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1、Thermodynamics-Chapter 3 The Properties of Ideal-Gas and Ideal-Gas Processes-Chapter 3 The Properties of Ideal-Gas and Ideal-Gas Processes ContentsvThe Ideal-Gas Equation of Statevc，cv，cp，u，h，s of Ideal-GasvIdeal-Gas Processes（Four Basic Processes,Polytropic Processes)3-1-1 Ideal-Gas 1.Real Gas（实际气体
2、）实际气体）cant neglect the volume of moleculesthere are attraction among molecule of real gas 2.Ideal-Gas（理想气体）（理想气体）An Ideal gas is an imaginary substance that obeys these conditions:The molecules of ideal-gas have no volumeThere are no attraction among molecules of ideal-gas3-1 The Ideal-Gas Equation
3、of State （理想气体状态方程式）（理想气体状态方程式）N2,H2,O2,CO2,CO（1）The behavior of real gas:high temperature,low pressure,low density(high specific volume),far from a state of condensation-treated as an ideal gas.（2）The behavior of real gas:low temperature,high pressure,high density(low specific volume),not far from
4、a state of condensation-treated as a real gas.Steam power plants-water vapor Refrigerator-ammonia,freonnotes（3）Water Vapor an Ideal Gas?Water vapor exists in the air Working substance in the steam power plant 3-1-2 Ideal-Gas Equation of State （理想气体状态方程式）理想气体状态方程式）1.Ideal-Gas Equation of State For Id
5、eal-Gas 1kg:pv=RgT For Ideal-Gas mkg:pV=mRgT（1）Clapeyrons equation(In 1834)（2）At any equilibrium state:the p-v-T behavior of ideal-gas（3）Where:pabsolute pressure T Kelvin temperature RgThe gas constantnotes 2.Alternative Forms of the Ideal-Gas Equation of State(不同物量的理想气体状态方程式不同物量的理想气体状态方程式)1kg:p v=R
6、g T mkg:pV=m Rg T 1mol:P Vm=R T nmol:p V=n R T Rg-gas constant 气体常数气体常数 R-universal gas constant 通用气体常数通用气体常数notes理想气体状态方程式的应用理想气体状态方程式的应用 某蒸汽锅炉燃煤需要的标准状况下，空气量某蒸汽锅炉燃煤需要的标准状况下，空气量 为为 qV=66000m3/h，若鼓风炉送入的热空气温度为，若鼓风炉送入的热空气温度为t1=250C，表压力，表压力 pg1=20.0kpa。当时当地的大气。当时当地的大气压力压力 pb=101.325kpa。求实际的送风量为多少？求实际的送风
7、量为多少？例题例题说明说明标准状态下，标准状态下，其中其中：故故实际实际的送的送风风量：量：简单求解过程简单求解过程：解：按理想气体状态方程式，解：按理想气体状态方程式，可得：可得：3-2 Specific Heat,Internal Energy,Enthalpy and Entropy of Ideal-Gas （理想气体的热容，热力学能，焓和熵）理想气体的热容，热力学能，焓和熵）3-2-1 Heat Capacity（热容的定义）热容的定义）1.Heat Capacity：The energy required to raise the temperature of substance
8、by one degree.Denoted by C,unit J/K.1kg Iron20C-30C4.5kJ1kg Water20C-30C41.8kJ（3）Volume Heat（体积热容）（体积热容）Relationship among C，c，Cm，CV 3.Influence Factors（1）Substances（2）Thermal processes（3）Temperature（2）Molar heat（摩尔热容）摩尔热容）2.C,c,Cm,CV（1）Specific heat（比热容）（比热容）3-2-2 The Specific Heat at Constant Volu
9、me and at Constant Pressure（比定容热容和比定压热容）（比定容热容和比定压热容）1.Specific heat at constant volume cv:The energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant.Specific heat at constant pressure cp:The energy required to raise the temperatur
10、e of the unit mass of a substance by one degree as the pressure is maintained constant.2.Specific Heats cv,cp in Reversible ProcessesHow to express du,dh?For constant volume ConditionReversible ProcessesAny Substance For constant pressure=the change in internal energy with temperature at constant vo
11、lume=the change in enthalpy with temperature at constant pressure 3.Specific Heats cv,cp of Ideal-Gases（理想气体的（理想气体的cv,cp）For ideal gasConditionAny process for ideal-gasConstant volume process for real gasConstant pressure process for real gas For ideal gas,the specific heats cv,cp depend on temperat
12、ure only.notes 4.Specific-Heat Relations of Ideal Gases （理想气体的理想气体的 c cv v 和和 c cp p 的关系）的关系）（1）迈耶公式迈耶公式：（2）Specific heat ratio （比热容比）比热容比）How to obtain?Why cp cv at the same temperature?Analysis Specific heats cp,cv,cp-cv,cp/cv depend on substance or state?思考思考题题 How to determine heat transferred b
13、y virtue of specific heats?(利用比热容，如何求解热量利用比热容，如何求解热量?)cp，cv，cp-cv，cp/cv与物质的种类是否有关？与物质的种类是否有关？与气体的状态是否有关？与气体的状态是否有关？思考思考题题 利用比热容，如何求解热量？利用比热容，如何求解热量？解：解：cp，cv：与物质的种类和状态都有关系；：与物质的种类和状态都有关系；cp-cv=Rg：与物质的种类有关，与状态无关；：与物质的种类有关，与状态无关；cp/cv=：与物质的种类和状态都有关系。与物质的种类和状态都有关系。3-2-3 The calculation of Heat （利用理想气体的
14、比热容计算热量）利用理想气体的比热容计算热量）1.The real specific heat capacity（真实比热容）真实比热容）The specific heat in the state of instantaneous temperature.HeatSo Specific heat at constant pressure cp Specific heat at constant volume cv Heat transferred in a constant pressure process Heat transferred in a constant volume proc
15、ess 2.The average specific heats（平均比热容）平均比热容）Average value of specific heat at a temperature interval t1 to t2.（某一温度间隔内比热容的平均值）某一温度间隔内比热容的平均值）Heat:c0tt1t2dtqGeometric significancy Table of average specific heats（平均比热容图表）平均比热容图表）Where average specific heat from 0 to t temperature interval.Same as 平均比
16、定压热容图表：平均比定压热容图表：Table2（P360）平均比定容热容图表：平均比定容热容图表：Table 3（P361）3.Specific Heats Vary Linearly with Temperature （平均比热容直线关系式）平均比热容直线关系式）Heat Average specific heatc0tt1t2dtq 4.Constant Specific Heats（定值比热容）定值比热容）In the cursory calculation of specific heat,if the precision required is not strict the chan
17、ge in temperature is not big Principles Kinetic theory of gases（气体分子运动论）（气体分子运动论）Principle of Equipartition of Energy （能量按自由度均分）（能量按自由度均分）Principle of equipartition of energy:The gases of having the same amount of atomics have the same value molar specific heat capacity.The internal energy for ideal
18、 gas per unit molar The molar specific heat at constant volume Specific heat ratioI-degrees of freedom（分子运动自由度）分子运动自由度）Kinetic theory of gases The molar specific heat at constant pressure Heat1.291.401.67Cp,mCV,m753ipolyatomicdiatomicmonatomicThe fixed value molar specific heats of ideal gases例题例题理想
19、气体的比热容理想气体的比热容 在燃气轮机动力装置的回热器中，将空气在燃气轮机动力装置的回热器中，将空气从从150C定压加热到定压加热到350C，试按下列比热容值，试按下列比热容值计算对每公斤空气所加入的热量。计算对每公斤空气所加入的热量。（1）按真实比热容计算；）按真实比热容计算；（2）按平均比热容表计算（附表）按平均比热容表计算（附表2，3）；）；（3）按定值比热容计算；）按定值比热容计算；（4）按空气的热力性质表计算（附表）按空气的热力性质表计算（附表4）；）；3-2-4 Internal energy,Enthalpy of ideal-gas （理想气体的热力学能，焓）理想气体的热力学
20、能，焓）1.Internal Energy,Enthalpy For any substance u/h is a property u/h is a function of two independent propertiespv0T1T212V2p2 u1-2 and h1-2 depend on the ends states only,no relevantwith processes.notes For ideal gas u/h is a function of temperature alone（1）u1-2,h1-2 depend on temperature only,no
21、relevant with processes and states.（2）Three lines have the same meaning for ideal gases constant temperature line constant internal energy line constant enthalpy linepv0T1T212v2p2notesConditionAny process for ideal-gasConstant volume process for real gasConstant pressure process for real gas 2.Calcu
22、lation of u,h According to cv,cp Three ways for calculating u and h Table4（P362）3-2-5 Entropy of Ideal-Gas（理想气体的熵）理想气体的熵）1.Definition（1）Specific entropy change is a property of substance（2）Condition:reversible processes （3）To indicate the direction of heat transfer for a reversible processnotesIf ds
23、 0,then q 0,heat is added into the systemIf ds 0,then q 0,heat is rejected from the systemIf ds=0,then q=0,the process is an adiabatic process,also called constant entropy process.2.The Calculation of Entropy Changes（熵变的计算）熵变的计算）The first law of thermodynamics Ideal gas Entropy change in reversible
24、process The calculation of entropy changes Condition Ideal gas Any process Fixed value specific heat思考思考题题Statement the valid conditions for each stepof the equations 任意气体任意微元过程任意气体任意微元过程 任意气体微元可逆过程任意气体微元可逆过程 任意气体微元定压过程任意气体微元定压过程 任意气体微元定压过程任意气体微元定压过程 焓的定义焓的定义 理想气体任意过程理想气体任意过程例题例题 已知某理想气体的比定容热容已知某理想气
25、体的比定容热容cv=a+bT,其中其中a，b为常数，试导出其热力学能，焓为常数，试导出其热力学能，焓和熵变的计算式。和熵变的计算式。解：解：3-4 The Process of the Ideal Gas （理想气体的基本热力过程）理想气体的基本热力过程）3-4-1 Purpose and Method 1.Purpose 2.Basic Task （1）To determine the change of properties during the process （2）To analysis the relationship of energy transfer 3.Theories （1
26、）The first law of thermodynamics （2）The Ideal-Gas Equation of State （3）Expressions of reversible process 4.Assumptions and idealizations （1）Irreversible processes-Reversible processes （2）Complex processes-Simple processes Constant volume process（定容过程）（定容过程）Constant pressure process（定压过程）（定压过程）Isothe
27、rmal process（定温过程）（定温过程）Reversible adiabatic process（可逆绝热过程）（可逆绝热过程）5.Contents（研究内容）研究内容）（1）Process equation（过程方程式）过程方程式）（2）Relationships between the initial and final states （3）Determine the change of internal energy,enthalpy,entropy:u,h,s （4）Analysis of p-v diagram,T-s diagram （5）Determine the wor
28、k transferred:w，wt （6）Determine the heat transferred:q 3-4-2 The Basic Thermal Processes of Ideal-Gases （理想气体的基本热力过程）（理想气体的基本热力过程）1.The Constant-Volume Processes（定容过程）定容过程）(Isochoric Processes)（1）Process equation（2）Relationships of p,v,T（3）u,h,s（4）p-v and T-s diagrams p-v diagram:vertical straight l
29、ines T-s diagram:Exponential shape curves（指数曲线）（指数曲线）p2210vvvT2210s.vv（5）Work W Wt（6）Heat At constant volume processes:q=u,T.The preparation process that heat convert into work.notes 2.The Constant-Pressure Processes（定压过程）定压过程）(Isobaric processes)（2）Relationships of p,v,T（1）Process equation（3）u,h,sv
30、v p-v diagram：Horizontal straight lines T-s diagram：Exponential shape curves（指数曲线）（指数曲线）vcontrast（4）p-v and T-s diagramsp210v2ppT2210s.pp（5）Work W Wt（6）Heat At constant pressure processes,the heat absorbed equals to the change of enthalpy.notes 一容积为一容积为 0.15m3 的储气罐，内装氧气，其初始的储气罐，内装氧气，其初始压力压力 p1=0.55M
31、Pa，温度，温度 t1=38C。若对氧气加热，。若对氧气加热，其温度，压力都升高。储气罐上装有压力控制阀，其温度，压力都升高。储气罐上装有压力控制阀，当压力超过当压力超过 0.7MPa 时，阀门便自动打开，放走部时，阀门便自动打开，放走部分氧气，即储气罐中维持的最大压力为分氧气，即储气罐中维持的最大压力为 0.7MPa。问当罐中氧气温度为问当罐中氧气温度为285C时，对罐中氧气共加入时，对罐中氧气共加入了多少热量？设氧气的比热容为定值。了多少热量？设氧气的比热容为定值。例题例题简单求解过程简单求解过程：这一问题包含了两个过程这一问题包含了两个过程：过程过程1-2:是由压力为是由压力为 被定容加
32、热被定容加热到到 ，该过程中氧气质量不变；，该过程中氧气质量不变；过程过程2-3:是压力由是压力由 被定压加热到被定压加热到 ，该过程是一个质量不断变化，该过程是一个质量不断变化的定压过程。的定压过程。（1）首先分析）首先分析1-2定容过程：定容过程：温度变化：温度变化：1-2中吸收热量：中吸收热量：在在2-3过程中的吸热量：过程中的吸热量：（3）共加入热量为：）共加入热量为：微元变化过程吸热量：微元变化过程吸热量：（2）2-3为变质量的定压过程：为变质量的定压过程：3.The Isothermal Processes（定温过程）定温过程）（Hyperbolic Processes）（1）Pr
33、ocess equation（2）Relationships of p,v,T（3）u,h,s p-v diagram:rectangular hyperbola（等边双曲线）等边双曲线）T-s diagram:horizontal lines（4）p-v and T-s diagrams.p2210vTTT2210sTT（5）Work w wt Relationship of w,wt（6）HeatAnalysis Can we use to calculate heat at isothermal process?T2210sTs During an isothermal expansio
34、n,all the heat transferred is converted into external work.Conversely,during an isothermal compression,all the work done on the gas is rejected by the gas as heat transfer.notes 4.The Reversible Adiabatic Process（可逆绝热过程）可逆绝热过程）Derivation Entropy ds Reversible adiabatic process Assume=constant k:Adia
35、batic index（绝热指数）（绝热指数）（1）Process equation Hence（2）Relationships of p,v,T（3）u,h,s Negligible k change with temperature k=1.67（monatomic gases）k=1.40（diatomic gases）k=1.29（polyatomic gases）TT p-v diagram:Hyperbolas of higher order（高次双曲线）高次双曲线）T-s diagram:Vertical lines（4）p-v and T-s diagrams.p2210vss
36、Tcontrast1T220sss（5）Heat（6）Work W Any substances,adiabatic processes Ideal gases,cv=constant,adiabatic processes Ideal gases,reversible adiabatic processes Wt Any substances,any adiabatic processes Ideal gases,cv=constant,adiabatic processes Ideal gases,reversible adiabatic processes Relationship of
37、 W,Wt During adiabatic processes,no heat transfer,the work done by system is from the total energy of substance.notes例题例题已知已知2kg空气分别经过定温膨胀和绝热膨胀的空气分别经过定温膨胀和绝热膨胀的可逆过程，从初态压力为可逆过程，从初态压力为 p1=9.807bar，t1=300 C膨胀到终态容积为初态容积的膨胀到终态容积为初态容积的5倍。倍。试计算不同过程中空气的终态参数，对外界所作的试计算不同过程中空气的终态参数，对外界所作的功和交换的热量，过程中热力学能，焓和熵的变化
38、功和交换的热量，过程中热力学能，焓和熵的变化量。量。(设空气的设空气的cp=1.004kJ/(kg.K)，Rg=0.287kJ/(kg.K)，k=1.4)简单求解过程简单求解过程：取空气作热力系取空气作热力系（1）可逆定温过程）可逆定温过程1-2:由参数间的相互关系得：由参数间的相互关系得：按理想气体状态方程式得按理想气体状态方程式得:定温过程：定温过程：T1=T2=573K，气体对外所作的膨胀功及交换的热量：气体对外所作的膨胀功及交换的热量：过程中热力学能，焓，熵的变化为：过程中热力学能，焓，熵的变化为：过程中热力学能，焓，熵的变化为过程中热力学能，焓，熵的变化为：（2）可逆绝热过程）可逆绝
39、热过程：按可逆绝热过程参数间关系可得：按可逆绝热过程参数间关系可得：气体对外所作膨胀功及交换的热量气体对外所作膨胀功及交换的热量：3-4-3 The polytropic Processes（多变过程）多变过程）n:Polytropic exponent（多变指数）（多变指数）In theory,the range of the value of the index n is from-to+.Different polytropic process has different polytropic exponent.For four basic processes:（1）Process eq
40、uation The calculation of the index n For any polytropic processes Integral above Then If the properties of initial(p1,v1)and final(p2,v2)states are known,the value of index n can be calculated in terms of the formula above.（2）Relationships of p,v,T（3）u,h,s p-v diagram Slope（多变过程的斜率）（多变过程的斜率）Where（4
41、）p-v and T-s diagrams-isobaric processes-isothermal processes-constant entropy processes-isochoric processes.p2210vTsvp绘制空气绘制空气n=1.6的膨胀过的膨胀过程，判断程，判断q,w,u的正的正负负.p0v-n=kn=1n=0+n=1.6n=1.6膨胀过程膨胀过程 Determine the value of n Determine q,w,u Distribution ruleThe value of index n：Starting from constant volum
42、e process Increase clockwise Value of n:-00 1 k+.T-s diagram Slope Cn:Polytropic specific heat capacity（多变比热）（多变比热）Heat at polytropic processes Definition of entropy Entropy change of ideal gases Where:-isobaric processes-isothermal processes-constant entropy processes-isochoric processes绘制空气绘制空气n=1
43、.6的膨胀过的膨胀过程，判断程，判断q,w,u的正的正负负T20s.pv21TsT20s.21n=kn=1n=0-+n=1.6n=1.6膨胀过程膨胀过程 Starting from constant volume process:n:-00 1 k+Increase clockwise Starting from constant entropy process:Cn:0ccV V cp Increase clockwise Distribution rule（5）Work W Wt（6）Heat When n=1,Isothermal processes When n1,and cn=con
44、stant（7）Coordinate diagram analysis of polytropic processesT20s.pv21Ts.p2210vTsvp Work-isochoric processes Heat-constant entropy processes Internal energy,enthalpy-isothermal processes Wt-isobaric processes1.将满足下列要求的理想气体多变过程表示将满足下列要求的理想气体多变过程表示2.在在p-v图和图和T-s图上。图上。（1）工质又升压，又升温，又放热的过程。）工质又升压，又升温，又放热的过
45、程。（2）工质又膨胀，又降温，又放热的过程。）工质又膨胀，又降温，又放热的过程。2.试分析多变指数在试分析多变指数在 1nk 范围内的膨胀范围内的膨胀 过程特点。过程特点。思考思考题题（1）工质又膨胀，又升温，又吸热的过程。）工质又膨胀，又升温，又吸热的过程。（2）工质又膨胀，又降温，又放热的过程。）工质又膨胀，又降温，又放热的过程。2.试分析多变指数在试分析多变指数在 1nk 范围内的膨胀范围内的膨胀 过程特点。过程特点。思考题思考题答：答：过程特点为过程特点为 吸热，降温，膨胀，压力减小。吸热，降温，膨胀，压力减小。.p2210vTsvp若上例改为若上例改为 n=1.3 的多变过程，则各项结果的多变过程，则各项结果 又如何？将定温过程，定熵过程及此多变又如何？将定温过程，定熵过程及此多变 过程分析比较。过程分析比较。例题例题