First Law of Thermodynamics

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1 First Law of Thermodynamics
Objective First Law of Thermodynamics Energy Forms and Energy Transfer …Work, Heat and Mass Energy Balance for Closed System Energy Balance for Open System Energy Balance for Steady-Flow System Objective

2 First Law of Thermodynamics
Conservation of Energy Principle Energy can be neither created nor destroyed ;It can only change forms. Energy can change many forms, but its total value keeps constant. The perpetual-motion machine of the first kind can never be made.

3

4 First Law of Thermodynamics
In 1843, at the age of 25, James Prescott Joule did a series of careful experiments to prove the equivalence of heat and work.

5 Forms of Energy Transfer
{Energy Entering CV } － {Energy Leaving CV} ={Energy Accumulating in CV} There are three forms of energy transfer: Work:caused by a force acting through a distance volume work: moving boundary work Heat: caused by temperature difference between the system and surroundings. Mass flow: mass flow in or out of the system serves as an energy transfer.

6 Energy Transfer by Work
• Definitions – Energy transfer associated with force acting through a distance – Energy crossing the boundary of a closed system that is not heat must be work 系统与外界相互作用而传递的能量，其全部效果为使外界物体改变宏观运动状态。 包含两个必要条件： 确定有力作用在边界上，即系统与外界有相互作用，且力的不平衡势差无限小； 2) 系统边界发生位移，即外界物体改变宏观运动状态。 二者缺一不可

7 Energy Transfer by Work
例1： p 拔掉销钉后，系统向真空膨胀， 系统是否做功？ Vacuum 例2： 刚性容器，加热后系统内压力 升高，系统是否向外做功？ p Q

8 Energy Transfer by Work
Units – or J; usually kJ in SI • Work examples – Boundary work (e.g., a gas working against a moving piston) closed system – Shaft work (e.g., a rotating crankshaft on a motor) open system

9 Energy Transfer by Work
示功图 Work－Process function 功是过程量

10 Energy Transfer by Work
• Work per unit mass • Sign Convention W > 0: work done by the system W < 0: work done on the system W = 0: no work

11 例： Q 解： 如图所示，气缸初始状态下： 现对气缸加热，使气体膨胀至： 已知：初始状态下，弹簧与气缸接触但不受力，弹簧刚度 大气压力
活塞面积 求：1) 气缸内终了压力和气体做的功？ 2) 若活塞与气缸间有摩擦力 存在， 气体做的功是多少？ Q 解： 取气缸内的气体为系统，气缸内壁为边界 准静态过程 终了状态为平衡状态，活塞两侧的受力相等

12 利用示功图求解

13 若活塞与气缸间存在摩擦，不可逆因素出现在系统外，为外部不可逆过程，这时气体
需要抵抗外力 做功。 在工程实际中，将活塞和气缸作为整体考虑，注重整套装置，即活塞的有效输出功， 这时，摩擦成为内部不可逆因素，整套装置对外做功 但是由于内部摩擦的存在，使得气体必须克服摩擦做功，因此气体必须做功

14 Energy Transfer by Heat
• Definitions — Energy transfer only by a temperature difference －Heat is energy in transformation；path function. • Units • Heat per unit mass

15 Energy Transfer by Heat
• Sign Convention Q > 0: heat transfer to the system Q < 0: heat transfer from the system Q = 0: adiabatic T 1(T1,s1) 2(T2,s2) S － Entropy 熵 S ds

16 Comparison Between Heat and Work
Heat and work are energy transfer mechanisms between a system and its surroundings. Similarities: Both are recognized at boundaries of a system as they cross the boundaries. That is ,both are boundary phenomenon. System processed energy, but not heat and work. Both are associated with a process, not a state. Heat and work have no meaning at a state. Both are path functions. Difference:

17 Forms of Energy Total Energy =Internal+External =U+Ek+Ep
Although energy can take a large number of forms, we can consider only: External and Internal. Kinetic External Total Energy =Internal+External =U+Ek+Ep Potential Physical Internal (U) State Property Chemical Nuclear Note: U is state property

18 Energy Balance for Closed System
{Energy Entering CV } － {Energy Leaving CV} ={Energy Accumulating in CV} For Closed System difference 注意过程量和状态量数学表达的区别！

19 推动质量进出系统的推动功，即保证质量流动耗费的功
Energy Balance for Open System For Open System Difference with closed system: Energy in mass to and from system Mass equilibrium Including volumetric work and flow work with surrounding 推动质量进出系统的推动功，即保证质量流动耗费的功

20 Flow Work Total Energy of Flowing Fluid 焓－工质流动时具有的与其热力状态有关的总能量。
1 １kg : dx1 2 Total Energy of Flowing Fluid 焓－工质流动时具有的与其热力状态有关的总能量。 Enthaply 焓

21 Energy Balance for Open System
{Energy Entering CV } － {Energy Leaving CV} ={Energy Accumulating in CV} Energy Entering CV: Heat-in energy with mass in flow work by upfluid

22 Energy Balance for Open System
Energy Leaving CV: Shaftwork energy with mass out flow work by fluid

23 Energy Balance for Open System
适用于任何工质的任何流动过程

24 Energy Balance for Open System
For Steady Flow System — fluid properties remain constant during the entire process — mass equilibrium — energy in=energy out

25 Energy Balance for Open System
=0 适用于任何工质的任何稳定流动过程

26 Analysis of Energy Balance
Shaft Work w 技术功

27 Analysis of Energy Balance
稳定流动系统能量方程 适用于任何工质的任何稳定流动过程

28 Analysis of Energy Balance
对可逆过程 1 p v 2 vdp dp > 0,wt<0 work done on the system dp < 0,wt>0 work done by the system dp = 0, wt=0

29 Mechanical Energy Conservation
For Reversible Process For Quasi-Equilibrium Process with Friction =0 Bonulli Equation

30 Example 例2-2 解： q 绝热压缩 3 换热器 2 4 换热器吸热 喷管 压缩机 绝热膨胀 气轮机 空气流量 5 空气 1
过程中忽略位能变化 求:1)压缩机功率；2)喷管出口流速；3)气轮机功率；4)整套装置功率 解： 工质在整套装置内的流动为稳定流动，应用稳定流动能量方程求解。

31 1)压缩机功率 2)喷管出口流速,流经换热器和喷管 1 2 3 4 5 空气 压缩机 换热器 喷管 气轮机 q 3)气轮机功率

32 4)整套装置功率 或 将整套装置取为系统

33 End of 1st Law of Thermodynamics
Well done!

34 2.6 Second Law of Thermodynamics
Natural process is directive. 2nd Law is used to determine the direction condition limitation of thermal process. *** We will use the 2nd Law as a tool to evaluate whether a process is possible. ***

35 Spontaneous Process Mechanical Heat Heat Transfer Others:
Transferring heat to a paddle wheel will not cause it to rotate. A cup of hot coffee does not get hotter in a cooler room. Others: Gas Free Expansion Mixture Process Combustion and Reaction Process

36 2.6 Second Law of Thermodynamics
一切实际的热力过程都具有方向性，只能单独自动地朝一个方向进行，这类过程称为自发过程；而其逆方向地过程不能单独自动进行，这类过程称为非自发过程。若要非自发过程得以实现，必须附加某些补充条件，付出一定的代价。

37 Statement of 2nd Law Clausius:
It is impossible to transfer energy from a cooler to a hotter body as a sole effect (requires a heat pump or other device, which needs energy input) 2. Kelvin-Planck: It is impossible to operate a thermodynamic cycle to produce work with only heat transfer from a single reservoir (requires both heat addition and heat rejection)

38 Statement of 2nd Law Clausius: High-Temperature Reservoir TH
Low-Temperature Reservoir TL High-Temperature Reservoir TH Reservoir: a “large” body that can absorb or supply heat without a “noticeable” temperature change

39 High-Temperature Reservoir TH
Statement of 2nd Law 2. Kelvin-Planck High-Temperature Reservoir TH Heat Engine

40 Statement of 2nd Law 3. 2nd perpetual-motion machine NEVER be made

41 Second Law of Thermodynamics
热力学第二定律的实质: 自发过程是不可逆的； 若要非自发过程得以实现，必须伴随一个适当的 自发过程作为补充条件。

42 Heat Transfer 若要实现热量由低温向高温的非自发过程，必须消耗功， 即配合功变热这个自发过程作为补充条件。
Spontaneous Process 若要实现热量由低温向高温的非自发过程，必须消耗功， 即配合功变热这个自发过程作为补充条件。

43 Mechanical Heat Mechanical Heat 自发过程 非自发过程
1.若要实现热转化功的非自发过程，必须配合向低温热源放热 这一自发过程，因此热机的效率必定小于1。 2.“功可以全部转化成热，但热不能完全变为功” 理想气体的等温膨胀 3.热变功的最高效率－Carnot Cycle

44 Carnot Cycle(Reversible)
Ⅰ Isothermal Expansion Ⅱ Adiabatic Expansion Ⅲ Isothermal Compression T S TH TL S1 S2 Ⅳ Adiabatic Compression

45 Thermal Efficiency Thermal Efficiency = Desired Output Required Input
Net Work Output Total Heat Input WO Qin =1－ Qout Definition: Q1—heat transfer between cycle device and high-temerature medium at TH Q2—heat transfer between cycle device and Low-temerature medium at TL

46 Thermal Efficiency For Heat Engine(卡诺热机) Heat Engine
High-Temperature Reservoir TH Heat Engine 区别？ Low-Temperature Reservoir TL

47 Thermal Efficiency For Refrigerator(卡诺制冷机) Refrig- erator 制冷系数
High-Temperature Reservoir TH Refrig- erator 制冷系数 Low-Temperature Reservoir TL

48 Thermal Efficiency For Heat Pump(卡诺热泵) Heat Pump 供暖系数
High-Temperature Reservoir TH Heat Pump 供暖系数 Low-Temperature Reservoir TL

49 Carnot Principles 1. All reversible cycles operating between TL
and TH have the same efficiency (ηrev) . 2. For same TL and TH, reversible cycles have higher efficiencies than any irreversible cycles.

50 Conclusions of Carnot Principles
Carnot 循环热效率只取决于高温热源和低温热源的温度，即工质吸热和放热的温度。提高T1或降低T2均可提高热效率。 任何循环的热效率均小于1。 T1=T2时，循环热效率为零，说明只从单一热源吸热是不可能把热转变成功的。 要提高实际装置的热效率，必须尽可能减少摩擦等不可逆功损失。

51 Carnot Cycle with Multi-reservoir
Why is the Carnot cycle the highest efficiency of heat to work? A B C D

52 Example 解： 例2-5 冬季室外温度-10℃，为保持室内温度20℃，需要室内供热 7200kJ/h。试计算：
(1)若采用电暖气供暖，需要电功率为多少？ 解： (1) 取室内的空气为系统

53 (2)若采用逆向卡诺循环机供暖，则供暖机功率为多少？
Outside Heat Pump ？ Carnot Cycle

54 Outside (3)若该供暖机由另一正向卡诺热机带动，其高温热源温度500K， 低温热源为大气，则正向卡诺热机的供热率为多少？
Heat Pump High-Temperature Heat Engine Atmosphere

55 Entropy 熵 Objective Definition of Entropy Clausius Inequality
Entropy Change in Irreversible Process Entropy Generation Increase of Entropy Principal of Isolated System

56 Entropy A B P Q M N Carnot Cycle 代数值 for any reversible cycle

57 Entropy p A B P Q M N for whole cycle v State Property

58 Entropy (J/K) or (kJ/K) Entropy per Unit Mass
Entropy indicates the direction and magnitude of heat transfer during reversible process. heat into system, entropy increasing heat out system, entropy decreasing adiabatic system, entropy constant

59 Clausius Inequality If part of cycle is irreversible process.According to the Carnot Principle: For same TL and TH,reversible cycles have higher efficiencies than irreversible cycles. 代数值

60 Clausius Inequality “=” Reversible Process “<” Irreversible Process
－Heat from surrounding －Temperature of heat reservoir “=” Reversible Process “<” Irreversible Process

61 Entropy Change in Irreversible Process
1 2 A B p v 1A2 Reversible Process 1B2 Irreversible Process

62 Entropy Change in Irreversible Process
特别需要理解： 因为熵是状态参数，所以无论是可逆过程还是不可逆过程，如果过程的起始和终了状态相同，那么系统的熵变相同。 该熵变等于(可逆过程)或大于(不可逆过程)过程中系统与外界交换的热量与热源温度比值的积分。 如果初始状态相同，且过程 相同，则系统经过不可逆过程的熵变大于可逆过程熵变。

63 Entropy Change in Irreversible Process
Adiabatic Process Adiabatic + Reversible 定熵过程 Isentropic Process Adiabatic + Irreversible Caused by irreversibility T s 1 2 2’

64 Entropy Flow and Entropy Generation
Irreversible Process Entropy Flow Entropy Generation Entropy change by heat transfer Entropy increase by irreversibility for reversible process for irreversible process

65 Entropy Flow and Entropy Generation
Friction Irreversible Process Work by Friction Reversible Process 吸收摩擦热的热源温度 Friction Work Heat

66 Entropy Flow and Entropy Generation
Heat Transfer by Temperature Difference Temperature of Heat Reservoir

67 2.10 Increase of Entropy Principal of Isolated System
System+Surroundings 孤立系统内所进行的一切实际过程（不可逆过程） 都朝着使系统熵增加的方向进行；在极限情况下 （可逆过程），系统的熵维持不变；任何使系统熵 减小的过程都是不可能的。

68 Increase of Entropy Principal of Isolated System
1.熵增原理是对孤立系统而言的，系统内的某个物体可与系统内的其它物体相互作用，其熵可增、可减、也可以维持不变。 是指系统内各部分熵变的代数和。既可以按照组成物体计算，也可按照不可逆因素计算。 要想使 的过程得以实现，必须寻找一个使熵增加的过程与原孤立系统伴随进行。而且必须使两者组成的新孤立系统的熵变大于零。 4.当熵不断增加直至达到某个最大值时，系统处于平衡状态，过程即告停止。

69 Example 11.某热机工作于1000K和400K的两恒温热源之间，每kg工质从高温热源吸热200kJ， (1)试计算其最大循环功；
(4)上述三种循环中的熵产各为多少？

70 (1)计算其最大循环功 解： (2)若工质吸热时与高温热源温差为150K，放热时与低温热源温差20K，则最大输出功是多少？ (3)如果上述过程中不仅存在温差传热，由于摩擦又使循环功减少40J，则热机效率又是多少？

71 (4)上述三种循环中的熵产各为多少？ 1.∵为可逆过程 ∴ 熵产为零 2.∵温差传热产生不可逆性 ∴

72 温差传热 温差传热 摩擦损耗功 3.∵温差传热和摩擦产生不可逆性
解法一：将高温热源、低温热源(Surroundings)和热机(System)取为孤立系统,按照不可逆因素计算孤立系统熵增 温差传热 温差传热 200-w 摩擦损耗功 T2’

73 请思考： 1.物体熵变计算中， 高温热源 的物理意义？ 低温热源 2.热机的不可逆摩擦损耗 如何处理？ 热机
3.∵温差传热和摩擦产生不可逆性 解法二：将高温热源、低温热源(Surroundings)和热机(System)取为孤立系统,按照组成物体计算孤立系统熵增 请思考： 1.物体熵变计算中， 的物理意义？ 高温热源 低温热源 2.热机的不可逆摩擦损耗 如何处理？ 热机

74 1.对于热源 与外界交换的热量,包括数量和方向 换热温度 高温热源 低温热源 2.工质经历循环后熵变 热机摩擦损耗全部转变成热，被低温热源吸收。

75 热机 高温热源 低温热源

76 本章小结 热力学第二定律的实质和表达 卡诺循环的组成及图形表示 卡诺定理 判断热力过程方向和限度的方法 Clausius 不等式
可逆过程与不可逆过程的熵变 熵流与熵产 孤立系统熵增原理和计算方法

77 Laws of Thermodynamics completed!
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