4讲能量方程及例题

nullnull2012年3月29日复习：无论开口、闭口系统，热力学第一定律是能量守恒问题 分析 定性数据统计分析pdf销售业绩分析模板建筑结构震害分析销售进度分析表京东商城竞争战略分析 过程总是从系统图开始分析内容在系统本身、边界作用两方面的能量情况边界作用有物质和能量两方面物质作用考虑携带的焓、动能和位能边界作用考虑热量和功记住这个图null工质系统的能量守恒方程课下准备的问题讨论：1）技术功膨胀功区别，如何在示功图上表示？对机器作功＋动能差＋位能差注意差异！对机器作功不等于工质的膨胀功null可逆（准平衡）过程，工质膨胀功即工质膨胀功与技术功的差异是流动功！示功图表示：膨胀功过程线下面积技术功过程线右面积注：只有可逆或准平衡过程才有图示关系null2）推动功与流动功区别？3）若工质流入一容器，带入容器能量如何表示？4）理想气体是某一类气体吗？流动功是进出口推动功差假若进入工质状态恒定若进入工质状态是变化的，最好的表达是按理想气体模型，不是。null今天课程内容：讨论第一定律各种情况下的应用（例题）工质的热力学能确定null例题2.2已知汽轮机进口水蒸气参数为p1＝9MPa，t1＝500℃，流速c1＝50m/s；出口水蒸气参数为p2＝0.5MPa，t2＝180℃，流速c2＝120m/s。蒸气的质量流量qm＝330t/h 。蒸气在汽轮机中进行稳定的绝热流动，求汽轮机的功率。解：取系统如下图所示 （稳态稳流）分析：null略去其中位能差，可得：查：水蒸气参数注意：此处系统所作之功并不仅仅是工质膨胀而得，还有流动净功和动能差所带来的功若还将动能差忽略，结果是：null可见，忽略动能差后，计算结果相差不大，1.94%Example 2.3: A gas within a piston-cylinder device is compressed at a constant pressure of 0.5MPa from 1000 to 400cm3. The frictional force at the piston-cylinder interface is 200N, the piston surface area is 100cm3, and the atmospheric pressure is 0.1MPa. Determine the work transferred by the piston to the gas and the work supplied by the connecting rod.另外，技术功和工质对机器作功当忽略动能差和位能差后在数量上是相同的。nullSolution:Given: a gas is compressed in a piston-cylinder device, with friction occurring at piston-cylinder interface. The systems of interest and known data are shown in the fig. right.Find: Wcom , WrodModel: Quasi-equilibrium process， frictional force and atmospheric pressure are constant. Closed system.Strategy: according the definition of work to evaluate the work transfer by the piston. nullAnalysis: 1)compressing work of gas ; 2)work supplied by the rod.1) compressing work of gas as definition:The work transferred direction is into the system2)work supplied by the rod as definition of forces equilibrium:nullRearrangemented as:Work:The work transferred to gas is －300 J, the work transferred from atmosphere is 60 J, and the work against friction is －12 J.Comments:the work transferred of 252 J is supplied by the piston rod.wihout any friction, only 240 J of piston work would be required to compress the gas.the work of the atmosphere helps reduce the work required(60 J).nullExample2.4: Air initially at 0.1 MPa and 290 K is compressed in steady state to 0.5 MPa and 450 K. The power input to the air under steady-flow conditions is 5kW, and a heat loss of 5 kJ/kg occurs during the process. If the changes in potential and kinetic energies are neglected, determine the mass flow rate. (the enthalpies for air at 290 K and 450 K are 290.2 and 451.8 kJ/kg respectively)Solutions:Find: Mass flow rateModel:steady state flow neglect Ek and Ep nullStrategy: Known-initial and final states, q Unknown-mass flow rateUsing: mass and energy balance equationsAnalysis:Mass eq.Energy eq.Comment: Note here q is 3% less then enthalpy changenullExample 2.5:A small nuclear reactor is cooled by passing liquid sodium（钠） through it. The liquid sodium leaves the reactor at 0.2 MPa and 400℃. It is cooled to 320 ℃ by passing through a heat exchanger before returning to the reactor. In the heat exchanger, heat is transferred from the liquid sodium to water, which enters the exchanger at 10 MPa and 49 ℃, and leaves at the same pressure as a saturated vapor. The mass flow rate of sodium is 10,000kg/h, its specific heat is constant at 1.25kJ/kg·K, and its pressure drop is negligible. Determine (a) the mass flow rate of water evaporated in the heat exchanger, and (b) the heat transfer rate between the two fluids. Neglect kinetic and potential energy changes.nullSolution:Given: shown in fig.Find: mass flow rate of water or vapor; heat transfer rateModel: steady state; incompressible ;W=0 Ek=0 Δp=0Strategy: energy balanceAnalysis:Sodium enthalpy change:nullEnthalpy of water in: Hwater,1=176.4kJ/kgEnthalpy of vapor out: Hvap.,2=2724.7kJ/kgBalance equation:null例题2.6解：模型：取容器内空间为开口系统方法：能量守恒，质量守恒null假设：充气过程很快，以至于过程中容器内气体与环境之间的传热可以忽略；刚性容器，内部气体对外作功为0null评述：假如充气过程非常缓慢，结果又如何？充气完毕后，容器内气体的温度比充气管道内气体温度高，如何解释？此时，容器内气体温度应该与环境的温度相同，如果环境温度为30℃，那么null假如有一容器，内部气体通过打开放气阀放气，应该如何分析计算？上述例题中，均出现焓（或热力学能）的确定问题，确定的原理是什么？这涉及到：null习题：P58~60 2-12 2-13 2-18 2-19 2-21课下:阅读:第二章内容书写作业：思考题: P56 2－6 2-10 2-11 2-13预习：（3－1、3－2、3－3） 问题：1）比热与过程有关吗？用什么原理得出你的结论？2）给定状态1和2的焓值和热力学能值，如何计算平均比定容热容和比定压热容？