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# ChE 142 ebook (final version)

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### ChE 142 ebook (final version)

1. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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.