总复习 (Overall Review).

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1 总复习 (Overall Review)

2 Basic Concept 系统 (Thermodynamic System): A quantity of matter or a region in space chosen for study. （系统就是指被选做研究对象的物体或空间。） 2. 平衡状态(Equilibrium State) A system in equilibrium experiences no changes with time when it is isolated from its surroundings. 所谓平衡状态就是指在没有外界影响的情况下，系统的状态不随时间而发生变化。

3 在无限小势差推动下的由连续平衡态组成的过程，就是准静态过程。
3.准静态过程(Quasi-static Process ) 在无限小势差推动下的由连续平衡态组成的过程，就是准静态过程。 When a process proceeds in such a manner that the system remains infinitesimally close to an equilibrium state at all times, it is called a quasi-static or quasi-equilibrium process.

4 4.可逆过程(Reversible process)
系统经历某一过程后，如果能使系统与外界同时恢复到初始状态，而不留下任何痕迹，则此过程为可逆过程。 A process that can reversed without leaving any trace on the surroundings. That is, both the system and the surroundings are returned to their initial states at the end of the reverse process.

5 自发过程(Spontaneous Process)
不需要任何外界作用而自动进行的过程。 A process which can happy naturally without the interference from the surroundings.

6 5.稳态稳定流动(Steady-flow process)
A process during which a fluid flows through a control volume steadily. That is, the properties remain the same at a fixed point during the entire process.

7 6.循环 (Cycle) If a system returns to its initial state at the end of the process, it is said a system have undergone a cycle. 系统经历了一系列状态变化过程，又回到了原来状态，称为热力循环。

8 7.比热容（Specific Heat) 将单位物量的物质温度升高1K所需加入的热量。
The energy required to raise the temperature of a unit quantity of a substance by one degree

9 8.分容积 ( component volume )
假定混合气体中组成气体具有与混合气体相同的温度及压力时，单独存在所占有的容积。 9.分压力（Component pressure） 假定混合气体中组成气体单独存在，并具有与混合气体相同的温度及容积时的压力

10 10.焓（Enthalpy） Enthalpy, a state function, is defined as follows, h= u + pv This energy is composed of two parts: the internal energy of the fluid (u) and the flow work (pv) associated with pushing the mass of fluid across the system boundary. 对开口系统而言，焓指流动工质所携带能量的一部分，这部分能量取决于热力状态

11 11.饱和状态 Saturation state 饱和状态：汽化与凝结的动态平衡 12.蒸发（Evaporize）和沸腾（Boiling)
13.干度（Dryness) 14.汽化潜热（Latent heat)

12 15. 未饱和湿空气(Unsaturated Air)
Dry air+ Superheated water vapor 干空气和过热水蒸汽的混合物。 饱和湿空气(Saturated Air) Dry air+ Saturated water vapor 干空气和饱和水蒸汽的混合物。

13 17.相对湿度φ (Relative Humidity of Air)
16.露点温度(Dew-point Temperature) It is defined as the temperature at which condensation begins when the air is cooled at constant pressure. 17.相对湿度φ (Relative Humidity of Air) It is defined as the ratio of the amount of moisture in the air to the maximum amount of moisture the air can hold at the same temperature. 湿空气中实际包含的水蒸汽量与同温度下最多能包含的水蒸汽量的百分比。

14 18.含湿量(Humidity ratio) 含湿量d：以干空气为基准，包含1千克干空气的湿空气中中所含有的水蒸汽的质量。 The mass of water vapor present in a unit mass of dry air is called humidity Ratio.

15 19.马赫数：（The Mach number） M,
It is the ratio of the flow speed, c, to the velocity of sound in the same fluid at the same state. It is denoted as M. (流体在某一点的运动速度和该点当地声速之比, 以M表示)

16 20.滞止参数(Stagnation Properties)
气体速度为零时的状态称为滞止状态,该状态的参数称为滞止参数. Stagnation properties represent the enthalpy of a fluid when it is brought to rest adiabatically

17 21.Critical Pressure Ratio(临界压力比)　　

18 22. 绝热节流 (Adiabatically Throttling) 　
流速为 的气流,由于局部阻力使流体降压膨胀的现象称为节流, 因流速高,时间短,与外界换热少,可视作绝热,故称绝热节流.

19 Critical Point(关键点）: 1. Pressure (压力） Absolute pressure （绝对压力）
Relative pressure （相对压力） Gage pressure （表压力） Vacuum pressure （真空度）

20 pe p pv pb p absolute pressure 表压力 pe 当 p > pb Gage pressure
绝对压力与相对压力 absolute pressure relative pressure 表压力 pe 当 p > pb Gage pressure 当 p < pb 真空度 pv Vacuum pressure pe p pv pb p

21 Exercise.1 (练习1）---压力的测量
As shown in the following figure，it is known that pb=101325Pa, the height difference is H=300mm for mercury liquid. The gauge pressure of B is MPa, Then what is the pressure for side A, and what is the value of Pg,A? 已知大气压pb=101325Pa，U型管内 汞柱高度差H=300mm，气体表B读数为0.2543MPa，求：A室压力pA及气压表A的读数pgA

22 解：

23 2. Equation of State for Ideal Gas （理想气体的状态方程的应用）
Pv=RT Example. Determine the mass of air in a room whose dimensions are 4m× 5m× 6m at 100kPa and 25℃

24 3.平均比热 Mean specific heat
(cp ,cv) c=f (t) 求O2在 ℃平均定压热容 t1 t2 t

25 4.理想气体u、 h和s的计算 理想气体，任何过程

26 Point function---Exact differentials--- d
5.闭口系统能量方程 Point function---Exact differentials--- d Path function---Inexact differentials---  一般式  Q  W Q = dU + W Q = U + W q = du + w q = u + w 单位工质 适用条件： 1）任何工质 2) 任何过程

27 w = pdv q = du + pdv q =  u +  pdv  q = Tds Tds = du + pdv
可逆闭口系能量方程 w = pdv q = du + pdv 任何工质、可逆过程 q =  u +  pdv 简单可压缩系可逆过程  q = Tds Tds = du + pdv 热力学恒等式  Tds =  u +  pdv

28 理想气体、可逆过程 理想气体，任何过程

29 6. Corollaries of the First Law （热力学第一定律的推论）
(1)  Work done in any adiabatic (Q=0) process is path independent. (2) For a cyclic process heat and work transfers are numerically equal or

30 7. 稳态稳定流动系统能量方程 q = dh-vdp

31 Exercise (练习) Correct the error in the following equations. (纠正下列各式中的错误) (1) (2) (3) (4) (5)

32 2. Make a statement of suitable conditions under which the following equations can be adopted. ( 叙述下列各公式的适用条件) (1) (2) (3) (4)

33 3. Initially, there is m working fluid in an adiabatic and rigid container, its state is denoted as ‘1’.Then m working fluid of state ‘2’ is charged into the container. The process finished until the state of the working fluid reaches ‘3’ . Write out the energy equation for the process.

34 8.The procedures adopted (所采取的步骤)
Pv=RT

35 35

36 9.单级压气机 (1) Work consumption (耗功量）
If it is isothermal compression, then (若压缩过程为可逆定温过程，则：) If it is isentropic compression, then (若压缩过程为可逆绝热压缩，则: )

37 （2） Discharge Temperature (排气温度）
If it is polytropic compression, then (若压缩过程为可逆多变过程，则 : ) （2） Discharge Temperature (排气温度）

38 （2） The influence on compression work（对耗功量的影响）
10. The influence of residual volume of inter-space. (余隙容积的影响) （1）The influence on discharge volume( 对排气量的影响） （2） The influence on compression work（对耗功量的影响） There is no influence on compression work.(对耗功量无影响)

39 11. Multistage Compression with inter-cooling （多级压缩、中间冷却）

40 不可能将热从低温物体传至高温物体而不引起其它变化。
12. 克劳修斯表述 Clausius statement 不可能将热从低温物体传至高温物体而不引起其它变化。 It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body.

41 12.Carnot‘s Theorems (卡诺定理)
在相同的高温热源和相同的低温热源间工作的可逆热机的热效率恒高于不可逆热机的热效率； The efficiency of an irreversible heat engine is always less than that of a reversible one operating between the same two thermal reservoirs.

42 在相同的高温热源和相同的低温热源间工作的一切可逆热机有相同的热效率，而与工质无关。
The efficiencies of all reversible heat engines operating between the same two thermal reservoirs are the same.

43 13. The significance of the Carnot Theorems (卡诺定理的意义)
(1)卡诺定理指明了热变功的最高效率 The Carnot Theorem indicates the maximum thermal efficiency of heat engine, which converts heat into work.

44 (2)卡诺定理指明可以通过提高高温热源的温度,降低低温冷源的温度或减少过程的不可逆因素等方式来提高热效率
The Carnot Theorem point out thermal efficiency can be improved by means of raising the temperature of high temperature thermal reservoir, lowering the temperature of lower temperature reservoir, or reducing irreversibilities.

45 (3) 卡诺热效率表明了热量的最大可用能 The Carnot thermal efficiency value reveals the maximum amount of high temperature thermal energy which can be converted to work. (4)卡诺定理表明能量不仅有数量的差别，还有品质的高低 The Carnot Theorem indicates that energy has quality as well as quantity.

46 热源温度越高，热量的品质就越高,其可转化为的可用能就越大。
The higher the temperature, the higher the quality of thermal energy. (5) 基于卡诺定理,才证明熵是一个状态参数 It is based on Carnot theorem that entropy is investigated to be a property.()

47 A 热机是否能实现 1000 K 2000 kJ 1200 kJ A 可能 1500 kJ 800 kJ 500 kJ 300 K 不可能
例题: A 热机是否能实现 1000 K 卡诺定理举例 2000 kJ 1200 kJ A 可能 1500 kJ 800 kJ 如果：W=1500 kJ 500 kJ 300 K 不可能

48 14.克劳修斯不等式 ∴ 对任意循环 克劳修斯 不等式 热源温度 = 可逆循环 < 不可逆循环 > 不可能 热二律表达式之一

49 克劳修斯不等式例题 A 热机是否能实现 1000 K 可能 2000 kJ A 1200 kJ 1500 kJ 800 kJ 500 kJ
如果：W=1500 kJ A 1200 kJ 1500 kJ 800 kJ 500 kJ 不可能 300 K 注意： 热量的正和负是站在循环的立场上

50 (4). > , for irreversible process (不可逆过程)。
=, for reversible process (可逆过程)

51 15.The increase principle of Entropy (孤立系统的熵增原理)
For isolated system The entropy of an isolated system during a process always increases or, in the limiting case of a reversible process remains constant. In other word, it never decreases. (孤立系统中的过程总是向着熵增大的过程进行,若为可逆过程,则熵不变。换句话说，即孤立系统的熵不会减小）

52 16. Entropy generation and Entropy flow (熵产与熵流)
Entropy generation is caused by any irreversibility. ( 熵产是由不可逆因素引起的熵变) (2) Entropy Flow/Transfer (熵流) Entropy transfer by heat transfer (热量熵流) Entropy transfer by mass flow (质量熵流)

53 For steady-flow system (对于稳态稳定流动系统）
For closed system (对于闭口系统) For steady-flow system (对于稳态稳定流动系统） For single stream, adiabatic, steady flow （对于单股绝热的稳态稳定流动系统）

54 17. Significance of Entropy and its application
(熵的意义及应用) （1） Heat absorption and heat rejection during a reversible processcan be calculated by resorting to Entropy. (可逆过程中的吸热或放热量可借助熵来计算） （2） Entropy generation indicates the direction of process in isolated System.(熵产是孤立系统中过程进行方向的标志）

55 T T0 s heat transfer across a finite temperature difference（温差传热）
Free or unrestrained expansion（自由膨胀） （3）Entropy is a measurement of the amount of thermal energy which can not be converted to work. (熵是热量不可用能大小的量度） (4) Entropy Generation indicates the amount of loss in energy which can be converted to work. (熵产是做功能力损失的量度） T TH T0 s s1 s2

56 (1) Saturated water Tables (饱和水和饱和蒸汽表)
18. Types of Property Tables (水蒸汽表的分类) (1) Saturated water Tables (饱和水和饱和蒸汽表) Saturated water---T Tables(以温度t为独立变量排列) Saturated water---p Tables(以压力p为独立变量排列) (2) Subcooled water and Superheated vapor Table (未饱和水和过热水蒸汽表) P and t are dependent variables (以p和t为自变量)

57 （3）

58 （4） Consulting property tables of water vapor to determine the state of each point and their h，s，x.(利用水蒸气表判断下列各点的状态，并确定其h，s，x的值。)  

59 19. Thermodynamic Processes of Water Vapor (水蒸气的热力过程)
(1) Based on the given conditions, determine the initial state and its properties . (根据已知条件，确定初始状态，查出其余参数。) (2) Based on the characteristics of the process and one of the properties of the final state, determine the final state and its properties. ( 根据过程特点和一个终态参数，确定终态，再查出其余终态参数。) (3) Based on the initial and final state, calculate the q、Δu、w during the process。 (根据初、终态参数，计算q、Δu、w等。)

60 (1) Constant Volume Process ( 定容过程, v=定值。)

61 (2) Constant Pressure Process (定压过程 p=定值)

62 (3) Constant Temperature Process (定温过程 T=定值)
Isothermal Process (定温过程)

63 (4) Adiabatic Process ( 绝热过程)
Isentropic Process (定熵过程)

64 Example: 1 kg water vapor, initially, ℃ undertakes an isentropic expansion process and reaches the final state of Calculate the amount of work done by the water vapor during the process.

65 The atmospheric air is at ℃ .
20. Example: The atmospheric air is at ℃ . （１）If is relative humidity is equal to 60%, then calculate such properties as （２）If ℃, then calculate such properties as 　　Discussion: To determine the state of moist air, it need three state properties at least.(为了确定湿空气的状态，至少需要３个状态参数）

66 The Psychrometric Chart (h-d图)
21. 湿空气的焓湿图 The Psychrometric Chart (h-d图) Saturation Line (饱和湿空气线） In addition, there are other lines(还有下列曲线：） 1. Lines of constant enthalpy (定焓线，与y轴成45oC角向下) 2. Lines of constant humidity ratio (定含湿量线，垂直直线) 3.Lines of constant temperature (定温线) 4. Lines of constant volume ( 定容线) 5. Lines of constant relative humidity line( 定相对湿度线。) 6. Lines of constant pv和d. (水蒸汽分压力pv和d关系曲线 )

67 不同的pb 不同的h-d图 tw td

68 （Thermodynamic Process of Moist Air ）
22. 湿空气的热力过程 （Thermodynamic Process of Moist Air ） (1) Sensible Heating Process (加热过程) 加热过程 B A The load on the heater is:

69 (若冷却介质温度没有降到露点以下，则过程前后)：
(2) Sensible Cooling Process at Constant Moisture Content (单纯冷却过程) If the temperature of the cooling medium is still above its dew point temperature, then (若冷却介质温度没有降到露点以下，则过程前后)： 冷却过程 A B

70 (3) Sensible Cooling with Dehumidification (冷却去湿)
If the air is continuously cooled to a temperature which below its dew-point temperature, then water vapor will condense. (温度下降到露点以后，继续降温，析出水分，减小含湿量。) A C B

71 (4) Humidification Process (加湿过程)
Adiabatic Humidification Process (绝热加湿过程) 由于水的焓值很小，可以忽略不计，所以可以看作加湿之后焓不变，在h-d图上是沿着定焓线向d增大的方向进行（过程1-2）。

72 (2) Isothermal Humidification Process (等温加湿)

73 (5) Mixing of Two Air Streams (混合过程)

74 例题(Example) 已知空气的t1=20oC， p1=0.1MPa
将其加热至t2=50oC，后送入干燥室，从干燥室排出时t3=30oC，求：1） 2）从干燥室每吸收1kg水分所需空气量和加热量 h2 h 2 h1 3 1 d

75 23.朗肯循环 T h 1 1 4 4 2 3 2 3 s s 12 汽轮机 s 膨胀 23 凝汽器 p 放热 34 给水泵 s 压缩

76 朗肯循环热效率的计算 h 1 一般很小，占0.8~1%，忽略泵功 4 2 3 s

77 t1 , p2不变，p1 T 1 5 6 缺点： 不利于汽轮机安全。一般要求出口干度大于0.85~ 0.88 4 2 3 s
A.蒸汽初压对朗肯循环热效率的影响 优点： ，汽轮机出口尺寸小 t1 , p2不变，p1 T 1 5 6 缺点： 对强度要求高 不利于汽轮机安全。一般要求出口干度大于0.85~ 0.88 4 2 3 s

78 p1 , p2不变，t1 T 1 5 6 4 3 2 s B.蒸汽初温对朗肯循环热效率的影响 优点： ，有利于汽机安全。 缺点：
对耐热及强度要求高，目前初温一般在550℃左右 汽机出口尺寸大 5 6 4 3 2 s

79 p1 , t1不变，p2 T 1 缺点： 5 6 4 2 3 s C.乏汽压力对朗肯循环热效率的影响 优点：
受环境温度限制，现在大型机组p2为0.0035~0.005MPa，相应的饱和温度约为27~ 33℃ ，已接近事实上可能达到的最低限度。冬天热效率高 5 6 4 2 3 s

80 24. 蒸汽抽汽回热循环 T 1 1kg a 6  kg 5 4 (1- )kg 3 2 s a  kg 4 (1- )kg 5
由于T-s图上各点质量不同，面积不再直接代表热和功

81 T 1 1kg a 6  kg 5 4 (1- )kg 3 2 a  kg 4 (1- )kg 5 1kg s
(2)抽汽回热循环的抽汽量计算 T 1 以混合式回热器为例 热一律 1kg a 6  kg 5 4 (1- )kg 3 2 a  kg 4 (1- )kg 5 1kg s

82 (3)抽汽回热循环热效率的计算 吸热量： T 1 1kg a 放热量： 6  kg 5 4 (1- )kg 3 2 净功： s 热效率：

83 25. 蒸汽再热循环 (reheat cycle) 1 T 6 b 5 4 3 s

84 蒸汽再热循环的定量计算 吸热量： 1 T 6 b 放热量： 5 4 净功（忽略泵功）： 3 热效率： s

85 26.Otto循环(定容加热循环）_ 1’ - 2 Compress mixture quasi-statically and adiabatically Ignite and burn mixture at constant volume (heat is added) 3 (4)- 5 Expand mixture quasi-statically and adiabatically 5 - 1’’ Cool mixture at constant volume

86 Thermal Efficiency of Otto Cycle
(奥托循环的热效率) 提高循环的压缩比 提高循环的最高温度

87 27.The Diesel Cycle (狄塞尔循环)

88 Thermal Efficiency of Disel Cycle 狄塞尔循环的热效率：
提高循环的压缩比 降低循环的预胀比 提高循环的最高温度

89 28. The Dual Cycle (混合加热循环 )
其中,1-2是定熵压缩过程, 2-3是可逆定容加热过程, 3-4是可逆定压加热过程, 4-5是定熵膨胀过程, 5-1是可逆定容放热过程.

90 混合加热循环的热效率: 几个定义： 压缩比: 　　 定容升压比: 定压预胀比: :

91 提高热效率方法：提高循环的压缩比 降低循环的预胀比 提高循环的最高温度
　　由上面的三个定义,得: 　　                           讨论: 　　  随  ,   ,   的升高而升高; 　　  随  的升高而降低. 提高热效率方法：提高循环的压缩比 降低循环的预胀比 提高循环的最高温度

92 29.空气压缩制冷循环 冷却水 2 3 冷却器 冷藏室 膨胀机 压缩机 4 1

93 Reversed Brayton Cycle
p-v图和T-s图 p 逆勃雷登循环 T 3 2 2 3 T0 T2 1 4 1 4 v s s 绝热压缩 绝热膨胀 s 等压冷却 p 等压吸热 p

94 制冷系数 T 2 3 1 4 s

95 2-4：定压放热过程 4-5：绝热节流过程 5-1：定压吸热过程
30.蒸气压缩制冷循环装置 1-2：绝热压缩过程 4 2-4：定压放热过程 4-5：绝热节流过程 5-1：定压吸热过程 5

96 lnp-h图及计算 T lnp h 2 3 4 2 3 4 1 5 1 5 s

97 压焓图 P-h diagram

98 过冷措施 工程上常用 lnp h T 2 3 4’ 4 2 3 4 4’ 1 5’ 5 1 5’ 5 s 不变