1.Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystemEnergy & Pressure Work BasicsChapter 1Basics Topics Reviewed The topic menu above allows you to move directly to any of the four sections for each topic. The 1. System, State, Process and Cycle sections are: 2. Pressure, Temperature and theCase Intro: To help introduce and understand theZeroth Law of Thermodynamics basic principles, a case study is presented. 3. Heat and WorkTheory: This section will review the basic principles 4. Energy, Specific Heat, and and equations that you should know to answer theEnthalpy exam questions. It does not give detailed derivations of the theory. Case Solution: The case study is solved in detail in this section. Graphics, narrations, animations, and equations are used to help you understand how the problem was solved. Simulation: You can adjust several parameters of a given problem and learn how they affect the results.

2. Ch 1. BasicsMultimedia Engineering ThermodynamicsSystem TemperatureHeat andEnergy& PressureWorkSystem Case IntroTheoryCase Solution SimulationTHERMODYNAMICS - CASE STUDY Introduction Jacks mom bought a new pressure cooker. Jack is interested in the petcock, which is a small piece of mass, sits on top of the only opening in the middle of the lid and prevents steam from escaping until the pressure force overcomes the weight of the petcock. He just wonders the weight of the petcock which can maintain a high pressure inside the cooker. What is known: The operation pressure is 100 kPa gagePressure CookerThe opening cross-sectional area is 4 mm2 Atmospheric pressure is 101 kPa QuestionsWhat is the mass of the petcock? Approach Solve the problem using the basic steps in engineering problem solving: 1) read 2) drawPressure Cookerdiagrams 3) write equations 4) solve and 5)Click here to view movie check solution. For step 3, use the force balance in the base of the petcock: F - ( F a+ G ) = 0 3. Ch 1. Basics Multimedia Engineering ThermodynamicsSystem TemperatureHeat and Energy& PressureWorkSystemCase IntroTheoryCase SolutionSimulationTHERMODYNAMICS - THEORYSystemsAll thermodynamic systems contain three basicelements:System boundary: The imaginary surface that bounds the system.System volume: The volume within the imaginary surface.System and SurroundingsThe surroundings: The surroundings is everything external to the system.Systems can be classified as being closed,open, or isolated. Closed system: Mass cannot cross theboundaries, but energy can. Open system (control volume): Both massand energy can cross the boundaries. Isolated system: Either mass nor energy cancross its boundaries.Closed System Click here to view movie Open System Isolated System Click here to view movieClick here to view movie 4. Ch 1. Basics Multimedia Engineering ThermodynamicsSystem Temperature Heat and Energy& Pressure WorkSystemCase IntroTheoryCase SolutionSimulationTHERMODYNAMICS - THEORY Property, Equilibrium and State A property is any measurable characteristic of a system. The common properties include: pressure (P) temperature (T) volume (V) velocity (v) mass (m) enthalpy (H) entropy (S) Properties can be intensive or extensive. Intensive properties are those whose values are independent of the mass possessed by the system, such as pressure, temperature, and velocity. Two States of a System Extensive properties are those whose values are dependent of the mass possessed by the system, such as volume, enthalpy, and entropy (enthalpy and entropy will be introduced in following sections). Extensive properties are denoted by uppercase letters, such as volume (V), enthalpy (H) and entropy (S). Per unit mass of extensive properties are called specific properties and denoted by lowercase letters. For example, specific volume v = V/m, specific enthalpy h = H/m and specific entropy s = S/m (enthalpy and entropy will be introduced in following sections). Note that work and heat are not properties. They are dependent of the process from one state to another state. 5. Ch 1. BasicsMultimedia Engineering ThermodynamicsSystem Temperature Heat and Energy& Pressure WorkSystemCase IntroTheory Case Solution SimulationTHERMODYNAMICS - THEORY Property, Equilibrium and State (continuation) When the properties of a system are assumed constant from point to point and there is no change over time, the system is in a thermodynamic equilibrium. The state of a system is its condition as described by giving values to its properties at a particular instant. For example, gas is in a tank. At state 1, its its mass is 2 kg, temperature is 20oC, and volume is 1.5 m3. At state 2, its mass is 2 kg, temperature is 25oC, and volume is 2.5 m3. A system is said to be at steady state if none of its properties changes with time. Process, Path and Cycle The changes that a system undergoes from one equilibrium state to another is called a process. The series of states through which a system passes during a process is called path. In thermodynamics the concept of quasi-equilibrium processes is used. It is a sufficiently slow process that allows the system to adjust itself internally so that its properties in one part of the system do not change any faster than those at other parts. When a system in a given initial state experiences a series of quasi-equilibrium processes and returns to the initial state, the system undergoes a cycle. For example, the piston of car engine undergoes Intake stroke, Compression stroke, Combustion stroke, Exhaust stroke and goes back to Intake again. It is a cycle. 6. Ch 1. BasicsMultimedia Engineering ThermodynamicsSystem TemperatureHeat andEnergy& PressureWorkSystem Case Intro Theory Case Solution SimulationTHERMODYNAMICS - CASE STUDY SOLUTION The weight of the petcock Using the basic force-equilibrium equation on the base of the petcock:F - ( G + Fa ) = 0 F is the force given by the pressure inside the cooker and G is the weight of the petcock. Fa is the force acting on the base by the atmosphere. Then the mass can be expressed as PA - ( mg + PaA ) = 0 Force Diagram m = ( P - Pa )(A)/g= PgA/g Put the given values into previous equation yields m = 0.0408 kg The result shows that a small mass of petcock can handle a high pressure inside the cooker. What the Pressure Cooker Tells Us? Before the pressure in the cooker reaches the operation pressure, no mass is transferred between the cooker and the surroundings. So the cooker is a closed system. When the pressure approaches the operation pressure, the petcock moves upward and stream can leak out the cooker. Then the system becomes an open system. Pressure Cooker The atmospheric pressure always exists for Click here to view movieobjects which are explored in the air. Be sure the pressure given is absolute, gage or vacuum.The concept about pressure will be introduced later. 7. Ch 1. Basics Multimedia Engineering ThermodynamicsSystem TemperatureHeat and Energy& PressureWorkSystemCase IntroTheory Case Solution SimulationRun Simulation Suggested Help Questions Why the base area of the peckcock is small? What are the forces acting on the base of the petcock? Technical Help This simulation demonstrates the relation between the operation pressure and the weight of the petcock. Use this slider to change the operation pressure. It is the gage pressure. The range is from 0 to 100 kPa. Use this slider to change the base area of the petcock. The range is from 0 to 10 mm2. This window gives the solution of this problem: the weight of the petcock. 8. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy& Pressure WorkTemperature and PressureCase IntroTheory Case Solution SimulationTHERMODYNAMICS - CASE STUDYIntroductionA nurse is taking care of three children. Sheneeds to measure their temperatures andcompare the temperatures with the standardizedbody temperature to determine if someone got afever.What is known:The standardized body temperature for differentage people is shown in the table. Temperature Taking Age Temperature ( o F) 0 - 3 month 99.4 3 - 6 month 99.5 6 -12 month 99.7 1 - 3 year99.0 3 - 5 year98.6 5 - 9 year98.3Thermometer 9 - 13 year 98.0 > 13 year 97.8 - 99.1The temperature of the three kids are: Age Temperature ( o C) 3 month 37.0 6 month 38.5 3 year37.2QuestionsIs there someone who got a fever?ApproachThe relation between oC, oF and K are:T (K) = T (oC) + 273.15T (oF) = 1.8 T(oC) + 32.0 9. Ch 1. Basics Multimedia Engineering Thermodynamics TemperatureHeat andSystemEnergy& PressureWorkTemperature and Pressure Case Intro TheoryCase SolutionSimulation THERMODYNAMICS - THEORY Units for Mass, Length, Time and Force There are two widely used systems of units: the International System (or Systeme International dUnites in French), S.I.; and the English System. Unit S.I. EnglishLengthmeter (m)foot (ft) The base units in the S.I. system are meters (m) Timesecond (s) second (s) for length, second (s) for time, and kilogram (kg)Masskilogram (kg) slug (slug)for mass. The force unit is derived using NewtonsForcenewton (N) pound (lb) 2nd Law: blue = derived units F = ma = 1 kg (1 m/s2) = 1 kg m/s2 = 1 N The base units in the English system are foot (ft) for length, second (s) for time, and pound-force (lbf) for force. The mass unit is derived using Newtons 2nd Law: m = F/a = 1 lb/(ft/s2) = 1 lb s2/ft= 1 slug = 32.174 lbm The table to the left compares the two systems. All the units in thermodynamics can be derived from these base units. Details of the thermodynamic units will be introduced in the following section. Pressure The absolute pressure (P) is a force acting on a unit area. Definition of PressureIn the SI system, the unit for pressure is Pa,Pascal. In the English system, it is psi.Pa = N/mpsi = lbf/in2Since Pa is a small unit in the SI system, other unitsare also used in thermodynamics, such as: 10. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystemEnergy& Pressure WorkTemperature and PressureCase Intro Case Intro Theory Case Case Solution Theory Solution SimulationTHERMODYNAMICS - THEORY Pressure (continuation)1 bar = 105 Pa1 kPa = 103 Pa1 MPa = 106 Pa1 atm = 101325 Pa The air surrounding the earth can be treated as a homogeneous gas, called atmosphere. Atmospheric pressure (Pa) is the pressure due to the force by the atmosphere mass. Standard atmospheric pressure is Atmosphere101325 Pa. Barometer is a device used to measure the atmospheric pressure. Pa = g h where = The density of the working liquid, kg/m3 g = The acceleration of gravity, 9.8 m/s2 h = The height of the working liquid in Barometer the tube, m Gage pressure (Pg) is the difference between the absolute pressure and the atmospheric pressure if the difference is positive. If the difference is negative, it is called vacuum pressure (Pv).Pg = P - Pa (P > Pa)Pv = Pa- P (P < Pa) Absolute pressure is used in thermodynamicGage Pressure and Vacuum Pressurerelations and tables. 11. Ch 1. BasicsMultimedia Engineering Thermodynamics Temperature Heat andSystem Energy& Pressure WorkTemperature and PressureCase IntroTheoryCase Solution SimulationTHERMODYNAMICS - THEORYPressure (continuation)U-Tube Manometer is used to measure pressuredifference. One end of it is open to theatmosphere and the other end is connected tothe equipment whose pressure is needed to bemeasured.At the right side,P1 = Pgas + gas g h1 U-Tube ManometerUsually, the second term on the right hand of Click here to view moviethe previous equation is negligible since thedensity of the work fluid is much larger than thedensity of the gas. P1 = PgasAt the left side,P1 = Pa + working fluid g h1Combine the two equations above, thepressure in the gas tank can be determined asPgas = Pa + working fluid g h1Temperature and the Zeroth LawThe measurement of the degree of hotness orcoolness is temperature.If two bodies at different temperatures arebrought together, the hot body will warm up thecold one. At the same time, the cold body willcool down the hot one. This process will endwhen the two bodies have the sameThe Zeroth Lawtemperatures. At that point, the two bodies aresaid to have reached thermal equilibrium. 12. Ch 1. BasicsMultimedia Engineering Thermodynamics TemperatureHeat andSystemEnergy& PressureWorkTemperature and PressureCase Intro TheoryCase Solution Simulation THERMODYNAMICS - THEORY Zeroth Law (continuation)The Zeroth Law of thermodynamics states: Two bodies each in thermal equilibrium with a third body will be in thermal equilibrium with each other. The Zeroth Law of thermodynamics is a basis for the validity of temperature measurement. Temperature ScalesTo establish a temperature scale, two fixed, easyduplicated points are used. The intermediatepoints are obtained by dividing the distancebetween into equal subdivisions of the scalelength.Temperature ScaleFixed Point 1Fixed Point 2 Fahrenheit Scale ( o F) Freezing Point of Water = 32.0 Boiling Point of Water = 212.0Celsius Scale ( o C) Freezing Point of Water = 0.0Boiling Point of Water = 100.0 Thermodynamic The pressure of an ideal gas is zero The Triple Point of Water = Temperature Scale (K) = 0.0273.16The relations between the above temperature scalesare: T (K) = T(oC) + 273.15 T (oF) = 1.8T(oC) + 32.0 T (oF) = 1.8 (T(K)-273.15) + 32.0 The thermodynamic temperature scale in the English system is the Rankine scale. The temperature unit on this scale is the rankine, which is designated by R. The thermodynamic Relations between Different temperature scale in S.I. system (K) and EnglishTemperature Scales system (R) are related by T(R) = 1.8 T(K) 13. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystemEnergy& Pressure WorkTemperature and Pressure Case Intro Theory Case Solution SimulationTHERMODYNAMICS - THEORY Thermometers Thermometers measure temperature, by using materials that change in some way when they are heated or cooled.In a mercury or alcohol thermometer the liquid expands as it is heated and contracts when it is cooled, so the length of the liquid column becomes longer or shorter depending on the temperature.ThermometerModern thermometers are calibrated in standardClick here to view movie temperature units such as Fahrenheit or Celsius. Three practical points for using thermometer are:1. The thermometer should be isolated toeverything except the body which temperatureis measured. The general method is toimmerse the thermometer in a hole in a solidbody, or directly in a fluid body.2. When thermal equilibrium is reached, thethermometer can indicate its own temperatureas well as the body measured. TheDigital Thermometer thermometer should be small relative to thebody so that it only has a small effect uponthe body.3. The thermometer must not be subject toeffects such as pressure changes, whichmight change the volume independently oftemperature.Digital thermometers almost replace the mercuryones in nowadays because they are more accuracyand more easy to use. 14. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystemEnergy& Pressure WorkTemperature and Pressure Case IntroTheoryCase Solution SimulationTHERMODYNAMICS - CASE STUDY SOLUTIONThe standardized body temperatures unit isoF but the temperature taken by the nurse isoC. So use the following relations to convertthe standardized body temperature into oC T (K) = 5.0 (T(oF) - 32.0)/9.0 + 273.15 T (oC) = 5.0 (T(oF) - 32.0)/9.0 The standardized body temperature in oF, oC and K: Age Temperature(o F) Temperature(o C) Temperature(K) 0 - 3 month 99.437.4310.5 3 - 6 month 99.537.5310.6 6 -12 month 99.737.6310.7 1 - 3 year99.037.2310.3 3 - 5 year98.637.0310.1 5 - 9 year98.336.8309.9 9 - 13 year 98.036.7309.8 > 13 year 97.8 - 99.1 36.6 - 37.3 309.7 -310.4 Comparing the temperature of each kid to the standardized body temperature in oC shows the 6 month child has a fever.AgeTemperature(oC) standardized(oC) conditions3 month 37.037.56 month 38.537.6 fever3 year37.2 37.2 15. Ch 1. Basics Multimedia Engineering Thermodynamics TemperatureHeat andSystem Energy& PressureWorkTemperature and PressureCase IntroTheory Case SolutionSimulation Run SimulationSuggested Help QuestionsMarys 1 year old girl feels not good. She takes her temperature.It shows the girls temperature is 99.0 oF. Does she run a fever? Technical Help This simulation shows different temperature scales of a mercury thermometer. The red flag indicates the standardized body temperatures for different ages. Use this menu to choose the age of the people whose temperature is taken. Use this menu to choose the temperature scale used. Use this slider to change the temperature value. 16. Ch 1. BasicsMultimedia Engineering Thermodynamics Temperature Heat andSystemEnergy & PressureWork Heat and Work Case IntroTheoryCase SolutionSimulationTHERMODYNAMICS - CASE STUDY Introduction Before ironing, an iron needs to be warmed up to a certain temperature. The iron temperature is different for different fabric. The resistance wires are used to generate the energy needed to warm the iron up. Known: The power of the iron is = 100 W and 85% of the heat generated in the resistance wires is transferred to the iron base plate Iron and Iron Board The materials of the base plate is aluminum alloy 2024-76 ( = 2770 kg/m and Cp= 875 J/(kg- C)) 3o and the base plate thickness = 0.5 cm, and the area is A = 0.02 m2 Initially the iron is in equilibrium with the ambient air at Tinitial = 20 oC The convection heat transfer coefficient h = 35 W/(m2-oC) The emissivity of the base plate to the ambient = 0.6 Questions 1.Determine the minimum time needed for the base plate temperature to reach Tfinal = 140oC. 2.After the temperature reaches 140oC, how much energy is needed to keep the base at 140oC. 17. Ch 1. BasicsMultimedia Engineering Thermodynamics Temperature Heat andSystem Energy & PressureWork Heat and WorkCase IntroTheory Case SolutionSimulationTHERMODYNAMICS - CASE STUDY Approach 1. The heat needed to increase the base plates temperature from Tinital to Tfinal is: where m = mass of the base Cp = specifric heat 2. The energy balance of the base plate (Energy balance will be introduced in the following section):Energy in: 85% of the heat generated bythe wires.Energy out: convection, radiation heat lossto the ambient air.Energy storage: the energy in the baseto increase the bases temperature. Energy Balance Diagram 18. Ch 1. Basics Multimedia Engineering ThermodynamicsTemperatureHeat andSystem Energy& Pressure WorkHeat and Work Case IntroTheory Case SolutionSimulationTHERMODYNAMICS - THEORYPath Function and Point FunctionPath function and Point function are introduced toidentify the variables of thermodynamics. Path function: Their magnitudes depend on the path followed during a process as well as the end states. Work (W), heat (Q) are path functions. Process A: W A = 10 kJ Process B: W B = 7 kJ Point Function: They depend on the state only, and not on how a system reaches that Path Function and Point Functionstate. All properties are point functions. Process A: V2 - V 1 = 3 m3 Process B: V2 - V 1 = 3 m3HeatHeat is energy transferred from one system toanother solely by reason of a temperaturedifference between the systems. Heat exists onlyas it crosses the boundary of a system and thedirection of heat transfer is from higher temperatureto lower temperature.For thermodynamics sign convention, heat Heat Transfer Directiontransferred to a system is positive; Heat transferred Click here to view movie from a system is negative. 19. Ch 1. BasicsMultimedia Engineering Thermodynamics Temperature Heat andSystem Energy & PressureWork Heat and WorkCase Intro TheoryCase Solution SimulationTHERMODYNAMICS - THEORYHeat (continuation)The heat needed to raise a objects temperature fromT1 to T2 is: Q = cp m (T2 - T 1) where cp = specific heat of the object (will be introduced in the following section) m = mass of the objectUnit of heat is the amount of heat required to cause aunit rise in temperature of a unit mass of water atatmospheric pressure. Btu: Raise the temperature of 1 lb of water 1oF Cal: Raise the temperature of 1 gram of water 1 oCJ is the unit for heat in the S.I. unit system. Therelation between Cal and J is 1 Cal = 4.184 JNotation used in this book for heat transfer: Q : total heat transfer: the rate of heat transfer (the amount of heat transferred per unit time) Q: the differential amounts of heat q: heat transfer per unit mass q: heat transfer per unit mass 20. Ch 1. Basics Multimedia Engineering Thermodynamics TemperatureHeat andSystemEnergy & Pressure Work Heat and Work Case IntroTheoryCase Solution SimulationTHERMODYNAMICS - THEORY Modes of Heat Transfer Conduction: Heat transferred between two bodies in direct contact. Fouriers law: Conduction If a bar of length L was put between a hot object TH and a cold object TL , the heat transfer rate is: where kt = Thermal conductivity of the bar A = The area normal to the direction of heattransfer Convection: Heat transfer between a solid surface and an adjacent gas or liquid. It is the combination of conduction and flow motion. Heat transferred from a solid surface to a liquid adjacent is conduction. And then heat is brought away by the flow motion. Newtons law of cooling:Convection where h = Convection heat transfer coefficient Ts = Temperature of the solid surface Tf = Temperature of the fluid The atmospheric air motion is a case of convection. In winter, heat conducted from deep ground to the surface by conduction. The motion of air brings the heat from the ground surface to the high air. 21. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & PressureWork Heat and WorkCase IntroTheoryCase SolutionSimulationTHERMODYNAMICS - THEORY Modes of Heat Transfer (continuation) Radiation: The energy emitted by matter in the form of electromagnetic waves as a result of the changes in the electronic configurations of the atoms or molecules.Stefan - Boltzmann law:where Radiation = Stefan - Boltzmann constant = emissivityTs = Surface temperature of the objectSolar energy applications mainly use radiationenergy from the Sun.The three modes of heat transfer always existsimultaneously. For example, the heat transferassociated with double pane windows are:Conduction: Hotter (cooler) air outside eachpane causes conduction through solid glass.Convection: Air between the panes carries heatfrom hotter pane to cooler pane.Radiation: Sunlight radiation passes throughglass to be absorbed on other side. Double Pane WindowClick here to view moviePlease view heat transfer books for details ofmodes of heat transfer. 22. Ch 1. Basics Multimedia Engineering Thermodynamics TemperatureHeat andSystem Energy & Pressure Work Heat and WorkCase IntroTheoryCase SolutionSimulation THERMODYNAMICS - THEORYWorkWork is the energy transfer associated with aforce acting through a distance.In thermodynamics sign convection, worktransferred out of a system is positive withrespect to that system. Work transferred in isnegative. Definition of Work Click here to view movie Dot product means the distance along the forcesdirection. For example, if a car runs at a flat road,its weight does zero work because the weightand the moving distance have a 90o angle.Like heat, Work is an energy interactionbetween a system and its surroundings andassociated with a process.Units of work is the same as the units of heat.Notation:W : total workW: differential amount of workw: work per unit mass : Power, the work per unit time 23. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & PressureWork Heat and WorkCase Intro TheoryCase SolutionSimulation THERMODYNAMICS - THEORYExpansion and Compression WorkA system without electrical, magnetic, gravitationalmotion and surface tension effects is called a simplecompressible system. Only two properties areneeded to determine a state of a simplecompressible system.Considering the gas enclosed in a piston-cylinderdevice with a cross-sectional area of the piston A.Initial State:Pressure P1Volume V1Expansion and Compression WorkClick here to view movieFinial State:Pressure P2Volume V2 Then a work between initial and final states is: Pressure P, Volume V. Let the piston moving ds in a quasi-equilibrium manner. The differential work done during this process is: W = F ds = P A ds = P dV The total work done during the whole process (from state (P1,V1) to state (P2,V2)) is: This quasi-equilibrium expansion process can be shown on a P-V diagram. The differential area dA is equal to P dV. So the area under the process curve on a P-V diagram is equal, in magnitude, to the work done during a quasi-equilibrium expansion or compression process of a closed system. 24. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & PressureWork Heat and WorkCase IntroTheoryCase SolutionSimulationTHERMODYNAMICS - CASE STUDY SOLUTION A certain time will be needed to warm up the iron to a certain temperature before ironing. The energy generated by the wires partly lost to the ambient by heat transfer. For a given ironing temperature, the warm up time and the quantity of heat loss need to be determined. Energy balance of this warming up process: Ein - E out = Estorage Energy Balance Diagram During this process, the temperature of the base becomes higher than the ambient air temperature. So convection and radiation heat transfer exist. Ein = Egenerate by wires Eout= Econvection + Eradiation1. If all the energy transferred to the base is stored to increase the temperature of the base, the time will be shortest. In another word, assuming no heat loss to the ambient by convection and radiation.Ein - E out = Estorage Estorage = Ein - E outThe heat generated by the wires is: 25. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & PressureWork Heat and WorkCase IntroTheoryCase Solution SimulationTHERMODYNAMICS - CASE STUDY SOLUTION (cont.) Energy stored in the base is: Estorage = cpm (Tfinal - T initial) m = APlug the numbers, the solution for time t is: t = 342 s 2. Consider the convection and radiation heat lossfrom the iron after the temperature reaches 140oC. Because temperature keeps at 140 oC, noenergy is stored.Comparing the heat loss and heat input. 98.8/100 = 0.98 = 98.8%The power of this iron is not enough to keep thetemperature at 140 oC during ironing because theclothes will absorb more heat than the ambient. 26. Ch 1. Basics Multimedia Engineering Thermodynamics TemperatureHeat andSystemEnergy & Pressure Work Heat and WorkCase Intro Theory Case Solution Simulation Run Simulation Suggested Help Questions If a silk shirt needs to be ironed. How long time will it take to warm the iron up? Technical Help This simulation shows at least how long time it will take the iron to warm up to a certain temperature before ironing. Also the heat loss to the ambient air at certain temperature is compared with the electric power of the iron.Use this button to choose thefabric. Default is synthetic. Use this slider to change the convection heat transfer coefficient. The range is from 0 to 100W/(m2-oC). Default is 50 W/(m2-oC). 27. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & PressureWork Heat and WorkCase IntroTheory Case Solution Simulation Simulation (continuation) Use this slider to change the area of the iron base plate. The range is from 0.01 to 0.05 m2. Default is 0.03 m2. Use this slider to change the input power. The range is from 100 to 1600 W. Default is 850 W. Use this slider to change the emissivity of the iron base. The range is from 0 to 1.0. Default is 0.5. This window gives the warm up time in seconds. 28. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & PressureWorkEnergy, Specific Heat & EnthalpyCase IntroTheory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction Bob went home with a red print on his face yesterday. He was involved in a fight in school. Someone slapped his face which caused the temperature of the affected area of his face to rise. How fast the slapping hand is? Known:Bobs Bad Day The mass of the slapping hand Click here to view movie mhand = 1.5 kg The mass of the affected tissue maffected tissue = 0.2 kg The specific heat of the tissue c = 3.8 kJ/(kg-oC) Temperature of face rise 1.5oC QuestionsDetermine the velocity of the hand just before impact. Approach Take the hand and the affected portion of the face as a system. The energy equation: Ein - E out = Esystem = (U + KE + PE)affected tissue + (U + KE + PE)handSystem 29. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystemEnergy & PressureWorkEnergy, Specific Heat & Enthalpy Case IntroTheory Case SolutionSimulation THERMODYNAMICS - THEORY Energy Energy is the capacity for doing work. It may exist in a variety of forms such as thermal, mechanical, kinetic, potential, electric, magnetic, chemical, and nuclear. It may be transferred from one type of energy to another. For example, Heating water by gas: Chemical energy ---> thermal energy Heating water by electricity: electric energy ---> thermal energy Chemical Energy Transfers to Running nuclear power plant:Kinetic Energy in Rocket Nuclear energy ---> electric energy Flying rocket: Chemical energy ---> thermal Energy ---> Kinetic Energy Forms of Energy Kinetic Energy (KE): The energy that a system possesses as a result of its motion.KE = mv2/2 where m = mass of the systemv = velocity of the systemKinetic Energy and If an object of mass m changes velocity from v1 Gravitational Potential Energy to v2. thus the change of its kinetic energy is: Click here to view movieKE = 1/2 (v2 2- v 1 2) 30. Ch 1. BasicsMultimedia Engineering Thermodynamics Temperature Heat andSystemEnergy & PressureWorkEnergy, Specific Heat & Enthalpy Case IntroTheoryCase SolutionSimulationTHERMODYNAMICS - THEORY Forms of Energy (continuation) Potential Energy (PE): The energy that a system possesses as a result of its elevation in a gravitational field or change of configurations. Gravitational potential energy (elevation in a gravitational field):PE = mgz wherem = mass of the systemz = height relative to a reference frame Moving an object from location A to B, its gravitational potential energy change is:PE = mg (ZB - Z A) Elastic potential energy (change of configurations):PE = 1/2 kx2wherek = spring constant x = change in spring length If a spring elongates from L1 to L2, the elastic potential energy stored in the spring is :ElasticPotentialEnergyPE = 1/2 k L2 2- 1/2 k L1 2Click here to view movie 31. Ch 1. BasicsMultimedia Engineering Thermodynamics Temperature Heat andSystemEnergy & PressureWorkEnergy, Specific Heat & Enthalpy Case Intro Theory Case SolutionSimulation THERMODYNAMICS - THEORY Forms of Energy (continuation) Internal energy (U): The energy associated with the random, disordered motion of molecules. It is the sum of the kinetic and potential energies of all molecules. Experience has shown that for most substances with no phase change involved, internal energy strongly depends on temperature. Its dependence on pressure and volume is relatively small. Molecules Random Movement It is not possible to calculate the absolute value Click here to view movieof the internal energy of a body. Only internal energy change of a system can be determined. Internal energy is a property. Total Energy (E): The sum of all forms of energy exist in a system. The total energy of a system that consists of kinetic, potential, and internal energies is expressed as:E = U + KE + PE = U + mv2/2 + mgz The change in the total energy of a system is:E = U + KE + PE Enthalpy (H) Enthalpy is a thermodynamics property of a substance and is defined as the sum of its internal energy and the product of its pressure and volume. H = U + PV 32. Ch 1. BasicsMultimedia Engineering Thermodynamics Temperature Heat andSystemEnergy & PressureWorkEnergy, Specific Heat & Enthalpy Case Intro Theory Case SolutionSimulation THERMODYNAMICS - THEORY Specific Heat (c) Experiment shows that the temperature rise of liquid water due to heat transfer to the water is given by Q = m c (T2 - T 1) where Q = heat transfer to the water m = mass of water T2 - T 1 = temperature rise of the water c = specific heat, an experiment factor In general, the value of specific heat c depends on the substance in the system, the change of state involved, and the particular state of the system at the time of transferring heat. Specific heat of solids and liquids is only a function of temperature but specific heat of gaseous substances is a function of temperature and process. Specific Heat at Constant Volume (cv) Specific heat at constant volume is the change of specific internal energy with respect to temperature when the volume is held constant (Isochoric process).Isochoric Process 33. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystemEnergy & PressureWorkEnergy, Specific Heat & Enthalpy Case IntroTheory Case SolutionSimulation THERMODYNAMICS - THEORY Specific Heat at Constant Volume (cv) (continuation) For constant volume process: Specific Heat at Constant Pressure (Cp) Specific heat at constant pressure is the change of specific enthalpy with respect to temperature when the pressure is held constant (Isobaric process). For constant pressure processIsobaric Process 34. Ch 1. BasicsMultimedia Engineering Thermodynamics Temperature Heat andSystemEnergy & PressureWorkEnergy, Specific Heat & Enthalpy Case Intro TheoryCase Solution SimulationTHERMODYNAMICS - CASE STUDY SOLUTION The velocity of a 1.2 kg hand needs to cause the face temperature to rise 1.8 oC when slapped. The kinetic energy of the hand decreases during the process, due to a decrease in velocity from the initial value to zero. At the same time, the internal energy of the affected area increase, due to an increase in the temperature. Assumptions: The hand is brought to a complete stopafter the impact and the face does notmove. No heat is transferred from the affectedarea to the surroundings. No work isdone to or by the system. The potential energy change is zero. System The energy balance equation:Ein - E out = Esystem= (U + KE + PE)affected tissue + (U + KE + PE)hand With all these assumptions, the equations can be simplified as: 35. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystemEnergy & PressureWorkEnergy, Specific Heat & Enthalpy Case IntroTheory Case SolutionSimulationTHERMODYNAMICS - CASE STUDY SOLUTION Solution (continuation) Plug in the numbers, the solution for velocity v is: 0 = (mcDT)affected tissue + [m(0 - v2)/2]hand This is a very fast hand. 36. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystemEnergy & PressureWorkEnergy, Specific Heat & Enthalpy Case IntroTheory Case SolutionSimulationRun Simulation Suggested Help Questions How fast the hand is to cause the temperature in the affected area increase 1 oC if the mass of the hand is 1.0 kg, the affected mass of face is 0.5 kg, and the specific heat of the tissue is 1.0 kJ/(kg-oC)? Technical Help This simulation shows a face is slapped by a hand and and cause the temperature of the face to rise. Push this button to slap after setting all the numbers. Use this slider to change the face temperature rise. The range is from 0 to 3 oC. 37. Ch 1. BasicsMultimedia Engineering Thermodynamics Temperature Heat andSystemEnergy & PressureWorkEnergy, Specific Heat & Enthalpy Case IntroTheoryCase Solution Simulation Simulation (continuation)Use this slider to change the mass of theaffected tissue of the face. The range isfrom 0.00 to 1.00 kg.Use this slider to change mass of theslapping hand. The range is from 0.00 to1.00 kg.Use this slider to change the specificheat of the face tissue. The range isfrom 0.00 to 5.00 kJ/(kg-oC).This window gives the result handvelocity in m/s. 38. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase PropertyProperty IdealDiagrams Tables GasPure Substances Chapter 2Pure SubstancesTopics ReviewedThe topic menu above allows you to move directly to any of the four sections for each topic. The sections are:1.Phase and Phase Change of Pure Substance2.Property Diagrams for Phase-change Case Intro: To help introduce and understand the basic principles,Processesa case study is presented.3.Property Tables for Pure Substance4.Ideal GasTheory: This section will review the basic principles and equations that you should know to answer the exam questions. It does not give detailed derivations of the theory. Case Solution: The case study is solved in detail in this section. Graphics, narrations, animations, and equations are used to help you understand how the problem was solved. Simulation: You can adjust several parameters of a given problem and learn how they affect the results. 39. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property IdealDiagramsTables Gas Case Intro Theory SimulationPhase and Phase Change Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDYIntroduction A student is working on his thermodynamicsexperiment about phase change. He boils room Phase Change Experimenttemperature water using a pan placed on the top of aelectric heater at room pressure. A thermometer isused to measure the temperature of the substance inthe pan.Known:The mass of the water in the pan is m = 1.0 kgThe power (energy per unit time) of the electric heater is = 2.0KWThe temperature of the room is Troom = 20 oC and the pressureis P = 1 atmThe specific heat of water is c = 4.184 kJ/(kg-K)The latent heat of vaporization of water is L = 2257.0 kJ/kgQuestions The student needs to record the temperature over a period of time. The records should include all phases such as subcooled liquid, saturated liquid, saturated mixture, saturated vapor, and superheated vapor. Please determine the time intervals between each two records.Approach Take the water in the pan as a system The process is shown on the T-v diagraPhase Change Process on T-v DiagramClick to View Movie (4 kB 40. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGasPhase and Phase Change Case IntroTheory Case Solution Simulation THERMODYNAMICS - THEORYPure Substance A substance that has a fixed chemical composition throughout is called a pure substance. Examples of pure substances: water mixture of ice and water Examples of non-pure substances: mixture of water and oilPure Substance: Ice and Watermixture of liquid air and gaseous air Solid, Liquid, and GasSubstances exist in different phases. A phase isidentified as having a distinct molecular arrangementthat is homogeneous throughout and separated fromother phases by easily identifiable boundary surface.The three principal phases are solid, liquid, and gas.Solid: The large attractive forces of molecules on eachother keep the molecules at fixed position. Ice is thesolid phase of water.Liquid: Chunks of molecules float about each other.The molecules maintain an orderly structure withineach chunk and remain their original positions withrespect to one another. Water in room temperature and1 atm pressure is in liquid phase.Gas: Molecules are far apart from each other and moveabout at random. Air is in gaseous phase in roomtemperature and 1 atm pressure. 41. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property PropertyIdealDiagramsTablesGas Phase and Phase Change Case Intro Theory Case SolutionSimulationTHERMODYNAMICS - THEORY Latent Heat When a material changes from a solid to liquid, or from a liquid to a gas, an amount of energy is involved in the change of phase. This energy must be supplied or removed from the system to cause the molecular rearrangement. This energy is called the latent heat. Latent heat relative to melting a solid is calledLatent Heatthe latent heat of fusion (LF). Latent heat relative to vaporizing a liquid is called the latent heat of vaporization (LV). For example, when ice at 1 atm is melted to water at 0 oC, the latent heat of fusion is 333 kJ/kg.The same quantity of heat will be removed for freezing a pound of water to ice. Liquid water boils into vapor at 100 oC, the latent heat of vaporization is 2257 kJ/kg. Also the same quantity of heat will be removed when condensing a pound of water vapor to liquid water at this condition.Phase-change Processes Consider a piston-cylinder device containing liquid water at 20 oC and 1 atm. At this state, the water is in liquid phase and is called compressed liquid or subcooled liquid.Subcooled LiquidWhile keeping the pressure constant which is 1.0atm, add heat to the piston-cylinder device tillthe temperature reaches 100 oC. If additionalheat is added to the water, vapor will appear. TheSaturated Liquidliquid water at this state is called saturated liquid. 42. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGasPhase and Phase Change Case Intro TheoryCase SolutionSimulation THERMODYNAMICS - THEORY Phase-change Processes (Continued) Continuing to add heat to the piston-cylinder device, the liquid will vaporize. The piston- cylinder contains both liquid water and vapor. It is called saturated liquid-vapor mixture or saturated mixture.Saturated MixtureThe temperature remains at 100 oC if theliquid and vapor coexist. The vapor iscalled saturated vapor just when all liquidbecomes vapor. Saturated vapor The temperature of the vapor will rise if more heat is added to the piston-cylinder system after it reaches the saturated vapor state. The vapor for which temperature is higher than that of saturated vapor is called superheated vapor.Superheated vapor Click to View Movie (112 The entire process can be described on a T-v diagram shown on the left. 1 = Subcooled Liquid 2 = Saturated Liquid 3 = Saturated Mixture 4 = Saturated Vapor 5 = Superheated Vapor T-v Diagram 43. Ch 2. Pure SubstancesMultimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGas Phase and Phase ChangeCase Intro Theory Case Solution SimulationTHERMODYNAMICS - CASE STUDY SOLUTIONA student heats 1 kg of water of 20 oC in an experimentto learn the phase change of water. Temperature needsto be recorded for each phase.Assumptions: The pressure in the pan stays at 1 atm duringthe process and thus the saturated temperature is100 oC.Phase Change Process No water or vapor leaks from the pan. So the system is a Click to View Movieclosed system.If the time when the water becomes saturated liquid andsaturated vapor are known, then the recording timeschedule can be determined.Qwater is the energy needed to heat the water fromsubcooled liquid (room temperature) to saturated liquid(saturated temperature)whereTwater = T saturated - Troom = The power of the electric heater, is given as 2.0 kWQvapor is the total energy needed to heat the roomtemperature water to saturated vapor.Heat, Temperature and Time Relation 44. Ch 2. Pure SubstancesMultimedia Engineering ThermodynamicsPhase PropertyProperty IdealDiagrams Tables GasPhase and Phase ChangeCase IntroTheory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION(Continued) Using known values, Twater = 100 - 20 = 80 oC tsaturated liquid = (1.0) (4.184) (80)/2.0 = 167.36 s tsaturated vapor = 2257.0/2.0 + 167.36 = 1295.86 s The time range for each phase is shown in the table. StatesSubcooled LiquidSaturated Liquid Saturated LiquidSaturated vapor Superheated VaporTime (s) 0.0 ~ 167.36167.36167.36 ~ 1295.861295.86 > 1295.86 45. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGasPhase and Phase Change Case Intro Theory Case SolutionSimulation Run Simulation Suggested Help Questions What is the phase of the water if the heating time reaches 10 minute? Technical Help This simulation shows the whole process about water phase change from subcooled liquid water to superheated vapor. Temperature and energy needed are given and the process is shown on the T-v diagram.Use this slider to change theheating time.The range is from0 to 1500 s.This window shows thetemperature of the water orvapor.This window shows the totalheat added. This window shows the phase of the water or vapor. 46. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGasProperty DiagramsCase IntroTheoryCase SolutionSimulation THERMODYNAMICS - CASE STUDYIntroductionA mechanical engineering student gets an internship in apower plant. His first task is to design a piston-cylinderdevice which is used to vaporize the saturated water tosaturated vapor under a given pressure. The volume of thedevice and how much heat needed for the vaporizationprocess are the first thing he has to figure out.What is known:The mass of water m = 10.0 kgOperation pressure P = 100.0 kPaSaturated Water in A Piston- cylinderClick to view movie (48 kB)Questions What is the volume change from liquid water to vapor How much heat is added for the vaporization processApproach Latent heat of vaporization (LV) of water under100 kPa is 2257.0 kJ/kg. Specific volume (Will be introduced in nextsection) of saturated water (vf) under 100 kPa is0.001 m3/kg. Specific volume of saturated vapor (vg) under100 kPa is 1.673 m3/kg. 47. Ch 2. Pure SubstancesMultimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGasProperty DiagramsCase Intro Theory Case Solution Simulation THERMODYNAMICS - THEORYT-v DiagramThe phase-change process of water under 1 atm describedin the previous section can be repeated for differentpressures. Put all the processes in the T-v diagram. A lineconnected by all the saturated liquid states is calledsaturated liquid line. All the saturated vapor states are Phase Change Process Under Constantconnected to create the saturated vapor line. WhenPressurepressure becomes as high as Pcr(critical pressure), theClick to view Movie (112 kB)saturated liquid state and the saturated vapor state becomea single point in T-v diagram. This point is called the criticalpoint. For water, Pcr equals 22.09 MPa, Tcr (criticaltemperature) equals 374.14 oC. At pressure above thecritical pressure, there will not be a distinct phase.The saturated liquid line and saturated vapor line divide theregion on the T-v diagram into three regions: subcooledliquid region, saturated liquid-vapor region and superheatedvapor region.The Process of Creating T-v Diagram Click to View Movie (112 kB)The Construction of T-v Diagram 48. Ch 2. Pure SubstancesMultimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGasProperty DiagramsCase IntroTheory Case SolutionSimulationP-v DiagramThe water vaporization process can also be described inP-v diagram. The method of creation the P-v Diagram ismuch like the method for the T-v diagram. By consideringthe piston-cylinder device again, the temperature keepsconstant by heat transfer The pressure is changed by .Piston-cylinder Device for P-v Diagramremoving the weight. Like the process in previoussection, the water will become saturated liquid,saturated mixture, saturated vapor and superheatedvapor if enough heat added. The process is repeatedunder several different temperatures and plot them in a P-v diagram. By connecting all the saturated liquid statesunder different temperatures, one can create thesaturated liquid line and connecting all the saturatedvapor states, one can create the saturated vapor line onthe P-v diagram. There are three regions on the P-vdiagram: subcooled liquid region, saturated liquid-vaporregion, and superheated vapor region. The Process of Create P-v DiagramThe Construction of P-v DiagramP-v Diagram Including Solid Phase The above diagram can be extended to include solid phase as well as the solid-liquid and solid-vapor regions. Under some conditions, all three phase can coexist in equilibrium. On a P-v diagram, it forms the triple line and on a T-v diagram, it forms only a point called the triple point. 49. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGasProperty DiagramsCase IntroTheoryCase Solution SimulationP-v Diagram Including SolidPhase (Continued)The Construction of P-v Diagram of a Substance That Contracts (left) or Expends (right) on FreezingP-T DiagramThe P-T diagram is called the phase diagram because threephases are separated from each other by three lines(sublimation, melting, and vaporization). The three linesmeet at the triple point where three phases coexist inequilibrium. The Construction of P-T Diagram 50. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGasProperty DiagramsCase IntroTheory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTIONA piston-cylinder device is used to vaporize saturated liquidto saturated vapor under 100 kPa. The volume change andthe amount of energy added need to be determined.Assumptions: The pressure is kept as a constant during theprocess(1)Volume changeThe total volume of the saturated liquidVf = vfmThe total volume when all the water becomes saturatedvapor.Vg = vgmVolume Change Vg - Vf The volume difference is:Vdifference = vgm vfm = 10 (1.673 - 0.001) = 16.72 m3(2) Amount of energy addedLatent heat of vaporization of water under 100 kPa is LV =2257.0 kJ/kg. It is the same amount of energy needed tochange 1 kg saturated water to saturated vapor under 100kPa.Q = m Lv = 10 (2257.0) = 22570 kJ 51. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase PropertyProperty IdealDiagrams Tables GasProperty Diagrams Case Intro TheoryCase SolutionSimulation Run SimulationSuggested Help QuestionsWhat is the temperature if the pressure in the device is 200 kPa? If 2kg water needs to vaporize, how much heat needed?Technical HelpThis simulation shows the evaporating of the saturated liquid tosaturated vapor under different pressures. Saturated temperature,volume changed and heat added are calculated.Use this menu to change thepressure.Use this slider to change themass of the water to be heated.This window shows thesaturation temperature.This window shows the volumechange.This window shows the energyadded. 52. Ch 2. Pure SubstancesMultimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGas Property Tables Case Intro Theory Case Solution Simulation THERMODYNAMICS CASE STUDYIntroductionJohn got a new job in a power plant. His job is tocontrol a boiler that is used to produce steam andwhich will be sent to the turbine for powergeneration. The only property that can be readdirectly from the monitor provided is the pressure. Inorder to monitor the vaporization process, he needsto know the temperature of the water and how muchliquid water left at some pressure.What is known: The pressure in the boiler is 5 MPa The volume of the boiler is 10 m3 and the mass of water is 500 kg Boiler in the Power PlantQuestions Determine the temperature of water in the boiler Determine the total enthalpy Determine the mass of each phase of waterApproach Consider water in the boiler as a closed system Use the specific volume and pressure to locate the state on the P-v diagramSteps to Determine the States of WaterClick to View Movie (68 kB) 53. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGas Property Tables Case Intro TheoryCase SolutionSimulation THERMODYNAMICS THEORYFor most substances, thermodynamic properties arepresented in the form of tables because they are toocomplex to be expressed by simple equations.In thermo-systems, many working fluids can be used.Water is one of the most common working fluidsinvolved. It is the only liquid presented in this section.The thermodynamic properties that are commonly usedare: Temperature (T), oC or oF Pressure (P), Pa or psia Specific volume (v) (the volume per unitmass), m3/kg or ft3/lb Specific internal energy (u), kJ/kg or Btu/lb Specific enthalpy (h), kJ/kg or Btu/lb Specific entropy (s) (Which will be introducedin the following section), kJ/(kg-K) or Btu/(lb-R)Saturated Liquid and Saturated VaporAt a given pressure, the temperature at which a puresubstance changes phase is called the saturationtemperature. At 1 atm, the saturation temperature of wateris 100 oC. At a given temperature, the pressure at which apure substance changes phase is called the saturationpressure. At 100 oC, the saturation pressure of water is 1atm. The saturation temperature and saturation pressuredepends on each other.There are two types of tables for saturated water. Bothtables give the same information. The only difference isthat the properties are listed either as a function oftemperature or pressure.Saturated Water Table The properties listed in the tables includeClick to View Movie (120 kB) Saturation pressure (Psat) for a given temperature or saturation temperature (Tsat) for a given pressureSaturated Water Temperature Table Specific volume of saturated liquid (vf) and saturated vapor (vg)Saturated Water Pressure Table 54. Ch 2. Pure SubstancesMultimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGas Property Tables Case IntroTheory Case SolutionSimulation THERMODYNAMICS THEORYSaturated Liquid and Saturated Vapor(Continued) Specific internal energy of saturated liquid (uf), saturated vapor (ug), and vaporization (ufg = ug - uf ) Specific enthalpy of saturated liquid (hf), saturated vapor (hg), and vaporization (hfg = hg - hf) or latent heat of vaporization. Specific entropy of saturated liquid (sf), saturated vapor (sg), and vaporization (sfg = sg - sf)Saturated MixtureDuring the vaporization process of water, the substance is amixture of saturated liquid and saturated vapor. Quality (x)is defined to describe the fraction of saturated vapor in themixture. x = mvapor/mtotalwheremvapor = mass of vapor in the mixturemtotal = total mass of the mixturemtotal = mliquid + mvapor = mf + mgQuality has a value between 0 and 1. x equals 1 forsaturated vapor and x equals 0 for saturated liquidaccording to its definition.Saturated mixture is a two-phase system. For convenience,it can be treated as a homogenous mixture and theproperties of this mixture is simply the average propertiesof the saturated liquid and saturated vapor. For example,the specific volume of the mixture can be determined by Vav = Vf + Vg mtotal vav = mf vf + mg vg vav = mf vf/mtotal + mg vg/mtotal = vf (1 - mg /mtotal) + vg mg /mtotal = vf (1 - x) + vg x = vf + vfg xQuality is related to the horizontal distance on T-v diagram 55. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGas Property Tables Case IntroTheoryCase Solution Simulation THERMODYNAMICS THEORYSaturated Mixture (Continued)Rearranging the above equation to give an expression forthe quality x as x = (vav - vf)/vfgBased on this equation, quality can be determined from thehorizontal distance on the T-v or P-v diagram as shown inthe figure.The above process can be repeated for u and h. uav = uf + ufg x hav = hf + hfg xSuperheated VaporSince the superheated vapor is a single phase substance,the temperature and pressure are independent. In thesuperheated vapor tables, the properties are listed as afunction of both the temperature and pressure. Thetabulated properties include v, u, h and s. The saturationtemperature corresponding to the pressure is given in theparentheses following the pressure value.Superheated vapor can be characterized by P < Psat at a given T T > Tsat at a given P Superheated Water Table v > vg at a given P or TClick to view Movie (112 kB) u > ug at a given P or T h > hg at a given P or TSuperheated Steam Table Subcooled LiquidThe format of the subcooled liquid water tables is similarto the format of the superheated vapor tables. In theabsence of compressed liquid data, a common practiceis to use the saturated liquid data based on the giventemperature to estimate the properties of compressedliquid. For h, the error can be reduced using the followingapproximation 56. Ch 2. Pure SubstancesMultimedia Engineering ThermodynamicsPhase PropertyProperty IdealDiagrams Tables Gas Property TablesCase Intro Theory Case Solution Simulation THERMODYNAMICS THEORYSubcooled Liquid (Continued) Subcooled liquid can be characterized by P > Psat at a given T T < Tsat at a given P v < vf at a given P or T u < uf at a given P or T h < hf at a given P or TSubcooled Water Table 57. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGas Property Tables Case Intro Theory Case Solution SimulationTHERMODYNAMICS CASE STUDY SOLUTION500 kg of water is in a 10 m3 boiler under 5 Mpa. Thetemperature, enthalpy, and the mass of each phase ofwater are needed to monitor the vaporization process.(1) Determining the state of waterThe specific volume of the water is vav = V/m = 10/500 = 0.02 m3/kgFrom the saturated water table, find the specific volume forsaturated liquid vapor at 5 Mpa. vf = 0.001286 m3/kg vg = 0.03944 m3/kgvf = 0.001286 < vav = 0.02< vg = 0.03944 m3/kgSince vf < vav< vg, the water is in saturated mixture state. Steps to Determine the States(2) TemperatureClick to view Movie (68 kB)Since the water is a saturated mixture, the temperature isthe saturation temperature. From the saturated watertable, the saturation temperature at 5 Mpa is 263.99oC.(3) Quality xThe quality of the saturated mixture is given by x = (vav - vf)/vfg vfg = vg - vf = 0.038154 m3/kg x = (0.02 - 0.001286)/0.038154 = 49%(4) EnthalpyOnce again, from the saturated water table, hf and hgcan befound at P = 5 Mpa. hf = 1,154.23 kJ/kg 58. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGas Property Tables Case IntroTheoryCase Solution Simulation THERMODYNAMICS CASE STUDY SOLUTION (Continued) hg = 2,794.3 kJ/kg hfg = hg - hf = 1,640.1 kJ/kg hav = hf + xhfg= 1,154.23 + (49%)(1,640.1) = 1,958 kJ/kg(5) Mass of each phaseThe definition of quality is given by x = mg/mtotalThe mass for each phase can be calculated as mg = x mtotal = ( 49%)(500) = 245 kg ml = mtotal - mg = 500 - 245 = 255 kg 59. Ch 2. Pure SubstancesMultimedia Engineering ThermodynamicsPhase PropertyPropertyIdealDiagrams TablesGas Property Tables Case IntroTheoryCase SolutionSimulationTHERMODYNAMICS SIMULATIONRun SimulationSuggested Help QuestionsUse steam tableDetermine the state and calculate quality x firstTechnical HelpThis simulation contains 5 cases. Two properties are given and other two need to bedetermined for each case. For the quality x column, if the state is subcooled liquid orsuperheated vapor, put "" in the window Use this window to input data. After input all data, push this button to submit the results. If the result is right, the data remains no change. If it is not correct, the right solution will be shown in the window as red. Push this button to link the steam tables used in these calculations. 60. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property PropertyIdealDiagramsTablesGas Ideal GasCase Intro Theory Case Solution SimulationTHERMODYNAMICS - CASE STUDY Introduction Ballooning is a sport and hobby for many people around the world. Three people were staring ballooning in a cool spring day. Alex, a 11 year old son of one of the ballooning person, was one of the audience. He is full of curiosity about the huge color ball. After his dad came home, he asked some questions about the ballooning. What is known: The balloon is approximately a sphere with a diameter of D = 20 m Local pressure Pa = 90 kPa Local air temperature Ta = 15 oC The mass of the empty balloon and the cage mb= 80 kgHow Hot Air Balloon Works The mass of each person mp = 65 kgClick to view movie (116 kB) Questions Determine the mass of the hot air in the balloon when the balloon is still in the air Determine the average temperature in the balloon in order to keep the balloon still in the air What will happen if the local temperature increases to 30 oC?Approach The force that lifts the balloon upward is: FB = cool air g Vballon The forces that will be balanced by the force (FB) are : (1)The weight of the cage, the ropes, and the balloon materials (2) The mass of the hot air in the balloon (3) The weight of the people and other load in the cageForce Balance on The Balloon 61. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhaseProperty PropertyIdeal DiagramsTables Gas Case Intro Theory Case SolutionSimulationIdeal GasCase Intro Theory Case SolutionSimulationTHERMODYNAMICS - THEORYThe Ideal-gas Equation of StateAny equation that relates the pressure, temperature, andspecific volume of a substance is called the equation ofstate. The following equation is the ideal-gas equation ofstate. A gas that obeys this relation is called an ideal gas. Pv = RTR is the gas constant, which is determined fromR = Ru/MwhereRu = universal gas constant, 8.314 kJ/(kmol-K)M = molar mass, the mass of one mole of asubstance in gramsThe ideal-gas equation of state can also be expressed asPV = mRT or PV = nRuTwherem = mass of the gasn = mole of the gasFor a fixed mass system (m = constant), the properties of an idealgas at two different states can be related asP1v1/T1 = P2v2/T2Equations of State for a Non-ideal GasThe ideal-gas equation of state is very simple, but its applicationrange is limited. The following three equations which are based onassumptions and experiments can give more accurate result over alarger range.Van der Waals Equation of State:The Van der Waals equation of state was proposed in 1873,and it states that(P + a/v2)(v-b) = RTa = 27 R2(Tcr)2/(64Pcr)b = RTcr/(8Pcr)where Tcr = critical temperature Pcr = critical pressureVan der Waals equation of state is the first attempt to modelthe behavior of a real gas. However, it is only accurate over alimited range.Beattie-Bridgeman Equation of State:The Beattie-Bridgeman equation of state was proposed in1928. It has five experimentally determined constant. 62. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase PropertyProperty IdealDiagrams Tables Gas Ideal GasCase Intro Theory Case SolutionSimulationTHERMODYNAMICS THEORY(Continued) The properties with a bar on top are molar basis. Per unit massPer unit moleThe five constants can be found in the table below where P is in kPa, is in m3/kmol, T is in K and Ru is equal to 8.314 (kPa-m3)/(Kmol-K). v (m3/kg)(m3/kmol)The Beattie-Bridgeman equation of state is valid when < 2.5 cr(critical density).u (kJ/kg) (kJ/kmol)h (kJ/kg) (kJ/kmol)Properties Per Mass and Per MoleGasA0 a B0b cAir 131.8441 0.01931 0.04611-0.001101 43400Argon, Ar 30.78020.02328 0.039310.0 59900Carbon dioxide, CO2 507.2836 0.07132 0.10476 0.07235 660000 Helium, He2.18860.05984 0.014000.0 40 Hydrogen, H2 20.0117-0.005060.02096 -0.04359504 Nitrogen, N2 136.2315 0.02617 0.05046 -0.00691 42000 Oxygen, O2 151.0857 0.02562 0046240.004208 48000Source: Gordon J. Van Wylen and Richard E. Sonntag, ClassicalThermodynamics, Fundamentals of 3rd ed., p46, table 3.3 63. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhaseProperty PropertyIdeal DiagramsTablesGasIdeal Gas Case IntroTheoryCase SolutionSimulationTHERMODYNAMICS THEORY(Continued)Benedict-Webb-Rubin Equation of State:Benedict, Webb, and Rubin raised the number of experimentallydetermined constants in the Beattie-Bridgeman Equation of State toeight in 1940.The constants appearing in the equation are given in the table belowwhere P is in kPa, is in m3/kmol, T is inK and Ru is 8.314 (kPa-m3)/(Kmol-K). This equation ofstate is accurate when < 0.8 cr. GasaA0 bB0 c10-C010-5 1054n-Butane, C4H10190. 102 0.03990.124 32051006 110.10.0340681.6 9836Carbon dioxide,13.8 277.0.00720.049 151.1 140.48.4700.0054 CO2630 1091 Carbon monoxide,3.71 135.0.00260.054 10.54 8.67313.500.0060 CO87 3254 Methane, CH45.00 187.0.00330.042 25.78 22.8612.440.0060 91 8060Nitrogen, N2 2.54 106.0.00230.040 7.379 8.16412.720.0053 73 2874 Source: Kenneth Wark, Thermodynamics, 4th ed., p.141. 64. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhaseProperty Property Ideal DiagramsTables GasIdeal Gas Case IntroTheoryCase SolutionSimulation THERMODYNAMICS THEORY (Continued) Compressibility Factor The compressibility factor (Z) is a dimensionless ratio of the product of pressure and specific volume to the product of gas constant and temperature.Z = (Pv)/(RT) orZ = vactual/videal The compressibility factor (Z) is a measure of deviation from the ideal-gas behavior. For ideal gas, Z is equal to 1. Z can be either greater or less than 1 for real gases. The further away Z is from unity, the more the gas deviates from the ideal-gas behavior. The generalized compressibility chart is developed to be used for all gases. They are plotted as a function of the reduced pressure and reduced temperature, which are defined as follows: PR = P/Pcr and TR= T/Tcr where PR = the reduced pressureThe Generalized Compressibility ChartTR = the reduced temperature Click to view Movie (84 kB) The gases behave as an ideal gas regardleFrom the generalized compressibility chart, the following observations can be made. ss of temperature for very low pressure (PR 2). In the vicinity of the critical point, the gases deviate from ideal gas greatly. The animation on the left shows the error involved in assuming steam to be an ideal gas. The red region wherePercentage of error involved in assuming steam steam can be treated as an ideal gas has error of less thanto be an ideal gas 1%. Click to view Movie (68 kB) percentage of error = (|vtable - videal|/vtable) (100%) 65. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhaseProperty PropertyIdeal DiagramsTablesGasIdeal GasCase IntroTheory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION While the hot air balloon is still in the air. The mass of the hot air and the average temperature in the balloon needs to be determined. Assumptions: The air is ideal gas with gas constant R = 287 J/(kg-K). (1) The mass of air in the balloon Since the air in the balloon is ideal gas, it obeys the ideal-gas equation of state. PV = mairRT whereR = gas constantV = volume of the balloon P =pressure in the balloonmair = mass of the air in the balloonT = temperature of the air in the balloon Since the balloon is open to the air, the pressure P in the balloon is the same as the ambient pressure. According to the force diagram shown on the left, the sum of all forces should be zero when the balloon is still in the air.FB - (Gb + Gp + Gair) = 0FB = cool airgVballoon = cool airg (4/3)(D/2)3cool air = P/(RT)= 90(103)/(287(15+273.15)) = 1.089 kg/m3 Rearranging the equation to give the expression of mass of the hot air.mair = FB/g - mb - 3(mp) Force Balance on The BalloonWith all the data knownmair = (1.089)(9.8)(4/3)(10)3 /9.8 - 80 - 3(65) = 4287 kg (2) The average temperature in the balloon After the mass of hot air is determined, the average temperature of the hot air can be determined using the ideal-gas equation of state.PV = mairRT T = PV/ (mairR) = 90(103)(4/3)(3.14)(10)3/ ((4287)(287)) = 306.5 K = 33.34 oC 3) The movement of the balloon if the local air temperature is 30 oC The density of the air decreases with the temperature increases. Hence the buoyancy force will decrease and the balloon will move downward. Repeating the solution above, the temperature of the hot air equals 50.38 oC if the balloon keeps still again. 66. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhaseProperty PropertyIdeal DiagramsTablesGasIdeal GasCase Intro TheoryCase Solution SimulationRun Simulation Suggested Help Questions For a certain ambient pressure, if the ambient temperature increases, to which direction will the balloon move? Technical Help This simulation shows the movement of a hot air balloon at a specific temperature and pressure. The wind direction is used to movement the balloon horizontally.Use this menu to choose the directionof movement of the balloon inhorizontal level.Use this slider to change theambient temperature. Therange is from 0 to 30 oC. Use this slider to change the ambient pressure. The range is from 90 to 120 kPa. This window shows the average temperature of the hot air in the balloon. This window shows the mass of the hot air in the balloon. 67. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservationConservationSolids and Ideal Gasof Massof Energy LiquidsFirst Law of Thermodynamics Chapter 3 First LawTopics ReviewedThe topic menu above allows you to move directly to any of the four sections for each topic. The sections are:1.Conservation of Mass2.Conservation of Energy Case Intro: To help introduce and understand the basic3.Energy Analysis of solids and liquidsprinciples, a case study is presented.4.Energy Analysis of Ideal Gas Theory: This section will review the basic principles and equations that you should know to answer the exam questions. It does not give detailed derivations of the theory. Case Solution: The case study is solved in detail in this section. Graphics, narrations, animations, and equations are used to help you understand how the problem was solved. Simulation: You can adjust several parameters of a given problem and learn how they affect the results. 68. Ch 3. First Law of ThermodynamicsMultimedia Engineering ThermodynamicsConservation Conservation Solids andIdeal Gasof Mass of EnergyLiquidsConservation of Mass Case Intro TheoryCase SolutionSimulation THERMODYNAMICS - CASE STUDY IntroductionRobert has bought some land, and plans to plant vegetablewhich will need to be watered everyday in the summer. Thewater used to irrigate the land is from a well. A pond isneeded to store the water. He has an old pump which has avolume flow rate of 20 L/s.Known: Roberts Pond Robert will plant 45 acres of vegetables. It is estimated that each acre needs 2,000 liters of water per day. QuestionsHow big is the pond?Determine how long the pump will be operated daily.ApproachConsider the water in the pond as a system.Consider the Pond as a SystemThe water stored in the pond equals msystem of the fillingprocess.At 30 oC and 1 atm, water = 996 kg/m3 69. Ch 3. First Law of ThermodynamicsMultimedia Engineering ThermodynamicsConservation ConservationSolids and Ideal Gasof Mass of Energy LiquidsConservation of MassCase Intro Theory Case SolutionSimulation THERMODYNAMICS - THEORY Mass and Volume Flow Rate The mass flow rate () is defined as the amount of mass flowing through a cross-section per unit time. The mass flow rate of a fluid flowing in or out of a pipe or duct is proportional to the cross-sectional area (A) of the pipe or duct, the density of the fluid (), and the velocity of the flow (V). The flow rate through a differential area dA is:Mass Flow Rate Through a Ductd= VndA WhereVn = the velocity component normal to the area dA Integrating the above equation to get the total mass flow rate. The volume flow rate () is the volume of the fluid flowing through a cross-sectional area per unit time.The Normal Velocity Component The mass and volume flow rate are related by Conservation of Mass Principle The conservation of mass principle states the following:Net mass transfer to or from asystem during a process is equalto the net change in the totalmass of the system during thatprocess. In an equation format, the conservation of mass principle is:(Total mass entering the system)System Used for Conservation of -Mass Equation(Total mass leaving the system)= (Net change in mass within the system) 70. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gasof Mass of EnergyLiquidsConservation of MassCase Intro Theory Case Solution Simulation Conservation of Mass Principle(continuation) or, min - mout = msystem where msystem = msystem@final - msystem@initial The rate form of the conservation of mass principle is: Filling and Emptying Bathtub is an (Rate at which mass entering the system) Example of Mass Conservation-(Rate at which mass leaving the system)=(time rate of change in mass within the system) or,Conservation of Mass for Closed System A closed system is defined as a system which mass can not cross its boundaries, but energy transfer is allowed. Since no mass flows in or out of the system, the mass of the closed system remains constant during a process. Mass Remains Constant for a Closed System 71. Ch 3. First Law of ThermodynamicsMultimedia Engineering ThermodynamicsConservationConservation Solids andIdeal Gasof Massof EnergyLiquidsConservation of MassCase IntroTheory Case SolutionSimulationTHERMODYNAMICS - CASE STUDY SOLUTION A pond will be built to store water to irrigate vegetables. The volume of the pond and the time to fill the pond need to be determined. Assumptions:No irrigation during the filling processThe flow of the pump is steady, m = t (Steady flowprocess will be introduced in the following sections)Considering the Pond as a System (1) Determine the volume of the pondThe amount of water needed for 45 acres per day: Vtotal = (45)(2,000) = 90,000 LAt the end of the filling process, the pond should have enough water to irrigate the 45 acres for the day. Hence the volume of the pond needs to be at least 90,000 L. (2) Determine the filling time Considering the water in the pond as a system. During the filling process, no water is going out of the system for irrigation. The mass balance on this system during the filling process is:At the end of the filling process, the pond has the water Considering the Pond as a System needed for one days irrigation. That givesmsystem@final = mtotal At the beginning of the filling process, the pond is empty. msystem@initial = 0 72. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation ConservationSolids and Ideal Gasof Mass of Energy LiquidsConservation of MassCase Intro Theory Case SolutionSimulationTHERMODYNAMICS - CASE STUDY SOLUTION(Continuation)From the mass balance, the mass imported to the pondcan be determined.min = msystem = msystem@final - msystem@initial = mtotal - 0 = Vtotal = (90,000/1,000)(996) = 89,640 kgmin= 89,640 kgThe flow rate of the pump is 20 L/s, or = 20 L/s.According to the relation between the mass flow rateand volume flow rate, the mass flow rate can bedetermined.= = (20/1,000)(996) = 19.9 kg/sThe total time needed for the pump to beoperated is:min = tt = min/=(89,640)/(19.9) = 4505 s = 1.25 h 73. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids andIdeal Gasof Mass of EnergyLiquidsConservation of MassCase IntroTheory Case SolutionSimulation Run Simulation Suggested Help Questions If Robert wants to fill his pond in 1 hour, what is the inlet flow rate? If Roberts pump has a flow rate of 25 L/s, and the exit valve leaks water at a flow rate of 1 L/s during the filling process, what is the filling time?Technical HelpThis simulation shows the process of filling or emptying a pond. Thefilling time is calculated.Use this slider to change the inlet flow rate. The range isfrom 0 to 30 L/s.Use this slider to change the exit flow rate. The range isfrom 0 to 30 L/s. This window shows time needed to fill the empty pond at the selected inlet and outlet flow rate.After selecting the flow rate at inlet and exit, push "Run"button to view the filling or empty process. Push "Reset"button to stop the process and reset the flow rates. 74. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservationConservation Solids andIdeal Gasof Massof EnergyLiquidsConservation of Energy Case Intro Theory Case Solution SimulationTHERMODYNAMICS - CASE STUDYIntroductionLee is a new AME student at the University ofOklahoma. He rents a small apartment which has awindow air conditioner. Lee spends most of his time atschool, hence he only wants to run the air conditionerwhen he is at home. But he is worrying that it will taketoo long for the room temperature to reach thetemperature setting. Also, he is interested in knowinghow much heat will be released to the ambient duringthe cooling process. How Window Air What is known: Conditioner Works Coefficient of performance (the ratio of the heat removed from the room to the net work required for the air conditioner) of the air conditioner COP = 2.5 Power input of the air conditioner= 900 W Room temperature Troom = 90 oF Temperature setting of the air conditioner Tsetting= 75 oF 10 kJ heat needs to be removed for the temperature of the room to go down 1 oFQuestions How long time will it take to cool the room down to the temperature setting? How much heat will be released to the ambient?Approach Take the window air conditioner as a system The energy balance equation is: (Qin - Qout ) + (Win - Wout) + (Emass,in -Emass,out) = Esystem 75. Ch 3. First Law of ThermodynamicsMultimedia Engineering ThermodynamicsConservation ConservationSolids and Ideal Gasof Mass of Energy LiquidsConservation of EnergyCase Intro Theory Case Solution SimulationTHERMODYNAMICS - THEORY The First Law of Thermodynamics The first law of thermodynamics, also known as the conservation of energy principle, states:Energy can be neither created nordestroyed; it can only change forms.The first law is based on experimentalobservations. It can not be provedmathematically, but no known process innature violates the first law. Conservation of Energy for Energy Balancean Air Conditioner During a process, the first law can be expressed as: the net change in the total energy during a process is equal to the difference between the total energy entering and leaving the system during that process. In an equation format, the energy balance for a system is: (Total energy entering the system) -(Total energy leaving the system)= Energy Balanced for a (Net change in the total energy of the system)Piston-cylinderor, Spring Device Ein- Eout = Esystem The rate form of the energy balance of a system is:( Rate at which energy entering the system)- (Rate at which energy leaving the system)= (time rate of change in the total energy of the system) or, 76. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservationConservation Solids and Ideal Gasof Massof EnergyLiquidsConservation of EnergyCase Intro Theory Case Solution Simulation Mechanisms of Energy Transfer Ein and EoutEnergy is transferred to or from a system by three forms:heat, work and mass flow. For a closed system, the onlytwo forms are heat and work since no mass crosses itsboundaries. 1. Heat transfer (Q)Heat transfer is the energy interaction caused by atemperature difference between a system and itssurroundings. Heat transfer to a system (heat gain) willcause the internal energy of the system to increase, andheat transfer from a system (heat loss) will cause theinternal energy of the system to decrease. Mode of heat transfer2. Work (W)The energy interaction that is not caused by atemperature difference between a system and itssurroundings is work. Work transferred to a system willincrease the energy of the system, and work transferredfrom a system will decrease the energy of the system.3. Mass flowWhen mass enters a system, the energy of the systemSome Forms of Workincreases because of the energy accompanied by mass.Also the energy of the system decreases when massleaves the system.Heat and work have been introduced in the previoussections. Systems that involve energy balance with massflow are considered in the following section.Note that net transfer of a quantity is equal to thedifference between the amounts transferred in and out,the energy balance can be written as:(Qin - Qout ) + (Win - Wout) + (Emass,in - Emass,out) =EsystemAll the quantities of the six terms in the left are positiveamounts since the direction of the energy transfer isdescribed by the subscripts "in" and "out". 77. Ch 3. First Law of ThermodynamicsMultimedia Engineering ThermodynamicsConservation ConservationSolids and Ideal Gasof Mass of Energy LiquidsConservation of EnergyCase Intro Theory Case SolutionSimulation Energy Change of a System Esystem The energy change of a system during a process is: Energy change = Energy at final state - Energy at initial state or, Esystem = Esystem@final - Esystem@initial From the discussion in the previous section, it is known that the change in the total energy of a system during a process is the sum of the changes in its internal, kinetic, and potential energies. E = U + KE + PE 78. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gasof Mass of EnergyLiquidsConservation of MassCase Intro Theory Case SolutionSimulation THERMODYNAMICS - CASE STUDY SOLUTION A window air conditioner is running to reduce the room temperature to the temperature setting. The operation time and the total heat released to the ambient need to be determined. Assumption: Assume energy of the system remains constant (1) Heat transferred to the air conditioner Qin In order to cool the room to the setting temperature, heat needs to be removed from the room, which is transferred to the air conditioner, and it is the energy entering the system. Qin = (10)(1000) (90 - 75) = 150,000 J (2) Total work input for the air conditioner Win According to the definition of COP, the amount of work input needed for the air conditioner is given by COP = Qin / Win Win = Qin /COP = 150,000/2.5 = 60,000 J The power input for the air conditioner is 900 W. So the operation time is: t = Win /= 60,000/900 = 66.67 s = 1.1 minute (3) The total heat released to the ambient Qout The energy equation can be simplified according to the assumptions. Since energy of the system remains constant,Esystem is 0. No work is done by the air conditioner to its ambient , hence Wout = 0. The system is a closed system, hence no mass crossed its boundaries. That is, Emass,in - Emass,out = 0. Energy Balance of the Air Conditioner Qout = Qin + Win = 150,000 + 60,000 = 210,000 J = 210 kJ 79. Ch 3. First Law of ThermodynamicsMultimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gasof Mass of EnergyLiquids Solids and Liquids Case Intro TheoryCase Solution SimulationRun SimulationSuggested Help QuestionsHow long time will it take to cool the room down from 98 oF to 76 oF?Technical HelpThis simulation shows the cooling process of a room by a window airconditioner. For different room temperature, temperature setting andCOP, the cooling time is calculated.Use this slider to change COP of the air conditioner. Use this slider to change the temperature setting of the air conditioner. Use this slider to change the room temperature.Use this slider to change power iinput to the airconditioner. This window shows operation time in minutes. 80. Ch 3. First Law of ThermodynamicsMultimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gasof Mass of EnergyLiquids Solids and Liquids Case Intro Theory Case Solution SimulationTHERMODYNAMICS - CASE STUDY Introduction Noel is an electric engineer who is designing an elevator for the new wing of the library. The elevator needs to carry new books to the fourth floor where the university bookstore is located. The elevator is always fully loaded when it travels upwards and empty when it travels downwards. The main power line was installed when the library was built, hence the power for this elevator is limited. Noel needs to make sure his design is feasible for the power requirement. How the Elevator WorksWhat is known: The mass of the empty car mcar = 150 kg The maximum load is mload = 650 kg The mass of the counterweight is 50% of the total mass of the elevator when it is fully loaded The operating velocity of the elevator is v = 2.0 m/s Questions Determine the power required when the elevator is fully loaded and moving upwards Determine the power required when the elevator is empty and moving downwards How much energy can be saved by using counterweight for a single delivery? ApproachConsider the car and the counterweighttogether as a systemConsider the energy change for a time spanfrom t1 to t2 81. Ch 3. First Law of ThermodynamicsMultimedia Engineering ThermodynamicsConservation ConservationSolids and Ideal Gasof Mass of Energy Liquids Solids and Liquids Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Energy Balance for Closed SystemThe energy balance for a system, which has beenpreviously introduced, is: (Qin - Qout ) + (Win - Wout) + (Emass,in - Emass,out) = Esystem = (U + KE + PE)systemFor a closed system, the only forms of energy that canbe supplied or removed from a system are heat andwork.Energy Balance for Closed System(Qin - Qout ) + (Win - Wout) = (U + KE + PE)system If the adopted sign convention is such that the heat entering the system is positive, and the work done by the system is positive for a process from state 1 to state 2, then the energy balance for a closed system becomes:Q - W = E2 - E1 = Esystem= (U + KE + PE)system For a stationary system, in which no velocity andSign Convention for Heat Transferelevation changes during a process, the change of the total energy of the system is due to the change of the internal energy only. That is, Q- W = U2 - U1 Specific Heats of Solids and LiquidsThe definitions of constant volume and constantpressure specific heats have been introducedpreviously. They are Sign Convention for Work 82. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation ConservationSolids and Ideal Gasof Mass of Energy Liquids Solids and Liquids Case Intro Theory Case SolutionSimulationSpecific Heats of Solids and Liquids(Continuation)For most solids and liquids, they can beapproximated as incompressible substances,hence the constant volume and constantpressure specific heats are the same. That is, cP = cv = cThe specific heats of incompressible substancesdepend on the temperature only. Hence thespecific heats are simplified as: cv = du/dTcP = dh/dTInternal Energy and EnthalpyDifference of Solids and LiquidsFrom the definition of specific heat, the change ofinternal energy becomesdu = cvdT = c(T) dTFor a process from state 1 to state 2, the change ofinternal energy is obtained by integrating the aboveequation from state 1 to state 2. For small temperature intervals, an average specific heat (c) at the average temperature is used and treated as a constant, yielding Enthalpy is another temperature dependent variable. The definition of enthalpy is: H = U + PV It can be rewritten in terms of per unit mass as follows: h = u+ Pv 83. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gasof Mass of EnergyLiquids Solids and Liquids Case Intro Theory Case Solution SimulationInternal Energy and EnthalpyDifference of Solids and Liquids(Continuation)Note that v is a constant, so the differential form ofthe above equation is:Integrating from state 1 to state 2 yields h = u + v P + vPInternal Energy and Enthalpy of Solids For solids, the term vP is insignificant.and Liquidsh = u For liquids, two cases are encountered. They are:Constant pressure process, P = 0, h = uConstant temperature process, T = 0, h = vP 84. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gasof Mass of EnergyLiquids Solids and Liquids Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION An elevator is designed to deliver books for the bookshop located on the fourth floor of the library. The power required to run the elevator when it is fully loaded and empty needs to be determined.Assumptions:The guide rails and the pulleys are frictionlessThe ropes can not be stretchedConsider the car and the counterweight as a system. Itis a closed system because the car (with or without aload) and the counterweight are solids, hence themass remains constant during the process. Consider the Car and theCounterweight as a System The energy balance for a closed system is:Q - W = (U + KE + PE)systemIt can be simplified according to the followingconditions:No heat is transferred in or out, Q = 0No internal energy change, U = 0The velocity remains constant during the process, KE=0The energy balance equation becomes: - W =(PE)system(1) Determine the power needed when the elevator isfully loaded and moving upwardsFrom the analysis above, the work needed to run thefully loaded elevator upwards is:W = - (PE)systemThe mass of the counterweight is:mcounter = (mcar + mload )(50%)= (150 + 650)(50%) = 400 kgAt time t1, the total potential energy of the system is:(PE)1 = (mcar + mload )g x1 + mcounterg y1 Locations of the Elevator at Time t1and t2when the Elevator Moves Upwards 85. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation ConservationSolids and Ideal Gasof Mass of Energy Liquids Solids and Liquids Case Intro TheoryCase SolutionSimulation THERMODYNAMICS - CASE STUDYSOLUTION(Continuation) At time t2, the total potential energy of the system changes to: (PE)2 =(mcar + mload )g x2 + mcounterg y2 (PE)system = (PE)2 - (PE)1= (mcar + mload )g ( x2 - x1) + mcounterg ( y2 - y1) Since the ropes cannot be stretched, the distance that the car moves upwards is the same as the distance that the counterweight moves downwards.(x2 - x1) = - (y2 - y1) = (t2 - t1)vHence,(PE)system= (mcar + mload- mcounter)g (t2 - t1)vW = - (mcar + mload- mcounter)g (t2 - t1)vPower is work per unit time, and it is given by = W/(t2 - t1) = - (mcar + mload- mcounter)gv = - (150 + 650 - 400)(9.81)(2.0) = -