高等工程热力学作业
高等工程热力学试题-(07-1-17)[1].
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姓名： ；学号：高等工程热力学试题 (2006.9～2007.1研)一、 基本概念题(共52分＋4分)1、 简述热力学研究的基本内容和建立经典热力学体系的基本研究路线是什么？(3分、3分)2、 对于以下各例，说明所规定的系统作的功、热、内能、熵的变化（a ）一质量为M 重物（系统）在真空重力场中下落过程。
(2分)(b) 一木质雪橇（系统）从无摩擦的水平冰层滑到粗糙金属板上而停止（雪橇为刚性绝热材料制成，取地面为参考坐标）。
(4分)3、 写出特征函数焓h 的基本表达式和熵函数()v T s ,的微分方程式。
(2分、3分)4、 写出利用饱和液体焓的通用对应比方程()r r T f h ∆='∆和已知物性的A 物质(例如R22)，求温度为B物质(例如R134a)在温度2T 的比焓值的步骤，并简要说明正确选择通用对比参数的方法。
(6分)5、 简述偏摩尔参数与化学势的异同点，以及偏摩尔参数与偏离函数的区别(3分、3分)6、 说明熵变、熵增、熵流、熵源强度的概念及其区别 。
(5分)7、 写出太阳辐射能的总体和光谱辐射能的有效能表达式。
(6分)试说明光谱等效温度与黑体温度本质的异同点(加分：4分)8、 给出理论最高燃烧温度和平衡火焰温度的定义，需要什么方程或条件才可求出平衡火焰温度。
(6分)9、 以磁系统的热力能()H S U ˆ,函数的基本微分方程为例，说明处理特殊系统热力系统的关键点是什么？(6分)二、 已知1mol 液态苯的在标准大气压P 0下凝固点Tm ＝268.2K ，凝固热9874S =∆h J/mol ，熔点K 7.278=f T ，熔解热 9916=f h J/mol ， 固体和液体比热分别为61.122=ps c J/(K ·mol) 和80.126=pf c J/(K ·mol) 。
求熔化时熵变，并判断熔化过程能否自发进行。
(12分)三、 请把下面的T-s 图转为流程图，并说明各部件的名称，标识热、功的进出。
高等工程热力学试卷和答案
1True or false(20pts)1) Mass can cross the system boundary in the Open system .(T)2) Heat and work can not cross in the Closed system. (F)3)Intensive property is a property that is independent of the extent or mass of the system.(T)4)Heat exchangers are devices that transfer energy between fluids at different temperatures.(T)5) Heat engine can have a thermal efficiency of 100%. (F)6)Heat Engines are Devices that are used to convert heat to work.(T)7)The decrease in quality is always accompanied by an increase in entropy.(T)8)Reversible processes actually can occur in nature.F)9)Energy can be neither created or destroyed;it can only change forms.（T）10)Both Heat and Work are associated with a process and a state.(F)11)Once the temperature equality is established,the heat transfer stops.(T)12)The direction of heat transfer is always from the lower temperature body to the higher temperature.body.(F)13)Work is not a property of a system or its surroundings.(T)14)Closed Feedwater Heaters require a separate pump for each heater.(F)15)The reheat process in general can significantly change the cycle efficiency.(T)2Chosing a correct answer(30pts)1) A thermodynamic system generally has a boundary, this boundary （b）a)can be personally defined from your own interestsb)can only be defined following certain rolesc)is depend on the physical boundariesd)is independent on the physical as well as personal wills2) A system is called a closed system when （c）a)there is only heat exchange between the system and its surroundingsb)there is only work interactions between the system and its surroundingsc)there are both heat exchange and work interactoins between the system and its surroundingsd)there are neither heat exchange nor work interactions between the system and its surroundings3)The specific internal energy,u,other than U,of a system is determined by the（c）of the systema) volume b) pressure c) temperature d) mass4) An example of internal property of a system is （c？）a)mass b)total volume c)pressure d)enthalpy5) A steady state process is related to （a ？）a) a closed system with invarible propertiesb) a control volume with its property changesc) an open system keeping its properties constantd) a closed system having its properties changed6) A good approximation for a stream flowing through a porous plug is （a ）a) h in = h out b) P in = P out c)T in = T out d) a and b e) b and c7) Δ T= 0 is true （d ）a) for an adiabatic process. b) if no work is done.c) for an isentropic process. d) for a reversible process.8) Characteristics of Heat Engines include （e ）a) They receive heat from a high-temperature source.1b)They convert part of this heat to work.c)They reject the remaining waste heat to low-temperature sink.d) They operate on a cycle.e)all of above9) Quality （a）a) is defined by M /M + M ) liq liq vapb) is defined by v = x v + (1-x) v f gc)equals zero for a saturated vapor.d)is only defined in the two phase region.e)is the fraction of mass that is liquid.10) A good approximation for a stream flowing through a heat exchanger is （e？）a) h = h in out b) P = P in out c) T = T in outd) a and b e) b and c11)In an adiabatic process:（b）a)temperature cannot change.b)there is no heat flow.c)internal energy cannot change.d) a and be)a,b and c12) Which of the following is an extensive property ? （b）a)temperature b)weight c)compositiond)pressure e)none of the above13) What is the efficiency of an ideal Carnot heat engine cycle operating between 38 C and 482 C ? （c）a) 90 % b) 89 % c) 59 % d) 16 % e) 11 %14) Which of the following does not have units of energy? （d）a) N-m b) kw-hr c) Pa-m3 d) All have units of energy15)Extensive properties（d）a)apply at a point.b) do not depend on the size of the system.c)depend on position within a system.d)are proportional to the mass of a system.e)are printed in the tables in the back of your book.3Provide definition for the following terms(10pts)1)Internal energy:Internal energy is really the kinetic and potential energy of the atoms in the molecules.2)Heat:Energy crossing the system boundary,not associated with mass,due to a temperature difference.3)Work:Work is the energy transfer associated with a force acting through a distance.4) Closed system: No mass crosses the boundary. Heat and work can cross.5) Open system: Mass can cross the system boundary.4Answering and calculations(40pts)1) Feedwater Heater:Inlet 1 T1 = 200 ºC, p1 = 700 kPa, m1 40 kg/s2Inlet 2 T2 = 40 ºC, p2 = 700 kPa,Exit sat. liquid, p3 = 700 kPa,Find m 2 ?, m 3 ? andV 2 ? A2 = 25 cm2 Steady State m i m e m 1 m 2 m 3AVvInlet 2: compressed liquidTable A-4, v2 = 0.001008 m3/kgExit: saturated liquid Table A-5, v3 = 0.001108 m3/kgm 3 (AV)3v 3 0.060.00110854.15 kg/s m 2 m 3 m 1= 54.15 – 40 = 14.15 kg/sV m 2v 2 (14.15)(0.001008) 5.7 m/s2) A rigid tank contains 10 kg of water at 90 ºC. If 8 kg of the water is in the liquid form and the rest is in the vapor form. Determine the pressure in the tank and the volume of the tank . Table A-4, Psat = 70.14 kPa, P = Psat = 70.14 kPaTable A-4, vf = 0.001036 m3/kg, vg = 2.361 m3/kg v = vf + x(vg -vf) = 0.001036 + 0.2(2.361 – 0.001036) = 0.473 m3/kg V = mv = 10(0.473) = 4.73 m33)Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output and the thermal efficiency. W Q H Q L 80 50 30 MWQ H 80 MW, Q L 50 MW0.375Q H 80 m A V 2 A 0.00252(AV)3 0.06 m 3 /sW304) A heat engine receives 600 kJ of heat from a high-temperature source at 1000 K during a cycle. It converts150kJ of this heat to work and rejects the remaining450kJ to a low-temperature sink at300 K. Determine if this heat engine violates the 2nd law of thermodynamics on the basis of(a)the Clausius inequality.(b)the Carnot principle.(a)Clausius inequalityQ QH QL600450T T T10003003H L= - 0.9 kJ/K < 0(b) Carnot principleth 1 Q Q H L 1 0.25T 300ηth < ηrev5)A heat pump maintains a house at a specified temperature. The rate of heat loss of the house and the power consumption of the heat pump are given. The exterior temperature outside of the house is T L , 20C. The interior temperature e inside of the house is T H ,220C. The rate of heat loss of the house is Q H , 110.000kj/h. It is to be determined if this heat pump can do thejob.Assumptions: The heat pump operates steadily.Analysis: The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only, and is determined fromCOP HPjev =11（T /T ）LH = 11（2273K ）/(22273K ) 14.75The required power input to this reversible heat pump is determined from the definition of the coefficient of performance to beW net.jev.min= Q H = 110.000kj /h ( 1h ) =2.07kwThis heat pump is powerful enough since 8 kw >2.07 kw.6) 0.5 kg of air undergoes a Carnot cycle with η = 0.5.Given the initial pressure p1 = 700 kPa, initial volume V1 = 0.12 m3 and heat transfer during the isothermal expansion process Q12 = 40 kJ, Find the highest and the lowest temperatures in the cycle.p 1V 1 (700)(0.12)mR (0.5)(0.287)T H T 1 585.4 K H COP P 14.75 3600srev 1 T L 1 10000.7 Hth T T Q T，T L = 292.7 K4 H1 L 1 L Q H T H 1 1 0.5 0.5 L。
工程热力学大作业
1、根据班级序号自己计算参数利用通用压缩因子图确定氧气在温度为313K（113+10×20号），比体积为0.0074m3/kg时的压力。
2、工程热力学中为什么要引入“可逆过程”?3、以空调制热（班级序号为双号）为例，画出工作原理图、工热关系图，指出其中的代价、收益和经济性指标分别是什么。
4、在热力学发展历史中，有哪些科学家做出贡献？至少列出5人及其成就。
答：（1）J.R.von迈尔：他提出了能量守恒理论，认定热是能的一种形式，可与机械能互相转化，并且从空气的定压比热容与定容比热容之差计算出热功当量。
（2）焦耳：他的实验结果已使科学界彻底抛弃了热质说，公认能量守恒、而且能的形式可以互换的热力学第一定律为客观的自然规律。
热力学的形成与当时的生产实践迫切要求寻找合理的大型、高效热机有关。
（3）法国人S.卡诺：提出著名的卡诺定理，指明工作在给定温度范围的热机所能达到的效率极限（4）开尔文(即W.汤姆森)根据卡诺定理制定了热力学温标。
（5）克劳修斯根据卡诺定理提出并发展了熵5、为什么要引入“焓”和“熵”的概念？它们是如何定义的？答：“焓”：研究流动能量方程中，为了工程应用的方便，才引入焓。
（因为在流动过程中，工质携带的能量除热力学能之外，总伴有推动功，所以为了工程应用的方便起见，把U和PV组合起来，引入焓。
）“熵”6、下面的图是如何得到的？有何作用？7、一台清水离心泵，若泵中压力最低的点为K点，那么K点的压力与此泵工作时所处温度对应的饱和蒸汽压之间满足何种大小关系时，离心泵中压力最低点的水会汽化为气泡？8、一压缩机将1kg温度为T1、压力为p1的空气压缩到温度T2和压力p2，已知压缩过程是多变过程，多变指数为m。
（1）完整推导出T2与T1、p1、p2和m之间的关系式。
（2）空气在压缩机中可以分为三个阶段：首先空气通过吸气阀进入压缩机，此时吸气阀开启、排气阀关闭；接着吸排气阀均关闭，空气被压缩；最后排气阀开启，此时吸气阀仍关闭，空气通过排气阀排出压缩机。
高等工程热力学习题
气，每秒钟的抽气容积为 0.00025m3，假定在抽气过程中容器中气体的温度保持 300K 不变。试确定：
（a） 需要多少时间能使容器中压力降低到 0.35bar；
（b） 容器与环境之间的传热量；
（c） 被抽出氮的质量。
3-2 条件如上题，但进行的是绝热抽气，仍使压力降低到 0.35bar，问需要多少抽气时间，并求抽出
(d) df = −(βpv + s)dT + μpvdp
(e)
dg
=
(
βv μ
−
s)dT
−
dv μ
4-2
式中，β和μ分别是（定压）容积膨胀系数和定温压缩系数，其定义为：
β = 1 ⎜⎛ ∂v ⎟⎞ v ⎝ ∂T ⎠ p
μ
=
-
1 v
⎜⎜⎝⎛
∂v ∂p
⎟⎟⎠⎞T
6-1 试推证伯特洛（Berthelot）方程 的维里展开式为，
其中
β
=
B0 RT
−
A0
−
Rc T2
γ
=
−B0 RTb +
A0a −
B0 Rc T2
δ = B0 Rbc T2
而，
R=208.14 J/kg/K a=5.8390×10-4 m3/kg b=0.0 m3/kg
A0=81.99 m4/kg2 B0=9.8451×10-4 m3/kg c=1498.0m3K3/kg 。
6-9 试编制用贝蒂-布里奇曼状态方程计算流体 p-v-T 关系的通用程序（方程中的有关参数按蒋
汉文与邱信立编著的《热力学原理与应用》表 6-2 确定），并计算 215K 、3.7kg 的 CO 在 0.03m3
的刚性容器中所具有的压力。如果是氩气，则结果又会如何？
工程热力学第三章热力学第一定律作业
第3章 热力学第一定律3-1 一辆汽车 1 小时消耗汽油 34.1 升, 已知汽油发热量为 44000kJ/kg , 汽油密度 0.75g/cm3 。
测得该车通过车轮出的功率为 64kW, 试求汽车通过排气, 水箱散热等各种途径所放出的热量。
解: 汽油总发热量Q = 34.1×10-3m3 ×750kg/m3 ×44000kJ/kg =1125300kJ汽车散发热量Qout = Q-W ×3600 = (1125300-64×3600)kJ/h = 894900kJ/h3-2 气体某一过程中吸收了 50J 的热量, 同时, 热力学能增加 84J, 问此过程是膨胀过程还是压缩过程？ 对外作功是多少 J ？解 取气体为系统, 据闭口系能量方程式 Q = ΔU +WW = Q -ΔU = 50J -84J = -34J所以过程是压缩过程, 外界对气体作功 34J 。
3-3 1kg 氧气置于图 3-1 所示气缸内, 缸壁能充分导热, 且活塞与缸壁无磨擦。
初始时氧气压力为 0.5MPa, 温度为 27℃, 若气缸长度 2l , 活塞质量为 10kg 。
试计算拔除钉后, 活塞可能达到最大速度。
解：由于可逆过程对外界作功最大, 故按可逆定温膨胀计算：w = RgT ln V2/ V1 = 0.26kJ/(kg•K)×(273.15+ 27)K图3-1 图3-2×ln（A×2h）/ （A×h）= 54.09kJ/kgW =W0 + m'/2*Δc2 = p0(V2 -V1)+ m'/2*Δc2 （a）V1 =m1RgT1/ p1=1kg×260J/(kg•K)×300.15K/0.5×106Pa= 0.1561m3V2 = 2V1 = 0.3122m3代入(a)c2 = （2×(54.09J/kg×1kg×103-0.1×106Pa×0.1561m3)/10kg）1/2 = 87.7m/s3-4 有一飞机的弹射装置, 如图 3-2, 在气缸内装有压缩空气, 初始体积为 0.28m3 , 终了体积为0.99m3, 飞机的发射速度为61m/s, 活塞、连杆和飞机的总质量为 2722kg。
工程热力学习题及答案
工程热力学习题及答案
工程热力学学习题及答案
热力学是工程学习中的重要一环，它涉及到能量转化、热力循环等方面的知识。
在学习热力学的过程中，我们常常会遇到各种各样的学习题，下面就来看一些
典型的热力学学习题及答案。
1. 问题：一个理想气体在等压过程中，从初始状态到终了状态，其内能增加了
多少？
答案：在等压过程中，内能的增加量等于热量的增加量，即ΔU = q。
因此，
内能增加量等于所吸收的热量。
2. 问题：一个气缸中的气体经历了一个等温过程，温度为300K，初始体积为
1m³，末了体积为2m³，求气体对外界所做的功。
答案：在等温过程中，气体对外界所做的功等于PΔV，即气体的压强乘以体
积的变化量。
因此，气体对外界所做的功为PΔV = nRTln(V₂/V₁)。
3. 问题：一个理想气体经历了一个绝热过程，初始温度为400K，初始体积为
1m³，末了体积为0.5m³，求末了温度。
答案：在绝热过程中，气体的内能保持不变，即ΔU = 0。
根据理想气体的状
态方程PV = nRT，我们可以得到P₁V₁^γ = P₂V₂^γ，其中γ为绝热指数。
利用这个关系式，可以求得末了温度。
通过以上几个典型的热力学学习题及答案，我们可以看到热力学知识的应用和
计算是非常重要的。
只有通过不断的练习和思考，我们才能更好地掌握热力学
的知识，为今后的工程实践打下坚实的基础。
希望大家在学习热力学的过程中
能够勤加练习，不断提高自己的能力。
高等工程热力学试卷
一、单选题。
（每题2分，共16分）1.若已知工质的绝对压力P=0.18MPa，环境压力Pa=0.1MPa，则测得的压差为( )A.真空p v=0.08MpaB.表压力p g=0.08MPaC.真空p v=0.28MpaD.表压力p g=0.28MPa2.简单可压缩热力系的准平衡过程中工质压力降低，则( )A.技术功为正B.技术功为负C.体积功为正D.体积功为负3.理想气体可逆定温过程的特点是( )A.q=0B. W t=WC. W t>WD. W t<W4.若从某一初态经可逆与不可逆两条途径到达同一终态，则不可逆途径的△S必( )可逆过程△S。
A.大于B.等于C.小于D.大于等于5.饱和蒸汽经历绝热节流过程后( )A.变为饱和水B.变为湿蒸汽C.作功能力减小D.温度不变6.绝热节流过程中节流前、后稳定截面处的流体( )A.焓值增加B.焓值减少C.熵增加D.熵减少7.空气在渐缩喷管内可逆绝热稳定流动，其滞止压力为0.8MPa，喷管后的压力为0.2MPa，若喷管因出口磨损截去一段，则喷管出口空气的参数变化为( )A.流速不变，流量不变B.流速降低，流量减小C.流速不变，流量增大D.流速降低，流量不变8.把同样数量的气体由同一初态压缩到相同的终压，经( )过程气体终温最高。
A.绝热压缩B.定温压缩C.多变压缩D.多级压缩二、填空题。
（每空2分，共30分）1、温度是衡量_____________的一个物理量，热力学温标的计量单位是_________。
2、物系的温度是用以判别物系间是否处于_________状态的参数。
3、理想气体的逸度等于______，逸度和逸度系数都是物质_______的函数。
4、压缩因子、偏差函数和逸度系数等表示实际流体对理想气体的____________。
5、热力学能和体积不变的简单封闭系统就将等同于________系统。
6、__________称为克劳修斯不等式，对______循环，上式等号成立。
工程热力学作业
1-1 一立方形刚性容器，每边长1m ，将其中气体的压力抽至1000Pa ，问其真空度为多少毫米汞柱？容器每面受力多少牛顿？已知大气压力为0.1MPa 。
解：p = 1 000 Pa = 0.001 MPa真空度mmHg Pa MPa MPa MPa p p p b V 56.74299000099.0001.01.0===-=-= 容器每面受力F ＝p V A = 9 900 Pa×1m 2 =9.9×104 N1-2 试确定表压力为0.01 MPa 时U 形管压力计中液柱的高度差。
(1)U 形管中装水，其密度为1 000 kg/m 3；(2) U 形管中装酒精，其密度为789 kg/m 3。
解： 因为表压力可以表示为p g =ρgΔz ，所以有gp z gρ=∆既有（1）mm m s m m kg Pa g p z g72.101901972.1/80665.9/10001001.0236==⨯⨯=∆＝水ρ（2） mm m sm m kg Pag p z g34.129729734.1/80665.9/7891001.0236==⨯⨯=∆＝酒精ρ 1-7 从工程单位制热力性质查得，水蒸气在500℃、100at 时的比体积和比焓分别为v =0.03347m 3/kg 、h =806.6kcal/kg 。
在国际单位制中，这时水蒸气的压力和比热力学能各为多少？解： 水蒸气压力p =100at×9.80665×104Pa/at = 9.80665×106Pa=9.80665MPa 比热力学能u=h-pv=806.6kcal ×4.1868kJ/kcal)/kg-9806.65kPa ×0.03347m 3/kg = 3377.073kJ-328.228kJ =3048.845kJ2-1 冬季，工厂某车间要使室内维持一适宜温度。
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高等工程热力学作业（编程）第三章实际气体状态方程第四章实际气体导出热力学性质与过程题目：一、用PR方程计算制冷剂R290、R600a和混合制冷剂R290/R600a：50/50wt%的PVT性质。
二、用PR方程计算制冷剂R290、R600a和混合制冷剂R290/R600a的导出热力学性质焓和熵。
源程序：1、牛顿迭代法求Zfunction Z=newton(A,B,Z)err=1e-6;for n=0:1000f=Z^3-(1-B)*Z^2+Z*(A-2*B-3*B^2)-(A*B-B^2-B^3);Z=Z-f/(3*Z^2-2*(1-B)*Z+(A-2*B-3*B^2));if(abs(f)<err)breakendend %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%2、求a、b、Z、v等参数函数function [v,Z,a,b,beta]=vv(p,T)R=8.31451;N1=[44.096 369.89 4.2512 0.1521];N2=[58.122 407.81 3.6290 0.1840];k1=0.37464+1.54226*N1(4)-0.26992*N1(4)^2;alpha1=(1+k1*(1-(T/N1(2))^0.5))^2;a1=0.45724*alpha1*R^2*N1(2)^2/N1(3)/10^6;aa1=0.45724*R^2*N1(2)^2/N1(3)/10^6*2*sqrt(alpha1)*(-k1/(2*sqrt(N1(2)*T))); b1=0.07780*R*N1(2)/N1(3)/10^6;k2=0.37464+1.54226*N2(4)-0.26992*N2(4)^2;alpha2=(1+k2*(1-(T/N2(2))^0.5))^2;a2=0.45724*alpha2*R^2*N2(2)^2/N2(3)/10^6;aa2=0.45724*R^2*N2(2)^2/N2(3)/10^6*2*sqrt(alpha2)*(-k2/(2*sqrt(N2(2)*T))); b2=0.07780*R*N2(2)/N2(3)/10^6;a3=0.25*a1+0.5*(1-0.01)*sqrt(a1*a2)+0.25*a2;aa3=0.25*aa1+0.5*(1-0.01)*1/2/sqrt(a1*a2)*(a1*aa2+a2*aa1)+0.25*aa2;b3=0.5*(b1+b2);a=[a1 a2 a3];b=[b1 b2 b3];beta=[aa1 aa2 aa3];for i=1:3;A(i)=a(i)*p*10^6/(R^2*T^2);B(i)=b(i)*p*10^6/(R*T);Z(i)=newton(A(i),B(i),1);vv(i)=R*T*Z(i)/p/10^6;digits(5);v(i)=vpa(vv(i),5);i=i+1;enda=[a1 a2 a3];b=[b1 b2 b3];beta=[aa1 aa2 aa3];end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%3、余函数法求ar、sr、hrfunction [ar,sr,hr]=as(p,T)[v,Z,a,b,beta]=vv(p,T);R=8.31451;for i=1:3;sr(i)=-R*log((v(i)-b(i))/v(i))+beta(i)/(2*sqrt(2)*b(i))*log((v(i)-0.414*b(i))/(v(i) +2.414*b(i)))-R*log(v(i)/(R*T/p/10^6));ar(i)=R*T*log((v(i)-b(i))/v(i))-a(i)/(2*sqrt(2)*b(i))*log((v(i)-0.414*b(i))/(v(i)+2 .414*b(i)))+R*T*log(v(i)/(R*T/p/10^6));hr(i)=ar(i)+T*sr(i)+R*T*(1-Z(i));endend %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%4求绝对焓熵（以0℃饱和液体为标准)/(1satl ℃,0K kg kJ s ⋅=）function [h,s]=hs(p,T)M1=44.096;M2=58.122;x1=(1/M1)/(1/M1+1/M2);x2=(1/M2)/(1/M1+1/M2);Mm=M1*x1+M2*x2;M=[M1 M2 Mm]; ps=[0.015696 0.32979 0.47446];T0=273.15;R=8.31451;c1=[-95.80 6.945 -3.597*10^(-3) 7.290*10^(-7)];c2=[-23.91 6.605 -3.176*10^(-3) 4.981*10^(-7)];c3=[-64.79 6.798 -3.415*10^(-3) 6.294*10^(-7)];cps1=inline('-95.80./t+6.945-3.597*10^(-3)*t+7.290*10^(-7)*t.^2');cps2=inline('-23.91./t+6.605-3.176*10^(-3)*t+4.981*10^(-7)*t.^2');cps3=inline('-64.79./t+6.798-3.415*10^(-3)*t+6.294*10^(-7)*t.^2');cph1=inline('-95.80+6.945*t-3.597*10^(-3)*t.^2+7.290*10^(-7)*t.^3');cph2=inline('-23.91+6.605*t-3.176*10^(-3)*t.^2+4.981*10^(-7)*t.^3');cph3=inline('-64.79+6.798*t-3.415*10^(-3)*t.^2+6.294*10^(-7)*t.^3');Is1=quad(cps1,273.15,T)/1000;Is2=quad(cps2,273.15,T)/1000;Is3=quad(cps3,273.15,T)/1000;Ih1=quad(cph1,273.15,T)/1000;Ih2=quad(cph2,273.15,T)/1000;Ih3=quad(cph3,273.15,T)/1000;Is=[Is1 Is2 Is3];Ih=[Ih1 Ih2 Ih3];[ar,sr,hr]=as(p,T);for i=1:3[ar1,sr1,hr1]=as(ps(i),T0);ar0(i)=ar1(i);sr0(i)=sr1(i);hr0(i)=hr1(i);s(i)=1*M(i)+sr0(i)+Is(i)*M(i)-R*log(p/ps(i))-sr(i);h(i)=200*M(i)+hr0(i)+Ih(i)*M(i)-hr(i);endend %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5、主程序求v、h、sclearP=input('输入R600a工质压力:P/MPa:\n');T=input('输入R600a工质温度:T/K:\n');[v]=vv(p,T)[h,s]=hs(p,T) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% R290、R600a、R290/R600a的比体积v/(m^3/mol);R290、R600a、R290/R600a的焓h/(J/mol);R290、R600a、R290/R600a的熵s/(J/(mol.K); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%运行结果：单位制：SI第六章气液相平衡题目：试用Peng-Robinson方程计算纯质R290 P-T相图和溶液R290/R600a分别在p=1atm和p=10atm下的T-X相图。
纯质R290 P-T相图：源程序：1、求纯质R290逸度系数函数function [phi1]=phi(T1,P1,Z);R=8.3145;M1=44.096e-3;Tc1=369.89;Pc1=4.2512e6;w1=0.1512;Tr1=T1/Tc1;k1=0.37464+1.54226*w1-0.26992*w1^2;alpha1=(1+k1*(1-Tr1^0.5))^2;a1=0.45724*alpha1*(R^2)*(Tc1^2)/Pc1;b1=0.07780*R*Tc1/Pc1;A1=a1*P1/((R^2)*(T1^2));B1=b1*P1/(R*T1);Z=newton(A1,B1,Z);phi1=exp(Z-1-log(Z-B1)-A1*log((Z+2.414*B1)/(Z-0.414*B1))/(2*sqrt(2)*B1)); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%2、主程序p1=3e5;dp=100;N=20000;err=1e-8;for T1=200:0.1:369.89for n=1:Nphi1v=phi(T1,p1,1.1);phi1L=phi(T1,p1,0.001);if abs(phi1v-phi1L)<=errbreakelsep1=p1+dp;endendif n==N+1fprintf('error!')break;elseplot(T1,p1/10^6,'r-');hold on;endendgrid;title('R290工质p-T相图');xlabel('T/K');ylabel('p/MPa');运行结果：溶液R290/600a分别在P=1atm和10atm下的T-X相图源程序：1、求R290/R600a逸度系数函数function phimix=phimix(type,x1,T,p,Z)R=8.3145;x2=1-x1;N1=[44.096 369.89 4.2512 0.1521];N2=[58.122 407.81 3.6290 0.1840];k1=0.37464+1.54226*N1(4)-0.26992*N1(4)^2;alpha1=(1+k1*(1-(T/N1(2))^0.5))^2;a1=0.45724*alpha1*R^2*N1(2)^2/N1(3)/10^6;b1=0.07780*R*N1(2)/N1(3)/10^6;k2=0.37464+1.54226*N2(4)-0.26992*N2(4)^2;alpha2=(1+k2*(1-(T/N2(2))^0.5))^2;a2=0.45724*alpha2*R^2*N2(2)^2/N2(3)/10^6;b2=0.07780*R*N2(2)/N2(3)/10^6;a=x1*x1*a1+2*x1*x2*(1-0.01)*sqrt(a1*a2)+x2*x2*a2; b=x1*b1+x2*b2;A=a*p*10^6/(R^2*T^2);B=b*p*10^6/(R*T);Z=newton(A,B,Z);if type==1bi=b1;sai=2*(x1*a1+x2*0.99*sqrt(a1*a2));else if type==2bi=b2;sai=2*(x2*a2+x1*0.99*sqrt(a1*a2));endendphimix=exp(bi/b*(Z-1)-log(Z-B)-A*(sai/a-bi/b)*log((Z+2.414*B)/(Z-0.414*B))/(2*sqrt( 2)*B));end%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%2、1atm下R290/R600a的T-x图clearx1=0:0.01:1;x2=1-x1;t=length(x1);y1=x1;y2=1-y1;P=0.1;n=0;for i=1:tT=220;while 1n=n+1;fug1_l=phimix(1,x1(i),T,P,0.01);fug1_v=phimix(1,y1(i),T,P,1.1);fug2_l=phimix(2,x1(i),T,P,0.01);fug2_v=phimix(2,y1(i),T,P,1.1);k1=fug1_l/fug1_v;k2=fug2_l/fug2_v;y1(i)=k1*x1(i);y2(i)=k2*x2(i);sumy=y1(i)+y2(i);sumy1=sumy;if n==1y1(i)=k1*x1(i)/sumy;y2(i)=k2*x2(i)/sumy;fug1_v=phimix(1,y1(i),T,P,1.1); fug2_v=phimix(2,y1(i),T,P,1.1);k1=fug1_l/fug1_v;k2=fug2_l/fug2_v;y1(i)=k1*x1(i);y2(i)=k2*x2(i);sumy=y1(i)+y2(i);endwhile 1if abs((sumy-sumy1)/sumy1)<1e-4breakendsumy1=sumy;y1(i)=k1*x1(i)/sumy;y2(i)=k2*x2(i)/sumy;fug1_v=phimix(1,y1(i),T,P,1.1); fug2_v=phimix(2,y1(i),T,P,1.1);k1=fug1_l/fug1_v;k2=fug2_l/fug2_v;y1(i)=k1*x1(i);y2(i)=k2*x2(i);sumy=y1(i)+y2(i);endif abs(sumy-1)<=1e-4q(i)=T;breakendT=T+0.01;endR(:,i)=[x1(i),y1(i),T];ends=0;for i=1:tif R(3,i)<265s=s+1;L(:,s)=R(:,i);endendL(:,1)=[0,0,261];L(:,s+1)=[1,1,230.61];plot(L(1,:),L(3,:),'r'); hold on; plot(L(2,:),L(3,:));legend('泡点线'，’露点线’);xlabel('R290摩尔分数');ylabel('混合工质温度/K');title('p=1atm下，R290/R600a的T-x图');grid on %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 3、10atm下R290/R600a的T-x图clearx1=0:0.01:1;x2=1-x1;t=length(x1);y1=x1;y2=1-y1;P=1;n=0;for i=1:tT=290;while 1n=n+1;fug1_l=phimix(1,x1(i),T,P,0.01);fug1_v=phimix(1,y1(i),T,P,1.1);fug2_l=phimix(2,x1(i),T,P,0.01);fug2_v=phimix(2,y1(i),T,P,1.1);k1=fug1_l/fug1_v;k2=fug2_l/fug2_v;y1(i)=k1*x1(i);y2(i)=k2*x2(i);sumy=y1(i)+y2(i);sumy1=sumy;if n==1y1(i)=k1*x1(i)/sumy;y2(i)=k2*x2(i)/sumy;fug1_v=phimix(1,y1(i),T,P,1.1); fug2_v=phimix(2,y1(i),T,P,1.1);k1=fug1_l/fug1_v;k2=fug2_l/fug2_v;y1(i)=k1*x1(i);y2(i)=k2*x2(i);sumy=y1(i)+y2(i);endwhile 1if abs((sumy-sumy1)/sumy1)<1e-4breakendsumy1=sumy;y1(i)=k1*x1(i)/sumy;y2(i)=k2*x2(i)/sumy;fug1_v=phimix(1,y1(i),T,P,1.1); fug2_v=phimix(2,y1(i),T,P,1.1);k1=fug1_l/fug1_v;k2=fug2_l/fug2_v;y1(i)=k1*x1(i);y2(i)=k2*x2(i);sumy=y1(i)+y2(i);endif abs(sumy-1)<=1e-4q(i)=T;breakendT=T+0.01;endR(:,i)=[x1(i),y1(i),T];endplot(R(1,:),R(3,:),'r'); hold on;plot(R(2,:),R(3,:));legend('泡点线'，’露点线’);xlabel('R290摩尔分数');ylabel('混合工质温度/K');title('p=10atm下，R290/R600a的T-x图');grid on %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%运行结果：。