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1、2.3 热力学第一定律的一般表达式热力学第一定律的一般表达式 当系统由状态当系统由状态1 1经历一系列状态变化达到终状态经历一系列状态变化达到终状态2 2时，系时，系统总能量的变化为统总能量的变化为EsysEinEout（mineinQ）（mouteoutW）当系统处于宏观上静止时当系统处于宏观上静止时 U（mineinQ）（mouteoutW）看问题的角度看问题的角度: :进入系统的能量为正进入系统的能量为正, ,离开为负离开为负 moutminQWEsys 能量守恒12.4闭口系统能量方程闭口系统能量方程 A Conservation of Energy Principle for Clo
2、sed SystemsA closed system undergoing a process between state 1 and state 2, the energy equation will be: UQ W or QUW or for 1kg of mass quwfor a differential basis QdUW or qduw 2这里功的数量并不等于可以利用的功这里功的数量并不等于可以利用的功nFor a reversible process, the work can be calculated as: For a closed system, heat trans
3、fer and work transfer are the only mechanisms by which energy can be transferred across the boundary.If we need to express the general energy balance on a rate basis by a finite time interval. This yields:Then, in the limit as t approaches zero3例题例题例题例题2.12.1气缸内储有定量的气缸内储有定量的CO2气体，初态气体，初态p1300kPa，T12
4、00，V10.2m3。经历一可逆过程后温度下降至经历一可逆过程后温度下降至T2100。如果过程中压力和比容间的关系满足如果过程中压力和比容间的关系满足pv1.2常常数，试确定该过程中数，试确定该过程中CO2气体所作的功、比功、热力学气体所作的功、比功、热力学能变化量，气体与外界之间的热量交换。能变化量，气体与外界之间的热量交换。Solution：Given: initial and end states, process, CO2 in a cylinder Find: W, w, U, QModel: closed systemStrategy: apply the basic closed
5、 system energy balance to solve for the Q and U, apply the definition of work to calculation W and w State1 State 24Analysis:1)work：可逆：可逆wpdvWmw0.6719450063425 J 因气体体积变化，故此功称为膨胀功。因气体体积变化，故此功称为膨胀功。W0，为气体对，为气体对外做功。在终态，气体体积为外做功。在终态，气体体积为V20.656m32)系统内热力学能的变化：系统内热力学能的变化：(E)sysUmcV(T2T1)查表取查表取cv0.656kJ/k
6、gK5所以所以U0.671103(100200)44018J （说明系统的热力学能是减少的）（说明系统的热力学能是减少的）3)Q的的计计算算：QUW63425（44018）19407 JComment:该该例例题题中中气气体体对对外外作作出出的的功功量量大大于于气气体体热热力力学学能能的的减减少少量量，减减少少的的部部分分由由外外界界对对气气体体的的加加热热量量所所补补充充。另外此处能量的单位用焦耳显得不方便，用另外此处能量的单位用焦耳显得不方便，用kJ则好些。则好些。62.5开口系统的能量方程开口系统的能量方程 Conservation of Energy Principle for A O
7、pen System or Control VolumeTwo steps of analysis1st Conservation of Mass Principle for A Open System or Control Volume2nd Conservation of Energy Principle for An Open System or Control Volume1st: Conservation of Mass Principle for A Open System or Control VolumeIs defined as the mass flow rate kg/s
8、7Case 3 for both steady state and one-dimensional flow:Steady state: any system in steady state if the system properties are constant with time at every position within and on the boundaries of the systems.One-dimensional flow: if the properties at a flow boundary are uniform over the cross-sectiona
9、l area. Case 1 For steady state: Case 2 For one-dimensional flow:82nd Conservation of Energy Principle for A Open System or Control VolumeIts clear when regardless the kinetic and potential energy of a system , well have:If we take the assumption- Steady state: If we take the assumption- One-dimensi
10、onal flow: If we take both the assumptions- Steady state and One-dimensional flow: 9The energy in or out of the control volume ：When mass gets in the system, it will carry the several items of energy into the system1) internal energy of the mass;2) ?3) kinetic energy of the mass;4) and potential ene
11、rgy of the mass.1) internal energy of the mass (if m kg mass enters)2) how to let the mass across the boundary into a system?10According to the fig. shown below:For one-dimensional assumption, Work should be done as:We give this work the name of flow work or push work.As 1) and 2) discussion, we hav
12、e:When m kg of mass enters in a control volume :11A definition:orWe give h the name of enthalpy.This is a property of matter. (Combined property)3) kinetic energy of the mass (When m kg of mass enters in a control volume )4) and potential energy of the mass (When m kg of mass enters in a control vol
13、ume)12From above discussing, for each m kg mass enters in the system, the total energy enters the control volume should be:the total energy leaves the control volume should be:The control volume energy equation on one dimensional assumption:For mass rate: 13If we take another assumption of steady st
14、ate:There are numerous applications of the steady flow and steady state conservation of energy principle for which there is only one inlet (position 1) and one outlet (position 2) . In this circumstance above equation will be expressed as:On a unit mass basis, it is convenient to express as:14例题例题例题
15、例题2.22.2已知汽轮机进口水蒸气参数为已知汽轮机进口水蒸气参数为p19MPa，t1500，流速cf150m/s；出口水蒸气参数为出口水蒸气参数为p20.5MPa，t2180，流速流速cf2120m/s。蒸气的质量流量蒸气的质量流量qm330t/h 。蒸气在汽轮机中进行稳定的绝热流动，求汽蒸气在汽轮机中进行稳定的绝热流动，求汽轮机的功率。轮机的功率。解解：取系统如下图所示取系统如下图所示 （稳态稳流）（稳态稳流）分析：分析：Q = 0 U = 0einu1p1v1c12/2gz1einu2p2v2c22/2gz2 系统对外作功：系统对外作功：Wtm（u1u2）（p1v1p2v2） （c12/
16、2c22/2）（gz1gz2） m（h1h2）（c12c22）/2 g（z1z2） 例题2.2附图moutminws15续解：续解：对每对每kg工质，功可以表示为：工质，功可以表示为： wt（h1h2）（）（c12c22）/2g（z1z2）略去其中位能差，可计算得：略去其中位能差，可计算得：w wt t（h h1 1h h2 2）（）（c c1 12 2c c2 22 2）/2/2 (3386.4(3386.42812.1)2812.1)10103 3(50(502 21201202 2)/2)/2 568.4568.410104 4J/kgJ/kgP Pm wt 33033010103 35
17、68.4568.410104 4/3600/360052.152.110103 3 kW kW 16特殊情况特殊情况特殊情况特殊情况 忽略动能差和位能差时忽略动能差和位能差时q（h2h1）wtQ（H2H1）Wt绝热时的能量方程绝热时的能量方程 WtH1H2绝热节流过程绝热节流过程 H2H1 技术功技术功WtW（p2V2p1V1） 图2.6 技术功在p-v图上的表示17习题习题:阅读阅读:第二章内容第二章内容思考题思考题: P45 1. 3. 4. 5. 7.习题习题: p46 2-3 2-4 p47 2-9 2-1018SummaryMechanical work is defined as:
18、Expansion or compression work in a reversible process is defined as:Heat transfer of energy Q is the interaction across the boundaries under the temperature differenceTotal energy of a system includes internal energy, kinetic energy, and potential energyThe 1st law of thermodynamics for a closed sys
19、tem: QUW 19The 1st law of thermodynamics for a opened system:We have defined a new property-enthalpy For solving a control volume problem, we should take mention of two assumptions: steady state and one dimensional flow.This equation is used to steady state and one dimensional flow. 20部分资料从网络收集整理而来，供大家参考，感谢您的关注！
