第四讲热力学第一定律进一步的例题

1、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 pisto

2、n to the gas and the work supplied by the connecting rod.Solution: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 expansion frict

3、ional force and atmospheric pressure are constant. Closed system.Strategy: according the definition of work to evaluate the work transfer by the piston. Analysis: 1)compressing work of gas ; 2)work supplied by the rod.1) compressing work of gas as definition:31221cm )1000400(MPa 5 . 0)(VVppdVWJ 300c

4、m 10m 1MPa 1Pa 10cmMPa 300 36363The work transferred direction is into the system2)work supplied by the rod as definition of forces equilibrium:rodatmfrigasFFFFRearrangement as:ApFpAFFFFatmfriatmfrigasrodWork:mN 252101001060042004621AVVFhFWrodrodrodThe work transferred to gas is -300 J, the work tra

5、nsferred from atmosphere is 60 J, and the work against friction is -12 J.Comments:1) the work transferred of 252 J is supplied by the piston rod.2)wihout any friction, only 240 J of piston work would be required to compress the gas.3)the work of the atmosphere helps reduce the work required(60 J).N

6、4200 20010) 1 . 05 . 0(10100200)(64atmppAExample2.4: 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 en

7、ergies 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:111,Tpm q222,Tpm Wsteady state flowneglect Ek and Ep Strategy: Known-initial and final states, q Unknown-mass flow rateUsing: mass

8、 and energy balance equationAnalysis:Mass eq.mmm21021hmhmWqmEnergy eq.kg/s 0.030kJ/kg )2 .2908 .451(kJ/kg 5kJ/s 5)(12hhqWmComment: Note here q is 3% less then enthalpy changeExample 2.5:A small nuclear reactor is cooled by passing liquid sodium（钠）（钠） through it. The liquid sodium leaves the reactor

9、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 r

10、ate of sodium is 10,000kg/h, its specific heat is constant at 1.25kJ/kgK, 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 tow fluids. Neglect kinetic and potential energy changes.Solution:Give

11、n: shown in fig.Find: mass flow rate of water or vapor; heat transfer rateModel: steady state; incompressible ;W=0 Ek=0 p=0Strategy: energy balanceAnalysis:C4910MPa water10MPa vaporsaturatedC4000.2MPa sodium liquid C320QQkg/h000,10 KkJ/kg25. 1 sodium mcSodium enthalpy change:kJ/h 1010)400320(25. 110

12、000)(512.1.,2.,TTcmHHsodsodsodsodEnthalpy of water in: Hwater,1=176.4kJ/kgEnthalpy of vapor out: Hvap.,2=2724.7kJ/kg2.,2.,1 ,1 ,sodvapsodwaterHHHHkJ/h 101051.,2.,1 ,2.,sodsodwatervapHHHHkg/h 392.4kJ/kg )7 .2724(176.4kJ/h1010 5.vapm kJ/h 1010)4 .1767 .2724(4 .392)(51.,2.,.watvapvaphhmQ第第3章章 理想气体的性质理想

13、气体的性质n3.1 物质的三态及相变过程物质的三态及相变过程 热力学面由六个面（固、液、气；固热力学面由六个面（固、液、气；固液、固液、固气、气、液液气），一条线（固气），一条线（固液液气三相）组成气三相）组成 在三相线上，体积变化，表明气固液的比例不同，但温在三相线上，体积变化，表明气固液的比例不同，但温度和压力是不变的，换句话说，三相点的温度与外界压度和压力是不变的，换句话说，三相点的温度与外界压力无关。力无关。 相变：物态由固态转变为液态或气态，或相反的过程相变：物态由固态转变为液态或气态，或相反的过程 潜热：物质在相变过程中放出或吸收的热量潜热：物质在相变过程中放出或吸收的热量 汽化潜

14、热：液体在汽化过程中放出的热量汽化潜热：液体在汽化过程中放出的热量 融化热：固体在融化过程中吸收的热量。融化热：固体在融化过程中吸收的热量。 升华：固体直接汽化的过程升华：固体直接汽化的过程 凝华：气体直接转变为固体的过程凝华：气体直接转变为固体的过程 相变过程中可以放出或吸收大量的热量；相变过程中可以放出或吸收大量的热量； 相变过程中，温度变化与否取决于该过程的压相变过程中，温度变化与否取决于该过程的压力是否变化，当压力不变时，温度也不变。力是否变化，当压力不变时，温度也不变。 对比：潜热的数值比显热大对比：潜热的数值比显热大 本节学习目的： 理想气体状态方程 比热容 热力学能 焓及熵的计算

15、 模型假定： 1） 理想气体分子为弹性小球且不占居体积的质点； 2）分子之间无作用力，分子两次碰撞之间为直线运动， 碰撞过程为弹性碰撞且无能量损失 理想气体是一种实际上并不存在的假象气体 当气体状态处于温度比较高、压力比较低的范围内时，气体一般可以处理成为理想气体 不符合上述假设的气体称为实际气体 实际气体特点： 1）临界温度较高 2）内聚力较大 实际气体状态参数间的关系繁杂Ax理想气体压强推导分子作用于面积分子作用于面积A上的作上的作用力（总压强）由两部分组用力（总压强）由两部分组成，分子力和分子运动，由成，分子力和分子运动，由模型，分子力可以忽略模型，分子力可以忽略 单个分子：作用于面积单

16、个分子：作用于面积A 上的动量为上的动量为 mvi 传递给传递给A的动量的动量 PinivixtAmvix 力的定义：力的定义：Fipi /t 大量分子对壁面产生的平均效果是均匀而持续的压力大量分子对壁面产生的平均效果是均匀而持续的压力 piFi/Ani m vix2 iiximvnp2)(iiximvnp2)(由左向右的作用由左向右的作用由右向左的作用由右向左的作用两者合成两者合成 iiximvnppp2)()(2xvnmp 231vnmp nKTp32nRTpV KTvm221微观方法：微观方法：实验方法：实验方法：通用气体常数通用气体常数 R8.314J/molK气体常数气体常数Rg与通

17、用气体常数的关系与通用气体常数的关系MRRg 对对n mol气体气体 pVnRT对对m kg气体气体 pVmRgT对对1 kg气体气体 pvRgT标准状态标准状态：p0101325 Pa；T0273.15 K；Vm22.414103 m3/mol例题例题3.1:已知体积为已知体积为0.03m3的钢瓶内装有氧气，初始的钢瓶内装有氧气，初始压力压力p17105Pa，温度，温度t1=20。因泄漏，后压。因泄漏，后压力降至力降至p24.9105Pa ，温度未变。问漏去多少氧，温度未变。问漏去多少氧气？气？ 解：取钢瓶的容积为系统（控制容积），泄漏过程看成是解：取钢瓶的容积为系统（控制容积），泄漏过程看

18、成是一个相当缓慢的过程，以至于过程中钢瓶中气体于外界一个相当缓慢的过程，以至于过程中钢瓶中气体于外界之间有充分的传热，使之温度不变。初终态均已知。假之间有充分的传热，使之温度不变。初终态均已知。假定瓶内氧气为理想气体定瓶内氧气为理想气体 111TRVpmg122TRVpmg0827. 0)20273(32/831403. 010)9 . 47()(512121TRVppmmmgkg 例题例题3.2:已知空气在下表所列的温度和压力下测得的比已知空气在下表所列的温度和压力下测得的比体积，试用理想气体状态方程计算表中给定条件下的比体积，试用理想气体状态方程计算表中给定条件下的比体积。体积。 T，Kp，atm实测值v，m3/kg计算值v，m3/kg相对误差3000.