lake to a storage tank at a specified rate. The overall efficiency of the pump-motor unit and the pressure difference between the inlet and the exit of the pump are to be determined.Assumptions 1 The elevations of the tank and the lake remain constant. 2 Frictional losses in the pipes are negligible. 3 The changes in kinetic energy are negligible. 4 The elevation difference across the pump is negligible.Properties We take the density of water to be ρ = 1000 kg/m 3. Analysis (a ) We take the free surface of the lake to be point 1 and the free surfaces of the storage tank to be point 2. We also take the lake surface as the reference level (z 1 = 0), and thus the potential energy at points 1and 2 are pe 1 = 0 and pe 2 = gz 2. The flow energy at both points is zero since both 1 and 2 are open to the atmosphere (P 1 = P 2 = P atm ). Further, the kinetic energy at both points is zero (ke 1 = ke 2 = 0) since the water at both locations is essentiallystationary. The mass flow rate of water and its potential energy at point 2 are k g /s70/s)m 070.0)( kg/m 1000(33===V ρmk J /k g 196.0/s m 1000 kJ/kg 1m) 20)(m/s(9.8122222=⎪⎭⎫⎝⎛==gz pe Then the rate of increase of the mechanical energy of water becomeskW13.7kJ/kg) 6kg/s)(0.19 70()0()(22in mech,out mech,fluid mech,===-=-=∆pe m pe m e e m EThe overall efficiency of the combined pump-motor unit is determined from itsdefinition,67.2%or 0.672 kW20.4 kW 7.13inelect,fluid mech,motor-pump==∆=W E η(b ) Now we consider the pump. The change in the mechanical energy of water as itflows through the pump consists of the change in the flow energy only since the elevation difference across the pump and the change in the kinetic energy arenegligible. Also, this change must be equal to the useful mechanical energy supplied by the pump, which is 13.7 kW:P P P me e m E ∆=-=-=∆V ρ12in mech,out mech,fluid mech,)(Solving for ∆P and substituting,kPa 196=⎪⎪⎭⎫⎝⎛⋅=∆=∆ kJ 1m kPa 1/s m 0.070 kJ/s13.733fluidmech,VE PTherefore, the pump must boost the pressure of water by 196 kPa in order to raise its elevation by 20 m.Discussion Note that only two-thirds of the electric energy consumed by the pump-motor is converted to the mechanical energy of water; the remaining one-third is wasted because of the inefficiencies of the pump and the motor .2-112The work required to compress a gas in a gas spring is to be determined.Assumptions All forces except that generated by the gas spring will be neglected.Analysis When the expression given in the problem statement is substituted into the work integral relation, and advantage is taken of the fact that the force and displacement vectors are collinear, the result is[]Btu0.0228=⎪⎭⎫⎝⎛⋅⋅=⋅=⎪⎭⎫ ⎝⎛--⋅=--===----⎰⎰ft lbf 169.778Btu 1ft)lbf 74.17(ftlbf 74.17in 12ft 1 in)1(in)4(4.11inlbf 200)(1Constant Constant0.40.41.411122121k k kx x k dxxFds W。
工程热力学与传热学(英文) 第9章 热传导
y
t y
dxdz
dфy+dy
dz dx
dфx
dz dфx+dx
✓ In the Z direction
z
t z
dxdy
dфy
y
zx
The rate of heat flow out of the element by conduction
（导出微元体的总热流量dфout）
dV V dxdydz
The rate of increase of internal energy
dфy+dy
dz dx
（单位时间热力学能的增加 du）
dU c t dxdydz
dфx dфz+dz
dy dфv dU
dфy
Thus
dфz dфx+dx
y zx
d in
dV
d out
dU
x
y
z
• According to: Fourier’s law and energy balance
1. Heat conduction differential equation in the rectangular coordinate system （直角坐标系下的导热微分方程）
Assume: isotropic medium（各向同性介质）, λ,ρ,сkeep const， have heat sources（nuclear reactors, electrical conductors, and chemically reacting systems）。
motion of molecules and collisions each other. （气体分子不规则的热运动和相互碰撞而产生的热量传递）
工程热力学与传热学(英文) 绪论
deals with the principles of energy conversion among thermal energy and mechanical energy, as well as other form energies.
Thermodynamics and Heat Transfer
Thermodynamics and Heat Transfer
--- Introduction ---
Introduction
❖ Energy and Energy Resources ❖ The Contents of Thermodynamics
Mechanical energy Thermal energy
Electrical energy Chemical energy
Nuclear energy
Radiation energy
0-1-2 Thermal Energy and Its utilization
1. Energy Resources（能源）:
such as: cooking, warming, drying, smelting. etc
厨房用热
太阳能热水器
熔炼炉
➢ The powered utilization：
Converting heat into other form energy.
• Steam power equipment • Gas power equipment • Thermoelectricity power generation • Nuclear electric power generation • Solar power
(最新整理)高等工程热力学英文课件：Chapter1conceptsanddefinition
Everything external to the thermodynamic system is the environment or surrounding. The exchanges of mass or energy take place between the environment and the system.
environment. The boundary can be a real/physical interface or an imaginary one. It
can be mobile or fixed. It can be rigid or deformable.
2021/7/26
13
2021/7/26
Related concepts
(1) Working substance or working medium The medium for the inter-transformation between thermal energy and mechanical energy. Requirements: expansibility, mobility, thermal capacity, stability, safety, friendly to environment, low cost. Typical working media: combustion gas and steam. The state of working medium can be static or flowing.
5. Basic thermodynamic processes
6. Single phase systems
工程热力学课后答案 英文版 电子工业出版社
1-11 A barometer is used to measure the height of a building by recording reading at the bottom and at the top of the building. The height of the building is to be determined. Assumptions The variation of air density with altitude is negligible.Properties The density of air is given to be ρ = 1.18 kg/m 3density of mercury is 13,600 kg/m 3.Analysis building arekPa100.70N/m 1000kPa1m/s kg 1N1m) )(0.755m/s )(9.807kg/m (13,600)(kPa97.36m/s kg 1N1m) )(0.730m/s )(9.807kg/m (13,600)(2223bottombottom 223toptop =⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅===⎪⎪⎭⎫⎝⎛⋅==h g P h g ρP ρTaking an air column between the top and the bottom of the building and writing a force balance per unit base area, we obtainkPa 97.36)(100.70N/m 1000kPa 1m/s kg 1N1))(m/s )(9.807kg/m (1.18)(/2223topbottom air topbottom air -=⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅-=-=h P P gh P P A W ρIt yields h = 288.6 mwhich is also the height of the building.1-21 The air pressure in a duct is measured by an inclined manometer. For a given vertical level difference, the gage pressure in the duct and the length of the differential fluid column are to be determined.Assumptions The manometer fluid is an incompressible substance.Properties The density of the liquid is given to be ρ = 0.81 kg/L = 810 kg/m 3. Analysis The gage pressure in the duct is determined fromPa636=⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅==-=2223atm abs gage N/m 1Pa1m/s kg 1N1m) )(0.08m/s )(9.81kg/m (810ghP P P ρ The length of the differential fluid column is730 mmHgcm 13.9=︒==35sin /)cm 8(sin /θh LDiscussion Note that the length of the differential fluid column is extended considerably by inclining the manometer arm for better readability.2-4 No. This is the case for adiabatic systems only.2-6 A classroom is to be air-conditioned using window air-conditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. The number of 5-kW window air conditioning units required is to be determined.Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. Analysis The total cooling load of the room is determined fromQ Q Q Q cooling lights people heat gain =++whereQ Q Q lights people heat gain 10100W 1kW40360kJ /h 4kW 15,000kJ /h 4.17kW=⨯==⨯=== Substituting,.Q cooling 9.17kW =++=14417Thus the number of air-conditioning units required is units 2−→−=1.83kW/unit 5kW9.172-18 The flow of air through a flow channel is considered. The diameter of the wind channel downstream from the rotor and the power produced by the windmill are to be determined. Analysis The specific volume of the air is/k g m 8409.0k P a100K) K)(293/kg m kP a 287.0(33=⋅⋅==P RT v The diameter of the wind channel downstream from the rotor ism 7.38===−→−=−→−=m/s9m/s10m) 7()4/()4/(21122221212211V V D D V D V D V A V A ππ The mass flow rate through the wind mill iscool·kg/s 7.457/kg)m 4(0.8409m/s)(10m) 7(3211===πvV A mThe power produced is thenkW 4.35=⎪⎭⎫⎝⎛-=-=22222221/s m 1000kJ/kg 12)m/s 9()m/s 10(kg/s)7.457(2V V m W 2-19 The available head, flow rate, and efficiency of a hydroelectric turbine are given. The electric power output is to be determined.Assumptions 1 The flow is steady. 2 Water levels at the reservoir and the discharge site remain constant. 3 Frictional losses in piping are negligible. Properties We take the density of water to beρ = 1000 kg/m 3 = 1 kg/L.Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely. Therefore, the power potential of water is its potential energy,which is gz per unit mass, and gz m for a given mass flow rate.kJ/kg 177.1/s m 1000kJ/kg 1m ) 120)(m /s (9.81222mech =⎪⎭⎫ ⎝⎛===gz pe e The mass flow rate iskg/s ,000100/s)m 0)(10kg/m 1000(33===V ρmThen the maximum and actual electric power generation becomeMW 7.117kJ/s 1000MW 1kJ/kg) 7kg/s)(1.17 000,100(mech mech max =⎪⎭⎫ ⎝⎛===e mE WMW 94.2===MW) 7.117(80.0max overall electric W W η Discussion Note that the power generation would increase by more than 1 MW for each percentage point improvement in the efficiency of the turbine –generator unit.3-9 A rigid container that is filled with R-134a is heated. The temperature and total enthalpy are to be determined at the initial and final states.Analysis This is a constant volume process. The specific volume is/kg m 0014.0kg10m 014.03321====m Vv vR-134a 300 kPa 10 kgThe initial state is determined to be a mixture, and thus the temperature is the saturation temperature at the given pressure. From Table A-12 by interpolation C 0.61︒==kPa 300 @sat 1T TandkJ/kg52.54)13.198)(009321.0(67.52009321.0/kgm )0007736.0067978.0(/kg m )0007736.00014.0(113311=+=+==--=-=fg f fgf h x h h x v v vThe total enthalpy is thenkJ 545.2===)kJ/kg 52.54)(kg 10(11mh HThe final state is also saturated mixture. Repeating the calculations at this state,C 21.55︒==kPa 600 @sat 2T TkJ/kg64.84)90.180)(01733.0(51.8101733.0/kgm )0008199.0034295.0(/kg m )0008199.00014.0(223322=+=+==--=-=fg f fgf h x h h x v v vkJ 846.4===)kJ/kg 64.84)(kg 10(22mh H3-22 rigid tank contains an ideal gas at a specified state. The final temperature is to be determined for two different processes.Analysis (a ) The first case is a constant volume process. When half of the gas is withdrawn from the tank, the final temperature may be determined from the ideal gas relation as()K 400=⎪⎭⎫ ⎝⎛==K) 600(kP a 300kP a 1002112212T P P m m T (b ) The second case is a constant volume and constant mass process. The ideal gas relation for this case yields kPa 200=⎪⎭⎫ ⎝⎛==kP a) 300(K 600K 4001122P T T P3-32 Complete the following table for H 2 O :P , kPa T , ︒C v , m 3 / kgu , kJ/kg Phase description 200 30 0.001004 125.71 Compressed liquid 270.3130--Insufficient informationv200 400 1.5493 2967.2 Superheated steam 300133.520.5002196.4Saturated mixture, x=0.825500 473.1 0.6858 3084 Superheated steam4-14 Oxygen is heated to experience a specified temperature change. The heat transfer is to be determined for two cases.Assumptions 1 Oxygen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 154.8 K and 5.08 MPa. 2 The kinetic and potential energy changes are negligible, 0pe ke ≅∆≅∆. 3 Constant specific heats can be used for oxygen.Properties The specific heats of oxygen at the average temperature of (25+300)/2=162.5︒C=436 K are c p = 0.952 kJ/kg ⋅K and c v = 0.692 kJ/kg ⋅K (Table A-2b ).Analysis We take the oxygen as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for a constant-volume process can be expressed as)(12in energiesetc. potential, kinetic, internal,in Change system massand work,heat,by nsfer energy tra Net out in T T mc U Q E E E -=∆=∆=-vThe energy balance during a constant-pressure process (such as in a piston-cylinder device) can be expressed as)(12in out ,in out ,in energiesetc. potential, kinetic, internal,in Change system massand work,heat,by nsfer energy tra Net outin T T mc H Q U W Q UW Q E E E p b b -=∆=∆+=∆=-∆=-since ∆U + W b = ∆H during a constant pressure quasi-equilibrium process. Substituting for both cases, kJ 190.3=-⋅=-==K )25K)(300kJ/kg 692.0(kg) 1()(12const in,T T mc Q v VkJ 261.8=-⋅=-==K )25K)(300kJ/kg 952.0(kg) 1()(12const in,T T mc Q p P4-25 A rigid tank filled with air is connected to a cylinder with zero clearance. The valve is opened, and air is allowed to flow into the cylinder. The temperature is maintained at 30︒C at all times. The amount of heat transfer with the surroundings is to be determined.Assumptions 1 Air is an ideal gas. 2 The kinetic and potential energy changes are negligible,∆∆ke pe ≅≅0. 3 There are no work interactions involved other than the boundary work. Properties The gas constant of air is R = 0.287 kPa.m 3/kg.K (Table A-1).Analysis We take the entire air in the tank and the cylinder to be the system. This is a closed system since no mass crosses the boundary of the system. The energy balance for this closed system can be expressed asoutb,in 12out b,in energiesetc. potential, kinetic,internal,in Change system massand work,heat,by nsfer energy tra Net out in 0)(W Q u u m U W Q E E E ==-=∆=-∆=-since u = u (T ) for ideal gases, and thus u 2 = u 1 when T 1 = T 2 . The initial volume of air is33112212222111m 0.80)m (0.41kP a200kP a 400=⨯⨯==−→−=V V V V T T P P T P T P The pressure at the piston face always remains constant at 200 kPa. Thus the boundary work done during this process iskJ 80m kPa 1kJ 10.4)m kPa)(0.8 (200)( 312221out b,=⎪⎪⎭⎫⎝⎛⋅-=-==⎰V V V P d P W Therefore, the heat transfer is determined from the energy balance to bekJ 80==in out b,Q W4-27 An insulated cylinder is divided into two parts. One side of the cylinder contains N 2 gas and the other side contains He gas at different states. The final equilibrium temperature in the cylinder when thermal equilibrium is established is to be determined for the cases of the piston being fixed and moving freely.Assumptions 1 Both N 2 and He are ideal gases with constant specific heats. 2 The energy stored in the container itself is negligible. 3 The cylinder is well-insulated and thus heat transfer is negligible. Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m 3/kg.K is c v = 0.743 kJ/kg·°C for N 2, and R = 2.0769 kPa.m 3/kg.K is c v = 3.1156 kJ/kg·°C for He (Tables A-1 and A-2)Analysis The mass of each gas in the cylinder is()()()()()()()()kg0.808K 298K /kg m kPa 2.0769m 1kPa 500kg 4.77K 353K /kg m kPa 0.2968m 1kPa 50033He111He33N 111N 22=⋅⋅=⎪⎪⎭⎫ ⎝⎛==⋅⋅=⎪⎪⎭⎫ ⎝⎛=RT P m RT P m V VTaking the entire contents of the cylinder as our system, the 1st law relation can be written as()()He12N 12HeN energiesetc. potential, kinetic, internal,in Change system massand work,heat,by nsfer energy tra Net out in )]([)]([0022T T mc T T mc U U U E E E -+-=∆+∆=∆=∆=-v vSubstituting,()()()()()()0C 25C kJ/kg 3.1156kg 0.808C 80C kJ/kg 0.743kg 4.77=︒-︒⋅+︒-⋅f fT TIt givesT f = 57.2︒Cwhere T f is the final equilibrium temperature in the cylinder.The answer would be the same if the piston were not free to move since it would effect only pressure, and not the specific heats.Discussion Using the relation P V = NR u T , it can be shown that the total number of moles in the cylinder is 0.170 + 0.202 = 0.372 kmol, and the final pressure is 510.6 kPa.6-9 An inventor claims to have developed a heat engine. The inventor reports temperature, heat transfer, and work output measurements. The claim is to be evaluated.Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from42%or 42.0K500K 29011C th,max th,=-=-==H L T T ηη The actual thermal efficiency of the heat engine in question is42.9%or 0.429kJ700kJ300net th ===H Q W η which is greater than the maximum possible thermal efficiency. Therefore, this heat engine is a PMM2 and the claim is false .6-11 A heat pump maintains a house at a specified temperature. The rate of heat loss of the house and the power consumption of the heat pump are given. It is to be determined if this heat pump can do the job.Assumptions The heat pump operates steadily.Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only, and is determined from ()()()14.75K 27322/K 273211/11COP rev HP,=++-=-=H L T TThe required power input to this reversible heat pump is determined from the definition of the coefficient of performance to bekW 2.07=⎪⎪⎭⎫ ⎝⎛==s 3600h 114.75kJ/h 110,000COP HP min in,net,H Q W This heat pump is powerful enough since 5 kW > 2.07 kW.7-8A reversible heat pump with specified reservoir temperatures is considered. The entropy change of two reservoirs is to be calculated and it is to be determined if this heat pump satisfies the increase in entropy principle.Assumptions The heat pump operates steadily. Analysis Since the heat pump is completely reversible, the combination of the coefficient of performance expression, first Law, and thermodynamic temperature scale gives73.26)K 294/()K 283(11/11COP rev HP,=-=-=HL T T The power required to drive this heat pump, according to the coefficient of performance, is thenkW 741.326.73kW 100COP rev HP,in net,===HQ WAccording to the first law, the rate at which heat is removed from the low-temperature energy reservoir iskW 26.96kW 741.3kW 100in net,=-=-=W Q Q H L The rate at which the entropy of the high temperature reservoir changes, according to the definition of the entropy, iskW/K 0.340===∆K294kW 100H HHT Q S and that of the low-temperature reservoir iskW/K 0.340-=-==∆K283kW 26.96L L L T Q S The net rate of entropy change of everything in this system iskW/K 0=-=∆+∆=∆340.0340.0total L H S S Sas it must be since the heat pump is completely reversible.net7-11 Steam is expanded in an isentropic turbine. The work produced is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 The process is isentropic (i.e., reversible-adiabatic).Analysis There is one inlet and two exits. We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form aso u tin energiesetc. potential, kinetic, internal,in change of Rate (steady) 0systemmassand work,heat,by nsfer energy tra net of Rate out in 0E E E E E==∆=-332211out out332211h m h m h m W W h m h m h m --=++=From a mass balance,kg/s 75.4)kg/s 5)(95.0(95.0kg/s 25.0)kg/s 5)(05.0(05.01312======m mm mNoting that the expansion process is isentropic, the enthalpies at three states are determined as follows:6)-A (Table KkJ/kg 6953.7kJ/kg4.2682C 100 kPa 503333 ⋅==⎭⎬⎫︒==s h T P 6)-A (Table kJ/kg 3.3979K kJ/kg 6953.7 MPa 41311=⎭⎬⎫⋅===h s s P 6)-A (Table kJ/kg 1.3309K kJ/kg 6953.7 kPa 7002322=⎭⎬⎫⋅===h s s P Substituting,kW6328=--=--=kJ/kg) .4kg/s)(2682 75.4(kJ/kg) .1kg/s)(3309 25.0(kJ/kg) .3kg/s)(3979 5(332211outh m h m h m W7-13 The entropy change relations of an ideal gas simplify to∆s = c p ln(T 2/T 1) for a constant pressure processand ∆s = c v ln(T 2/T 1) for a constant volume process.Noting that c p > c v , the entropy change will be larger for a constant pressure process.7-22 Air is compressed in a piston-cylinder device. It is to be determined if this process is possible.Assumptions 1 Changes in the kinetic and potential energies are negligible. 4 Air is an ideal gas with constant specific heats. 3 The compression process is reversible.Properties The properties of air at room temperature are R = 0.287 kPa ⋅m 3/kg ⋅K, c p = 1.005 kJ/kg ⋅K (Table A-2a).Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed asin,out 12out in ,12out in ,12out in ,energiesetc. potential, kinetic, internal,in Change system massand work,heat,by nsfer energy tra Net out in )(since 0)( )(b b p b b W Q T T Q W T T mc Q W u u m U Q W E E E ===--=--=∆=-∆=-The work input for this isothermal, reversible process iskJ/kg 8.897kP a100kP a 250K)ln K)(300kJ/kg 287.0(ln12in =⋅==P P RT w That is,kJ/kg 8.897in out ==w qThe entropy change of air during this isothermal process isK kJ/kg 0.2630kP a100kP a250K)ln kJ/kg 287.0(ln ln ln121212air ⋅-=⋅-=-=-=∆P P R P P R T T c s p The entropy change of the reservoir isK kJ/kg 0.2630K300kJ/kg 89.78R R ⋅===∆R T q s Note that the sign of heat transfer is taken with respect to the reservoir. The total entropy change (i.e., entropy generation) is the sum of the entropy changes of air and the reservoir:K kJ/kg 0⋅=+-=∆+∆=∆2630.02630.0R air total s s sNot only this process is possible but also completely reversible.8-1 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).Analysis (b ) From the ideal gas isentropic relations and energy balance,()()K 579.2k P a 100k P a1000K 3000.4/1.4/11212=⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=-kk P P T THeatv()()()K3360==−→−-⋅=-=-=3max 32323in 579.2K kJ/kg 1.005kJ/kg 2800T T T T T c h h q p(c )()K 336K 3360kP a1000kP a1003344444333===−→−=T P P T T P T P v v ()()()()()()()()21.0%=-=-==-⋅+-⋅=-+-=-+-=+=k J /k g2800k J /k g 221211k J /k g2212K 300336K k J /k g 1.005K 3363360K k J /k g 0.718in out th14431443out 41,out 34,out q q T T c T T c h h u u q q q p ηvDiscussion The assumption of constant specific heats at room temperature is not realistic in this case the temperature changes involved are too large.8-5 An ideal Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and temperature in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effective pressure are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a ) Process 1-2: isentropic compression.()()()()kP a 2338kP a 100K 308K 757.99.5K757.99.5K 3081122121112220.412112=⎪⎪⎭⎫⎝⎛==−→−===⎪⎪⎭⎫ ⎝⎛=-P T T P T P T P T T k v v v v vvProcess 3-4: isentropic expansion.()()K 1969==⎪⎪⎭⎫ ⎝⎛=-0.4134439.5K 800k T T vvProcess 2-3: v = constant heat addition.()kPa 6072=⎪⎪⎭⎫⎝⎛==−→−=kPa 2338K 757.9K 19692233222333P T T P T P T P v v (b ) ()()()()kg 10788.6K 308K /kg m kPa 0.287m 0.0006kPa 100433111-⨯=⋅⋅==RT P m V vs()()()()()kJ0.590=-⋅⨯=-=-=-K 757.91969K kJ/kg 0.718kg 106.78842323in T T mc u u m Q v(c) Process 4-1: v = constant heat rejection.()()()()kJ0.2 40K 308800K kJ/kg 0.718kg 106.788)(41414out =-⋅⨯-=-=-=-T T mc u u m Q v kJ 0.350240.0590.0out in net =-=-=Q Q W59.4%===kJ0.590kJ0.350inout net,th Q W η(d ) ()()kPa 652=⎪⎪⎭⎫⎝⎛⋅-=-=-===kJm kPa 1/9.51m 0.0006kJ0.350)/11(MEP 331outnet,21outnet,max 2min r W W rV V V V V V8-7 An ideal diesel cycle has a a cutoff ratio of 1.2. The power produced is to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The specific volume of the air at the start of the compression is/kg m 8701.0kPa95K)288)(K /kg m kPa 287.0(33111=⋅⋅==P RT vThe total air mass taken by all 8 cylinders when they are charged is kg 008665.0/kgm 8701.0m)/4 12.0(m) 10.0()8(4/3212cyl 1cyl===∆=ππv v VS B N N mThe rate at which air is processed by the engine is determined fromkg/s 1155.0rev/cycle2rev/s) 1600/60kg/cycle)( (0.008665rev ===N nm msince there are two revolutions per cycle in a four-stroke engine. The compression ratio is2005.01==r At the end of the compression, the air temperature is()K 6.95420K) 288(14.1112===--k r T ToutApplication of the first law and work integral to the constant pressure heat addition giveskJ/kg 1325K )6.9542273)(K kJ/kg 005.1()(23in =-⋅=-=T T c q pwhile the thermal efficiency is6867.0)12.1(4.112.12011)1(1114.111.41th =---=---=--c k c k r k r r ηThe power produced by this engine is thenkW105.1====kJ/kg) 67)(1325kg/s)(0.68 (0.1155in th net netq m w m W η8-12 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).Analysis For the isentropic compression process,K .1527K)(10) 273(0.4/1.4/)1(12===-kk p r T TThe heat addition iskJ/kg 500kg/s1kW 500in in===m Q qApplying the first law to the heat addition process,K 1025KkJ/kg 1.005kJ/kg500K 1.527)(in 2323in =⋅+=+=-=p p c q T T T T c q The temperature at the exit of the turbine isK 9.530101K) 1025(10.4/1.4/)1(34=⎪⎭⎫⎝⎛=⎪⎪⎭⎫⎝⎛=-kk p r T TApplying the first law to the adiabatic turbine and the compressor produceskJ/kg 6.496K )9.5301025)(K kJ/kg 1.005()(43T =-⋅=-=T T c w pkJ/kg 4.255K )2731.527)(K kJ/kg 1.005()(12C =-⋅=-=T T c w pThe net power produced by the engine is thenkW 241.2=-=-=kJ/kg )4.2556kg/s)(496. 1()(C T netw w m W Finally the thermal efficiency is0.482===kW 500kW241.2innet thQ W η。
工程热力学英文版9
A + B AB.
(10.3)
Suppose we introduce NA,0 moles of A and NB,0 moles of B into a container with volume V at time t = 0. We allow the system to evolve with no interaction with the surroundings (isolated). As time proceeds, we find that the mole numbers change, due to formation of AB molecules. If we wait long enough, the mole numbers come to steady-state values, which we will denote NA∗ , NB∗ , and NA∗B. Evidently, once steady state is reached, the rate at which AB is formed by reaction (10.1) just balances the rate at which it is destroyed by reaction (10.2).
When this condition is reached, we say the system is in chemical equilibrium. Depending on the nature of the chemical reaction, and also on the temperature
10.3 Reactions in Ideal Gases
If the mixture of A, B, and AB is an ideal gas mixture, then we have already
工程热力学(英文版)第10单元课件
Cogeneration
(热电合供循环)
SHANDONG UNIVERSITY
Chapter 10 Vapor Power Cycles (蒸汽动力循环)
3
SHANDONG UNIVERSITY
• Steam is the most common working fluid used in vapor power cycles because of its many desirable characteristics: low cost, availability, and high hfg. • This chapter is mostly devoted to the discussion of steam power plants. Steam power plants are commonly referred to as coal plants, nuclear plants, or natural gas plants, depending on the type of fuel used to supply heat to the steam.
过热蒸汽 过冷水
湿饱和蒸汽
饱和水
10
SHANDONG UNIVERSITY
Energy Analysis of the Ideal Rankine Cycle
• All four components associated with the Rankine cycle (the pump, boiler, turbine, and condenser) are steady-flow devices • Rankine cycle can be analyzed as steady-flow processes. • The kinetic and potential energy changes of the steam are usually small and can be neglected. • So, steady-flow energy equation per unit mass of the steam is