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- 1. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Basics Chapter 1 Basics 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 the Case Intro: To help introduce and understand the Zeroth Law of Thermodynamics basic principles, a case study is presented. 3. Heat and Work Theory: This section will review the basic principles 4. Energy, Specific Heat, and and equations that you should know to answer the Enthalpy 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. Basics Multimedia Engineering ThermodynamicsSystem Temperature Heat and Energy & Pressure Work System Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 gage Pressure Cooker The opening cross-sectional area is 4 mm2 Atmospheric pressure is 101 kPa Questions What is the mass of the petcock? Approach Solve the problem using the basic steps in engineering problem solving: 1) read 2) draw Pressure Cooker diagrams 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 Temperature Heat and Energy & Pressure Work System Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Systems All thermodynamic systems contain three basic elements: • System boundary: The imaginary surface that bounds the system. • System volume: The volume within the imaginary surface. System and Surroundings • The surroundings: The surroundings is everything external to the system. Systems can be classified as being closed, open, or isolated. • Closed system: Mass cannot cross the boundaries, but energy can. • Open system (control volume): Both mass and energy can cross the boundaries. • Isolated system: Either mass nor energy can cross its boundaries. Closed System Click here to view movie Open System Isolated System Click here to view movie Click here to view movie
- 4. Ch 1. Basics Multimedia Engineering ThermodynamicsSystem Temperature Heat and Energy & Pressure Work System Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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. Basics Multimedia Engineering ThermodynamicsSystem Temperature Heat and Energy & Pressure Work System Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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. Basics Multimedia Engineering ThermodynamicsSystem Temperature Heat and Energy & Pressure Work System Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 movie objects 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 Temperature Heat and Energy & Pressure Work System Case Intro Theory Case Solution Simulation Run 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 Work Temperature and Pressure Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction A nurse is taking care of three children. She needs to measure their temperatures and compare the temperatures with the standardized body temperature to determine if someone got a fever. What is known: The standardized body temperature for different age 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 year 99.0 3 - 5 year 98.6 5 - 9 year 98.3 Thermometer 9 - 13 year 98.0 > 13 year 97.8 - 99.1 The temperature of the three kids are: Age Temperature ( o C) 3 month 37.0 6 month 38.5 3 year 37.2 Questions Is there someone who got a fever? Approach The relation between oC, oF and K are: T (K) = T (oC) + 273.15 T (oF) = 1.8 T(oC) + 32.0
- 9. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Temperature and Pressure Case Intro Theory Case Solution Simulation 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. English Length meter (m) foot (ft) The base units in the S.I. system are meters (m) Time second (s) second (s) for length, second (s) for time, and kilogram (kg) Mass kilogram (kg) slug (slug) for mass. The force unit is derived using Newtons Force newton (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 Pressure In the SI system, the unit for pressure is Pa, Pascal. In the English system, it is psi. Pa = N/m psi = lbf/in2 Since Pa is a small unit in the SI system, other units are also used in thermodynamics, such as:
- 10. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Temperature and Pressure Case Intro Case Intro Theory Case Case Solution Theory Solution Simulation THERMODYNAMICS - THEORY Pressure (continuation) 1 bar = 105 Pa 1 kPa = 103 Pa 1 MPa = 106 Pa 1 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 Atmosphere 101325 Pa. Barometer is a device used to measure the atmospheric pressure. Pa = ρg h where ρ = 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 thermodynamic Gage Pressure and Vacuum Pressure relations and tables.
- 11. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Temperature and Pressure Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Pressure (continuation) U-Tube Manometer is used to measure pressure difference. One end of it is open to the atmosphere and the other end is connected to the equipment whose pressure is needed to be measured. At the right side, P1 = Pgas + ρgas g h1 U-Tube Manometer Usually, the second term on the right hand of Click here to view movie the previous equation is negligible since the density of the work fluid is much larger than the density of the gas. P1 = Pgas At the left side, P1 = Pa + ρworking fluid g h1 Combine the two equations above, the pressure in the gas tank can be determined as Pgas = Pa + ρworking fluid g h1 Temperature and the Zeroth Law The measurement of the degree of hotness or coolness is temperature. If two bodies at different temperatures are brought together, the hot body will warm up the cold one. At the same time, the cold body will cool down the hot one. This process will end when the two bodies have the same The Zeroth Law temperatures. At that point, the two bodies are said to have reached thermal equilibrium.
- 12. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Temperature and Pressure Case Intro Theory Case 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 Scales To establish a temperature scale, two fixed, easy duplicated points are used. The intermediate points are obtained by dividing the distance between into equal subdivisions of the scale length. Temperature Scale Fixed Point 1 Fixed Point 2 Fahrenheit Scale ( o F) Freezing Point of Water = 32.0 Boiling Point of Water = 212.0 Celsius Scale ( o C) Freezing Point of Water = 0.0 Boiling Point of Water = 100.0 Thermodynamic The pressure of an ideal gas is zero The Triple Point of Water = Temperature Scale (K) = 0.0 273.16 The relations between the above temperature scales are: 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 English Temperature Scales system (R) are related by T(R) = 1.8 T(K)
- 13. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Temperature and Pressure Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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. Thermometer Modern thermometers are calibrated in standard Click here to view movie temperature units such as Fahrenheit or Celsius. Three practical points for using thermometer are: 1. The thermometer should be isolated to everything except the body which temperature is measured. The general method is to immerse the thermometer in a hole in a solid body, or directly in a fluid body. 2. When thermal equilibrium is reached, the thermometer can indicate its own temperature as well as the body measured. The Digital Thermometer thermometer should be small relative to the body so that it only has a small effect upon the body. 3. The thermometer must not be subject to effects such as pressure changes, which might change the volume independently of temperature. Digital thermometers almost replace the mercury ones in nowadays because they are more accuracy and more easy to use.
- 14. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Temperature and Pressure Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION The standardized body temperatures unit is oF but the temperature taken by the nurse is oC. So use the following relations to convert the 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 99.4 37.4 310.5 3 - 6 month 99.5 37.5 310.6 6 -12 month 99.7 37.6 310.7 1 - 3 year 99.0 37.2 310.3 3 - 5 year 98.6 37.0 310.1 5 - 9 year 98.3 36.8 309.9 9 - 13 year 98.0 36.7 309.8 > 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. Age Temperature(oC) standardized(oC) conditions 3 month 37.0 37.5 6 month 38.5 37.6 fever 3 year 37.2 37.2
- 15. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Temperature and Pressure Case Intro Theory Case Solution Simulation Run Simulation Suggested Help Questions Marys 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. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Heat and Work Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 (ρ = 2770 kg/m and Cp= 875 J/(kg- C)) 3 o and the base plate thickness δ = 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 ε = 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. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Heat and Work Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 by the wires. Energy out: convection, radiation heat loss to the ambient air. Energy storage: the energy in the base to increase the bases temperature. Energy Balance Diagram
- 18. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Heat and Work Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Path Function and Point Function Path function and Point function are introduced to identify 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 Function state. All properties are point functions. Process A: V2 - V 1 = 3 m3 Process B: V2 - V 1 = 3 m3 Heat Heat is energy transferred from one system to another solely by reason of a temperature difference between the systems. Heat exists only as it crosses the boundary of a system and the direction of heat transfer is from higher temperature to lower temperature. For thermodynamics sign convention, heat Heat Transfer Direction transferred to a system is positive; Heat transferred Click here to view movie from a system is negative.
- 19. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Heat and Work Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Heat (continuation) The heat needed to raise a objects temperature from T1 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 object Unit of heat is the amount of heat required to cause a unit rise in temperature of a unit mass of water at atmospheric pressure. Btu: Raise the temperature of 1 lb of water 1oF Cal: Raise the temperature of 1 gram of water 1 oC J is the unit for heat in the S.I. unit system. The relation between Cal and J is 1 Cal = 4.184 J Notation used in this book for heat transfer: Q : total heat transfer : the rate of heat transfer (the amount of heat transferred per unit time) δQ: the differential amounts of heat q: heat transfer per unit mass q: heat transfer per unit mass
- 20. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Heat and Work Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 heat transfer 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 & Pressure Work Heat and Work Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 σ = Stefan - Boltzmann constant ε = emissivity Ts = Surface temperature of the object Solar energy applications mainly use radiation energy from the Sun. The three modes of heat transfer always exist simultaneously. For example, the heat transfer associated with double pane windows are: Conduction: Hotter (cooler) air outside each pane causes conduction through solid glass. Convection: Air between the panes carries heat from hotter pane to cooler pane. Radiation: Sunlight radiation passes through glass to be absorbed on other side. Double Pane Window Click here to view movie Please view heat transfer books for details of modes of heat transfer.
- 22. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Heat and Work Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Work Work is the energy transfer associated with a force acting through a distance. In thermodynamics sign convection, work transferred out of a system is positive with respect to that system. Work transferred in is negative. Definition of Work Click here to view movie Dot product means the distance along the forces direction. For example, if a car runs at a flat road, its weight does zero work because the weight and the moving distance have a 90o angle. Like heat, Work is an energy interaction between a system and its surroundings and associated with a process. Units of work is the same as the units of heat. Notation: W : total work δW: differential amount of work w: work per unit mass : Power, the work per unit time
- 23. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Heat and Work Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Expansion and Compression Work A system without electrical, magnetic, gravitational motion and surface tension effects is called a simple compressible system. Only two properties are needed to determine a state of a simple compressible system. Considering the gas enclosed in a piston-cylinder device with a cross-sectional area of the piston A. Initial State: Pressure P1 Volume V1 Expansion and Compression Work Click here to view movie Finial State: Pressure P2 Volume 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: δW = 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 & Pressure Work Heat and Work Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 η Eout= Econvection + Eradiation 1. 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 out The heat generated by the wires is:
- 25. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Heat and Work Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION (cont.) Energy stored in the base is: Estorage = cpm (Tfinal - T initial) m = ρAδ Plug the numbers, the solution for time t is: t = 342 s 2. Consider the convection and radiation heat loss from the iron after the temperature reaches 140 oC. Because temperature keeps at 140 oC, no energy 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 the temperature at 140 oC during ironing because the clothes will absorb more heat than the ambient.
- 26. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Heat and Work Case 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 the fabric. Default is synthetic. Use this slider to change the convection heat transfer coefficient. The range is from 0 to 100 W/(m2-oC). Default is 50 W/(m2-oC).
- 27. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure Work Heat and Work Case Intro Theory 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 & Pressure WorkEnergy, Specific Heat & Enthalpy Case Intro Theory 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 Questions Determine 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 = ΔEsystem = (ΔU + ΔKE + ΔPE)affected tissue + (ΔU + ΔKE + ΔPE)hand System
- 29. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure WorkEnergy, Specific Heat & Enthalpy Case Intro Theory Case Solution Simulation 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 system v = velocity of the system Kinetic 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 movie ΔKE = 1/2 (v2 2- v 1 2)
- 30. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure WorkEnergy, Specific Heat & Enthalpy Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 where m = mass of the system z = height relative to a reference frame Moving an object from location A to B, its gravitational potential energy change is: ΔPE = mg (ZB - Z A) Elastic potential energy (change of configurations): PE = 1/2 kx2 where k = spring constant x = change in spring length If a spring elongates from L1 to L2, the elastic potential energy stored in the spring is : ElasticPotentialEnergy ΔPE = 1/2 k L2 2- 1/2 k L1 2 Click here to view movie
- 31. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure WorkEnergy, Specific Heat & Enthalpy Case Intro Theory Case Solution Simulation 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 movie of 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: ΔE = ΔU + ΔKE + ΔPE 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. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure WorkEnergy, Specific Heat & Enthalpy Case Intro Theory Case Solution Simulation 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 andSystem Energy & Pressure WorkEnergy, Specific Heat & Enthalpy Case Intro Theory Case Solution Simulation 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 process Isobaric Process
- 34. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure WorkEnergy, Specific Heat & Enthalpy Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 stop after the impact and the face does not move. • No heat is transferred from the affected area to the surroundings. No work is done to or by the system. • The potential energy change is zero. System The energy balance equation: Ein - E out = Δ Esystem = (ΔU + ΔKE + ΔPE)affected tissue + (ΔU + ΔKE + ΔPE)hand With all these assumptions, the equations can be simplified as:
- 35. Ch 1. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure WorkEnergy, Specific Heat & Enthalpy Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 andSystem Energy & Pressure WorkEnergy, Specific Heat & Enthalpy Case Intro Theory Case Solution Simulation Run 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. Basics Multimedia Engineering Thermodynamics Temperature Heat andSystem Energy & Pressure WorkEnergy, Specific Heat & Enthalpy Case Intro Theory Case Solution Simulation Simulation (continuation) Use this slider to change the mass of the affected tissue of the face. The range is from 0.00 to 1.00 kg. Use this slider to change mass of the slapping hand. The range is from 0.00 to 1.00 kg. Use this slider to change the specific heat of the face tissue. The range is from 0.00 to 5.00 kJ/(kg-oC). This window gives the result hand velocity in m/s.
- 38. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Pure Substances Chapter 2 Pure SubstancesTopics Reviewed The 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,Processes a case study is presented.3.Property Tables for Pure Substance4.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.
- 39. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Case Intro Theory Simulation Phase and Phase Change Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction A student is working on his thermodynamics experiment about phase change. He boils room Phase Change Experiment temperature water using a pan placed on the top of a electric heater at room pressure. A thermometer is used to measure the temperature of the substance in the pan. Known: •The mass of the water in the pan is m = 1.0 kg •The power (energy per unit time) of the electric heater is = 2.0 KW •The temperature of the room is Troom = 20 oC and the pressure is P = 1 atm •The specific heat of water is c = 4.184 kJ/(kg-K) •The latent heat of vaporization of water is L = 2257.0 kJ/kg Questions 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 diagra Phase Change Process on T-v Diagram Click to View Movie (4 kB
- 40. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Phase and Phase Change Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Pure 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 oil Pure Substance: Ice and Water •mixture of liquid air and gaseous air Solid, Liquid, and Gas Substances exist in different phases. A phase is identified as having a distinct molecular arrangement that is homogeneous throughout and separated from other phases by easily identifiable boundary surface. The three principal phases are solid, liquid, and gas. Solid: The large attractive forces of molecules on each other keep the molecules at fixed position. Ice is the solid phase of water. Liquid: Chunks of molecules float about each other. The molecules maintain an orderly structure within each chunk and remain their original positions with respect to one another. Water in room temperature and 1 atm pressure is in liquid phase. Gas: Molecules are far apart from each other and move about at random. Air is in gaseous phase in room temperature and 1 atm pressure.
- 41. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Phase and Phase Change Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 called Latent Heat the 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 Liquid While keeping the pressure constant which is 1.0 atm, add heat to the piston-cylinder device till the temperature reaches 100 oC. If additional heat is added to the water, vapor will appear. The Saturated Liquid liquid water at this state is called saturated liquid.
- 42. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Phase and Phase Change Case Intro Theory Case Solution Simulation 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 Mixture The temperature remains at 100 oC if the liquid and vapor coexist. The vapor is called saturated vapor just when all liquid becomes 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 Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Phase and Phase Change Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION A student heats 1 kg of water of 20 oC in an experiment to learn the phase change of water. Temperature needs to be recorded for each phase. Assumptions: • The pressure in the pan stays at 1 atm during the process and thus the saturated temperature is 100 oC. Phase Change Process • No water or vapor leaks from the pan. So the system is a Click to View Movie closed system. If the time when the water becomes saturated liquid and saturated vapor are known, then the recording time schedule can be determined. Qwater is the energy needed to heat the water from subcooled liquid (room temperature) to saturated liquid (saturated temperature) where ΔTwater = T saturated - Troom = The power of the electric heater, is given as 2.0 kW Qvapor is the total energy needed to heat the room temperature water to saturated vapor. Heat, Temperature and Time Relation
- 44. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Phase and Phase Change Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION (Continued) Using known values, ΔTwater = 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. States Subcooled Liquid Saturated Liquid Saturated Liquid Saturated vapor Superheated Vapor Time (s) 0.0 ~ 167.36 167.36 167.36 ~ 1295.86 1295.86 > 1295.86
- 45. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Phase and Phase Change Case Intro Theory Case Solution Simulation 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 the heating time.The range is from 0 to 1500 s. This window shows the temperature of the water or vapor. This window shows the total heat added. This window shows the phase of the water or vapor.
- 46. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Property Diagrams Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction A mechanical engineering student gets an internship in a power plant. His first task is to design a piston-cylinder device which is used to vaporize the saturated water to saturated vapor under a given pressure. The volume of the device and how much heat needed for the vaporization process are the first thing he has to figure out. What is known: • The mass of water m = 10.0 kg • Operation pressure P = 100.0 kPa Saturated Water in A Piston- cylinder Click to view movie (48 kB) Questions • What is the volume change from liquid water to vapor • How much heat is added for the vaporization process Approach • Latent heat of vaporization (LV) of water under 100 kPa is 2257.0 kJ/kg. • Specific volume (Will be introduced in next section) of saturated water (vf) under 100 kPa is 0.001 m3/kg. • Specific volume of saturated vapor (vg) under 100 kPa is 1.673 m3/kg.
- 47. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Property Diagrams Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY T-v Diagram The phase-change process of water under 1 atm described in the previous section can be repeated for different pressures. Put all the processes in the T-v diagram. A line connected by all the saturated liquid states is called saturated liquid line. All the saturated vapor states are Phase Change Process Under Constant connected to create the saturated vapor line. When Pressure pressure becomes as high as Pcr(critical pressure), the Click to view Movie (112 kB) saturated liquid state and the saturated vapor state become a single point in T-v diagram. This point is called the critical point. For water, Pcr equals 22.09 MPa, Tcr (critical temperature) equals 374.14 oC. At pressure above the critical pressure, there will not be a distinct phase. The saturated liquid line and saturated vapor line divide the region on the T-v diagram into three regions: subcooled liquid region, saturated liquid-vapor region and superheated vapor region. The Process of Creating T-v Diagram Click to View Movie (112 kB) The Construction of T-v Diagram
- 48. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Property Diagrams Case Intro Theory Case Solution Simulation P-v Diagram The water vaporization process can also be described in P-v diagram. The method of creation the P-v Diagram is much like the method for the T-v diagram. By considering the piston-cylinder device again, the temperature keeps constant by heat transfer The pressure is changed by . Piston-cylinder Device for P-v Diagram removing the weight. Like the process in previous section, the water will become saturated liquid, saturated mixture, saturated vapor and superheated vapor if enough heat added. The process is repeated under several different temperatures and plot them in a P- v diagram. By connecting all the saturated liquid states under different temperatures, one can create the saturated liquid line and connecting all the saturated vapor states, one can create the saturated vapor line on the P-v diagram. There are three regions on the P-v diagram: subcooled liquid region, saturated liquid-vapor region, and superheated vapor region. The Process of Create P-v Diagram The Construction of P-v Diagram P-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 Property Property Ideal Diagrams Tables Gas Property Diagrams Case Intro Theory Case Solution Simulation P-v Diagram Including Solid Phase (Continued) The Construction of P-v Diagram of a Substance That Contracts (left) or Expends (right) on Freezing P-T Diagram The P-T diagram is called the phase diagram because three phases are separated from each other by three lines (sublimation, melting, and vaporization). The three lines meet at the triple point where three phases coexist in equilibrium. The Construction of P-T Diagram
- 50. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Property Diagrams Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION A piston-cylinder device is used to vaporize saturated liquid to saturated vapor under 100 kPa. The volume change and the amount of energy added need to be determined. Assumptions: • The pressure is kept as a constant during the process (1)Volume change The total volume of the saturated liquid Vf = vfm The total volume when all the water becomes saturated vapor. Vg = vgm Volume Change Vg - Vf The volume difference is: Vdifference = vgm – vfm = 10 (1.673 - 0.001) = 16.72 m3 (2) Amount of energy added Latent heat of vaporization of water under 100 kPa is LV = 2257.0 kJ/kg. It is the same amount of energy needed to change 1 kg saturated water to saturated vapor under 100 kPa. Q = m Lv = 10 (2257.0) = 22570 kJ
- 51. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Property Diagrams Case Intro Theory Case Solution Simulation Run Simulation Suggested Help Questions What is the temperature if the pressure in the device is 200 kPa? If 2 kg water needs to vaporize, how much heat needed? Technical Help This simulation shows the evaporating of the saturated liquid to saturated vapor under different pressures. Saturated temperature, volume changed and heat added are calculated. Use this menu to change the pressure. Use this slider to change the mass of the water to be heated. This window shows the saturation temperature. This window shows the volume change. This window shows the energy added.
- 52. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Property Tables Case Intro Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY Introduction John got a new job in a power plant. His job is to control a boiler that is used to produce steam and which will be sent to the turbine for power generation. The only property that can be read directly from the monitor provided is the pressure. In order to monitor the vaporization process, he needs to know the temperature of the water and how much liquid 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 Plant Questions • Determine the temperature of water in the boiler • Determine the total enthalpy • Determine the mass of each phase of water Approach • Consider water in the boiler as a closed system • Use the specific volume and pressure to locate the state on the P-v diagram Steps to Determine the States of Water Click to View Movie (68 kB)
- 53. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Property Tables Case Intro Theory Case Solution Simulation THERMODYNAMICS –THEORY For most substances, thermodynamic properties are presented in the form of tables because they are too complex to be expressed by simple equations. In thermo-systems, many working fluids can be used. Water is one of the most common working fluids involved. It is the only liquid presented in this section. The thermodynamic properties that are commonly used are: • Temperature (T), oC or oF • Pressure (P), Pa or psia • Specific volume (v) (the volume per unit mass), 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 introduced in the following section), kJ/(kg-K) or Btu/(lb- R) Saturated Liquid and Saturated Vapor At a given pressure, the temperature at which a pure substance changes phase is called the saturation temperature. At 1 atm, the saturation temperature of water is 100 oC. At a given temperature, the pressure at which a pure substance changes phase is called the saturation pressure. At 100 oC, the saturation pressure of water is 1 atm. The saturation temperature and saturation pressure depends on each other. There are two types of tables for saturated water. Both tables give the same information. The only difference is that the properties are listed either as a function of temperature or pressure. Saturated Water Table The properties listed in the tables include Click to View Movie (120 kB) • Saturation pressure (Psat) for a given temperature or saturation temperature (Tsat) for a given pressure Saturated Water Temperature Table • Specific volume of saturated liquid (vf) and saturated vapor (vg) Saturated Water Pressure Table
- 54. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Property Tables Case Intro Theory Case Solution Simulation THERMODYNAMICS –THEORY Saturated 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 Mixture During the vaporization process of water, the substance is a mixture of saturated liquid and saturated vapor. Quality (x) is defined to describe the fraction of saturated vapor in the mixture. x = mvapor/mtotal where mvapor = mass of vapor in the mixture mtotal = total mass of the mixture mtotal = mliquid + mvapor = mf + mg Quality has a value between 0 and 1. x equals 1 for saturated vapor and x equals 0 for saturated liquid according to its definition. Saturated mixture is a two-phase system. For convenience, it can be treated as a homogenous mixture and the properties of this mixture is simply the average properties of 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 x Quality is related to the horizontal distance on T-v diagram
- 55. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Property Tables Case Intro Theory Case Solution Simulation THERMODYNAMICS –THEORY Saturated Mixture (Continued) Rearranging the above equation to give an expression for the quality x as x = (vav - vf)/vfg Based on this equation, quality can be determined from the horizontal distance on the T-v or P-v diagram as shown in the figure. The above process can be repeated for u and h. uav = uf + ufg x hav = hf + hfg x Superheated Vapor Since the superheated vapor is a single phase substance, the temperature and pressure are independent. In the superheated vapor tables, the properties are listed as a function of both the temperature and pressure. The tabulated properties include v, u, h and s. The saturation temperature corresponding to the pressure is given in the parentheses 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 T Click to view Movie (112 kB) • u > ug at a given P or T • h > hg at a given P or T Superheated Steam Table Subcooled Liquid The format of the subcooled liquid water tables is similar to the format of the superheated vapor tables. In the absence of compressed liquid data, a common practice is to use the saturated liquid data based on the given temperature to estimate the properties of compressed liquid. For h, the error can be reduced using the following approximation
- 56. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Property Tables Case Intro Theory Case Solution Simulation THERMODYNAMICS –THEORY Subcooled 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 T Subcooled Water Table
- 57. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Property Tables Case Intro Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION 500 kg of water is in a 10 m3 boiler under 5 Mpa. The temperature, enthalpy, and the mass of each phase of water are needed to monitor the vaporization process. (1) Determining the state of water The specific volume of the water is vav = V/m = 10/500 = 0.02 m3/kg From the saturated water table, find the specific volume for saturated liquid vapor at 5 Mpa. vf = 0.001286 m3/kg vg = 0.03944 m3/kg vf = 0.001286 < vav = 0.02< vg = 0.03944 m3/kg Since vf < vav< vg, the water is in saturated mixture state. Steps to Determine the States (2) Temperature Click to view Movie (68 kB) Since the water is a saturated mixture, the temperature is the saturation temperature. From the saturated water table, the saturation temperature at 5 Mpa is 263.99oC. (3) Quality x The 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) Enthalpy Once again, from the saturated water table, hf and hgcan be found at P = 5 Mpa. hf = 1,154.23 kJ/kg
- 58. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Property Tables Case Intro Theory Case 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 phase The definition of quality is given by x = mg/mtotal The 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 Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Property Tables Case Intro Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions •Use steam table •Determine the state and calculate quality x first Technical Help This simulation contains 5 cases. Two properties are given and other two need to be determined for each case. For the quality x column, if the state is subcooled liquid or superheated 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 Property Ideal Diagrams Tables Gas Ideal Gas Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 kg How Hot Air Balloon Works •The mass of each person mp = 65 kg Click 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 = ρcool 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 cage Force Balance on The Balloon
- 61. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Case Intro Theory Case Solution Simulation Ideal Gas Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY The Ideal-gas Equation of State Any equation that relates the pressure, temperature, and specific volume of a substance is called the equation of state. The following equation is the ideal-gas equation of state. A gas that obeys this relation is called an ideal gas. Pv = RT R is the gas constant, which is determined from R = Ru/M where Ru = universal gas constant, 8.314 kJ/(kmol-K) M = molar mass, the mass of one mole of a substance in grams The ideal-gas equation of state can also be expressed as PV = mRT or PV = nRuT where m = mass of the gas n = mole of the gas For a fixed mass system (m = constant), the properties of an ideal gas at two different states can be related as P1v1/T1 = P2v2/T2 Equations of State for a Non-ideal Gas The ideal-gas equation of state is very simple, but its application range is limited. The following three equations which are based on assumptions and experiments can give more accurate result over a larger 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) = RT a = 27 R2(Tcr)2/(64Pcr) b = RTcr/(8Pcr) where Tcr = critical temperature Pcr = critical pressure Van der Waals equation of state is the first attempt to model the behavior of a real gas. However, it is only accurate over a limited range. Beattie-Bridgeman Equation of State: The Beattie-Bridgeman equation of state was proposed in 1928. It has five experimentally determined constant.
- 62. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Ideal Gas Case Intro Theory Case Solution Simulation THERMODYNAMICS – THEORY(Continued) The properties with a bar on top are molar basis. Per unit mass Per unit mole The 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 ρ < 2.5 ρcr(critical density). u (kJ/kg) (kJ/kmol) h (kJ/kg) (kJ/kmol) Properties Per Mass and Per Mole Gas A0 a B0 b c Air 131.8441 0.01931 0.04611 -0.001101 43400 Argon, Ar 30.7802 0.02328 0.03931 0.0 59900 Carbon dioxide, CO2 507.2836 0.07132 0.10476 0.07235 660000 Helium, He 2.1886 0.05984 0.01400 0.0 40 Hydrogen, H2 20.0117 -0.00506 0.02096 -0.04359 504 Nitrogen, N2 136.2315 0.02617 0.05046 -0.00691 42000 Oxygen, O2 151.0857 0.02562 004624 0.004208 48000 Source: Gordon J. Van Wylen and Richard E. Sonntag, Classical Thermodynamics, Fundamentals of 3rd ed., p46, table 3.3
- 63. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Ideal Gas Case Intro Theory Case Solution Simulation THERMODYNAMICS – THEORY (Continued) Benedict-Webb-Rubin Equation of State: Benedict, Webb, and Rubin raised the number of experimentally determined constants in the Beattie-Bridgeman Equation of State to eight in 1940. The constants appearing in the equation are given in the table below where P is in kPa, is in m3/kmol, T is in K and Ru is 8.314 (kPa-m3)/(Kmol-K). This equation of state is accurate when ρ < 0.8 ρcr. Gas a A0 b B0 c×10- C0×10-5 α ×105 γ 4 n-Butane, C4H10 190. 102 0.0399 0.124 3205 1006 110.1 0.0340 68 1.6 98 36 Carbon dioxide, 13.8 277. 0.0072 0.049 151.1 140.4 8.470 0.0054 CO2 6 30 10 91 Carbon monoxide, 3.71 135. 0.0026 0.054 10.54 8.673 13.50 0.0060 CO 87 32 54 Methane, CH4 5.00 187. 0.0033 0.042 25.78 22.86 12.44 0.0060 91 80 60 Nitrogen, N2 2.54 106. 0.0023 0.040 7.379 8.164 12.72 0.0053 73 28 74 Source: Kenneth Wark, Thermodynamics, 4th ed., p.141.
- 64. Ch 2. Pure Substances Multimedia Engineering ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Ideal Gas Case Intro Theory Case Solution Simulation 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) or Z = 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 pressure The Generalized Compressibility Chart TR = 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 << 1). •The gases can be assumed as an ideal gas with good accuracy regardless of pressure for high temperature (TR > 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 than to 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 ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Ideal Gas Case Intro Theory 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 where R = gas constant V = volume of the balloon P = pressure in the balloon mair = mass of the air in the balloon T = 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) = 0 FB = ρcool airgVballoon = ρcool airgπ (4/3)(D/2)3 ρcool 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 Balloon With all the data known mair = (1.089)(9.8)(4/3)π(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 ThermodynamicsPhase Property Property Ideal Diagrams Tables Gas Ideal Gas Case Intro Theory Case Solution Simulation Run 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 direction of movement of the balloon in horizontal level. Use this slider to change the ambient temperature. The range 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 ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids First Law of Thermodynamics Chapter 3 First Law Topics Reviewed The topic menu above allows you to move directly to any of the four sections for each topic. The sections are: 1.Conservation of Mass 2.Conservation of Energy Case Intro: To help introduce and understand the basic 3.Energy Analysis of solids and liquids principles, 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 Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Conservation of Mass Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction Robert has bought some land, and plans to plant vegetable which will need to be watered everyday in the summer. The water used to irrigate the land is from a well. A pond is needed to store the water. He has an old pump which has a volume flow rate of 20 L/s. Known: Robert’s Pond •Robert will plant 45 acres of vegetables. It is estimated that each acre needs 2,000 liters of water per day. Questions •How big is the pond? •Determine how long the pump will be operated daily. Approach •Consider the water in the pond as a system. Consider the Pond as a System •The water stored in the pond equals Δmsystem of the filling process. •At 30 oC and 1 atm, ρwater = 996 kg/m3
- 69. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Conservation of Mass Case Intro Theory Case Solution Simulation 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 (ρ), and the velocity of the flow (V). The flow rate through a differential area dA is: Mass Flow Rate Through a Duct d = ρVndA Where Vn = 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 a system during a process is equal to the net change in the total mass of the system during that process. 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 Gas of Mass of Energy Liquids Conservation of Mass Case Intro Theory Case Solution Simulation Conservation of Mass Principle(continuation) or, min - mout = Δmsystem where Δmsystem = 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 Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Conservation of Mass Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 process •The flow of the pump is steady, m = t (Steady flow process will be introduced in the following sections) Considering the Pond as a System (1) Determine the volume of the pond The amount of water needed for 45 acres per day: Vtotal = (45)(2,000) = 90,000 L At 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 gives msystem@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 Conservation Solids and Ideal Gas of Mass of Energy Liquids Conservation of Mass Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION(Continuation) From the mass balance, the mass imported to the pond can be determined. min = Δmsystem = msystem@final - msystem@initial = mtotal - 0 = Vtotalρ = (90,000/1,000)(996) = 89,640 kg min= 89,640 kg The flow rate of the pump is 20 L/s, or = 20 L/s. According to the relation between the mass flow rate and volume flow rate, the mass flow rate can be determined. = ρ = (20/1,000)(996) = 19.9 kg/s The total time needed for the pump to be operated is: min = t t = min/ =(89,640)/(19.9) = 4505 s = 1.25 h
- 73. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Conservation of Mass Case Intro Theory Case Solution Simulation 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 Help This simulation shows the process of filling or emptying a pond. The filling time is calculated. Use this slider to change the inlet flow rate. The range is from 0 to 30 L/s. Use this slider to change the exit flow rate. The range is from 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 ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Conservation of Energy Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction Lee is a new AME student at the University of Oklahoma. He rents a small apartment which has a window air conditioner. Lee spends most of his time at school, hence he only wants to run the air conditioner when he is at home. But he is worrying that it will take too long for the room temperature to reach the temperature setting. Also, he is interested in knowing how much heat will be released to the ambient during the 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 oF Questions •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) = ΔEsystem
- 75. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Conservation of Energy Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 nor destroyed; it can only change forms. The first law is based on experimental observations. It can not be proved mathematically, but no known process in nature violates the first law. Conservation of Energy for Energy Balance an 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-cylinder or, Spring Device Ein- Eout = ΔEsystem 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 ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Conservation of Energy Case Intro Theory Case Solution Simulation Mechanisms of Energy Transfer Ein and Eout Energy is transferred to or from a system by three forms: heat, work and mass flow. For a closed system, the only two forms are heat and work since no mass crosses its boundaries. 1. Heat transfer (Q) Heat transfer is the energy interaction caused by a temperature difference between a system and its surroundings. Heat transfer to a system (heat gain) will cause the internal energy of the system to increase, and heat transfer from a system (heat loss) will cause the internal energy of the system to decrease. Mode of heat transfer 2. Work (W) The energy interaction that is not caused by a temperature difference between a system and its surroundings is work. Work transferred to a system will increase the energy of the system, and work transferred from a system will decrease the energy of the system. 3. Mass flow When mass enters a system, the energy of the system Some Forms of Work increases because of the energy accompanied by mass. Also the energy of the system decreases when mass leaves the system. Heat and work have been introduced in the previous sections. Systems that involve energy balance with mass flow are considered in the following section. Note that net transfer of a quantity is equal to the difference between the amounts transferred in and out, the energy balance can be written as: (Qin - Qout ) + (Win - Wout) + (Emass,in - Emass,out) =ΔEsystem All the quantities of the six terms in the left are positive amounts since the direction of the energy transfer is described by the subscripts "in" and "out".
- 77. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Conservation of Energy Case Intro Theory Case Solution Simulation Energy Change of a System Δ Esystem The energy change of a system during a process is: Energy change = Energy at final state - Energy at initial state or, ΔEsystem = 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. ΔE = ΔU + ΔKE + ΔPE
- 78. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Conservation of Mass Case Intro Theory Case Solution Simulation 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,ΔEsystem 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 Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Solids and Liquids Case Intro Theory Case Solution Simulation Run Simulation Suggested Help Questions How long time will it take to cool the room down from 98 oF to 76 oF? Technical Help This simulation shows the cooling process of a room by a window air conditioner. For different room temperature, temperature setting and COP, 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 air conditioner. This window shows operation time in minutes.
- 80. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Solids and Liquids Case Intro Theory Case Solution Simulation THERMODYNAMICS - 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 Works What 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? Approach •Consider the car and the counterweight together as a system •Consider the energy change for a time span from t1 to t2
- 81. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Solids and Liquids Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Energy Balance for Closed System The energy balance for a system, which has been previously introduced, is: (Qin - Qout ) + (Win - Wout) + (Emass,in - Emass,out) = ΔEsystem = (ΔU + ΔKE + ΔPE)system For a closed system, the only forms of energy that can be supplied or removed from a system are heat and work. Energy Balance for Closed System (Qin - Qout ) + (Win - Wout) = (ΔU + ΔKE + ΔPE)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 = ΔEsystem = (ΔU + ΔKE + ΔPE)system For a stationary system, in which no velocity and Sign Convention for Heat Transfer elevation 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 Liquids The definitions of constant volume and constant pressure specific heats have been introduced previously. They are Sign Convention for Work
- 82. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Solids and Liquids Case Intro Theory Case Solution Simulation Specific Heats of Solids and Liquids (Continuation) For most solids and liquids, they can be approximated as incompressible substances, hence the constant volume and constant pressure specific heats are the same. That is, cP = cv = c The specific heats of incompressible substances depend on the temperature only. Hence the specific heats are simplified as: cv = du/dT cP = dh/dT Internal Energy and Enthalpy Difference of Solids and Liquids From the definition of specific heat, the change of internal energy becomes du = cvdT = c(T) dT For a process from state 1 to state 2, the change of internal energy is obtained by integrating the above equation 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 Gas of Mass of Energy Liquids Solids and Liquids Case Intro Theory Case Solution Simulation Internal Energy and Enthalpy Difference of Solids and Liquids (Continuation) Note that v is a constant, so the differential form of the above equation is: Integrating from state 1 to state 2 yields Δh = Δu + v ΔP + vΔP Internal Energy and Enthalpy of Solids For solids, the term vΔP is insignificant. and Liquids Δh = Δu For liquids, two cases are encountered. They are: •Constant pressure process, ΔP = 0, Δh = Δu •Constant temperature process, ΔT = 0, Δh = vΔP
- 84. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids 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 frictionless •The ropes can not be stretched Consider the car and the counterweight as a system. It is a closed system because the car (with or without a load) and the counterweight are solids, hence the mass remains constant during the process. Consider the Car and the Counterweight as a System The energy balance for a closed system is: Q - W = (ΔU + ΔKE + ΔPE)system It can be simplified according to the following conditions: •No heat is transferred in or out, Q = 0 •No internal energy change, ΔU = 0 •The velocity remains constant during the process, ΔKE =0 The energy balance equation becomes: - W =(ΔPE)system (1) Determine the power needed when the elevator is fully loaded and moving upwards From the analysis above, the work needed to run the fully loaded elevator upwards is: W = - (ΔPE)system The mass of the counterweight is: mcounter = (mcar + mload )(50%) = (150 + 650)(50%) = 400 kg At 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 Conservation Solids and Ideal Gas of Mass of Energy Liquids Solids and Liquids Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION(Continuation) At time t2, the total potential energy of the system changes to: (PE)2 =(mcar + mload )g x2 + mcounterg y2 (ΔPE)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)v Hence, (ΔPE)system= (mcar + mload- mcounter)g (t2 - t1)v W = - (mcar + mload- mcounter)g (t2 - t1)v Power is work per unit time, and it is given by = W/(t2 - t1) = - (mcar + mload- mcounter)gv = - (150 + 650 - 400)(9.81)(2.0) = - 7,840 W The negative sign means that work needs to be applied to the system when the elevator runs upwards with full load. (2) Determine the power needed when the elevator is empty and moving downwards At time t1, the total potential energy of the system is: (PE)1 = (mcar)g x1 + mcounterg y1 At time t2, the total potential energy of the system is: (PE)2 =(mcar)g x2 + mcounterg y2 (ΔPE)system = (PE)2 - (PE)1 = (mcar )g (x2 - x1) + mcounterg (y2 - y1) . Locations of the Elevator at Time t1and t2when the Elevator Moves Downwards
- 86. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Solids and Liquids Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION(Continuation) The counterweight moves upwards while the car moves downwards. y2 - y1 = - (x2 - x1) = (t2 - t1)v Hence, (ΔPE)system= (mcounter - mcar )g(t2 - t1)v Power can be determined by dividing the work by time. = W/(t2 - t1) = - (mcounter- mcar)gv = - (400 - 150)(9.81)(2.0) = - 4,900 W The result is negative which means that input power is required. (3) Determine the energy saved by using counterweight For a single delivery, the elevator runs upwards with full load and downwards with no load. Set mcounterequals to 0 in the above analysis. The energy needed for the elevator moving upwards and downwards are: up = W/(t2 - t1) = - (mcar + mload)gv = - (150 + 650)(9.8)(2.0) = - 15,680 W down = W/(t2 - t1) = - (- mcar)gv = - ( - 150)(9.8)(2.0) = 2,940 W The work is positive which means that no energy is needed to be applied to the system. The percent of energy saved by using the counterweight is (15,680 - 7,840 - 2,940)/15,680 = 31.3%
- 87. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Solids and Liquids Case Intro Theory Case Solution Simulation Run Simulation Suggested Help Questions •For a given load and car weight, how does the weight of the counterweight affect the power? •For a given counterweight, how does the load and car weight affect the power? Technical Help This simulation shows the process of an elevator used to send books to the bookstore which is located on the 4th floor. For different loads, mass of the car, operation velocity, and counterweight mass, the power needed to operate with a full load going up and with no load going down are calculated. Use this slider to change the load. Use this slider to change the mass of the car. Use this slider to change the operation velocity of the elevator. Use this slider to change the ratio of the mass of the counterweight to the mass of the fully loaded car. This window shows the power needed when the elevator is full and go upwards. This window shows the power needed when the elevator is empty and go downwards.
- 88. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Energy Analysis of Ideal Gas Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction Ann is living in a dormitory room, and she has a small fan in the room. On a summer day, she turns on her fan before leaving for classes, to cool the room before she comes back in the evening. Does it work? What is known: •The size of the room is 4 m x 6 m x 6 m •The power of the fan is 150 W •She leaves at 8:00 am and comes back at 6:00 pm Problem Description •The room temperature is 15 oC when she leaves in the morning Questions •What will the temperature of the room be when Ann comes back in the evening? Approach •Consider the air in the room as a system •Use specific heat values at the average room temperature •All the energy from the fan becomes internal energy of the air in the room
- 89. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsIdeal Gas ModeConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Energy Analysis of Ideal Gas Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Internal Energy, Enthalpy, and Specific Heats of Ideal Gas The ideal gas is defined as a gas which obeys the following equation of state: Pv = RT Ideal Gas Model The internal energy of an ideal gas is a function of Pv = PT temperature only. That is, u = u(T) u = u(T) h = h(T) = u(T) + RT Ideal Gas Model Using the definition of enthalpy and the equation of state of ideal gas to yield, Specific Heats of h = u + P v = u + RT Common Ideal Gases Since R is a constant and u = u(T), it follows that the enthalpy of an ideal gas is also a function of temperature only. h = h(T) Since u and h depend only on the temperature for an ideal gas, the constant volume and constant pressure specific heats cv and cp also depend on the temperature only. cv = cv (T) cP = cP (T) For an ideal gas, the definitions of cv and cp are given as follows: and they can be rewritten as cv = du/dT cP = dh/dT
- 90. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsIdeal Gas ModeConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Energy Analysis of Ideal Gas Case Intro Theory Case Solution Simulation Internal Energy, Enthalpy, and Specific Heats of Ideal Gas(Continuation) During a process from state 1 to state 2, the changes of internal energy and enthalpy are: There are three ways to determine the changes of internal energy and enthalpy for ideal gas. 1. By using the tabulated u and h data. This is the easiest and most accurate way. 2. By integrating the equations above if the relations of cv and cp as a function of temperature are known. This is for computerized calculations and very accurate. 3. By using the average values of specific heats. This is simple and the results obtained are reasonably accurate if the temperature interval is small. Specific Heat Relations of Ideal Gas Combing the definition of enthalpy and the equation of state for ideal gas to yield h = u + pv = u + RT Differentiating the above relation to give dh = du + RdT Replacing dh by cPdT and du by cvdT,
- 91. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsIdeal Gas ModeConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Energy Analysis of Ideal Gas Case Intro Theory Case Solution Simulation Specific Heat Relations of Ideal Gas (Continuation) This is a special relationship between cv and cP for an ideal gas. Also, the ratio of cP and cv is called the specific heat ratio, k = cP/cv The specific heat ratio is also a temperature dependent property. For air at T = 300 K, cP = 1.005 kJ/(kg-K) cv = 0.718 kJ/(kg-K) k = 1.4 The Polytropic Process Many processes which occur in practice can be described by an equation of the form Pvn = constant where n = constant For a process from state 1 to state 2, the relation is: P1v1 n = P2v2n Special Processes of Ideal Gas on a P-v Diagram By using the equation of state of ideal gas, the relations between P, v, and T are: T2/T1=(P2/P1)(n-1)/n P2/P1=(v1/v2)n For some specific values of n, the process becomes isobaric, isothermal, isometric, and adiabatic, and they are summarized as follows: Process n Isobaric (P = constant) 0 Isothermal (T = constant) 1 Isometric (V = constant) infinity adiabatic (Q = 0) k
- 92. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsIdeal Gas ModeConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Energy Analysis of Ideal Gas Case Intro Theory Case Solution Simulation Energy Analysis for Ideal Gas in a Closed System at Different Processes The energy balance for a stationary, closed system is: Q - W = ΔU For ideal gas, the internal energy can be determined by ΔU = U2 - U1 = cv(T2 - T1) where cv = the average value of constant volume specific heat If only boundary work is considered, then the work W can be determined by For a polytropic process, the work W is: Boundary Work For special processes such as the isobaric, isothermal, isometric, and adiabatic processes for ideal gas, using the average specific heats, the heat, work, and internal energy are given in the following table: Process Q W ΔU Isothermal mRT ln(P1/P2) mRT ln(P1/P2) 0 Isometric m cv ΔT 0 mcvΔT Isobaric mcP ΔT P(V2-V1) mcvΔT Adiabatic 0 -mcvΔT mcvΔT
- 93. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Energy Analysis of Ideal Gas Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION Ann turns on her fan trying to cool her room while she is gone. The room temperature when she comes back in the evening needs to be determined. Assumptions: •Consider the air as an ideal gas •All the doors and windows are tightly closed •Disregard any heat transfer through the walls and the windows •The room pressure remains at 1 atm The system is a closed system since all the doors and windows are tightly closed. It is a stationary system by disregarding the changes of the velocity and elevation of air during the cooling process. The energy balance for this system is: Q - W = U2 - U1 Also, the assumption indicates that the heat transfer through the walls and windows is disregarded, hence no Simplification of the heat transfer to or from this system. Energy Balance Equation Q=0 The electric energy is transferred into the system by the fan, hence this work is negative. W = - fant For an ideal gas, the internal energy is a function of the temperature. If the specific heat at the average room temperature is used, the difference of internal energy is: U2 - U1 = cv,av(T2 - T1) The energy equation for the air in the room becomes: fant = cv,avm(T2 - T1) T2 = fant/cv,avm + T1
- 94. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Energy Analysis of Ideal Gas Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION (Continuation) Ann leaves at 8:00 in the morning and returns at 6:00 in the evening. Hence the total time is 10 hours. The average value of specific heat is used to calculate the temperature. A room temperature is assume first. Then iteration is used till a reliable result obtained. Assume the room temperature is 55oC when she is back in the evening. Tav = (15 + 55)/2.0 = 35 oC At 1 atm and 35 oC, the density of the air is ρ = 1.166 kg/m3. The mass of the air in the room is: m = (4)(6)(6)(1.166) = 167.9 kg From table, at 35 oC, the constant volume specific heat is Specific Heats of Some Common cv,av = 0.718 kJ/(kg-K) = 718 J/(kg-K). Ideal Gases With all the data known, T2 = (150)(10)(3600)/((718)(167.9) + 15 = 59.8 oC Tav = (15 + 59.8)/2.0 = 37.4 oC The calculated average temperature is pretty close to 35 oC (the assumed temperature), hence no iteration is needed. As a result, it looks like Anns dream of a cooler room in the evening is impossible after all.
- 95. Ch 3. First Law of Thermodynamics Multimedia Engineering ThermodynamicsConservation Conservation Solids and Ideal Gas of Mass of Energy Liquids Energy Analysis of Ideal Gas Case Intro Theory Case Solution Simulation Run Simulation Suggested Help Questions •How high the room temperature can rise? •How does the fan power affect the room temperature? Technical Help This simulation shows the temperature rise in Anns room when she turns on the fan for the whole period while she is at school. Use this slider to change the time Ann leaves to school. Use this slider to change the time Ann comes back. Use this slider to change the power of the fan. Use this menu to setup all the parameters and run the clock and thermometer. This window shows the room temperature when Ann comes back from school. This window shows the operation time of the fan.
- 96. Ideal Gas Specific Heats of Some Common Gases (kJ/kg-K) Temp. cP cv k cP cv k cP cv k K Air Nitrogen, N 2 Oxygen, O 2 250 1.003 0.716 1.401 1.039 0.742 1.400 0.913 0.653 1.398 300 1.005 0.718 1.400 1.039 0.743 1.400 0.918 0.658 1.395 350 1.008 0.721 1.398 1.041 0.744 1.399 0.928 0.668 1.389 400 1.013 0.726 1.395 1.044 0.747 1.397 0.941 0.681 1.382 450 1.020 0.733 1.391 1.049 0.752 1.395 0.956 0.696 1.373 500 1.029 0.742 1.387 1.056 0.759 1.391 0.972 0.712 1.365 550 1.040 0.753 1.381 1.065 0.768 1.387 0.988 0.728 1.358 600 1.051 0.764 1.376 1.075 0.778 1.382 1.003 0.743 1.350 650 1.063 0.776 1.370 1.086 0.789 1.376 1.017 0.758 1.343 700 1.075 0.788 1.364 1.098 0.801 1.371 1.031 0.771 1.337 750 1.087 0.800 1.359 1.110 0.813 1.365 1.043 0.783 1.332 800 1.099 0.812 1.354 1.121 0.825 1.360 1.054 0.794 1.327 900 1.121 0.834 1.344 1.145 0.849 1.349 1.074 0.814 1.319 1000 1.142 0.855 1.336 1.167 0.870 1.341 1.090 0.830 1.313 Temp. Carbon Dioxide, CO 2 Carbon Monoxide, CO Hydrogen, H 2 K 250 0.791 0.602 1.314 1.039 0.743 1.400 14.051 9.927 1.416 300 0.846 0.657 1.288 1.040 0.744 1.399 14.307 10.183 1.405 350 0.895 0.706 1.268 1.043 0.746 1.398 14.427 10.302 1.400 400 0.939 0.750 1.252 1.047 0.751 1.395 14.476 10.352 1.398 450 0.978 0.790 1.239 1.054 0.757 1.392 14.501 10.377 1.398 500 1.014 0.825 1.229 1.063 0.767 1.387 14.513 10.389 1.397 550 1.046 0.857 1.220 1.075 0.778 1.382 14.530 10.405 1.396 600 1.075 0.886 1.213 1.087 0.790 1.376 14.546 10.422 1.396 650 1.102 0.913 1.207 1.100 0.803 1.370 14.571 10.447 1.395 700 1.126 0.937 1.202 1.113 0.816 1.364 14.604 10.480 1.394 750 1.148 0.959 1.197 1.126 0.829 1.358 14.645 10.521 1.392 800 1.169 0.980 1.193 1.139 0.842 1.353 14.695 10.570 1.390 900 1.204 1.015 1.186 1.163 0.866 1.343 14.822 10.698 1.385 1000 1.234 1.045 1.181 1.185 0.888 1.335 14.983 10.859 1.380
- 97. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Energy Analysis Chapter 4 The First Law of Thermodynamics: Control Volume Topics Reviewed The topic menu above allows you to move directly to any of the four sections for each topic. The sections 1. Steady-flow Process are: 2. Steady-flow Devices (1): Nozzles, Diffusers, and Throttling Devices Case Intro: To help introduce and understand the 3. Steady-flow Devices (2): Turbines, basic principles, a case study is presented. compressors, and Mixing Chambers Theory: This section will review the basic principles 4. Steady-flow Devices (3): Heat and equations that you should know to answer the Exchangers, Pipe and Duct Flow exam questions. It does not give detailed derivations 5. Unsteady-flow Process 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.
- 98. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady Flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction John has a power washer, and plans to clean the siding of his house this weekend. John can not find the instruction of the power washer, hence he does not know the power needs for running it. What is known: • Water enters the washer at 20 oC and 1 atm with a volumetric flow rate of 0.1 L/s through a 25- cm- diameter hose • Water leaves the jet at 23 oC, 1 atm, and a velocity of 50 m/s at an elevation of 5 m • The magnitude of the heat transfer rate from the John and his House power unit to the surroundings is 10% of the power input Question • Determine the power input to the motor of the power washer. • The power input is used to increase the enthalpy, kinetic energy, and potential energy of the water. Which one dominates power consumption? Approach • Consider the power unit and the delivery hose as a control volume. "1" denotes the water at inlet and "2" denotes the water at exit • The control volume is a one-inlet-one-exit system. So the energy balance equation is:
- 99. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady Flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY A control volume may involve one or more forms of work at the same time. If the boundary of the control volume is stationary, the moving boundary work is zero, and the work terms involved are shaft work and electric work. Another work form with the fluid is flow work. Flow Work (Flow Energy) Work is needed to push the fluid into or out of the boundaries of a control volume if mass flow is involved. This work is called the flow work (flow energy). Flow work is necessary for maintaining a continuous flow through a control volume. Consider a fluid element of volume V, pressure P, and cross-sectional area A as shown left. The flow immediately upstream will force this fluid element to enter the control volume, and it can be regarded as an imaginary piston. The force applied on the fluid element by the imaginary piston is: A Flow Element F = PA The work done due to pushing the entire fluid element across the boundary into the control volume is Wflow = FL = PAL = PV For unit mass, wflow = Pv The work done due to pushing the fluid element out of the control volume is the same as the work needed to push the fluid element into the control volume. Flow Work with Imaginary Piston Click to View Movie (68 kB)
- 100. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady Flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS – THEORY (Continuation) Total Energy of a Flowing Fluid The total energy of a simple compressible system consists of three parts: internal, kinetic, and potential energy. E = U + KE + PE For unit mass, e = u + ke + pe = u + v2/2 + gz where e = total energy u = internal energy v = velocity of the system Total Energy of a Flowing Fluid z = the elevation of the fluid Click to View Movie (52 kB) The fluid entering or leaving a control volume possess an additional energy, the flow work (Pv). Hence, the total energy of a flowing fluid becomes θ = Pv + u + v2/2 + gz where θ = methalpy, the total energy of a flowing fluid The definition of enthalpy gives h = Pv + u Replacing Pv + u by h yields θ = h + v2/2 + gz By using the enthalpy instead of internal energy, flow work is not a concern.
- 101. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady Flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS – THEORY (Continuation) The Steady-flow Process Steady flow process is a process where: the fluid properties can change from point to point in the control volume but remains the same at any fixed point during the whole process. A steady-flow process is characterized by the following: • No properties within the control volume change with time. That is mcv = constant Ecv = constant • No properties change at the boundaries with time. Thus, the fluid properties at an inlet or exit will remain the same during the whole process. They can be different at different Steady-flow Process opens. Click to View Movie (60 kB) • The heat and work interactions between a steady- flow system and its surroundings do not change with time.
- 102. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady Flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS – THEORY (Continuation) Mass and Energy Balance for Steady-flow Process The conservation of mass principle, which has been previously introduced, in rate format, is: During a steady-flow process, the total amount of mass contained within a control volume does not change with time. That is, dmsystem/dt = 0 Hence the conservation of mass principle gives the total Mass and Energy balance for Steady-flow amount of mass entering a control volume equal to the Process total amount of mass leaving it. In an equation format, it is (Total mass entering the control volume per unit time) = (Total mass leaving the control volume per unit time) or, where i = inlet e = exit Also, the energy balance for a process, which has been previously introduced, in rate format, is: For a steady-flow process, the total energy content of a control volume remains constant. That is, dEsystem/dt = 0
- 103. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady Flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS – THEORY (Continuation) Mass and Energy Balance for Steady- flow Process (Continuation) Thus, the amount of energy entering a control volume in all forms (heat, work, mass transfer) must be equal to the amount of energy leaving it for a steady-flow process. In an equation format, it is (Rate of net energy transfer in by heat, work and mass) = (Rate of net energy transfer out by heat, work and mass) Or, For a general steady-flow process, the energy balance can be written as: If the sign introduced previously for heat and work is used, the energy balance for a general steady-flow process can be rewritten as:
- 104. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady Flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION John plans to wash his house using a power washer. The power input to the power unit needs to be determined. Assumptions: • The control volume enclosing the power unit and the delivery hose is at steady state • The water is modeled as incompressible (1) Determine the power input The energy balance for the one-inlet, one-exit steady-flow process is: Introducing , and solving for The specific volume of water v at 20 oC is 1.0018(10-3) m3/kg (from water table). The mass flow rate can be evaluated using the given volumetric flow rate. = V/v = (0.1/1000)/(1.0018(10-3)) = 0.1 kg/s Since water is incompressible, the specific heat c is 4.18 kJ/kg-K. Hence, the enthalpy term can be determined. h1 - h2 = c(T1 - T2) + P(v1 -v2) = 4.18(20-23) + 0 = -12,540 J/kg
- 105. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady Flow Process Case Intro Theory Case Solution Simulation Run Simulation Suggested Help Questions •How does the temperature affect the enthalpy term and the power input? •How does the elevation of the hose affect the potential energy term and the power input? •How does the exit velocity of the hose affect the kinetic energy term and the power input? Technical Help This simulation gives the analysis of the energy for a control volume. The hose elevation, the volume velocity, the exit velocity, and the exit temperature are given and the input power is calculated. Also, the proportions of the enthalpy, kinetic energy, and potential energy to the power input are compared. Use this slider to change the hose elevation. Use this slider to change the volumetric flow rate of the power washer. Use this slider to change the water exit velocity at the hose. Use this slider to change the exit temperature of the water. This window shows the power input to the power washer. This window gives the data used in this calculation.
- 106. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (1) Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction Belinda is a member of the race car club at her university. The club has a trip to the wind tunnel laboratory to view the testing of a new race car. A diffuser is used to slow the air flow and exhaust it. The engineer working there told them that for safety considerations, the velocity at the exit of the diffuser should be less than 3 m/s. But the temperature has not been checked. If the temperature is higher than 35oC, it will be harmful to the environment. Is the air from the wind tunnel safe to exhaust after flowing through the diffuser? What is known: Consider the Diffuser as a Control Volume • Air temperature at diffuser inlet T 1 = 10 oC • Pressure of air at diffuser inlet P1 = 80 kPa • The mass flow rate = 200 kg/s • The inlet area of the diffuser A1= 1.5 m2 Questions • Determine the velocity of the air at the inlet of the diffuser • Determine the temperature of the air leaving the diffuser Approach Consider the diffuser as a control volume. It is a one-inlet- one-exit system. During the test process, the flow can be treated as steady.
- 107. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (1) Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Many engineering devices operate essentially under the same conditions for long periods of time. These devices can be treated as steady-flow devices. Recall, the energy balance for a control volume is: Nozzles and Diffusers Nozzles and diffusers are commonly utilized in jet engines, rockets, and spacecrafts. Any fluid-mechanical device designed to accelerate a flow is called a nozzle and any fluid-mechanical device designed to decelerate a flow is called a diffuser. For subsonic flow (velocity under the speed of sound), a converging duct is a nozzle and a diverging duct is a diffuser. Nozzle and diffuser are single-stream (one-inlet-one-exit) systems. If Nozzle the inlet is denoted by subscript 1 and exit is denoted by subscript 2, Click to View Movie (36 kB) the mass balance and energy balance for single-stream steady-flow systems become Diffuser There are several common assumptions that are made in Click to View Movie (36 kB) the energy analysis of nozzles and diffusers: The fluid has high velocity and thus not spending enough time in the device for any significant heat transfer to take place.
- 108. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (1) Case Intro Theory Case Solution Simulation THERMODYNAMICS – THEORY (Continuation) Nozzles and Diffusers (continuation) Nozzles and diffusers are properly shaped ducts and no shaft or electric work can be transferred in or out. As a fluid passes through a nozzle or diffuser, it experiences a large change in its velocity. Hence, the kinetic energy change must be accounted for. The fluid experiences small or no change in its elevation as it flows through the nozzle or diffuser. Summarizing the above analysis, the energy balance for nozzles and diffusers is: (h2 - h1) + ( v22 - v12)/2 = 0 Throttling Devices A significant reduction in pressure can be achieved by introducing a restriction into a line through which a gas or liquid flows. This is commonly done by means of an adjustable valve, a porous plug, or a capillary tube. They are called throttling devices. There are several common assumptions that are made in the energy analysis of throttling devices: • Throttling devices are usually small devices. There is neither sufficient Click to view movie time nor large enough area for significant heat transfer to take place. • No work is done to or from the devices. • The increase in kinetic energy is insignificant. • The change in potential energy is very small. Hence, the energy balance for throttling devices is: h2 = h1 When the flow through a valve or other restrictions is idealized in this way, the process is called a throttling process.
- 109. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (1) Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION Air temperature at the exit of the wind tunnel needs to be determined to see if it is safe to exhaust the air to ambient. Assumptions: • Air is an ideal gas • Heat transfer is negligible • No change in potential energy • No work interactions (1) Determine the velocity at the inlet of the diffuser To determine the velocity at the inlet, the specific volume of the air needs to be first determined. Since the air in the tunnel is an ideal gas, it obeys the ideal-gas equation of state. Pv = RT Consider the Diffuser as a Control where Volume R = 287 J/(kg-K) for air v = specific volume of the air The specific volume can be determined at the inlet conditions: v1 = RT1/P1 = 287(273+10)/80,000 = 1.015 m3/kg The velocity can be calculated using the following equation: v1 = v1/A1 =200(1.015)/1.5 =135.3 m/s (2) Determine the temperature at the exit of the diffuser Under the stated assumptions and observations, the energy balance for the steady-flow through the diffuser can be expressed as (h2 - h1) + ( v22 - v12)/2 = 0 h2 = h1 - ( v22 - v12)/2 The enthalpy of air at the diffuser inlet can be determined from the air table to be h1 = h@283 k = 283.14 kJ/kg Assume the velocity at the exit is 0, then the enthalpy reaches the maximum value. h2 = 283.14 - ( 02 - 135.32)/2/1,000 = 292.3 kJ/kg From the air table, the temperature corresponding to this enthalpy value is T2 = 292 K = 19oC < 35 oC which shows that the air is safe to exhaust to the outside environment.
- 110. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (1) Case Intro Theory Case Solution Simulation Run Simulation Suggested Help Questions •For a specific inlet temperature, what is the inlet area for exit temperature under 35 oC? •For a specific inlet area, what is the inlet temperature for exit temperature under 35 oC? Technical Help This simulation shows the velocity profile for a diffuser used to exit air from a wind tunnel. The exit temperature is calculated for given inlet area, inlet temperature and exit velocity. Use this slider to change the temperature at the inlet of the diffuser. Use this slider to change the area at the inlet of the diffuser. Use this slider to change the air velocity at the exit of the diffuser. This window shows the exit air temperature. This window shows the air velocity at the inlet of the diffuser. This window shows the air pressure at the inlet and the mass flow rate of the air used in this calculation.
- 111. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (2) Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY ___________________________________________________________________________________________________________________________________________ Introduction The pressure of a citys main water line is much higher than the pressure in the water line of apartments. It is suggested that a hydraulic turbine be installed to utilize the water pressure to run an electric motor which can be supplied to the lighting system of the apartments. Is the power generated by the turbine enough to light the apartments? What is known: • Water at room temperature ( 300 K) enters the The Water System turbine at P1 = 750 kPa • Water at room temperature ( 300 K) leaves the turbine at P2 = 170 kPa • One apartment uses 10,000 kg water per hour • Each apartment needs 500 W power for illumination Questions Determine the power that could be produced by this hydraulic turbine. Approach Consider the turbine as a control volume. It is a one-inlet- one-exit system. The flow is steady. The energy balance is: Control Volume
- 112. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (2) Case Intro Theory Case Solution Simulation THERMODYNAMICS – THEORY Many engineering devices operate essentially under the same conditions for long periods of time. Therefore, these devices can be treated as steady-flow devices. Recall, the energy balance for a control volume is: Turbines, Compressors, and Pumps In steam, gas or hydroelectric power plants, the device that drives the electric generator is the turbine. A turbine is a device in which work is developed as a result of a gas or liquid passing through a set of blades. The blades, which are attached to the shaft, force the shaft to rotate. The work is positive since it is done by the fluid. How Does Turbine Work Compressors are devices which raise the pressure of the gas Click to View Movie (56 kB) that passes through them. In pumps, the working fluid is a liquid instead of a gas. Work is supplied to these devices through a rotating shaft from an external source. There are several common assumptions that are made in the energy analysis of turbines and compressors: • If there is no intentional cooling, the heat transfer for these devices is small relative to the shaft work. • The work term is important for all these devices. For turbines, it is the power output; for compressors or pumps, it is the power input. A Reciprocating Compressor • Click to View Movie (64 kB) The fluid velocities in turbines are high and the fluid experiences a significant change in kinetic energy. However, it is small relative to the change of enthalpy. The velocities involved in compressors and pumps are too low to cause any significant change in kinetic energy. • The potential energy changes of the fluid passing through turbines, compressors, and pumps are small and are neglected.
- 113. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (2) Case Intro Theory Case Solution Simulation THERMODYNAMICS – THEORY (CONTINUATION) Mixing Chambers In engineering applications, the steady-flow mixing of two streams of the same fluid is another common process. The section where the mixing process takes place is referred to as a mixing chamber. An open feedwater heater is an example of mixing chamber. If 1 and 2 denote the inlets and 3 denotes the exit, a mass balance gives An Open Feedwater Heater is a Mixing Chamber There are several common assumptions that are made in Click to View Movie (32 kB) the energy analysis of mixture chambers: • Mixing chambers are usually well insulated. Although the temperatures of the flow streams may be quite different from the temperature of the environment, the most important energy transfer is between the two fluids and not between the fluids and the environment. • No work is involved in the mixing process. • The increase in kinetic energy is insignificant. • The change in potential energy is negligible. Hence, the energy balance for mixing chamber is:
- 114. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (2) Case Intro Theory Case Solution Simulation THERMODYNAMICS – CASE SOLUTION A hydraulic turbine is used to produce power using for the apartments lighting. The power generated needs to be determined. Assumptions: • Heat transfer is negligible • Potential energy change is negligible • Kinetic energy change is negligible The energy balance for one-inlet-one-exit system is: Applying the basic assumptions for turbines, the energy balance Control Volume can be rewritten as: Water enters the turbine at 300 K and 750 kPa. At 300 K and 750 kPa, water is compressed liquid. The enthalpy of compressed liquid can be approximated by h = hf@T + vf (P - Psat.) where hf@T = the enthalpy of saturated water at temperature given vf = the specific volume of saturated water at temperature given Psat. = the saturation pressure at temperature given From the water table, the saturated properties of water at 300 K are: hf@300 k = 104.89 kJ/kg vf = 0.001003 m3/kg Psat. = 3.169 kPa Hence, enthalpy at 300 K and 750 kPa is: h1 = 104.89 + 0.001003(750 - 3.169) = 105.64 kJ/kg At 300 K and 170 kPa, water is compressed liquid also. Using the same approximation, the enthalpy of water at 300 K and 170 kPa is h2 = 104.89 + 0.001003(170 - 3.169) = 105.06 kJ/kg Substituting these enthalpies and the given mass flow rate to the energy balance equation yields - = 10,000/3,600(105.06 - 105.64) = -1.6 kW = 1.6 kW The power is generated by the system, so it is positive. The power needs for lighting an apartment is 500 W, which is smaller than the power generated by the turbine. 500 W < 1.6 kW = 1,600 W Hence, the result shows the turbine can be used to generate power for the lighting system.
- 115. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (2) Case Intro Theory Case Solution Simulation THERMODYNAMICS – SIMULATIONS Run Simulation Suggested Help Questions • How does the inlet pressure affect the power output? • How does the exit pressure affect the power output? • How does the flow rate effect the power output? Technical Help This simulation shows the lighting power produced by a hydraulic turbine, which is run by the domestic water pressure of a small city Use this slider to change the pressure at the inlet of the turbine. Use this slider to change the pressure at the exit of the turbine. Use this slider to change the water flow rate. This window shows the power produced by the hydraulic turbine. This window shows conditions used in this calculation.
- 116. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY ___________________________________________________________________________________________________________________________________________ Introduction The engineering computer lab is planning to discard some old computer parts this summer. Alex plans to build a computer by himself using those parts but is worried about the cooling capacity of the fan he obtained since the electronic components of a computer can only work under a specific temperature range. For the given fan, he needs to determine how many printed circuit boards (PCBs) he can use in the computer before it overheats. What is known: Cooling Computer by a Fan • Each PCB dissipates 10 W of power Click to View Movie • The air enters the box at 20 oC and 1 atm • The temperature at the exit can not exceed 32oC • The power of the fan is 20 W. CPU and other components other than PCBs dissipates 60 W of power • The flow rate of the fan is 0.01 kg/s Questions How many PCBs can he install in this computer? Approach Method 1: Consider the whole enclosure as a control volume. Method 2: Consider the air as control volume A and the fan, the CPU, and PCBs as control volume B.
- 117. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (3) Case Intro Theory Case Solution Simulation THERMODYNAMICS – THEORY (continuation) Pipe Duct Flow Pipes and ducts are used to transfer fluid from one device to another. Flow through a pipe or duct can be treated as a steady-flow process since the start-up and shut-down periods, which are transient process; are excluded. When flow through pipes or ducts are analyzed, the following points should be considered: • An Open Feedwater Heater is a If the pipes or ducts are insulated, heat transfer from the Mixing Chamber pipes or ducts to the environments is negligible. Under Click to View Movie (32 kB) normal operating conditions, the pipes or ducts are not insulated, hence the heat gained or lost is large. Sometimes heat transfer is the main purpose of the flow, for example, in heat exchangers. In this case, heat transfer needs to be accounted for. Pipe Are Used to Transfer Fluid from • one Device to another If the control volume involves pumps or fans or other work devices, the work interaction terms should be considered. Otherwise, the work term is zero. • The change in kinetic energy is insignificant, particularly when the pipe or duct has a constant diameter. • The change in potential energy is large when the fluid undergoes a considerable elevation change.
- 118. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (3) Case Intro Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION The heat transfer between the printed circuit boards (PCBs) and Steady-flow Devices (3) the computer needs to be calculated to the air passing through determine how many PCBs can be installed in the computer. Assumptions: • Air is an ideal gas with cP = 1.005 kJ/kg-K • Heat transfer from the box to the surroundings is negligible • No change in potential energy • For noise control, the air velocity is relatively low. Hence the kinetic energy is negligible • The temperature change of the computer components is negligible Consider the Whole Enclosure as a Control Volume (1) Method 1: If the whole enclosure is a control volume, the heat transfer term is zero due to the assumptions. Thus, under the stated assumptions, the energy balance can be expressed as: If air enters at 20 oC and exits at 32 oC, the enthalpy change is: The power equals the sum of the power of the fan, the power of CPU and other components, and the power supplied to the PCBs. It is negative since it is done to the system. =PCBs - 80 W PCBs is negative because power needs to supply to the system. Each PCB can dissipate 10 W power. It means 10 W power needs to supply to each PCBs. The number of the PCBs is given by 40.6/10 = 4
- 119. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (3) Case Intro Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION (CONTINUATION) (2) Method 2: Consider air as control volume A and the computer components as control volume B. Control Volume A: Heat transfer occurs between the air and the electric components, Hence, the heat transfer term is not zero. No work is done by the air or to the air. The energy balance is: From method 1, the enthalpy change is determined to be 120.6 W. Hence, the heat transfer is given by = 120.6 W The positive sign means that the heat transfer is from the computer components to the air. Control Volume B: Heat transfer occurs between the air and the computer components. Hence, the heat transfer term is not zero. Work is done to the components, and given by the sum of the power of the fan, the CPU and other components, and the the PCBs. According to the assumptions, no enthalpy change occurs in control volume B. Thus, the energy balance is: The amount of heat transfer from the PCBs to the air is the same as the heat transfer from the air to the PCBs, hence, = -120.6 W The negative sign indicates that heat is transferred from the system. The power supplied to the PCBs can be determined by w PCBs = -120.6 - (-20 - 60) = -40.6 W Note that wfan and wCPU have negative sign since power is applied to them. The result ob is the obtained same as in method 1, hence the number of PCBs is 4. Based on the results from both methods, for the given fan, Alex can only install up to 4 PCBs.
- 120. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Steady-flow Devices (3) Case Intro Theory Case Solution Simulation THERMODYNAMICS – SIMULATIONS Run Simulation Suggested Help Questions If the computer has 5 PCBs installed and the room temperature is 20 oC, what are the exit temperature and the fan power? Technical Help This simulation shows the computer cooling process by a fan. For a given fan, air inlet temperature and exit temperature, the number of PCBs that can be used in the computer is calculated. Use this slider to change the temperature at the inlet of the fan. Use this slider to change the air temperature at the exit of the enclosure. Use this slider to change the power of the fan. This window shows the number of the PCBs that the computer can contain. This window shows other given conditions for this calculation
- 121. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Unsteady-flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction Minnie volunteers to help the new semester welcoming party. Her job is to fill the balloons with helium from a supply line. The balloons have some limitations that she needs to consider when she fills them. What is known: • The supply line of helium has a pressure of 120 kPa and a temperature of 25 oC. • Mass flow rate from the supply line is 0.01 kg/s. • The elasticity of the balloon material is such that the pressure inside varies with volume according to the Problem Description relation P = 10 V, where P is the absolute pressure in Click to view movie (68 kB) the balloon (kPa) and V is the volume of the balloon (m3). • The highest temperature the balloon material can withstand is 50 oC. • A cooling system is used to cool the balloon during the filling process. 6 kJ of heat is taken away by the cooling system per second Questions • Determine how long will it take to fill each balloon? • Is it safe to use this filling system to fill the balloons? Approach • Take the balloon as a control volume. The filling process belongs to unsteady-flow process. • Helium can be treated as an ideal gas with R = 2.0769 kJ/kg-K, cP = 5.1926 kJ/kg-K, and cv = 3.1156 kJ/kg-K.
- 122. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Unsteady-flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Mass and Energy Balance of Unsteady-flow Processes In the previous sections, it was noted that nozzles, diffusers, turbines, compressors and other devices undergo a steady- flow process because of their long-time running consideration. But their startup and shutdown periods undergo transient operations since their states change with time. The flow processes involved are called unsteady-flow processes, or transient-flow processes. Unlike steady-flow processes, unsteady-flow processes start and end over some finite time period (Δt). An additional example of an unsteady- flow process is filling or discharging a tank. Unsteady-flow Processes During an unsteady-flow process, the mass in the control volume changes with time. The mass balance for a system undergoing Click to View Movie (64 kB) any process, can be used for control volume as where i = inlet e = exit ΔmCV = mCV@final - mCV@initial mi = the mass flow into the control volume through one inlet me = the mass flow out of the control volume through one exit Or in rate form Mass Balance for Unsteady-flow Processes where i = the rate of mass flow into the control volume through an inlet e = the rate of mass flow out of the control volume through an exit = the rate of change of mass within the control volume
- 123. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Unsteady-flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS – THEORY (CONTINUATION) Also, the energy content of a control volume changes with time during an unsteady-flow process. The general energy balance can be used for the control volume as Ei - Ee = ΔECV where Ei = the total energy transferred into the control volume by heat, work, and mass Ee = the total energy transferred out of the control volume by heat, work, and mass ΔECV = energy change in the control volume in forms of internal, kinetic, potential, etc., energies or in rate form Where = rate of energy transferred into the control volume by by heat, work, and mass = rate of energy transferred out of the control volume by heat, work, and mass = rate of energy change in the control volume in forms of internal, kinetic, potential, etc., energies Noting that the energy can be transferred by heat, work, and mass only, the energy balance can be rewritten as where θ = h + v2/2 + gz , the total energy of a flowing fluid per unit mass ΔECV = (ΔU + ΔKE + ΔPE)CV
- 124. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Unsteady-flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS – THEORY (CONTINUATION) Uniform-flow Processes Uniform-flow processes are special cases of unsteady-flow processes. During a uniform-flow process, the state of the control volume changes with time, but it does so uniformly. That is, • At any instant during the process, the state of the control volume is the same throughout. Hence, at an instant, the state of the mass leaving from the exit is the same as the state of the mass in the control volume. Uniform-flow Processes • The fluid flows at an opening is uniform and steady. Click to View Movie (72 kB) That is, the properties do not change with time or position over the cross section of an inlet or exit. But they are different at different openings. With these identifications, the mass and energy balances for uniform-flow processes become where 2 = final state of the control volume 1 = initial state of the control volume i = inlet e = exit When no mass is entering or leaving the control volume, and the kinetic and potential energy changes associated with the control volume are negligible, the energy equation can be reduced to the first law relation for closed system. m2 = m1= m System Reduces to a Closed System when all the Inlets and Exits are Closed
- 125. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Unsteady-flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION Balloons need to be filled for a welcoming party. Determine the heat that has to be removed to prevent the balloon from being overheated. Also determine the time needed to fill a single balloon. Assumptions: • The supply line is steady • The state inside the balloon is uniform • Neglect the changes in kinetic energy and potential energy • No work is involved • The cooling system is steady Take the balloon as a control volume. The filling process belongs to unsteady-flow processes since the balloon is empty initially and contains some helium when the filling process ends. It has only one inlet and has no exit. The pressure in the balloon will reach the pressure in the supply line at the end of the filling process. If the temperature in the balloon at the end of the filling process is less than 50 oC, Minnie can use this supply line to fill her balloons. Hence, temperature in the balloon at the end of the filling process (T 2) needs to be determined. (1) Determine the temperature in the balloon when the filling process ends. Take the Balloon as a Control Volume At the beginning of the filling process, the balloon is empty. If 1 denotes the initial state and 2 denotes the final filled state, then the mass at state 1 is m1 = 0 Since the balloon has only one inlet and no exit (me = 0), the mass balance for the control volume is m2 - m1 = mi m2 - 0 = mi m2 = mi The energy balance for unsteady-flow process,
- 126. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Unsteady-flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION (CONTINUATION) can be simplified using the initial assumptions (v = 0, z = 0, ke = 0, pe = 0) and the mass terms, giving Q = - mihi + m2u2 It should be noted that work and potential energy are not really zero, but they offset each other. The work done to filling the balloon is assumed to be 100% converted into potential energy. Next, the enthalpy and the internal energy of Helium can be expressed as hi = cPTi u2 =cv T2 Also, the total heat removed from the balloon by the cooling system during the filling process can be expressed as where t = time needed to fill one balloon = heat removed per unit time by the cooling system Substittuting hi, u2 and Q (recall, mi = m2) gives or T2 = 31 oC < 50 oC00 It is safe to fill balloons using this supply line.
- 127. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Unsteady-flow Process Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION (CONTINUATION) (2 ) Determine the time needed to fill a single balloon. Helium is an ideal gas and thus obeys the ideal-gas equation of state. P2V2 = m2 RT2 where R = the gas constant, R = 2.0769 kJ/kg-K Since the relation between the pressure and the volume of the balloon is given as P = 10 V, the volume of the balloon at state 2 can be determined. That is P2 = 10V2 V2 = 120 /10 = 12 m3 Hence, the mass in the balloon at the end of the filling process is m2 = 120(1,000)(12)/(2.0769(1,000)(31+273)) = 2.28 kg The assumption states that the flow is steady from the supply line. Hence, = 2.28/0.01 = 228 s = 3.8 minute It will take 3.8 minutes to fill one balloon.
- 128. Ch 4. Energy Analysis Multimedia Engineering ThermodynamicsSteady-flow Steady-flow Steady-flow Steady-flow Unsteady-flow Process Devices (1) Devices (2) Devices (3) Process Unsteady-flow Process Case Intro Theory Case Solution Simulation Run Simulation Suggested Help Questions If the mass flow rate from the supply lines remains 0.01 kg/s, how large will the balloon becomes after 1 minute, 2 minutes, and 3 minutes? Technical Help This simulation shows the filling process of balloons. The filling time and heat removed from the filling process are calculated and the relations of P-v, P-t, and V-t of the filling process are plotted. Use this slider to change the mass flow rate flowing from the supply line to the balloon. Use this slider to change the amount of heat removed by the cooling system. This window shows final temperature in the balloon after the filling. This window shows the time required to fill one balloon. Buttons to choose which plot is shown in the graph window. This window gives other conditions used in the calculations.
- 129. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Second Law of Thermodynamics Chapter 5 The Second Law of Thermodynamics Topics Reviewed The topic menu above allows you to move directly to any of the 1. Heat Engine four sections for each topic. The sections are: 2. The Second Law 3. The Carnot Cycle Case Intro: To help introduce and understand the basic 4. Carnot Heat Engine principles, a case study is presented. 5. Carnot Refrigerator and Heat Pump 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.
- 130. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction A town located on an island near the equator needs a new power plant to generate additional electricity to meet the increasing demand for electrical power. An Ocean Thermal Energy Conversion (OTEC) power plant is considered since it operates between the ocean surface and the deep ocean near the equator of the earth. For economic consideration, the thermal efficiency of the power plant can not be lower than 5%. The ocean surface water, which flows through the evaporator (a heat exchanger) leaves with a temperature of 22 oC due to the vaporization process of the working fluid (propylene) used in the OTEC power plant. The deep ocean water used in the plant is exhausted at 10 oC. The 10 oC water will be used by a local bottling company to help refrigerate drink. In order to design the pipe used to draw water from the ocean, the flow rate of the ocean surface water and deep ocean water that the power plant consumes needs to be determined. What is known: • The average temperature at the ocean surface is Tsurface= 27 oC. • The water exhausted from the evaporator has a temperature of 22 oC. • The temperature in the ocean changes with the depth. Schematic of an OTEC Power Plant The relation is: T = -0.005 h + Tsurface where T is temperature (oC) and h is the depth (m). • Deep ocean water from the depth of 1,000 m is sent to the power plant and left with a temperature of 10 oC. • The power required from this power plant is 15,000 kW. • The thermal efficiency of the power plant is 5%.
- 131. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Question Determine the volumetric flow rate of the ocean surface water and the deep ocean water Approach • The power plant is a heat engine since it generates power at the expense of heat input. • The ocean surface is a source and the deep ocean is a heat sink.
- 132. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Thermal Reservoirs A thermal reservoir is a specific kind of system with a large thermal energy capacity that can supply or absorb finite amounts of heat and always remains at constant temperature. Such a system can be approximated in a number of ways: •Large land masses •Earths atmosphere •Large bodies of water: oceans, lakes, or rivers •Any physical body whose thermal energy capacity is large relative to the amount of energy it supplies or absorbs, for example, a large block of ice Atmosphere, Land , and Water in a Lake are Examples of Thermal Reservoirs A reservoir that supplies energy in the form of heat is called a source and one that absorbs energy in the form of heat is called a sink. For example, atmospheric air is a source for heat pumps and a sink for air conditioners. Energy Analysis of Cycles When a system in a given initial state experiences a series of quasi-equilibrium processes and returns to its initial state, the system undergoes a cycle. The energy balance for any system Source and Sink undergoing a cycle takes the form ΔEcycle = Qcycle - Wcycle where Qcycle = the net amount of energy transferred by heat for the cycle, Qcycle = Qin- Qout Wcycle = the net amount of energy transferred by work for the cycle, Wcycle = Wout - Win Notice that the directions of the heat and work are indicated by the subscripts in and out. Therefore, Qin, Qout, Wout, and Win are all positive numbers.
- 133. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Energy Analysis of Cycles (continuation) Since the system is returned to its initial state after the cycle, there is no net change in its energy. Therefore, ΔEcycle = 0 Then the equation reduces to Qcycle = Wcycle This expression can satisfy every thermodynamic cycle, regardless of the sequence of processes followed by the system undergoing the cycle or the nature of the substances making up the system. Power Cycle and Refrigeration and If the system undergoing cycles delivers a net work to its Heat Pump Cycle surroundings during each cycle, the cycle is called a power Click to View Movie (72 kB) cycle. Wcycle = Qin – Qout On the other hand, if the system needs work input from the surroundings to run each cycle, the cycle is called a refrigeration and heat pump cycle. Wcycle = Qout - Qin where Wcycle has a positive value. Heat Engine Most people understand that work can always be converted to heat directly and completely. But converting heat to work requires the use of special devices. These devices are called heat engines. Heat engines operate on a cycle and receive heat from a high- temperature source, convert part of this heat to work, and then reject the remaining waste heat to a low-temperature sink during the cycle. What is a Heat Engine A steam power plant is an example of heat engine. The schematic Click to View Movie (60 kB) of a basic steam power plant is shown on the left. The cycle is: •Heat (Qin) is transferred to the steam in the boiler from a furnace, which is the energy source.
- 134. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Heat Engine (continuation) • The turbine produces work (Wout ) when steam passes through it. • A condenser transfers the waste heat (Qout) from steam to the energy sink, such as the atmosphere. • A pump is used to carry the water from the condenser back to the boiler. Work (Win) is required to compress water to boiler pressure. The net work output from this power plant is the difference between the work output and the work input. Schematic of a Basic Power Plant Wnet, out = Wout - Win From the energy balance of the cycle, the net work output is Wnet, out = Qin - Qout Thermal Efficiency A heat engine can only convert part of the energy it received from the source to work. A certain amount of heat is dissipated to the sink as waste heat. The fraction of the heat input that is converted to net work output is a measure of the performance of a heat engine and is called the thermal efficiency(ηth). In general, the efficiency (or performance) can be expressed in terms of the desired output and the required input as Performance = Desired output/ Required input For heat engines, the desired output is the net work output (Wnet, out) and the required input is the heat input( Qin). Hence the thermal efficiency of a heat engine can be expressed as
- 135. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Thermal Efficiency (continuation) Thermal efficiency = Net work output/Heat input or ηth= Wnet, out/Qin Since Wnet, out = Qin - Qout , it can be rewritten as ηth= (Qin- Qout)/Qin = 1 - Qout/Qin To bring uniformity to the treatment of heat engines, refrigerators, and heat pumps (wIll be introduced in the following paragraph), QH and QL are defined as Thermal Efficiency of Heat Engine • QH equals the amount of heat transferred between the Click to View Movie(48 kB) device (heat engines, refrigerators, and heat pumps) and a thermal reservoir of high temperature TH . • QL equals the amount of heat transferred between the device (heat engines, refrigerators, and heat pumps) and a thermal reservoir of low temperature TL. Note that QH and QL are all positive numbers. Hence, the thermal efficiency for any heat engine is: ηth= Wnet, out/QH = (QH- QL)/QH = 1 - QL/QH Note that for heat engine, QL is always less than QH ,and ηth is always less than 1.
- 136. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Thermal Efficiency (continuation) For refrigerators or heat pumps, the efficiency is in terms of the coefficient of performance (COP). A subscript R is used to denote refrigerators (COPR) and HP for heat pumps (COPHP). A refrigerator is used to remove heat (QL) from a lower temperature space with an electric work input (Wnet,in), then dissipates the total energy from the heat input and the electric work (QH) to a higher temperature thermal reservoir. Hence, the desired output is QL and the required input is Wnet, in. So the COPR can be expressed as Coefficient of Performance of Refrigerators COPR = Heat removed /Net work input Click to View Movie (48 kB) = QL / Wnet, in = QL /( QH- QL) = 1/(QH/QL-1) A heat pump is a device which transfers heat from a low- temperature medium to a high-temperature one. For example, a heat pump is used to heat a room in winter, which transfer heat from the low-temperature outdoor air to the high-temperature air inside the room. Hence, the desired output is the heat transferred to the room (QH). Also, a net work input (Wnet, in) is necessary. The COPHP can be expressed as COPR = Heat delivered/Required work input = QH / Wnet, in Coefficient of Performance of Heat Pumps = QH /( QH- QL) Click to View Movie (52 kB) = 1/(1-QL/QH)
- 137. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION The volumetric flow rates of the ocean surface water and deep ocean water need to be determined for a new Ocean Thermal Energy Conversion (OTEC) power plant. Assumptions: Assume the ocean water has the same properties as pure water and is modeled as incompressible fluid. (1) Determine the heat transferred to the power plant from the surface water ( H) and the heat transferred from the power plant to the deep ocean water ( L). To operate an OTEC power plant involves both a heat source and a heat sink. Therefore, the hot ocean surface water serves as the heat source and the cold deep ocean water serves as a heat sink. The power plant generates net work. Hence, it is a heat engine. The definition of thermal efficiency for a heat engine is Schematic of an OTEC Power Plant wherenet,out is the power generated by the power plant, which is 15,000 kW in this case. If the ηth is given as 5%, heat transferred from the heat source to the power plant is The energy balance of the power plant is
- 138. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION Hence the heat transferred to the sink from the power plant is = 300,000 - 15,000 = 285,000 kW (2) Determine the volumetric flow rates of the ocean surface water and the deep ocean water The ocean surface water is sent to the evaporator (a heat exchanger) where the working fluid propylene vaporizes. Consider the ocean water in the heat exchanger as a system, the heat transfer between the ocean water and the propylene can be determined as: Evaporator: Heat Transferred to Propylene where ΔT is the temperature difference of the ocean water from the Ocean Surface Water between the inlet and exit, which is ΔT = 27 - 22 = 5 oC Hence the volumetric flow rate ( ) of the surface water is The deep ocean water is sent to the condenser (a heat exchanger) where the propylene vapor condenses to liquid. The flow rate of the deep ocean water can be determined using a similar way. That is, Condenser: Heat Transferred to the Deep Ocean Water from Propylene
- 139. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION At 1,000 meter depth of the sea, the temperature of the water is Tdeep = -0.022 h + Tsurface = -0.022(1,000) + 27 = 5oC The water leaves the condenser at 10 oC. Hence, the temperature increase of the deep ocean water is ΔT = 10 - 5 = 5 oC In heat exchangers, properties are determined using the average temperature at the inlet and exit. For the evaporator, the average temperature of the water is 24.5 oC.The density and the specific heat of water at that temperature are: ρsurface = 997 kg/m3 and cP,surface = 4.18 kJ/kg-oC Also, the average temperature for the condenser is 7.5 oC, the density and the specific heat of water are: ρdeep = 997.8 kg/m3 and cP,deep = 4.22 kJ/kg-oC Substituting all these numbers to the flow rate equations, the flow rates can be determined. They are: surface = 300,000/((997)(4.18)(5)) = 14.4 m3/s deep = 285,000/((997.8)(4.22)(5)) = 13.5 m3/s
- 140. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions •How does the ocean surface temperature affect the flow rate of the ocean surface water? •How does the depth of the ocean affect the flow rate of the deep ocean water if the ocean surface temperature is fixed as 27 oC and a power output of 20 MW? Technical Help This simulation shows the design of an Ocean Thermal Energy Conversion power plant. The thermal efficiency of the power plant and the volume flow rate of the cold medium are determined. Use this slider to change the ocean surface temperature. Use this slider to change the ocean depth where the power plant pump ocean water to the condenser. Use this slider to change the required output of the power plant. This window shows the flow rate of the ocean surface water. This window shows the flow rate of the deep ocean water.
- 141. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator The Second Law Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction Max is playing in the backyard, focuses the sunlight with a magnifying lens to burn peanuts in a sieve, which his mother left in the sunlight for drying. His mother shouts to him "Be careful! Do not burn yourself!" Max shouts back "Ok. How hot is it? Can I boil water with the lens?" Maxs mother knows that the solar system using the sunlight can boil water. But she has no idea how hot the focus can be. That is, what is the upper limit of the temperature the focusing sunlight system can reach? Can you help her? What is known: Problem Description Click to View Movie (64 kB) Focusing the sunlight with a magnifying lens Question Can this focus be made hotter than the sun itself? Approach Assume this focus can be hotter than the sun itself Use the second law of thermodynamics to verify the assumption
- 142. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator The Second Law Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY In nature, processes occur in a certain direction, and not in the reverse direction. For example, if a block of ice is put into a cup of hot water, the ice will melt and the temperature of the water will decrease as a result of heat transferred from the higher temperature water to the lower temperature ice. This process is satisfied the first law of thermodynamics, which requires the energy lost by the water equals the energy gained by the ice. Consider the reverse process. A block of ice is put into a cup of hot water. Heat is transferred from the ice to the water. As a result, the water temperature increases and the ice temperature decreases. This process still obeys the first law. That is, the energy lost by the block of ice equals the energy gained by the water. But this process never occurs. Ice Melt in Hot Water This example shows that if a process only satisfies the first law, it may not occur. Hence, another law must also be obeyed to guarantee a process to happen. This is the second law of thermodynamics. A process will not occur unless it satisfies both the first and the second laws of thermodynamics. Clausius Statement of the Second Law The Clausius statement of the second law states It is impossible for any system to operate in such a way that the sole result would be an energy transfer by heat from a cooler to a hotter body. Heat can transfer from a cooler body to a hotter body if other effects accomplishing the heat transfer occur within the system or its surroundings, or both. Air conditioners and refrigerators are devices to transfer heat from a cool space to its hot surroundings. But both of them need power input. The Clausius Kelvin-Plank Statement of the Second Law statement says that an air conditioner cannot cool a room without power input.
- 143. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator The Second Law Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Kelvin-Planck Statement of the Second Law A heat engine must reject some energy to a heat sink in order to run a cycle. That is, no heat engine can convert all the heat it received from a high-temperature source to work. It is the basis for the Kelvin-Planck statement of the second law of thermodynamics, which is It is impossible for any system to operate in a thermodynamic cycle and deliver an equivalent amount of work to its surroundings while receiving energy by heat transfer from a single thermal reservoir. Kelvin-Plank Statement of the Second Law The Kelvin-Planck statement puts forward the idea that no heat engine can have 100 percent efficiency. It must exchange heat with a low-temperature heat sink as well as a high-temperature source to complete the thermodynamic cycle. Equivalence of Two Statement The Clausius statement and the Kelvin-Planck statement of the second law of thermodynamics are equivalent in their consequences. Any device that violates the Kelvin-Planck statement also violates the Clausius statement. Consider two devices A and B working between a high- temperature reservoir and a low-temperature reservoir. Device A is assumed to transfer Heat (QA,H) to work ( WA, net) and have 100 percent efficiency. That is, WA, net = QA,H Device A violates the Kelvin-Planck statement. Device B is a heat pump which receives heat (QB,L) from the low-temperature reservoir, receives work (WA, net) from device A and rejects the total energy of WA, net and QB,L to the high- Any Device that Violates the temperature reservoir. Heat rejected by device C, which is the Kelvin-Plank Statement also Violates combination of devices A and B, to the high-temperature the Clausius Statement reservoir equals Click to View Movie (88 kB)
- 144. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator The Second Law Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Equivalence of Two Statement (continuation) (QB,L + WA,net) - QA,H = (QB,L + WA, net) - WA, net = QB,L Device C receives heat (QB,L) from the low-temperature reservoir and rejects the same amount of heat (QB,L) to the high-temperature reservoir. It violates the Clausius statement. It can also be shown in a similar way that if a device violates the Clausius statement, it also violates the Kelvin-Planck statement. Hence, the Clausius statement and the Kelvin- Planck statement are two equivalent expressions of the second law of thermodynamics. Perpetual-Motion Machines Any Device that Violates the Clausius Statement also Violates The previous discussion shows that a process will not occur the Kelvin-Plank Statement unless it satisfies both the first and the second laws of Click to View Movie (104 kB) thermodynamics. But before this was commonly understood, a lot of efforts were spent to create machines which violate the first law or the second law. This kind of machines are called perpetual-motion machines. Devices that violate the first law are called perpetual-motion machines of the first kind (PMM1). Devices that violate the second law are called perpetual-motion machines of the second kind (PMM2).
- 145. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator The Second Law Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION Determine how hot the focal point of a magnifying lens can be. Assuming that the focused sun beam can reach a temperature higher than the temperature of the sun itself. A machine can be designed like this: A focusing sunlight system is used to focus sunlight. The assumption gives that the beam can reach a higher temperature than the sun. Hence, the sun with the focusing system can be used as a heat source and the sun itself can be used as a heat sink. A heat engine receives heat (QH) from the heat source, outputs work (Wout) and rejects the waste heat (QL) to the sun. The heat engine needs to satisfy the first law first. That is, A Heat Engine working with the Focusing Sunlight QH = QL + Wout The combined system of the lens and the heat engine gives a new heat engine. In this design, the sun is the heat source and also the heat sink. The sun gives heat QH and has QL back. The net heat transfer to the new heat engine from the sun is: QH - QL = (QL + Wout) - QL = Wout Hence, the new heat engine works like this: It receives heat from the sun only and convert it totally to work. This design violates the Kelvin-Planck statement of the second law and the new heat engine is a perpetual-motion machine of the second kind. It means the assumption that the focused sun beam could become The New Heat Engine hotter than the sun itself is obviously wrong. This case gives an idea that a focusing system can not produce a temperature higher than the source itself.
- 146. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator The Second Law Case Intro Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions •In order to decide if the machine is a perpetual-motion machine of the first kind, what needs to be checked? •In order to decide if the machine is a perpetual-motion machine of the second kind, what needs to be checked? Technical Help This simulation contains a series of machines and ideas that are claimed by inventors and scientists. Determine whether each machine 1. Violates the First Law of Thermodynamics 2. Violates the Second Law of Thermodynamics 3. Violates both the First and the Second Law of Thermodynamics 4. Violates none of the First and the Second Law of Thermodynamics Use these buttons to select different questions. Use these check-boxes to indicate your answer for the questions.
- 147. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator The Second Law Case Intro Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Use the reset button to reset the conditions. Use the score button to check your score at any time. Use this menu to switch between the questions and the answers. This window shows the questions and answers.
- 148. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator The Carnot Cycle Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction In order to win a design contract for a power plant, a company is soliciting ideas from its three design teams. The power plant receives heat from a underground hot spring and rejects the waste heat to the surrounding air. The three teams are required to submit the schematic of their design and report the efficiency of the power plant. What is known: •The temperature of water in the underground hot spring is 140 oC A Power Plant •The temperature of the surrounding air is 30 oC •The thermal efficiencies reported by the three teams are: Team A: 18.0% Team B: 26.6% Team C: 28.0% Question Based on the Carnot Principles, determine which design is possible. Approach Compare the thermal efficiency reported by the three teams with the thermal efficiency of a Carnot Heat Engine (will be introduced in next section) working between 140 oC and 30 oC, which is: ηREV = 1 - (30 + 273)/(140 + 273) = 26.6%
- 149. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator The Carnot Cycle Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Reversible and Irreversible Process A process is reversible if, after it has been carried out, it is possible to restore both the system and its entire surroundings to exactly the same states they were in before the process. If the system and its surroundings cannot return to their initial states at the end of the reversed process, this process is an irreversible process. A system can be restored to its initial state following a process, regardless if the process is reversible or not. If the surroundings can also be restored to its initial state, the process is reversible. Otherwise, the process is irreversible. Examples of Reversible and Irreversible Reversible process does not occur in nature. It is the idealization of Processes actual process and serves as an idealized model to which actual Click to View Movie (52 kB) process can be compared. The factors that cause a process to be irreversible are called irreversibilities. They include: • heat transfers through a finite temperature difference • unrestrained expansion of a gas • mixing of two gases • friction • electric current flow through a resistance • inelastic deformation • chemical reactions The process is irreversible if any of these effects present. Internally and Externally Reversible Processes When a process is carried out, irreversibilities can be found within the system as well as in the systems surroundings. A process is called internally reversible if the system can be restored through exactly the same equilibrium states which the system goes through. No irreversibilities occur within the boundaries of the system as it goes through the process. If no irreversibilities occur outside the system boundaries during the process, the process is called externally reversible. A process is called totally reversible, or reversible, if it is both internally and externally reversible.
- 150. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator The Carnot Cycle Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY The Carnot Cycle Heat engine operates on a cycle. The efficiency of heat engine depends on how the individual processes are executed. The most efficient cycles are reversible cycles, that is, the processes that make up the cycle are all reversible processes. Reversible cycles cannot be achieved in practice. However, they provide the upper limits on the performance of real cycles. Carnot cycle is one of the best-known reversible cycles. The The Carnot Cycle (1-2): Reversible Isothermal Carnot cycle is composed of four reversible processes. Consider Expansion an adiabatic piston-cylinder device that contains gas. The four Click to View Movie (36 kB) reversible processes that make up the Carnot cycle are as follows: • Reversible Isothermal Expansion (process 1-2): Heat transfer between the heat source and the cylinder occurs with an infinitesimal temperature difference. Hence, it is a reversible heat transfer process. Gas in the cylinder expands slowly, does work to its surroundings, and remains at a constant temperature TH. The total amount of heat transferred to the gas during this process is QH. The Carnot Cycle (2-3): Reversible Adiabatic Expansion • Reversible adiabatic expansion (process 2-3): Click to View Movie (40 kB) The heat source is removed, and the gas expands in an adiabatic manner. Gas in the cylinder continues to expand slowly, do work to its surroundings till the temperature of the gas drops from TH to TL. Assuming the piston moves frictionless and the process to be quasi- equilibrium, the process is reversible as well as adiabatic. • Reversible isothermal compression (process 3-4): The cylinder is brought into contact with a heat sink at temperature TL. The piston is pushed by an external force and which does work on the gas. During the compression, the gas temperature maintains at TL and The Carnot Cycle (3-4): Reversible the process is a reversible heat transfer process. The Isothermal Compression total amount of heat rejected to the heat sink from the Click to View Movie (40 kB) gas during this process is QL.
- 151. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator The Carnot Cycle Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY The Carnot Cycle (continuation) Reversible adiabatic compression (process 4-1): The heat sink is removed and the gas is compressed in an adiabatic manner. Gas in the cylinder continues to be compressed slowly, accepting work from its surroundings till the temperature of the gas rises from TL to TH. The gas returns to its initial state, which completes the cycle. The Carnot Cycle (4-1): Reversible Adiabatic Compression Click to View Movie (44 kB) The Carnot Principles If the processes that make up the cycle of the heat engine are all reversible processes, the heat engine is a reversible heat engine. Otherwise, it is an irreversible heat engine. In practice, all heat engines are irreversible since no reversible process exists in nature. The Carnot Principles are conclusions of the second law of thermodynamics. They are expressed as follows: • The efficiency of an irreversible heat engine is always less than that of a reversible one operating between the same two reservoirs. The Carnot Principles • The efficiencies of all reversible heat engines operating between the same two reservoirs are the same.
- 152. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator The Carnot Cycle Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION Three designs for a power plant using the geothermal energy are proposed by a company. Use the Carnot principles to determine which design is possible. Power plant receives heat from a heat source (the hot spring), output net work, and rejects heat to a heat sink (the ambient air). Hence, it is a heat engine. All heat engines in practice are irreversible. The Carnot principles state that the efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs. The thermal efficiency of a Carnot heat engine working between 140 oC and Carnot Principles 30 oC is 26.6%. Team A B C efficiency 18.0% < 26.6% 26.6% = 26.6% 28.0% > 26.6% Possibility possible impossible impossible
- 153. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator The Carnot Cycle Case Intro Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions A scientist declared that a power plant which works between a hot spring at 150 oC and the environmental air at 30 oC has a thermal efficiency of 40%. Is it possible? Technical Help This simulation shows the possibility of a geothermal power plant according to its thermal efficiency. Use this slider to change the water temperature in the hot spring. Use this slider to change the environmental air temperature. Use this slider to change the thermal efficiency of the power plant. This window shows the possibility of the power plant according to the given thermal efficiency.
- 154. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Carnot Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction The bottom zone of a pond can be used as a heat source since that zone has a concentrated salt solution with a higher temperature from absorbing solar radiation. This type of pond is called a solar pond. Solar ponds have been used to produce hot water and have good efficiency. A company wants to use this concept to generate power. Hence, a model pond is built to test its performance. What is known: • The size of the pond is 10m x 10m x 3m. How a Solar Pond Traps Solar Energy • The ambient air temperature is 35 oC. Click to View Movie (80 kB) • The temperature at the bottom of the pond is 85 oC. • The salt water in the bottom layer absorbs solar radiation at the rate of 0.3 kW/m2. Question • Determine the maximum thermal efficiency that this power plant can have. • Determine the maximum power generated by each m2 of the bottom of the pond. Approach • The power plant is a heat engine since it generates power at the expense of heat input. Carnot heat engine has the highest thermal efficiency. •ηth,rev = 1- TL/TH • The thermal efficiency of a heat engine, regardless reversible or irreversible, is given by ηth = 1- QL/QH Heat Engine Works between the Solar Pond and the Ambient Air
- 155. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Carnot Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY The Thermodynamic Temperature Scale A temperature scale which is independent of the properties of the substances that are used to measure temperature is called a thermodynamic temperature scale. Its derivation is given below using some Carnot heat engines. The Carnot principle states that the reversible heat engines have the highest efficiencies when compared to irreversible heat engines working between the same two reservoirs. And the efficiencies of all reversible heat engines are the same if they work between the same two reservoirs. That is, the efficiency of a reversible heat engine is independent on the working fluid used and its properties, the way the cycle operates, and the type of the heat engine. The efficiency of a reversible heat engine is a function of the reservoirs temperature only. ηth = 1 - QL/QH = g(TH,TL) Carnot Principle or QH/QL = f(TH,TL) where QL = heat transferred to the low-temperature reservoir which has a temperature of TL QH = heat transferred from the high-temperature reservoir which has a temperature of TH g, f = any function Consider three reversible heat engines working between a high-temperature reservoir at temperature T1 and a low- temperature reservoir at temperature T3. Engine A receives Q1 from the high-temperature reservoir,and rejects Q2 to engine B at temperature T2. Engine B receives Q2 from engine A at temperature T2, and rejects Q3 to the low- temperature reservoir. Engine C receives Q1 from the high- temperature reservoir,and rejects Q3 to the low- temperature reservoir. Applying the equation above to the Relations between Reversible Heat three engines separately yields, Engines A, B , and C Click to View Movie (80 kB)
- 156. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Carnot Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY The Thermodynamic Temperature Scale (continuation) Engine A: Q1/Q2 = f(T1,T2) (1) Engine B: Q2/Q3 = f(T2,T3) (2) Engine C: Q1/Q3 = f(T1,T3) (3) Multiplying equations (1) and (2) gives (4) Comparing (4) and (3) yields, f(T1,T3) = f(T1,T2)f(T2,T3) (5) The left-hand side of equation (5) is only a function of T1 and T3, and thus the right-hand side of equation (5) is also a function of T1 and T3 only. This condition will be satisfied only if the function f has the following form: Substituting them to equation (5) gives, or, (6) Equation (6) provides a basis to define a thermodynamic temperature scale. There are several choices for the function φ. The Kelvin scale is obtained by making a simple choice as φ(T) = T
- 157. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Carnot Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY The Thermodynamic Temperature Scale (continuation) Hence, equation (6) becomes, Q1/Q3 = T1/T3 To bring uniformity to the treatment of heat engines, H is used to denote the high-temperature reservoir, and L to denote the low-temperature reservoir. The above equation becomes, QH/QL = TH/TL The Kelvin scale temperature is called absolute temperature. At the International Conference on Weights and Measures held in 1954, the triple-point of water was assigned the value 273.16 K. Then, if a reversible cycle is operated between a reservoir at 273.16 K and another reservoir at temperature T, temperature T can be expressed as T = 273.16(Q/Qtp)rev. cycle Determine Absolute Temperature by Measuring Heat Transfers QH and QL where Q and Qtp are heat transfer between the cycle and the reservoirs at temperature T and at temperature 273.16 K, respectively. The magnitude of a Kevin scale is defined as 1/273.16 of the temperature interval between the absolute zero and the tripe- point of water. The magnitude of temperature units on Kevin scale is the same as that of Celsius. 1 K 1 oC When numerical values of the thermodynamic temperature are to be determined, it is not possible to use a reversible cycle since that does not exist in practice. Absolute temperatures can be measured by other means, such as the constant-volume ideal gas thermometer.
- 158. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Carnot Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY The Carnot Heat Engine A heat engine is called a Carnot heat engine if it operates on the reversible Carnot cycle. The thermal efficiency of heat engine, regardless reversible or irreversible, is given as ηth = 1- QL/QH where QL and QH are heat transfer between the cycle and the low-temperature reservoir at temperature TL and the high- temperature reservoir at temperature TH, respectively. For reversible heat engines, the heat transfer ratio can be replaced by the absolute temperatures of the two reservoirs. ηth = 1- TL/TH This efficiency is called the Carnot efficiency since the Carnot heat engine is one of the well-known reversible engines. It is A Carnot Heat Engine has the Highest the highest efficiency that a heat engine working between a Efficiency of any Heat Engines Working high-temperature reservoir at temperature TH and a low- between the Same Source and Sink temperature reservoir at temperature TL can reach. All Click to View Movie (72 kB) irreversible heat engines working between the same two reservoirs have lower efficiencies. < ηth,rev irreversible heat engine ηth = ηth,rev reversible heat engine > ηth,rev impossible heat engine
- 159. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Carnot Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION A solar pond is built to test the possibility of generating power. The maximum thermal efficiency and the maximum power generation need to be determined. (1) Determine the maximum thermal efficiency Assuming the power plant is a Carnot heat engine, which works between a source at temperature 85 oC, and a sink at temperature 35 oC. ηth,rev = 1 - (35+273)/(85+273) = 14% (2) Determine the maximum power generated by the pond The definition of the thermal efficiency of a heat engine is ηth = Wnet,out/Qin A Heat Engine can Work between a Solar Rearranging this equation gives Pond and the Ambient Air Wnet,out = Qin ηth If the efficiency of this heat engine equals the Carnot efficiency, and the heat absorbed by the salt water is totally input to the heat engine, the output net work can achieve its maximum value. Wnet,max = QHηth,rev The bottom area of the pond is Apond = (10) (10) = 100 m2 Heat absorbed by the salt water in the pond bottom area is QH = (0.3) (100) = 30 kW Hence, the maximum net work output by the heat engine equals Wnet,out,max = Qin ηth,rev = ( 30)(0.14) = 4.2 kW
- 160. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine Refrigerator Carnot Heat Engine Case Intro Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions •How does the solar radiation rate effect the thermal efficiency? •How can you increase the power generated by this pond? Technical Help This simulation evaluates the design of an solar pond power plant. The maximum thermal efficiency of this solar pond and the maximum power generated by the pond are determined. Use this slider to change the solar pond surface temperature. Use this slider to change solar pond bottom temperature. Use this slider to change the solar radiation rate absorbed by the bottom of the solar pond. This window shows the maximum thermal efficiency of this solar pond. This window shows the maximum power generated by the pond.
- 161. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine RefrigeratorCarnot Refrigerator and Heat Pump Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY Introduction A refrigerator in a kitchen dissipates heat to the kitchen air through the fin coil unit located on the back of the refrigerator. If a heat engine is operating between the fin coil unit and the surrounding air, can it generate power? What is known: •The coefficient of performance (COP) of the refrigerator is 3.0. •The power input to the refrigerator is 1.0 kW. Problem Description •The average surface temperature of the fin coil unit Click to View Movie (56 kB) is 30 oC. •The kitchen air temperature is 20oC. Question Determine the maximum power generated by the heat engine operating between the the fin coil unit and the kitchen air. Approach •Determine the heat dissipated from the the fin coil unit, which is the heat input to the heat engine. + •Assume the heat engine is a Carnot heat engine. Its How a Refrigerator Works thermal efficiency is determined by Click to View Movie (68 kB) ηth,rev. = 1 - TL/TH
- 162. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine RefrigeratorCarnot Refrigerator and Heat Pump Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY The Reversed Carnot Cycle The Carnot cycle is a totally reversible cycle. The Carnot refrigeration cycle can be achieved if one reverses all the processes in the Carnot power cycle. In this case, heat in the amount of QL is received by the gas from a heat sink and heat in the amount of QH is rejected to a heat source, and a work input of Wnet,in is required to accomplish the cycle. The P-v diagram of the reversed Carnot cycle is shown on the left. The Carnot Refrigerator and Heat Pump P-v Diagram of the Reversed Carnot Cycle A refrigerator or a heat pump that operates on the reversed Carnot cycle is called a Carnot refrigerator or a Carnot heat pump. The coefficient of performance(COP) of reversible or irreversible refrigerator or heat pump is given by COPR = 1/((QH/QL)-1) COPHP = 1/(1-(QL/QH)) where QH = the amount of heat rejected to the high-temperature reservoir QL = the amount of heat received from the low-temperature reservoir For reversible refrigerators or heat pumps, such as Carnot refrigerators, or Carnot heat pumps, the COPs can be determined by replacing the heat transfer ratios in the above equations by the absolute temperature ratios. These are, COPR,rev = 1/((TH/TL)-1) COPHP,rev = 1/(1-(TL/TH)) where TH = the absolute temperature of the high-temperature reservoir TL = the absolute temperature of the low-temperature reservoir
- 163. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine RefrigeratorCarnot Refrigerator and Heat Pump Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY The Carnot Refrigerator and Heat Pump (continuation) COPR,rev (or COPHP,rev) is the highest COP a refrigerator (or a heat pump) which operates between a high-temperature reservoir at temperature TH and a low-temperature reservoir at temperature TL can reach. All irreversible refrigerators or heat pumps working between the same two reservoirs have lower COPs. < hR,rev irreversible refrigerator COP = hR,rev reversible refrigerator The Carnot Heat Pump Has the R Highest COP > hR,rev impossible refrigerator < hHP,rev irreversible heat pump COPHP = hHP,rev reversible heat pump > hHP,rev impossible heat pump Household Refrigerators The basic idea behind a household refrigerator is simple: it uses the evaporation of the refrigerant to absorb heat from the cooling space in the refrigerator. There are four basic parts to any refrigerator. • Compressor • Fin coil unit outside the refrigerator • Expansion valve • Heat-exchanging pipes inside the refrigerator How a Refrigerator Works Click to View Movie (68 kB)
- 164. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine RefrigeratorCarnot Refrigerator and Heat Pump Case Intro Theory Case Solution Simulation THERMODYNAMICS - THEORY Household Refrigerators (continuation) The basic mechanism of a refrigerator works like this: •The compressor compresses the refrigerant vapor, increasing the temperature and pressure of the refrigerant vapor. •The high-temperature and high-pressure refrigerant vapor dissipates heat to the ambient when it flow through the fin coil unit outside the refrigerator. The refrigerant vapor condenses into liquid form. •The cool liquid refrigerant flows through the expansion valve to reduce its pressure. •The cool liquid refrigerant vaporizes when it flows through the fin coil unit inside the refrigerator and receives heat from the inner space of the refrigerator, making the inner space of the refrigerator cold. Then the gas enters the compressor to be compressed again. The cycle repeats itself. A household refrigerator is a combination of refrigerator- A Household Refrigerator freezer. It is designed to maintain the freezer section at -18 oC and the refrigerator section at 3 oC.
- 165. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine RefrigeratorCarnot Refrigerator and Heat Pump Case Intro Theory Case Solution Simulation THERMODYNAMICS - CASE STUDY SOLUTION A heat engine is operating between a refrigerators outside fin coil unit and the kitchen air. Determine the maximum power generated by this heat engine. (1) Determine the heat dissipated from the refrigerator The coefficient of performance(COP) of a refrigerator is expressed as COPR = QL /Wnet,in = (QH - Wnet,in)/Wnet,in Rearranging this equation gives QH = (1 + COPR)Wnet,in = (1 + 3.0)(1.0) = 4.0 kW (2) Determine power generated by the heat engine The thermal efficiency of a heat engine is expressed as Problem Description ηth = Wnet,out/QH where Wnet,out = power generated by the heat engine QH = heat transferred to the heat engine Rearranging this equation yields Wnet,out = ηth QH where QH is the heat transferred from the heat-exchange pipes to the heat engine, which is the heat dissipated from the refrigerator, hence QH = 4.0 kW The thermal efficiency of a Carnot heat engine is ηth,rev = 1- (273 + 20)/(273 + 30) = 3.3% The work output from this Carnot heat engine is Wnet,out = ηth,rev QH = (3.3%)(4.0) = 0.12 kW The result shows the power generated from this Carnot heat engine is only enough to light a 100 W bulb. Hence, there is little benefit to use the heat dissipated by a refrigerator.
- 166. Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics The Second Carnot Carnot Heat CarnotHeat Engine Law Cycle Engine RefrigeratorCarnot Refrigerator and Heat Pump Case Intro Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions How can you increase the power generated by the heat engine? Technical Help This simulation tries to evaluate the idea toh use the heat dissipated by a household refrigerator to generate power. Use this slider to change the temperature of the outside heat-exchanging pipes. Use this slider to change the temperature of the kitchen air. Use this slider to change COP o of the refrigerator. Use this slider to change the power input to the refrigerator. This window shows the power generated by the heat engine which is working between the kitchen air and the refrigerator.
- 167. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Chapter 6 Entropy Topics Reviewed The topic menu above allows you to move directly to any of the four sections for each topic. The 1.Entropy sections are: 2.Tds Relations 3.Entropy Change Case Intro: To help introduce and understand the 4.Isentropic Process basic principles, a case study is presented. 5.Isentropic Efficiency 6.Entropy Balance Theory: This section will review the basic 7.Entropy Balance for Control Volume principles and equations that you should know to 8.Reversible Steady-flow Work 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.
- 168. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY Introduction According to a recent new law, refrigeration systems are no longer allowed to use refrigerant R-12 due to environmental concerns. A cryogenic manufacturing facility has to choose a new refrigerant to replace R-12 used in its refrigeration systems. Refrigerant R-134a is one choice which satisfies the energy requirement. To further evaluate this new refrigerant, the entropy changes need to be determined. What is known: •Refrigerant R-134a enters the evaporator of the Heat Exchange in the Evaporator refrigeration system as a saturated liquid-vapor mixture Click to View Movie (48 kB) at a pressure of 200 kPa. •Refrigerant R-134a leaves the evaporator as saturated vapor at a pressure of 200 kPa. •Refrigerant R-134a absorbs 150 kJ of heat from the cooled space. •The cooled space of the refrigeration system is maintained at -4oC. Question • Determine the entropy change of refrigerant R-134a. • Determine the entropy change of the cooled space. • Determine the total entropy change of the whole process. Approach • The evaporator is a heat exchanger which has one inlet and one exit. • Refrigerant R-134a enters the evaporator as a saturated liquid-vapor mixture and leaves the evaporator as saturated vapor. Hence the temperature in the evaporator remains constant at the saturation temperature of 200 kPa.
- 169. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Clausius Inequality The second law of thermodynamics is developed for systems undergoing cycles while exchanging heat with two reservoirs. One corollary of the second law is known as the Clausius inequality, which states that the cyclic integral of is always less than or equal to zero. That is, where delta;Q = differential heat transfer at the system boundary during a cycle T = absolute temperature at the boundary = integration over the entire cycle The Clausius inequality is valid for all cycles, reversible or irreversible. To demonstrate the Clausius inequality, consider two heat engines that work between the same two reservoirs. Heat engine A is an reversible heat engine and heat engine B is an irreversible one. Both of these two engines absorb heat QH from a heat source, which has a temperature of TH and reject heat to a heat sink at temperature TL. Applying the first law to both of the engines gives, WA,rev = QH - QA,L WB,irr = QH - QB,L Since a reversible engine has the highest efficiency of any heat engines working between the same source and sink, a reversible heat engine produces more work than an irreversible heat engine for the same heat input QH. WA,rev = QH - QA,L > WB,irr = QH - QB,L QA,L < QB,L
- 170. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Clausius Inequality (cont.) Since the temperature TH is constant during the heat transfer QH, and the temperature TL is constant during the heat transfer QA,L and QB,L, the cyclic integral of for both heat engines is Demonstration of the Clausius Inequality Comparing the above two equations with QA,L < QB,L yields, (1) A reversible heat engine holds the following relation: Therefore, the cyclic integral of for a reversible engine is zero. Combining the above equation with equation (1) gives, To summarize the above analysis, the Clausius inequality states for internally reversible cycles for irreversible cycles
- 171. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Entropy From the discussion above, , the Clausius inequality states, for an internally reversible cycle The cyclic integral of is zero. Therefore, is a property. Clausius in1865 named this new thermodynamic property entropy. The definition is The entropy change for process1-2 can be determined as Note that the above definition actually defines the change of entropy, instead of entropy itself. Because entropy is a The Entropy Change between two Specified property, the entropy change between two specified States are the Same Regardless the process states is the same whether the process is reversible or Is Reversible or Not not. The previous discussion only means that entropy is a useful property in the second law analysis of an engineering device. But what is entropy? Entropy can be viewed as a measure of molecular disorder. In other words, entropy is the amount of disorder in a system. A pure substance at absolute zero temperature is in perfect order, and its entropy is zero. This is the third law of thermodynamics. The entropy of a pure crystallite substance at absolute zero is zero.
- 172. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Entropy (cont.) If the substance is not a pure crystal like substance, its entropy at absolute zero is not zero. When the molecular disorder is produced, the ability of doing work is reduced. The Increase of Entropy Principle Consider a cycle as described in the diagram at the left. Peocess1-2 is an arbitrary (reversible or irreversible) process, Process 2-1 is an internally reversible process. From the Clausius inequality, Cycle 1-2-1 or where the equality holds for internally reversible process and inequality holds for irreversible process. For an isolated system, the heat transfer at the system boundary is zero. The above equation becomes ΔSisolated ≥ 0 This equation is known as the increase of entropy principle. It states The entropy of an isolated system during a process always increases, or remains constant if the process is reversible.
- 173. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY The Increase of Entropy Principle (cont.) Entropy is an extensive property, and thus the total entropy of a system is equal to the sum of the entropies of all parts of the system. The combination of a system and its surroundings is an isolated system. For this case, the increase of entropy principle is ΔSisolated = ΔSsys + ΔSsurr ≥ 0 The entropy change of a system can be negative. But the entropy change of an isolated system formed by this system and its surroundings can not be negative. Negative Entropy Change of a System The second law of thermodynamics states that processes Click to View Movie (48 kB) can occur only in a certain direction. But which direction? The increase of entropy principle answers this question: a process must proceed in the direction that complies with the increase of entropy principle. A process that violates this principle is impossible.
- 174. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION Refrigerant R-134a vaporizes in the evaporator and absorbs heat from the cooled space. Entropy changes of the Refrigerant R-134a, the cooled space and the whole process need to be determined. Saturated R-134a Temperature Table Refrigerant R-134a enters the evaporator as a saturated liquid-vapor mixture and leaves the evaporator as Saturated R-134a Pressure Table saturated vapor. Hence the temperature in the Superheated Steam Table evaporator remains constant at the saturation temperature of 200 kPa, which is, Tsat. @ 200 kPa = -10.09oC (1) Determine the entropy change of refrigerant R-134a Take refrigerant R-134a as a system (system A). It enters the evaporator as a saturated liquid-vapor mixture, which is state 1. P1 = 200 kPa T1 = Tsat. @ 200 kPa = -10.09oC Refrigerant R-134a leaves the evaporator as saturated Heat Transferred into System A vapor, which is state 2. P2 = 200 kPa T2 = Tsat. @ 200 kPa = -10.09oC During this process, refrigerant R-134 absorbs 150 kJ heat. Q = 150 kJ Hence, the entropy change for refrigerant R-134a from process 1-2 is Temperature T is a constant in this case, and the above equation can be integrated as Substituting Q and T into the above equation yields, ΔSA = 150(1000)/(-10.09 + 273) = 570.56 J/K
- 175. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION cont. (2) Determine the entropy change of the cooled space Take the cooled space as a system (system B). During the heat transfer process, the space is maintained at -4oC and it dissipates 150 kJ heat to refrigerant R-134a. The entropy change of the cooled space can be determined as ΔSB =Q/T = -150(1000)/(-4 + 273) = -557.6 J/K Heat Transferred out of System B (3) Determine the total entropy change of the process Take refrigerant R-134a and the cooled space together as a system (system C). Then refrigerant R-134a and the cooled space are subsystems of system C. The entropy change of system C is the sum of the entropy changes of its two subsystems. ΔSC = ΔSA+ ΔSB = 570.56 + (-557.6) = 12.44 J/K System C is an Isolated System System C is an isolated system since no heat and mass transfer out of its boundaries. The entropy change of system C is greater than 0, which satisfies the increase of entropy principle.
- 176. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Case Study Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions •What is the sign of the entropy change of the cooled space? •What is the sign of the entropy change of refrigerant R-134a? •What is the sign of the entropy changes of the sum of the cooled space and refrigerant R-134a? Technical Help This simulation shows the entropy changes of refrigerant R-134a used as a new refrigerant in the refrigeration systems. Use this menu to review three different systems. System A is the system involved refrigerant R-134a. System B is the system involved the cooled space. System C is the sum of system A and B. Use this slider to change the heat transferred from the cooled space to refrigerant R-134a. Use this slider to change the temperature at the cooled space. This window shows the entropy changes of the refrigerant R-134a, cooled space and the total system. This window gives the other conditions used in this calculations.
- 177. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Tds Relations Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY Introduction A manufacturing facility has a production line to fill its air conditioning systems with liquid refrigerant-134a. A throttling valve is used to reduce the pressure of refrigerant-134a before filling. If the throttling valve is replaced by a turbine, which can generate power, how mach can this turbine save in electricity cost? What is known: • Refrigerant-134a is at 0.3 MPa after throttling. • Refrigerant-134a is at -4oC and 1 MPa before throttling. • The mass flow rate of refrigerant-134a is 200 kg/s. Problem Description Click to View Movie (62 kB) Question • Determine the maximum power that the turbine can produce. • If the turbine operates 8,760 hr/yr, and electricity costs the facility $0.08/kWh, how much can the turbine save per year? Approach • Take the turbine as a system. This is a control volume. • Assume the turbine is adiabatic and the process is reversible. That gives, S1 = S2
- 178. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Tds Relations Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY In the second law analysis, it is useful to plot the process Saturated Water temperature Table on diagrams for which has one coordinate is entropy. The two diagrams commonly used in second law analysis are Saturated Water Pressure Table temperature-entropy (T-s) and enthalpy-entropy (h-s) Superheated Water Table diagrams. For some pure substance, like water, the entropy is tabulated with other properties. The T-s Diagram On a P-v diagram, the area under the process curve is equal, in magnitude, to the work done during a quasi- equilibrium expansion or compression process of a closed system. On a T-s diagram, the area under an internally reversible process curve is equal, in magnitude, to the heat transferred between the system and its surroundings. That is, The Total Heat Transfer Equals the Total Area under the Process Curve on the T-s Note the area has no meaning for irreversible processes. Diagram Click to View Movie (30 kB) The T-s diagram of a Carnot cycle is shown on the left. The area under process curve 1-2 (area 1-2-B-A-1) equals the heat input from a source (QH). The area under process curve 3-4 (area 4-3-B-A-4) equals the heat rejected to a sink (QL). The area enclosed by the 4 processes (area 1-2-3-4-1) equals the net heat gained during the cycle, which is also the net work output. T-s Diagram of a Carnot Cycle
- 179. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Tds Relations Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY The h-s Diagram The enthalpy-entropy (h-s) diagram is valuable in the analysis of steady-flow devices such as nozzle, turbine, and compressor. The h-s diagram is also called Mollier diagram after the German scientist R. Mollier. The Mollier diagram of water is shown on the left. The Mollier diagram for water contains constant-quality lines, constant-pressure lines, and constant-temperature lines. The temperature lines in the mixture region are straight. How to Use the Mollier Diagram Click to View Movie (125 kB) The Tds Relations Mollier Diagram for Water: SI Units Mollier Diagram for Water: US Units In the previous section, the definition of entropy is given by Rearranging the above equation gives (1) The entropy change during an internally reversible process (1-2) is Only when the relation between δQ and T is known, the entropy change can be determined. The relations between δQ and T can be found by considering the energy balance of a closed system. The Tds Relations for Closed System
- 180. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Tds Relations Case Study Theory Case Solution Simulation THERMODYNAMICS – THEORY The Tds Relations (cont.) The differential form of the energy balance for a closed system, which contains a simple substance and undergoes an internally reversible process, is given by dU = δQrev - δWrev (2) The boundary work of a closed system is δWrev = PdV (3) Substituting equations (1) and (3) into equation (2) gives dU = TdS- PdV TdS = dU + PdV or Tds = du +Pdv (4) where s = entropy per unit mass Equation (4) is known as the first relation of Tds, or Gibbs equation. The definition of enthalpy gives h = u + Pv differential the above equation yields dh = du +Pdv + vdP Replacing du + Pdv with Tds yields dh = Tds + vdP Tds = dh -vdP (5) Equation (5) is known as the second relation of Tds.
- 181. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Tds Relations Case Study Theory Case Solution Simulation THERMODYNAMICS – THEORY The Tds Relations (cont.) Although the Tds equations are obtained through an internally reversible process, the results can be used for both reversible or irreversible processes since entropy is a property. Rewriting equations (4) and (5) in the following form ds = du/T + Pdv/T ds = dh/T + vdP/T The entropy change during a process can be determined The Tds Relations for Open System by integrating the above equations between the initial and the final states.
- 182. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Tds Relations Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION A turbine replaces a throttling valve to reduce the pressure of liquid refrigerant-134a from 1 MPa to 0.3 MPa. The power generated by the turbine and the money saved from this energy conversion need to be determined. Assumptions: • The flow is steady. • The turbine is adiabatic. • The pressure reduction process is reversible. • Kinetic and potential energies are negligible. (1) Determine the power generated by the turbine Take the turbine as a control volume. The inlet is state 1 and the exit is state 2. Since the assumptions state that the turbine is adiabatic and the process is reversible, this means the process is an isentropic process (will be introduced in the following section). Hence, S1 = S2 Schematic of the Turbine The turbine has only one inlet and one exit. That gives s1 = s2 where s = entropy per unit mass The saturation temperature of the refrigerant R-134a at 1 MPa is 39.39 oC. Hence, the refrigerant R-134a at 1 MPa and -4oC is a compressed liquid. In the absence of compressed liquid data, the properties of compressed liquid can be approximated by the saturated liquid data. At -4oC, the saturated properties of refrigerant R-134 are Pressure Reducing process on the h-s Psat = 0.25274 MPa Diagram vf = 0.0007644 m3/kg hf = 44.75 kJ/kg s = 0.1777 kJ/(kg-K)
- 183. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Tds Relations Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION cont. The properties of state 1 are: Saturated R-134a h1 = 44.75 + 0.0007644( 1- 0.25274)(103) Temperature Table = 45.32 kJ/kg s1 = 0.1777 kJ/(kg-K) Saturated R-134a Pressure Table v1 = 0.0007644 m3/kg Superheated Steam Table At state 2, the pressure and the entropy are known, hence, enthalpy of state 2 can be determined as h2 = 44.75 + 0.0007644( 0.3- 0.25274)(103) = 44.75 kJ/kg The energy balance of the turbine is: substituting the mass flow rate and the enthalpies into the above equation yields Heat Transferred into System A = 200(45.32-44.75) = 114 kW (2) Determine how much the turbine can save the facility per year If the turbine operate 8,760 hour per year, the amount of power produced per year is Wyr = 114( 8760) = 998,640 kWh/yr If the price of the electricity is $0.08/kWh, then 998,640 kWh of electricity is worth of 998,640 (0.08) = $79,891 The facility can save $79,891 if the throttling valve is replaced by a turbine.
- 184. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Tds Relations Case Study Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions • How does pressure before throttling effect the power generated by the turbine? • How does pressure after throttling effect the power generated by the turbine? Technical Help This simulation shows the effect of replacing the throttling valve by a turbine. The power generated by this turbine and money saved per year are determined. Use this slider to change the mass flow rate of refrigerant R-134a. Use this slider to change the pressure of refrigerant-134a before throttling. Use this slider to change the pressure of refrigerant-134a after throttling. This window shows the power generated by the turbine. This window shows the money saved by this turbine per year. This window gives the other conditions used in this calculations.
- 185. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Change Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY Introduction A spot cooler is a device which can produce hot and cold streams of air from a single stream of air at an intermediate temperature without any work or heat transfer. This product can be used for emergency cooling or short-term cooling. A seller called Kay and said his spot cooler can work at normal atmospheric pressure and cost nothing to cool the air. Can Kay trust the salesperson? What is known: A Seller’s Call •The ambient air enters the spot cooler at a Click to View Movie (65 kB) temperature of 36oC and a pressure of 1 atm. •25% of the air exits as cooled air at 24oC and a pressure of 1 atm. •75% of the air exits as hot air at 40oC and a pressure of 1 atm. Question • Is the salesperson telling the truth? Approach •If the product satisfies both the first and the second law of thermodynamics, it can work. To verify if it satisfies the first law, check the energy balance of the system. To verify if it satisfies the second law, check the entropy change of the entire process.
- 186. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Change Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Entropy Change of Pure Substances The entropy of a pure substance is determined from tables in the same manner as other properties such as v, u, and h. The value of entropy at a specified state is determined just like any other property. • Compressed liquid: from tables • Saturated liquid (sf): from tables • Saturated mixture: s = sf + xsfg where x = quality sfg = sg - sf • Saturated vapor (sg): from tables • Superheated vapor: from tables If the compressed liquid data are not available, the The Entropy of a Pure Substance entropy of the compressed liquid can be approximated by Determined by Tables the saturated liquid at the same temperature. scompressed@T,P = sf@T Saturated Water Temperature Table Saturated Water Pressure Table Entropy Change of Solids and Liquids Superheated Steam Table For an incompressible substances, the specific volume Water Tables does not change during a process. That is, dv = 0 The internal energy of an incompressible substance is du = c(T) dT Most solids and liquids can be approximated as incompressible substances. The entropy change of solids and liquids can be obtained by applying the above two equations to the first Tds equation. The entropy change during a process 1-2 is
- 187. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Change Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Entropy Change of Solids and Liquids (cont.) If the temperature variation (T2 - T1) is small over the entire process, an average specific heat (cave) is used and treated as a constant (constant-specific-heats assumption). Then the entropy change can be evaluated by Constant Specific Heat Used in Small Entropy Change of Ideal Gases Temperature Interva The property relations for ideal gases are: pv = RT du = cvdT dh = cPdT The first relation of entropy change for ideal gases is obtained by replacing P by RT/v and du by cvdT in the first Tds equation: Tds = du + Pdv = cvdT + RTdv/v ds = cvdT/T + Rdv/v Integrating ds from state 1 to state 2 gives the first relation of entropy change for ideal gases. The second relation of entropy change for ideal gases is obtained by replacing v by RT/P and dh by cPdT in the second Tds equation: Tds = dh - vdP = cPdT - RTdP/P ds = cPdT/T - RdP/P Integrating ds from state 1 to state 2 gives the second relation of entropy change for ideal gases.
- 188. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Change Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Entropy Change of Ideal Gas (cont.) A common approximation is made by assuming that the specific heats of ideal gases are constants if the temperature variation is small (constant-specific-heats assumption). Replacing cv and cP with constants cv,av and cP,av in the first and second relations of entropy change and integrating from state 1 to state 2 gives, If the temperature variation is large, the variation of specific heats with temperature should be considered and relations between cv(T), cP(T) and temperature are needed to integrate the first and second relations. For convenience, a function s0, standard-state entropy, is defined as Function s0 is a Function of Temperature only Click to View Movie (49.0 kB) s0 is a function of temperature only and is tabulated with Ideal-gas Properties of Air Table temperature. Absolute zero is used as the reference temperature. Replacing T as T1 and T2 in the above definition gives
- 189. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Change Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Entropy Change of Ideal Gas (cont.) Then the entropy change of ideal gases during a process 1-2 is Approximate Analysis and Exact Analysis of Entropy Change during a Cooling Process Click to View Movie (54 kB)
- 190. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Change Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION A salesperson states that this spot cooler can work at the atmospheric pressure without any energy consumption. Evaluate his claim. Assumptions: •Air is an ideal gas and has constant specific heats. •The flow is steady, no work and heat transfer involved. •Kinetic and potential energy changes are ignored. Take the spot cooler as a control volume shown on the left. If the spot cooler satisfies both the first and the second law of thermodynamics, it will work. To verify if it satisfies the first law, check the energy balance of the system. To verify if it satisfies the second law, check the entropy change of the entire process. With all the assumptions, the mass and energy balance Control Volume of Spot Cooler of the control volume are: The mass balance is satisfied since The constant-specific-heat assumption is also used. So with constant specific heats, the energy balance becomes
- 191. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Change Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION cont. Substituting the temperatures into both sides of the above equation yields right side = 0.25(24 + 273) +0.75(40 + 273) = 309 left side = 36 + 273 = 309 = right side The result above shows the energy balance is satisfied. To check if the spot cooler satisfies the second law, calculate the entropy change of the entire process. If the entropy change is not negative, the process can happen. With constant specific heats, the entropy change equals The result shows that the spot cooler cannot work at atmospheric pressure. The air at the inlet needs to be compressed to a higher pressure to make the cooler work, which requires energy input.
- 192. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Change Case Study Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions •At what pressure can the cooler work? •If 50% of the ambient air at 35oC needs to be cooled to 25oC and sent to indoor, what is the exhausted hot air temperature? Technical Help This simulation uses the first and the second law of thermodynamics to verify if the spot cooler can work at the atmospheric pressure with no energy consumption. The ratio of the cooled air produced by the cooler is shown in a diagram. Use this slider to change the air pressure at the inlet of the cooler. Use this slider to change the temperature of the ambient air, which is the inlet air of the cooler. Use this slider to change temperature of the cooled air out of the cooler. Use this slider to change the temperature of the hot air out of the cooler. This window shows the entropy change of the whole cooler. If it is negative, the cooler can not work. This window shows the ratio of the cooled air out of the cooler to the ambient air sent to the cooler.
- 193. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Process Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY Introduction Jim is in charge of filling the compressed carbon dioxide bottles in his factory. A worker brought in a bottle, which needs to be refilled because of the leak from the broken spray probe. Before refilling the bottle, Jim has to empty the bottle by sending the remaining carbon dioxide through a recycle process. How much carbon dioxide is left in the bottle? What is known: •The well-insulated 200 liter bottle full of carbon dioxide at a pressure of 5 atm and a temperature of Problem Description 500 K initially. Click to View Movie (112 kB) •Carbon dioxide leaks slowly till the pressure in the bottle reaches 1 atm. Question How much carbon dioxide is left in the bottle? Approach •Take the bottle as a system. •Modeling this carbon dioxide leak process as an isentropic process.
- 194. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Process Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Isentropic Processes The term "isentropic" means constant entropy. A process during which the entropy remains constant is called an isentropic process, which is characterized by ΔS = 0 or s1 = s2 for a process 1-2 If a process is both reversible and adiabatic, then it is an isentropic process. A Reversible and Adiabatic Process is also an Isentropic Process An isentropic process is an idealization of an actual process, and serves as a limiting case for an actual process. An Isentropic Process and an Actual Process on a h-s Diagram Isentropic Processes for Ideal Gases The relations of entropy change for ideal gases are (1) and (2) An Isentropic Process of Ideal Gases on a T-s Diagram
- 195. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Process Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Isentropic Processes for Ideal Gases (cont.) By setting Δs to 0 in the above equations, the relations for an ideal gas which undergoes an isentropic process can be obtained. Setting equation (1) to zero gives, If the constant-specific-heats assumption is valid, the above equation can be integrated and rearranged to give where k = specific heat ratio, k = cP/cv and R = cP - cv The second relation can be obtained by setting equation (2) to zero. Also, if the constant-specific-heats assumption is valid, the above equation becomes Constant Specific Heat Used in Small Temperature Interval
- 196. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Process Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Isentropic Processes for Ideal Gases (cont.) The third relation can be obtained by combining the first and the second relations. That is, The three relations of an isentropic process for ideal gases with constant specific heats in compact form are Tvk-1 = constant TP(1-k)/k = constant Pvk = constant If the constant-specific-heats assumption is not valid, the entropy change of ideal gases during a process 1-2 is Setting the above equation to zero and rearranging, one obtains If exp(s0/R) is defined as the relative pressure Pr, then Using Pr Data to Determine Finial the above equation becomes Temperature during an Isentropic Process (P2/P1)s = constant = Pr2/Pr1 Click to View Movie (46.0 kB) Values of relative pressure are tabulated against temperature in tables.
- 197. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Process Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Isentropic Processes for Ideal Gases (cont.) In an automotive engine, the ratio v2/ v1 is used instead of the ratio P2/ P1. The ideal-gas relation gives v2/ v1 = T2P1/T1P2 Replacing P2/P1 by Pr2/Pr1 in the above equation gives T/Pr is defined as relative specific volume vr and its value is also tabulated with temperature. Thus, Using vr Data to Determine Finial (v2/v1)s = constant = vr2/ vr1 Temperature during an Isentropic Process Click to View Movie (49 kB) Ideal Gas Properties Table -- Air Isentropic Processes for an Incompressible Fluid or Solid The entropy change of incompressible fluids or solids during a process 1-2 with constant specific heats is Δs = caveln(T2/T1) Setting Δs to zero gives T2 = T 1 The Isentropic Process of an That is, an isentropic process of an incompressible Incompressible Substance is substance is also isothermal. also Isothermal
- 198. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Process Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION A leak develops in a well-insulated bottle initially containing compressed carbon dioxide. The mass of carbon dioxide left in the bottle needs to be determined. Assumptions: •No significant heat transfer occurs. •Since the carbon dioxide leaks slowly, the process is assumed reversible. •The carbon dioxide is modeled as an ideal gas with constant specific heats. Take the bottle as the system. Denote the state before leak as state 1 and the state after leak as state 2. From the assumptions made, carbon dioxide is modeled as an idea gas. With the ideal-gas equation of state, the mass remains in the bottle after leak is m2 = P2V2/RT2 where P2 and V2 are known. If the temperature T2 is determined, the mass remaining m2 can be obtained. The gas constant R of carbon dioxide is 188.9 J(kg-K). The leaking process is adiabatic and reversible, hence it is Take the mass of carbon dioxide an isentropic process. The second relation of isentropic remaining in the bottle after leak as a process for ideal gases is: system
- 199. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Process Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION cont. Assume the final temperature is 350 K and use the average value of the initial temperature and the final temperature to determine the value of k. Tav = (500 + 350 )/2 = 425 K From the table, k equals to 1.246 at 425 K. With all the data known, T2 can be obtained by substituting the values of T1, P1, and P2 into the above The Process on the T-s Diagram equation. T2 = 500(1/5)0.246/1.246 = 363.9 K Molar Mass, Gas Constant for Various Common Gases The calculated final temperature is close to 350 K (the assumed temperature), hence no iteration is needed. Specific Heats of Some Common Ideal Gases Inserting T2 , P2, and V2 into the expression for m2gives, m2 =(101,325)(200/1,000)/((188.9)(363.9)) = 295 g
- 200. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Process Case Study Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions •How does the initial pressure in the bottle effect the mass left in the bottle? •How does the final pressure in the bottle effect the mass left in the bottle? Technical Help This simulation shows the relation between the pressure and the mass of carbon dioxide left in a bottle which was initially full of compressed carbon dioxide. Use this slider to change the volume of the bottle. Use this slider to change the initial temperature in the bottle. Use this slider to change the initial pressure in the bottle. Use this slider to change the final pressure in the bottle. This window shows the mass of carbon dioxide left in the bottle.
- 201. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Efficiency Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY Introduction Mr. Williams has a small swimming pool in his backyard. The pump that pumps water from a well to the pool is broken. However, he found a used pump in a second-hand electric utility store. To insure that the pump matches his motor, he needs to figure out the power required to run the pump. What is known: •The pump is capable of increasing the pressure of water up to 300 kPa. •The flow rate through the pump is 20 kg/min. Problem Description •The pump has an isentropic efficiency of 85%. Click to View Movie (49 kB) Question What is the required power input to the pump? Approach •Take the pump as a control volume. •Model the pump as an adiabatic pump. •Use the definition of isentropic efficiency of pump to calculate the enthalpy change of the process. •Use the energy balance to calculate the power input to the pump.
- 202. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Efficiency Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Entropy Change of Pure Substance In engineering analysis, isentropic efficiency is a parameter to measure the degree of degradation of energy in steady-flow devices. It involves a comparison between the actual performance of a device and the performance that would be achieved under idealized circumstances for the same inlet and exit states. Although there exits heat transfer between the device and its surroundings, most steady-flow devices are intended to operate under adiabatic condition. Hence, normally an isentropic process is chosen to serve as the idealized process. Recall, if the inlet is denoted by subscript 1 and exit is denoted by subscript 2, the energy balance for a one-inlet- one-exit control volume is Isentropic Efficiency of Turbines For an adiabatic turbine which undergoes a steady-flow process, its inlet and exit pressures are fixed. Hence, the idealized process for turbine is an isentropic process between the inlet and exit pressures. The desired output from a turbine is the work output. Hence, the definition of isentropic efficiency of turbine is the ratio of the actual work output of the turbine to the work output of the turbine if the turbine undergoes an isentropic process between the same inlet and exit pressures. Schematic of a Turbine
- 203. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Efficiency Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Isentropic Efficiency of Turbines (cont.) ηT = Actual turbine work/Isentropic turbine work = wa/ws wa and ws can be obtained from the energy balance of the turbine. Usually the kinetic and potential energies associated with a process through a turbine is negligible compared with the enthalpy change of the process. In this case, the energy balance of the turbine is reduced to The isentropic efficiency of turbine can then be written as The h-s Diagram of the Actual and Isentropic ηT (h2a - h1)/(h2s - h1) Processes of an Adiabatic Turbine where Click to View Movie (50 kB) h1 = enthalpy at the inlet h2a = enthalpy of actual process at the exit h2s = enthalpy of isentropic process at the exit Isentropic Efficiency of Compressors and Pumps Compressors and pumps, when undergo a steady-flow process, consume power. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. ηC = Isentropic compressor (pump) work/Actual compressor (pump) work = ws/wa wa and ws can be obtained from the energy balance of the compressors or pumps. When the kinetic and potential energies associated with a gas flowing through a compressor are negligible compared with the enthalpy change of the gas, the energy balance of the compressor Schematic of a Compressor is reduced to
- 204. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Efficiency Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Isentropic Efficiency of Compressors and Pumps (cont.) The isentropic efficiency of a compressor becomes η C (h2s - h1)/(h2a - h1) Pumps are used to handle liquid instead of gas. Since liquid is incompressible and the process is an isentropic process, the enthalpy change from inlet to exit is The h-s Diagram of the Actual and Isentropic Processes of an Adiabatic Compressor or Pump If the kinetic and potential energies are negligible, the Click to View Movie (54 kB) isentropic efficiency of a pump is reduced to η P v(P2 - P1)/(h2a - h1) IN practice, compressors are intentionally cooled to minimize the work input. In those cases, a reversible isothermal process is served as the idealized process for compressors which are intentionally cooled, and then an isothermal efficiency is defined instead of the isentropic efficiency. The isothermal efficiency is defined as the ratio of the work input to the isothermal process, to the work input to the actual process between the same inlet and exit pressures. Schematic of a Pump ηC = Reversible isothermal compressor work/Actual compressor work = wt/wa
- 205. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Efficiency Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Isentropic Efficiency of Nozzles Nozzles are devices used to accelerate the fluid velocity at the cost of pressure. The isentropic efficiency of nozzles is defined as the ratio of the actual kinetic energy at exit to the kinetic energy at the exit when the process is isentropic for the same inlet and exit pressures. Schematic of a Nozzle ηN = Actual KE at exit/Isentropic KE at exit = There is no work interaction involved in nozzles and the potential energy change of the fluid is small. If the inlet velocity is small relatively to the exit velocity, the energy balance of a nozzle is reduced to Then the isentropic efficiency of nozzles becomes ηN (h1 - h2a)/(h1 - h2s) The h-s Diagram of the Actual and Isentropic Processes of an Adiabatic Nozzle Click to View Movie (61 kB)
- 206. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Efficiency Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION A used pump must match the motor in Mr. Williams swimming pool. The power required to run the pump needs to be determined. Assumptions: •Model the pump as an adiabatic one. •The flow process in the pump is steady. •The kinetic and potential energies associated with water flowing through the pump are negligible. •The pump operates at ambient temperature. Take the pump as a control volume, shown on the left. Denote the inlet by subscript 1 and exit by subscript 2. The definition of isentropic efficiency of pump is: η P v(P2 - P1)/(h2 - h1) It is assumed that the pump operates at ambient temperature. Hence the specific volume of water is v = 0.001 m3/kg The pump increases the pressure of the water up to 300 The Control Volume kPa, which means the pressure difference of water between the exit and inlet is 300 kPa. That is P2 - P1 = 300 kPa Substituting the given efficiency, which is 85%, and the pressure difference above to the definition of isentropic efficiency gives, h2 - h1 = v(P2 - P1)/η P = 0.001(300,000)/0.85 = 352.9 J/kg
- 207. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Efficiency Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION cont. With the assumptions that the pump is adiabatic and the kinetic and potential energies are negligible, the energy balance of the pump reduces to The mass flow rate is given as 20 kg/min. Substituting the enthalpy difference and the mass flow rate to the energy The Pumping Process on the h-s Diagram balance yields, = -20/60(352.9) = -117.6 W The negative sign means the pump needs power input.
- 208. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Isentropic Efficiency Case Study Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions •Can the entropy change of the actual process be negative? •What is the entropy change of the isentropic process? •How does the power required by the pump change with the isentropic efficiency? Technical Help This simulation shows the power required to run a swimming pool pump. Processes are shown on the h-s diagram. Use this slider to change mass flow rate of the pump. Use this slider to change the isentropic efficiency of the pump. Use this slider to change the required pressure increase of the water. This window shows the required power input to the pump.
- 209. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Balance Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY Introduction Pat is taking a thermodynamics course this semester and in the class, his professor states that person increase entropy every time they do anything in daily life. Pats mom cook eggs for his breakfast. He wonders how much entropy is generated when eggs are cooked. What is known: •6-cm-diameter spherical egg at 10oC is dropped into boiling water. Problem Description Click to View Movie (40 kB) •The boiling water temperature is 100oC. •Egg has a density of 1000 kg/m3 and a specific heat of 3.4 kJ/(kg-K). Question How much entropy is generated by boiling one egg to 100oC? Approach •Method A: Take the egg as a system. It is a closed system due to no mass transfer involved. The entropy transfer is only due to the heat transfer. Model the egg as incompressible, and the entropy change in the system is ΔS = mCaveln(T2/T1) where 1 denotes the initial state and 2 denotes the final state. •Method B: take the egg as a closed system. Consider the entropy change of the isolated system formed by the egg and the surrounding water.
- 210. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Balance Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Entropy Generation Consider a system undergoing a process 1-2 as shown in the diagram. The increase of entropy principle states The above equation can be expressed as Process 1-2 on T-s Diagram where Sgen is called entropy generation. Sgen measures the effect of irreversibilities present within the system during a process. Its value depends on the nature of the process and not solely on the start and end states. Hence, it is not a property. Rewriting the above equation gives the >0 irreversible process expression of entropy generation as internally reversible Sgen = 0 process <0 impossible Non-negative Entropy Generation According to the increase of entropy principle, entropy generation can only take a non-negative value. Entropy Balance The entropy balance is an expression of the second law of thermodynamics that is particularly convenient for thermodynamic analysis. The increase of entropy principle states that the entropy can only be created, and it is expressed as
- 211. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Balance Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Entropy Balance (cont.) (Entropy change in a system) = (Total entropy entering the system) - (Total entropy leaving the system) + (Total entropy generation in the system) or ΔSsystem = Sin - Sout + Sgen This expression is the entropy change of a system. It is applicable to any system undergoing any process. In unit- mass basis, it is Δssystem = sin - sout + sgen Entropy Balance for a System The first and second terms in the right hand side of the equation represent the entropy transfer between the system and its surroundings. The third term represents the entropy generation in the system. Entropy Transfer Entropy can be transferred to or from a system by two mechanisms: heat transfer and mass flow. Note that no entropy is transferred by work. Heat Transfer Heat transfer to a system increases the entropy of the system, and heat transfer out of a system decreases the entropy of the system.
- 212. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Balance Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Entropy Transfer (cont.) During a process 1-2 of a system as defined above, If heat transfer δQ happens at the system boundary which has an absolute temperature T, the entropy transfer associated with this heat transfer process is This expression is the entropy change of a system. It is applicable to any system undergoing any process. In unit- mass basis, it is Entropy Transfer Δssystem = sin - sout + sgen The first and second terms in the right hand side of the Entropy Transfer by Heat with Different equation represent the entropy transfer between the Boundary Temperatures system and its surroundings. The third term represents the entropy generation in the system. If the entire boundary temperature remains as a constant T, the above equation can be simplified as Sheat = Q/T where Q = total heat transfer during process 1-2 When the temperature T is not a constant, but a function of its location, the entropy change can be approximated as where Qk = heat transfer at boundary location k Tk = temperature at boundary location k
- 213. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Balance Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Entropy Transfer (cont.) Mass Transfer Mass contains entropy as well as energy. Mass m, which has a specific entropy s, contains a total entropy of ms. When a mass flows in a system, the entropy increase in the system is ms. In the same manner, mass m flowing out of a system will result in a decrease of entropy in amount of ms in the system. Smass = ms Entropy Flows in or out with Mass Click to View Movie (28 kB) Entropy Balance for Closed Systems A closed system involves no mass transfer between the system and its surroundings, and thus its entropy transfer is only from heat transfer. If the closed system undergoes an adiabatic process, then no heat transfer occurs at the boundary. The entropy Consider the Egg as a Closed System balance is reduced to Click to View Movie (52 kB) A system and its surroundings forms an isolated system. The entropy change of an isolated system formed by a closed system and its surroundings is ΔSisolated = ΔSsystem + ΔSsurr = Sgen Consider the Isolated System Formed by the Egg and the Surrounding Water Click to View Movie (57 kB)
- 214. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Balance Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION Entropy generated by cooking an egg from 10oC to 100oC in boiling water needs to be determined. Assumptions: •Model both the egg and the water as incompressible substance. •Constant-specific-heat assumption is valid for both water and egg. •Kinetic and potential energy changes of the egg are negligible. •No work interaction is involved. •Water remains 100oC during the cooking process. (1) Method A: Consider the egg as a closed system Take the egg as a system. It is a closed system since no mass flow in or out of the egg. The entropy balance for a closed system is: Rearranging the above equation gives the expression of the entropy generation. That is, Consider the Egg as a Closed System Click to View Movie (52 kB) The first two terms on the right hand side of the above equation corresponds to the entropy change in the system, and the third term is the entropy transfer by heat. The entropy change in the system: The egg is approximated as an incompressible substance. With constant-specific-heat assumption, its entropy change can be determined from ΔS = S2 - S1 = mCaveln(T2/T1)
- 215. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Balance Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION cont. where 1 denotes the initial state and 2 denotes the final state. Hence, T1 = 10 oC = 283 K T2 = 100 oC = 373 K The mass of the egg is m = Vρ = 4/3(3.14)(0.03)3(1000) = 0.113 kg The average specific heat is given as 3.4 kJ/(kg-K). Substituting all the data into the expression of entropy change yields S2 - S1 = 0.113(3,400) ln(373/283) = 106.1 J/K The entropy transfer to the egg by heat: The egg is cooked from 10oC to 100oC in 100oC water. Assume the boundary of the egg remains 100oC during the whole process and hence the entropy transfer to the egg by heat transfer can be determined by where Q is the heat transferred to the egg from the surrounding water, and T is 100oC. Q can be determined by the energy balance of the egg. Q - W = U2 - U1 Q = mcave (T2 - T1) = 0.113(3,400)(100 - 10) = 34,578 J
- 216. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Balance Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION cont. Substituting Q and T to the expression of entropy transfer gives, Q/T = 34,578/(373) = 92.7 J/K After calculating the entropy change in the system and the entropy transfer by heat, the entropy generation can be determined. It is Sgen = (S2 - S1) - Q/T = 106.1 - 92.7 = 13.4 J/K (2) Method B: Consider the isolated system formed by the egg and the water surrounding it First, consider the egg as a system. Then the egg and the surrounding water as an isolated system. The entropy generated during the cooking process equals the entropy change of the isolated system. Sgen = ΔSisolated = ΔSsystem + Δssurroundings The entropy change of the system has been calculated in Consider the Isolated System Formed by the Egg and the Surrounding Water method A, which is Click to View Movie (57 kB) ΔSsystem = 106.1 J/K
- 217. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Balance Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION cont. Consider the surrounding water. The water remains 100oC in the entire process. Hence, ΔSsurroundings = Q/T where Q equals the heat transferred to the system in value but has negative sign, which indicates that heat is transferred from the water. Q = -34,578 J The entropy change in the surrounding water is Q/T = -34,578/373 = - 92.7 J/K Summing up the entropy change in the system and the entropy change in the surrounding water yields the entropy generation during the egg cooking process. That is, Sgen = 106.1 - 92.7 = 13.4 J/K The result obtained is the same as that in method A.
- 218. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Entropy Balance Case Study Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions •Does the entropy generation increase or decrease if the initial temperature of the egg increases? •Does the entropy generation increase or decrease if the water temperature increases? Technical Help This simulation shows the entropy generation when eggs are cooked in hot water. Use this slider to change the water temperature. Use this slider to change the initial temperature of the egg. This window shows the entropy generated during the cooking process. This window gives other conditions used in this calculation.
- 219. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) WorkEntropy Balance for Control Volume Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY Introduction Austins dad likes to take a long shower in the morning. Austin needs to find some way to get him out more quickly. What is known: •Austins dad takes a 15-minute shower every morning. •The average flow rate through the shower head is 10 L/min. •An electric water heater heats the water from 15oC to Problem Description 60oC and then mixes it with cold water to 40oC at a T- Click to View Movie (42 kB) elbow before it is sent to the shower head. Question How much entropy is generated by Austins father when he takes his daily shower? Approach •Take the electric heater and the T-elbow pipe together as a control volume. •Consider the mass balance and entropy balance of the control volume simultaneously. T-elbow and the Electric Heater
- 220. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) WorkEntropy Balance for Control Volume Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Entropy Balance for Control Volumes A control volume permits both energy and mass to flow through its boundaries. The entropy balance for a control volume undergoing a process 1-2 can be expressed as or in the rate form, as Entropy Transferred by Heat and Mass where i and e denote inlet and exit, respectively. The above in Control Volume entropy balance relation states that the entropy change of a Click to View Movie (33 kB) control volume undergoing a process 1-2 equals the sum of the entropy transfer by heat, the net entropy transfer by mass, and the entropy generation in the control volume. Analysis of a Control Volume at Steady State Most control volumes encountered in practice, such as nozzles, turbines and compressors operate at steady state. Hence, there is no entropy change in the control volume. That is, The entropy balance in this case is In engineering analyses, the mass balance, energy balance, and the entropy balance often must be solved simultaneously. Propylene Flows Steadily through a Heat Recall, the mass balance and energy balance are, Exchanger
- 221. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) WorkEntropy Balance for Control Volume Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Analysis of a Control Volume at Steady State (cont.) One-inlet-one-exit Control Volumes Many applications involve a one-inlet-one-exit control volume at steady state, such as a nozzle. The three balance equations for this case are, Mass Balance: Energy Balance: A one-inlet-one-exit Nozzle Entropy Balance:
- 222. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Eficiency Balance (1) Balance (2) WorkEntropy Balance for Control Volume Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION The entropy generated when Austins father takes his daily shower needs to be determined. Assumptions: •The shower operates steadily at maximum flow condition. •Heat loss from the pipes is negligible. •Water is an incompressible substance with a density equals 1000 kg/m3 and a specific heat equals 4.18 kJ/(kg- K). •The kinetic and potential energies are negligible. Take the T-elbow and the electric heater together as a control volume shown on the left. Cold water at 15oC enters the electric heater (denoted as 1) and the T-elbow (denoted as 2), and leaves the T-elbow at 40oC. Since the flow is steady, the mass balance for this control volume is Take the T-elbow and the Electric Take the T-elbow and the electric heater together as a Heater together as a Control Volume control volume shown on the left. Cold water at 15oC enters the electric heater (denoted as 1) and the T-elbow (denoted as 2), and leaves the T-elbow at 40oC. Since the flow is steady, the mass balance for this control volume is The mass flow rate to the shower head is The entropy balance for a control volume at steady state is Simplify the Entropy Balance of the Control Volume Click to View Movie (44 kB)
- 223. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) WorkEntropy Balance for Control Volume Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION cont. It is assumed that there is no heat transfer between the pipes and the surroundings. That means no entropy flows in or out of the control volume associated with the heat transfer. The control volume has electric work input, but the work is entropy free. Hence, the entropy balance for this steady-flow control volume can be simplified as Since water enters the T-elbow and the electric heater at the same temperature, s1 = s2 Also, with the mass balance, the entropy generation can be determined by Hence, a 15-minute shower will generate entropy Sgen = (15)(60)(58.1) = 52,290 J/K = 52.3 kJ/K
- 224. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) WorkEntropy Balance for Control Volume Case Study Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions •How can Austens dad generate less entropy when he takes shower? Technical Help This simulation shows the entropy generated by preparing water for a shower for different cold mater temperature, shower water temperature, shower flow rate, and shower time. Use this slider to change the clod water temperature. Use this slider to change the shower water temperature. Use this slider to change the time of taking shower. Use this slider to change the flow rate of the shower. This window shows entropy generated by taking shower.
- 225. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Reversible Steady-flow Work Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY Introduction A hydroelectric power plant is to be built in a mountain region. Water flows from the elevated reservoir through pipes to a hydraulic turbine, whose shaft is connected to a electric generator. Thus, the mechanical energy of water is converted to electrical energy. A designer orders some 0.5-m-diameter-pipes but he needs the engineer to specify the exact number of intake pipes. Problem Description Click to View Movie (42 kB) What is known: •The water surface in the reservoir is 105 m higher than the turbine. •The entrance to the intake pipes is located 5 m below the water surface in the reservoir. •Water enters the intake pipe at 1 m/s and exhausts to a river directly after exiting the turbine with almost the same velocity. •The ambient pressure is 1 bar. •The pipe diameter is 0.5 m. Power Generated by a Hydraulic Turbine Click to View Movie (41 kB) •The power output is 1 MW. Question How many intake pipes are to be installed? Approach •Take the system from point 1 to 2 shown on the left as a control volume. •Consider the power plant works in reversible and steady flow conditions. Then the power output from the power plant equals the reversible steady-flow work of the turbine. A Hydroelectric Power Plant
- 226. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Reversible Steady-flow Work Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Reversible Steady-flow Work The energy balance for a steady-flow device (nozzle, compressor, turbine and pump) with one inlet and one exit is: Its differential form is: δq - δw = dh + dke + dpe If the device undergoes an internally reversible process, the heat transfer term δq can be replaced by dh - vdP since One-inlet-one-exit Nozzle δqrev = Tds Tds = dh - vdP (the second Tds relation) Then the energy balance becomes dh - vdP - δwrev = dh + dke + dpe By rearranging the above equation, the reversible steady- flow work can be expressed as - δwrev = vdP + dke + dpe Integrating it form location 1 to location 2 yields, The above equation is the relation for the reversible work output associated with an internally reversible process in a steady-flow device. When the changes in kinetic and potential energies are negligible, the relation reduces to Water Flowing through a Hydraulic Turbine
- 227. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Reversible Steady-flow Work Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Reversible Steady-flow Work (cont.) The above equation states that the larger the specific volume, the larger the reversible work produced or consumed by a steady-flow device. To minimize the work input during a compression process, one should keep the specific volume of the working fluid as small as possible. In the same manner, to maximize the work output during an expansion process, one should keep the specific volume of the working fluid as large as possible. One needs to know the relationship between the specific volume v and the pressure P for the given process to perform the integration in the above relation. If an incompressible fluid is used as the working fluid, the specific volume v is a constant. The relation for the reversible work output associated with an internally reversible process in a steady- flow device is simplified to give wrev = -v(P2 - P1) - Δke - Δpe Power Generated by Hydraulic Turbine Click to View Movie (41 kB) Hydraulic turbines used in hydroelectric power plants run in a steady-flow process with incompressible fluid, i.e., water, as the working fluid. If no work interactions are involved, like nozzle or pipe section, the above equation is reduced to where V is the velocity of the fluid. This equation is known as The Bernoulli Equation the Bernoulli equation in fluid mechanics. Click to View Movie (31 kB)
- 228. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Reversible Steady-flow Work Case Study Theory Case Solution Simulation THERMODYNAMICS - THEORY Reversible Steady-flow Devices produce Most and Consume Least Work The steady-flow devices deliver the most and consume the least work when it undergoes a reversible process. Consider two steady-flow devices, one is reversible and the other is irreversible (actual process), operating between the same inlet and exit states. The differential forms for the energy balance of these two devices are δqact - δwact = dh + dke + dpe δqrev - δwrev = dh + dke + dpe The right hand sides of these two equations are the same. It gives, qact - δwact = qrev - δwrev Rearranging this equation gives, δwrev - δwact = qrev - qact Since qrev = Tds, the above equation becomes, δwrev - δwact = Tds – qact the increase of entropy principle gives Thus, δwrev - δwact 0 or δwrev δwact Reversible Turbine Delivers more Work than Actual Turbine That is, for the same inlet and exit conditions, when the device undergoes a reversible process, a work-produce device like turbine produces the most work (w is positive), or a work-consuming device like compressor consumes the least work (w is negative).
- 229. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Reversible Steady-flow Work Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION A small power plant using hydraulic turbines is planned. The number of intake pipes which intake water from a reservoir to the turbines needs to be determined. Assumptions: •Model the water as incompressible substance with a density ρ = 1000 kg/m3. •Assume water flows through the pipes and turbines steadily, and the process is reversible. The schematic of the power plant is shown on the left. Take the pipes and the turbines together as a control volume. Denote the inlet of the intake pipes as 1 and the outflow of the turbine as 2.The reversible steady-flow work generated by this power plant is where Schematic of the Power Plant The first term on the left-hand side is the pressure term. Point 1 is located 5 meter under the water surface in reservoir. So pressure P1 is P1 = ρgh + Patm= 1000(9.8)(5) +100,000 =149,000 Pa = 1.49 bar Point 2 is located at the outflow of the power plant. Its pressure equals the ambient pressure, P2 = 1.0 bar
- 230. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Reversible Steady-flow Work Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION cont. The second term on the left-hand side is the kinetic energy term. Water enters the pipes with a velocity of 1 m/s and leaves the turbine as the same velocity. Hence, there is no kinetic energy change when water flows from point 1 to point 2. That is, ΔKE = 0 The third term on the left-hand side is the potential energy term. Water flowing from point 1 (100 m higher than point 2) to point 2 experiences a potential energy change of Substituting the three terms into the expression for reversible steady-flow work gives This power plant needs to generate 1 MW power, hence, Wrev= 1,000,000 W
- 231. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Reversible Steady-flow Work Case Study Theory Case Solution Simulation THERMODYNAMICS – CASE STUDY SOLUTION cont. Substituting this to the expression of the reversible steady- flow work gives 1,000,000 = 1,029,000 V V = 0.97 m3/s Water enters the intake pipe with a velocity of 1 m/s. The flow rate through each 0.5 m diameter pipe is Vpipe= (3.14)(0.25)2(1) = 0.196 m3/s In order to intake 0.97 m3 of water through the pipes per second, the number of the pipes needed is N = V/V = 0.97/0.196 = 5
- 232. Ch 6. Entropy Multimedia Engineering Thermodynamics Entropy Isentropic Isentropic Entropy Entropy ReversibleEntropy Tds Relations Change Process Efficiency Balance (1) Balance (2) Work Reversible Steady-flow Work Case Study Theory Case Solution Simulation THERMODYNAMICS – SIMULATION Run Simulation Suggested Help Questions If the power output is set at 1.2 MW, the elevation of the pipe inlet is set at 100 m, and only 4 pipes are allowed, what is the diameter of the pipes? Technical Help This simulation shows the design of hydropower plant by allowing user to control elevation, pipe diameter, and power output. The number of water intake pipes is needed to be determined Use this slider to change the elevation of the pipe inlet. Use this slider to change the pipe dismeter. Use this slider to change the required output of the power plant. This window shows the number of pipes needed in this power plant.

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