CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES


 Erin Rose
 2 years ago
 Views:
Transcription
1 Thermodynamics: An Engineering Approach 8th Edition in SI Units Yunus A. Çengel, Michael A. Boles McGrawHill, 2015 CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES Lecture slides by Dr. Fawzi Elfghi
2 Objectives Develop the conservation of mass principle. Apply the conservation of mass principle to various systems including steadyflow control volumes. Apply the first law of thermodynamics as the statement of the conservation of energy principle to control volumes. Identify the energy carried by a fluid stream crossing a control surface as the sum of internal energy, flow work, kinetic energy, and potential energy of the fluid and to relate the combination of the internal energy and the flow work to the property enthalpy. Solve energy balance problems for common steadyflow devices such as nozzles, compressors, turbines, throttling valves, Apply the energy balance to general unsteadyflow processes with particular emphasis on the uniformflow process as the model for commonly encountered charging and discharging processes. 2
3 CONSERVATION OF MASS Conservation of mass: Mass, like energy, is a conserved property, and it cannot be created or destroyed during a process. Closed systems: The mass of the system remain constant during a process. Control volumes: Mass can cross the boundaries, and so we must keep track of the amount of mass entering and leaving the control volume. Mass m and energy E can be converted to each other according to where c is the speed of light in a vacuum, which is c = m/s. The mass change due to energy change is negligible. 3
4 Conservation of Energy for Control volumes The conservation of mass and the conservation of energy principles for open systems or control volumes apply to systems having mass crossing the system boundary or control surface. In addition to the heat transfer and work crossing the system boundaries, mass carries energy with it as it crosses the system boundaries. Thus, the mass and energy content of the open system may change when mass enters or leaves the control volume. Thermodynamic processes involving control volumes can be considered in two groups: steadyflow processes and unsteadyflow processes. During a steadyflow process, the fluid flows through the control volume steadily, experiencing no change with time at a fixed position. 4
5 A pressurized water nuclear reactor steam power plant has many examples of open system operation. Some of these are the pressure vessel, steam generator, turbine, condenser, and pumps. Photo courtesy of Progress Energy Carolinas, Inc. 5
6 A heat exchanger, the heater core from a V8 Mustang, is another example of an open system. Cooling water flows into and out of the tubes and air is forced through the fin sturucture. 6
7 A hair drier is an example of a one entrance, one exit open system that receives electrical work input to drive the fan and power the resistance heater. 7
8 Let s review the concepts of mass flow rate and energy transport by mass. One should study the development of the general conservation of mass presented in the text. Here we present an overview of the concepts important to successful problem solving techniques. Mass Flow Rate Mass flow through a crosssectional area per unit time is called the mass flow rate m. Note the dot over the mass symbol indicates a time rate of change. It is expressed as where V n is the velocity normal to the crosssectional flow area. 8
9 If the fluid density and velocity are constant over the flow crosssectional area, the mass flow rate is Vave A m Vave A v where is the density, kg/m 3 ( = 1/v), A is the crosssectional area, m 2 ; and average fluid velocity normal to the area, m/s. Example 51 V ave is the Refrigerant134a at 200 kpa, 40% quality, flows through a 1.1cm inside diameter, d, tube with a velocity of 50 m/s. Find the mass flow rate of the refrigerant134a. At P = 200 kpa, x = 0.4 we determine the specific volume from table A12, Page 918 v v xv f fg ( ) 3 m kg 2 Vave A Vave d m v v 4 50 m / s (0.011 m) m / kg kg s 2 9
10 The fluid volume flowing through a crosssection per unit time is called the volume flow rate V. The volume flow rate is given by integrating the product of the velocity normal to the flow area and the differential flow area over the flow area. If the velocity over the flow area is a constant, the volume flow rate is given by (note we are dropping the ave subscript on the velocity) V VA 3 ( m / s) The mass and volume flow rate are related by Example 52 m V V v ( kg / s) Air at 100 kpa, 50 o C, flows through a pipe with a volume flow rate of 40 m 3 /min. Find the mass flow rate through the pipe, in kg/s. Assume air to be an ideal gas, so From table A1, page 898, we find the gas constant (R) v RT P kj kg K (50 273) K 100kPa 3 m kpa kj 3 m kg 10
11 m V v m / min 1min m / kg 60s Conservation of Mass for General Control Volume The conservation of mass principle for the open system or control volume is expressed as kg s or SteadyState, SteadyFlow Processes m m m ( kg / s) in out system Most energy conversion devices operate steadily over long periods of time. The rates of heat transfer and work crossing the control surface are constant with time. The states of the mass streams crossing the control surface or boundary are constant with time. Under these conditions the mass and energy content of the control volume are constant with time. 11
12 dm dt CV m CV 0 Steadystate, SteadyFlow Conservation of Mass: Since the mass of the control volume is constant with time during the steadystate, steadyflow process, the conservation of mass principle becomes or m m ( kg / s) in out 12
13 Flow work and the energy of a flowing fluid Energy flows into and from the control volume with the mass. The energy required to push the mass into or out of the control volume is known as the flow work or flow energy. The fluid up steam of the control surface acts as a piston to push a unit of mass into or out of the control volume. Consider the unit of mass entering the control volume shown below. As the fluid upstream pushes mass across the control surface, work done on that unit of mass is 13
14 FLOW WORK AND THE ENERGY OF A FLOWING FLUID Flow work, or flow energy: The work (or energy) required to push the mass into or out of the control volume. This work is necessary for maintaining a continuous flow through a control volume. 14
15 Total Energy of a Flowing Fluid h = u + Pv The flow energy is automatically taken care of by enthalpy. In fact, this is the main reason for defining the property enthalpy. The total energy consists of three parts for a nonflowing fluid and four parts for a flowing fluid. 15
16 Energy Transport by Mass When the kinetic and potential energies of a fluid stream are negligible When the properties of the mass at each inlet or exit change with time as well as over the cross section 16
17 ENERGY ANALYSIS OF STEADYFLOW SYSTEMS Steadyflow process: A process during which a fluid flows through a control volume steadily. 17
18 The term Pv is called the flow work done on the unit of mass as it crosses the control surface. The total energy of flowing fluid The total energy carried by a unit of mass as it crosses the control surface is the sum of the internal energy, flow work, potential energy, and kinetic energy. 2 V u Pv gz 2 2 h V gz 2 Here we have used the definition of enthalpy, h = u + Pv. Energy transport by mass Amount of energy transport across a control surface: 2 V Emass m mh gz 2 (kj) 18
19 Rate of energy transport across a control surface: 2 V Emass m mh gz ( kw ) 2 Conservation of Energy for General Control Volume The conservation of energy principle for the control volume or open system has the same word definition as the first law for the closed system. Expressing the energy transfers on a rate basis, the control volume first law is or E E E in out system ( kw) Rate of net energy transfer by heat, work, and mass Rate change in internal, kinetic, potential, etc., energies Considering that energy flows into and from the control volume with the mass, energy enters because net heat is transferred to the control volume, and energy leaves because the control volume does net work on its surroundings, the open system, or control volume, the first law becomes 19
20 where is the energy per unit mass flowing into or from the control volume. The energy per unit mass,, flowing across the control surface that defines the control volume is composed of four terms: the internal energy, the kinetic energy, the potential energy, and the flow work. The total energy carried by a unit of mass as it crosses the control surface is 2 V u Pv gz 2 2 h V gz 2 Where the time rate change of the energy of the control volume has been written as E CV 20
21 SteadyState, SteadyFlow Processes Most energy conversion devices operate steadily over long periods of time. The rates of heat transfer and work crossing the control surface are constant with time. The states of the mass streams crossing the control surface or boundary are constant with time. Under these conditions the mass and energy content of the control volume are constant with time. dm dt de dt CV CV m E Steadystate, SteadyFlow Conservation of Mass: CV CV m m ( kg / s) in out Steadystate, steadyflow conservation of energy Since the energy of the control volume is constant with time during the steadystate, steadyflow process, the conservation of energy principle becomes
22 or E E E in out 0 system ( kw) Rate of net energy transfer by heat, work, and mass Rate change in internal, kinetic, potential, etc., energies or Considering that energy flows into and from the control volume with the mass, energy enters because heat is transferred to the control volume, and energy leaves because the control volume does work on its surroundings, the steadystate, steadyflow first law becomes 22
23 Often this result is written as where Q Q Q net in out W W W net out in Steadystate, steadyflow for one entrance and one exit A number of thermodynamic devices such as pumps, fans, compressors, turbines, nozzles, diffusers, and heaters operate with one entrance and one exit. The steadystate, steadyflow conservation of mass and first law of thermodynamics for these systems reduce to 23
24 where the entrance to the control volume is state 1 and the exit is state 2 and mass flow rate through the device. When can we neglect the kinetic and potential energy terms in the first law? Consider the kinetic and potential energies per unit mass. For V = 45 m s V = 140 m s ke ke ke 2 V 2 2 ( 45m / s) 1kJ / kg m / s ( 140m / s) 1kJ / kg m / s pe gz 2 2 kj 1 kg 10 m 1kJ / kg kj For z 100m pe m s 1000m / s kg kj kg m 1kJ / kg kj z 1000m pe m s 1000m / s kg m is the 24
25 When compared to the enthalpy of steam (h 2000 to 3000 kj/kg) and the enthalpy of air (h 200 to 6000 kj/kg), the kinetic and potential energies are often neglected. When the kinetic and potential energies can be neglected, the conservation of energy equation becomes Q W m( h h ) ( kw) 2 1 We often write this last result per unit mass flow as Q where q and w W. m m Some SteadyFlow Engineering Devices q w ( h h ) ( kj / kg) 2 1 Below are some engineering devices that operate essentially as steadystate, steadyflow control volumes. 25
26 We may be familiar with nozzles, diffusers, turbines, compressors, compressors, heat exchangers, and mixing devices. However, the throttle valve may be a new device to many of us. The Throttle may be a simple as the expansion tube used in automobile air conditioning systems to cause the refrigerant pressure drop between the exit of the condenser and the inlet to the evaporator. 26
27 SOME STEADYFLOW ENGINEERING DEVICES Many engineering devices operate essentially under the same conditions for long periods of time. The components of a steam power plant (turbines, compressors, heat exchangers, and pumps), for example, operate nonstop for months before the system is shut down for maintenance. Therefore, these devices can be conveniently analyzed as steadyflow devices. A modern landbased gas turbine used for electric power production. This is a General Electric LM5000 turbine. It has a length of 6.2 m, it weighs 12.5 tons, and produces 55.2 MW at 3600 rpm with steam injection. 27
28 Nozzles and Diffusers Nozzles and diffusers are commonly utilized in jet engines, rockets, spacecraft, and even garden hoses. A nozzle is a device that increases the velocity of a fluid at the expense of pressure. A diffuser is a device that increases the pressure of a fluid by slowing it down. The crosssectional area of a nozzle decreases in the flow direction for subsonic flows and increases for supersonic flows. The reverse is true for diffusers. Energy balance for a nozzle or diffuser: 28
29 Deceleration of Air in a Diffuser 29
30 Acceleration of Steam in a Nozzle Steam 5 kg/s = 2.8 kj/kg MPa, 400C 0.02 m MPa 275 m/s 30
31 Turbines and Compressors Turbine drives the electric generator In steam, gas, or hydroelectric power plants. As the fluid passes through the turbine, work is done against the blades, which are attached to the shaft. As a result, the shaft rotates, and the turbine produces work. Compressors, as well as pumps and fans, are devices used to increase the pressure of a fluid. Work is supplied to these devices from an external source through a rotating shaft. A fan increases the pressure of a gas slightly and is mainly used to mobilize a gas. A compressor is capable of compressing the gas to very high pressures. Pumps work very much like compressors except that they handle liquids instead of gases. 31
32 Compressing Air by a Compressor 32
33 Power Generation by a Steam Turbine 33
34 Throttling valves Throttling valves are any kind of flowrestricting devices that cause a significant pressure drop in the fluid. What is the difference between a turbine and a throttling valve? The pressure drop in the fluid is often accompanied by a large drop in temperature, and for that reason throttling devices are commonly used in refrigeration and airconditioning applications. Energy balance 34
35 Expansion of Refrigerant134a in a Refrigerator 35
36 Mixing chambers In engineering applications, the section where the mixing process takes place is commonly referred to as a mixing chamber. The mixing chamber does not have to be a distinct chamber. An ordinary T elbow or a Yelbow in a shower, for example, serves as the mixing chamber for the cold and hotwater streams. The conservation of mass principle for a mixing chamber requires that the sum of the incoming mass flow rates equal the mass flow rate of the outgoing mixture. The conservation of energy equation is analogous to the conservation of mass equation. 36
37 60C Mixing of Hot and Cold Waters in a Shower 150 kpa 10C 45C 37
38 Heat exchangers Heat exchangers are devices where two moving fluid streams exchange heat without mixing. Heat exchangers are widely used in various industries, and they come in various designs. The heat transfer associated with a heat exchanger may be zero or nonzero depending on how the control volume is selected. 38
39 Cooling of Refrigerant134a by Water 39
40 40
41 Pipe and duct flow The transport of liquids or gases in pipes and ducts is of great importance in many engineering applications. Flow through a pipe or a duct usually satisfies the steadyflow conditions. 41
42 Electric Heating of Air in a House 42
43 Nozzles and Diffusers V 1 V V 2 1 V 1 V V 2 1 For flow through nozzles, the heat transfer, work, and potential energy are normally neglected, and nozzles have one entrance and one exit. The conservation of energy becomes 43
44 Solving for Example 54 V 2 V 2( h h ) V Steam at 0.4 MPa, 300oC, enters an adiabatic nozzle with a low velocity and leaves at 0.2 MPa with a quality of 90%. Find the exit velocity, in m/s. Control Volume: The nozzle Property Relation: Steam tables Process: Assume adiabatic, steadyflow Conservation Principles: Conservation of mass: For one entrance, one exit, the conservation of mass becomes m in m out m m m
45 Conservation of energy: According to the sketched control volume, mass crosses the control surface, but no work or heat transfer crosses the control surface. Neglecting the potential energies, we have Neglecting the inlet kinetic energy, the exit velocity is V 2( h h ) Now, we need to find the enthalpies from the steam tables Table A6, Page 908. Superheated Saturated Mix. kj T 300 C h P 0.2MPa h kg P1 0.4 MPa x o
46 At 0.2 MPa h f = kj/kg and h fg = kj/kg from table A5, Page 906. V 2 h = h + x h 2 f 2 fg = (0.90)(2201.6) = kj kg kj 1000 m / s 2( ) kg kj / kg m s
47 Turbines If we neglect the changes in kinetic and potential energies as fluid flows through an adiabatic turbine having one entrance and one exit, the conservation of mass and the steadystate, steadyflow first law becomes 47
48 Example 55 High pressure air at 1300 K flows into an aircraft gas turbine and undergoes a steadystate, steadyflow, adiabatic process to the turbine exit at 660 K. Calculate the work done per unit mass of air flowing through the turbine when (a) Temperaturedependent data are used. (b) C p,ave at the average temperature is used. (c) C p at 300 K is used. Control Volume: The turbine. Property Relation: Assume air is an ideal gas and use ideal gas relations. Process: Steadystate, steadyflow, adiabatic process 48
49 Conservation Principles: Conservation of mass: Conservation of energy: m in m out m m m 1 2 According to the sketched control volume, mass and work cross the control surface. Neglecting kinetic and potential energies and noting the process is adiabatic, we have W m ( h h ) 0 m1h 1 Wout m2h2 out
50 The work done by the air per unit mass flow is w out Wout h h m 1 2 Notice that the work done by a fluid flowing through a turbine is equal to the enthalpy decrease of the fluid. (a) Using the air tables, Table A17 (ideal gas properties of air), page 925 at T 1 = 1300 K, h 1 = kj/kg at T 2 = 660 K, h 2 = kj/kg w h h out 1 2 ( ) kj kg kj kg 50
51 (b) Using Table A2(b), page 900 by interpolation; at T ave = 980 K, C p, ave = kj/kgk wout h1 h2 Cp, ave( T1 T2 ) kj ( ) K kg K kj kg (c) Using Table A2(b) page 900, at T = 300 K, Cp = kj/kg K w h h C ( T T ) out 1 2 p 1 2 kj ( ) K kg K kj kg 51
52 Compressors and fans Compressor blades in a ground station gas turbine Photo courtesy of Progress Energy Carolinas, Inc. 52
53 Compressors and fans are essentially the same devices. However, compressors operate over larger pressure ratios (P 2 /P 1 ) than fans. Axial flow compressors are made of several fan blade like stages as shown on the pervious slide. If we neglect the changes in kinetic and potential energies as fluid flows through an adiabatic compressor having one entrance and one exit, the steadystate, steadyflow first law or the conservation of energy equation becomes 53
54 Example 56 Nitrogen gas is compressed in a steadystate, steadyflow, adiabatic process from 0.1 MPa, 25 o C. During the compression process the temperature becomes 125 o C. If the mass flow rate is 0.2 kg/s, determine the work done on the nitrogen, in kw. Control Volume: The compressor (see the compressor sketched above) Property Relation: Assume nitrogen is an ideal gas and use ideal gas relations Process: Adiabatic, steadyflow 54
55 Conservation Principles: Conservation of mass: Conservation of energy: m in m out m m m 1 2 According to the sketched control volume, mass and work cross the control surface. Neglecting kinetic and potential energies and noting the process is adiabatic, we have for one entrance and one exit 0 m 1( h1 0 0) ( W in) m 2( h2 0 0) W m ( h h ) in
56 The work done on the nitrogen is related to the enthalpy rise of the nitrogen as it flows through the compressor. The work done on the nitrogen per unit mass flow is w in Win m h h 2 1 Assuming constant specific heats at 300 K from Table A2(b) page 900, we write the work as w C ( T T ) in p kj ( ) K kg K kj kg 56
57 Throttling devices Consider fluid flowing through a oneentrance, oneexit porous plug. The fluid experiences a pressure drop as it flows through the plug. No net work is done by the fluid. Assume the process is adiabatic and that the kinetic and potential energies are neglected; then the conservation of mass and energy equations become 57
58 This process is called a throttling process. What happens when an ideal gas is throttled? When throttling an ideal gas, the temperature does not change. We will see later in Chapter 11 that the throttling process is an important process in the refrigeration cycle. A throttling device may be used to determine the enthalpy of saturated steam. The steam is throttled from the pressure in the pipe to ambient pressure in the calorimeter. The pressure drop is sufficient to superheat the steam in the calorimeter. Thus, the temperature and pressure in the calorimeter will specify the enthalpy of the steam in the pipe. 58
59 Example 57 One way to determine the quality of saturated steam is to throttle the steam to a low enough pressure that it exists as a superheated vapor. Saturated steam at 0.4 MPa is throttled to 0.1 MPa, 100 o C. Determine the quality of the steam at 0.4 MPa. Throttling orifice 1 2 Control Surface Control Volume: The throttle Property Relation: The steam tables Process: Steadystate, steadyflow, no work, no heat transfer, neglect kinetic and potential energies, one entrance, one exit Conservation Principles: Conservation of mass: m in m out m m m
60 Conservation of energy: According to the sketched control volume, mass crosses the control surface. Neglecting kinetic and potential energies and noting the process is adiabatic with no work, we have for one entrance and one exit, From table A6, page 908 we found h2 0 m ( h 0 0) 0 m ( h 0 0) m h h 1 2 m h o T2 100 C kj h P2 0.1MPa kg h 60
61 Therefore, as long as we have to determine the quality, means I have two phases (Vap+Liq). Then I have to use Table A5, Page 906 for 100 kpa h h kj kg h f x1h fg 0.4MPa x 1 h 1 h fg h f
62 Conservation Principles: Conservation of mass: m m m ( kg / s) in out system 0(steady) For one entrance, one exit, the conservation of mass becomes m in m out m m m
63 63
Chapter 5. Mass and Energy Analysis of Control Volumes. by Asst. Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn
Chapter 5 Mass and Energy Analysis of Control Volumes by Asst. Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics:
More informationChapter 5. Mass and Energy Analysis of Control Volumes
Chapter 5 Mass and Energy Analysis of Control Volumes Conservation Principles for Control volumes The conservation of mass and the conservation of energy principles for open systems (or control volumes)
More informationThe First Law of Thermodynamics. By: Yidnekachew Messele
The First Law of Thermodynamics By: Yidnekachew Messele It is the law that relates the various forms of energies for system of different types. It is simply the expression of the conservation of energy
More informationENT 254: Applied Thermodynamics
ENT 54: Applied Thermodynamics Mr. Azizul bin Mohamad Mechanical Engineering Program School of Mechatronic Engineering Universiti Malaysia Perlis (UniMAP) azizul@unimap.edu.my 0194747351 049798679 Chapter
More informationTHE FIRST LAW APPLIED TO STEADY FLOW PROCESSES
Chapter 10 THE FIRST LAW APPLIED TO STEADY FLOW PROCESSES It is not the sun to overtake the moon, nor doth the night outstrip theday.theyfloateachinanorbit. The Holy Qurān In many engineering applications,
More informationWeek 8. Steady Flow Engineering Devices. GENESYS Laboratory
Week 8. Steady Flow Engineering Devices Objectives 1. Solve energy balance problems for common steadyflow devices such as nozzles, compressors, turbines, throttling valves, mixers, heaters, and heat exchangers
More information1 st Law Analysis of Control Volume (open system) Chapter 6
1 st Law Analysis of Control Volume (open system) Chapter 6 In chapter 5, we did 1st law analysis for a control mass (closed system). In this chapter the analysis of the 1st law will be on a control volume
More informationfirst law of ThermodyNamics
first law of ThermodyNamics First law of thermodynamics  Principle of conservation of energy  Energy can be neither created nor destroyed Basic statement When any closed system is taken through a cycle,
More informationCLASS Fourth Units (Second part)
CLASS Fourth Units (Second part) Energy analysis of closed systems Copyright The McGrawHill Companies, Inc. Permission required for reproduction or display. MOVING BOUNDARY WORK Moving boundary work (P
More informationChapter 5: The First Law of Thermodynamics: Closed Systems
Chapter 5: The First Law of Thermodynamics: Closed Systems The first law of thermodynamics can be simply stated as follows: during an interaction between a system and its surroundings, the amount of energy
More informationThermodynamics ENGR360MEP112 LECTURE 7
Thermodynamics ENGR360MEP11 LECTURE 7 Thermodynamics ENGR360/MEP11 Objectives: 1. Conservation of mass principle.. Conservation of energy principle applied to control volumes (first law of thermodynamics).
More informationChapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn
Chapter 7 Entropy by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics: An Engineering Approach, 5th ed.,
More informationCHAPTER INTRODUCTION AND BASIC PRINCIPLES. (Tutorial). Determine if the following properties of the system are intensive or extensive properties: Property Intensive Extensive Volume Density Conductivity
More informationc Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 2 / 25 T145 = Q + W cv + i h 2 = h (V2 1 V 2 2)
Thermodynamic Processes & Isentropic Efficiency Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & Technology (BUET Dhaka1000, Bangladesh zahurul@me.buet.ac.bd
More informationEVALUATION OF THE BEHAVIOUR OF STEAM EXPANDED IN A SET OF NOZZLES, IN A GIVEN TEMPERATURE
Equatorial Journal of Engineering (2018) 913 Journal Homepage: www.erjournals.com ISSN: 01847937 EVALUATION OF THE BEHAVIOUR OF STEAM EXPANDED IN A SET OF NOZZLES, IN A GIVEN TEMPERATURE Kingsley Ejikeme
More informationI. (20%) Answer the following True (T) or False (F). If false, explain why for full credit.
I. (20%) Answer the following True (T) or False (F). If false, explain why for full credit. Both the Kelvin and Fahrenheit scales are absolute temperature scales. Specific volume, v, is an intensive property,
More informationDishwasher. Heater. Homework Solutions ME Thermodynamics I Spring HW1 (25 points)
HW1 (25 points) (a) Given: 1 for writing given, find, EFD, etc., Schematic of a household piping system Find: Identify system and location on the system boundary where the system interacts with the environment
More informationCHAPTER 7 ENTROPY. Copyright Hany A. AlAnsary and S. I. AbdelKhalik (2014) 1
CHAPTER 7 ENTROPY S. I. AbdelKhalik (2014) 1 ENTROPY The Clausius Inequality The Clausius inequality states that for for all cycles, reversible or irreversible, engines or refrigerators: For internallyreversible
More informationME Thermodynamics I. Lecture Notes and Example Problems
ME 227.3 Thermodynamics I Lecture Notes and Example Problems James D. Bugg September 2018 Department of Mechanical Engineering Introduction Part I: Lecture Notes This part contains handout versions of
More informationTHERMODYNAMICS, FLUID AND PLANT PROCESSES. The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial.
THERMODYNAMICS, FLUID AND PLANT PROCESSES The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial. THERMODYNAMICS TUTORIAL 2 THERMODYNAMIC PRINCIPLES SAE
More informationIntroduction to Chemical Engineering Thermodynamics. Chapter 7. KFUPM Housam Binous CHE 303
Introduction to Chemical Engineering Thermodynamics Chapter 7 1 Thermodynamics of flow is based on mass, energy and entropy balances Fluid mechanics encompasses the above balances and conservation of momentum
More informationWhere F1 is the force and dl1 is the infinitesimal displacement, but F1 = p1a1
In order to force the fluid to flow across the boundary of the system against a pressure p1, work is done on the boundary of the system. The amount of work done is dw =  F1.dl1, Where F1 is the force
More informationApplied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture No  03 First Law of Thermodynamics (Open System) Good afternoon,
More informationThermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am  10:00am & Monday 8:00am  9:00am MEP 261 Class ZA
Thermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am  10:00am & Monday 8:00am  9:00am MEP 261 Class ZA Dr. Walid A. Aissa Associate Professor, Mech. Engg. Dept. Faculty of Engineering
More informationIntroduction to Engineering thermodynamics 2 nd Edition, Sonntag and Borgnakke. Solution manual
Introduction to Engineering thermodynamics 2 nd Edition, Sonntag and Borgnakke Solution manual Chapter 6 Claus Borgnakke The picture is a false color thermal image of the space shuttle s main engine. The
More informationBMEA PREVIOUS YEAR QUESTIONS
BMEA PREVIOUS YEAR QUESTIONS CREDITS CHANGE ACCHA HAI TEAM UNIT1 Introduction: Introduction to Thermodynamics, Concepts of systems, control volume, state, properties, equilibrium, quasistatic process,
More informationR13 SET  1 '' ''' '' ' '''' Code No RT21033
SET  1 II B. Tech I Semester Supplementary Examinations, June  2015 THERMODYNAMICS (Com. to ME, AE, AME) Time: 3 hours Max. Marks: 70 Note: 1. Question Paper consists of two parts (PartA and PartB)
More informationExergy and the Dead State
EXERGY The energy content of the universe is constant, just as its mass content is. Yet at times of crisis we are bombarded with speeches and articles on how to conserve energy. As engineers, we know that
More informationME 2322 Thermodynamics I PRELECTURE Lesson 23 Complete the items below Name:
Lesson 23 1. (10 pt) Write the equation for the thermal efficiency of a Carnot heat engine below: T η = T 1 L H 2. (10 pt) Can the thermal efficiency of an actual engine ever exceed that of an equivalent
More informationKNOWN: Pressure, temperature, and velocity of steam entering a 1.6cmdiameter pipe.
4.3 Steam enters a.6cmdiameter pipe at 80 bar and 600 o C with a velocity of 50 m/s. Determine the mass flow rate, in kg/s. KNOWN: Pressure, temperature, and velocity of steam entering a.6cmdiameter
More informationChapter 17. For the most part, we have limited our consideration so COMPRESSIBLE FLOW. Objectives
Chapter 17 COMPRESSIBLE FLOW For the most part, we have limited our consideration so far to flows for which density variations and thus compressibility effects are negligible. In this chapter we lift this
More informationChapter 4. Energy Analysis of Closed Systems
Chapter 4 Energy Analysis of Closed Systems The first law of thermodynamics is an expression of the conservation of energy principle. Energy can cross the boundaries of a closed system in the form of heat
More informationESO201A: Thermodynamics
ESO201A: Thermodynamics First Semester 20152016 MidSemester Examination Instructor: Sameer Khandekar Time: 120 mins Marks: 250 Solve subparts of a question serially. Question #1 (60 marks): One kmol
More informationME Thermodynamics I
Homework  Week 01 HW01 (25 points) Given: 5 Schematic of the solar cell/solar panel Find: 5 Identify the system and the heat/work interactions associated with it. Show the direction of the interactions.
More informationChapter 1: Basic Definitions, Terminologies and Concepts
Chapter : Basic Definitions, Terminologies and Concepts . UThermodynamics:U It is a basic science that deals with: . Energy transformation from one form to another..
More information5/6/ :41 PM. Chapter 6. Using Entropy. Dr. Mohammad Abuhaiba, PE
Chapter 6 Using Entropy 1 2 Chapter Objective Means are introduced for analyzing systems from the 2 nd law perspective as they undergo processes that are not necessarily cycles. Objective: introduce entropy
More informationThermal Energy Final Exam Fall 2002
16.050 Thermal Energy Final Exam Fall 2002 Do all eight problems. All problems count the same. 1. A system undergoes a reversible cycle while exchanging heat with three thermal reservoirs, as shown below.
More informationChapter 6. Using Entropy
Chapter 6 Using Entropy Learning Outcomes Demonstrate understanding of key concepts related to entropy and the second law... including entropy transfer, entropy production, and the increase in entropy
More informationThermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGrawHill, Chapter 7 ENTROPY
Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGrawHill, 2011 Chapter 7 ENTROPY Copyright The McGrawHill Companies, Inc. Permission required for reproduction
More informationTwo mark questions and answers UNIT I BASIC CONCEPT AND FIRST LAW SVCET
Two mark questions and answers UNIT I BASIC CONCEPT AND FIRST LAW 1. What do you understand by pure substance? A pure substance is defined as one that is homogeneous and invariable in chemical composition
More informationMAE 11. Homework 8: Solutions 11/30/2018
MAE 11 Homework 8: Solutions 11/30/2018 MAE 11 Fall 2018 HW #8 Due: Friday, November 30 (beginning of class at 12:00p) Requirements:: Include T s diagram for all cycles. Also include p v diagrams for Ch
More informationThermodynamics: An Engineering Approach Seventh Edition in SI Units Yunus A. Cengel, Michael A. Boles McGrawHill, 2011.
Thermodynamics: An Engineering Approach Seventh Edition in SI Units Yunus A. Cengel, Michael A. Boles McGrawHill, 2011 Chapter 7 ENTROPY Mehmet Kanoglu University of Gaziantep Copyright The McGrawHill
More informationFirst Law of Thermodynamics Closed Systems
First Law of Thermodynamics Closed Systems Content The First Law of Thermodynamics Energy Balance Energy Change of a System Mechanisms of Energy Transfer First Law of Thermodynamics in Closed Systems Moving
More informationME 300 Thermodynamics II
ME 300 Thermodynamics II Prof. S. H. Frankel Fall 2006 ME 300 Thermodynamics II 1 Week 1 Introduction/Motivation Review Unsteady analysis NEW! ME 300 Thermodynamics II 2 Today s Outline Introductions/motivations
More informationReadings for this homework assignment and upcoming lectures
Homework #3 (group) Tuesday, February 13 by 4:00 pm 5290 exercises (individual) Thursday, February 15 by 4:00 pm extra credit (individual) Thursday, February 15 by 4:00 pm Readings for this homework assignment
More informationENTROPY. Chapter 7. Mehmet Kanoglu. Thermodynamics: An Engineering Approach, 6 th Edition. Yunus A. Cengel, Michael A. Boles.
Thermodynamics: An Engineering Approach, 6 th Edition Yunus A. Cengel, Michael A. Boles McGrawHill, 2008 Chapter 7 ENTROPY Mehmet Kanoglu Copyright The McGrawHill Companies, Inc. Permission required
More informationPrevious lecture. Today lecture
Previous lecture ds relations (derive from steady energy balance) Gibb s equations Entropy change in liquid and solid Equations of & v, & P, and P & for steady isentropic process of ideal gas Isentropic
More informationChapter 7. Entropy: A Measure of Disorder
Chapter 7 Entropy: A Measure of Disorder Entropy and the Clausius Inequality The second law of thermodynamics leads to the definition of a new property called entropy, a quantitative measure of microscopic
More informationThermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am  10:00am & Monday 8:00am  9:00am MEP 261 Class ZA
Thermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am  10:00am & Monday 8:00am  9:00am MEP 261 Class ZA Dr. Walid A. Aissa Associate Professor, Mech. Engg. Dept. Faculty of Engineering
More informationFirst Law of Thermodynamics
CH2303 Chemical Engineering Thermodynamics I Unit II First Law of Thermodynamics Dr. M. Subramanian 07July2011 Associate Professor Department of Chemical Engineering Sri Sivasubramaniya Nadar College
More informationLecture 35: Vapor power systems, Rankine cycle
ME 00 Thermodynamics I Spring 015 Lecture 35: Vapor power systems, Rankine cycle Yong Li Shanghai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Chuan Road Shanghai, 0040, P. R.
More informationLecture 44: Review Thermodynamics I
ME 00 Thermodynamics I Lecture 44: Review Thermodynamics I Yong Li Shanghai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Chuan Road Shanghai, 0040, P. R. China Email : liyo@sjtu.edu.cn
More informationAs Per New GTU Syllabus
As Per New GTU Syllabus Prepared by: Mr. Bhavin J. Vegada (M.Tech Thermal Engg.) Mr. Ketan C. Agola (M.Tech. Thermal Engg.) Darshan Institute of Engineering &Technology BRIEF CONTENTS Chapter No. Title
More informationToday lecture. 1. Entropy change in an isolated system 2. Exergy
Today lecture 1. Entropy change in an isolated system. Exergy  What is exergy?  Reversible Work & Irreversibility  SecondLaw Efficiency  Exergy change of a system For a fixed mass For a flow stream
More informationME 201 Thermodynamics
ME 0 Thermodynamics Solutions First Law Practice Problems. Consider a balloon that has been blown up inside a building and has been allowed to come to equilibrium with the inside temperature of 5 C and
More informationChapter 3 PROPERTIES OF PURE SUBSTANCES. Thermodynamics: An Engineering Approach, 6 th Edition Yunus A. Cengel, Michael A. Boles McGrawHill, 2008
Chapter 3 PROPERTIES OF PURE SUBSTANCES Thermodynamics: An Engineering Approach, 6 th Edition Yunus A. Cengel, Michael A. Boles McGrawHill, 2008 Objectives Introduce the concept of a pure substance. Discuss
More informationLecture 38: Vaporcompression refrigeration systems
ME 200 Termodynamics I Lecture 38: Vaporcompression refrigeration systems Yong Li Sangai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Cuan Road Sangai, 200240, P. R. Cina Email
More informationChapter Four fluid flow mass, energy, Bernoulli and momentum
41Conservation of Mass Principle Consider a control volume of arbitrary shape, as shown in Fig (41). Figure (41): the differential control volume and differential control volume (Total mass entering
More informationMAE 320 HW 7B. 1e. For an isolated system, please circle the parameter which will change with time. (a) Total energy;
MAE 320 HW 7B his comprehensive homework is due Monday, December 5 th, 206. Each problem is worth the points indicated. Copying of the solution from another is not acceptable. Multichoice, multianswer
More informationThe exergy of asystemis the maximum useful work possible during a process that brings the system into equilibrium with aheat reservoir. (4.
Energy Equation Entropy equation in Chapter 4: control mass approach The second law of thermodynamics Availability (exergy) The exergy of asystemis the maximum useful work possible during a process that
More informationIII. Evaluating Properties. III. Evaluating Properties
F. Property Tables 1. What s in the tables and why specific volumes, v (m /kg) (as v, v i, v f, v g ) pressure, P (kpa) temperature, T (C) internal energy, u (kj/kg) (as u, u i, u f, u g, u ig, u fg )
More informationFINAL EXAM. ME 200 Thermodynamics I, Spring 2013 CIRCLE YOUR LECTURE BELOW:
ME 200 Thermodynamics I, Spring 2013 CIRCLE YOUR LECTURE BELOW: Div. 5 7:30 am Div. 2 10:30 am Div. 4 12:30 am Prof. Naik Prof. Braun Prof. Bae Div. 3 2:30 pm Div. 1 4:30 pm Div. 6 4:30 pm Prof. Chen Prof.
More informationUNIT I Basic concepts and Work & Heat Transfer
SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road 517583 QUESTION BANK (DESCRIPTIVE) Subject with Code: Engineering Thermodynamics (16ME307) Year & Sem: IIB. Tech & IISem
More informationME 354 THERMODYNAMICS 2 MIDTERM EXAMINATION. Instructor: R. Culham. Name: Student ID Number: Instructions
ME 354 THERMODYNAMICS 2 MIDTERM EXAMINATION February 14, 2011 5:30 pm  7:30 pm Instructor: R. Culham Name: Student ID Number: Instructions 1. This is a 2 hour, closedbook examination. 2. Answer all questions
More informationIn the next lecture...
16 1 In the next lecture... Solve problems from Entropy Carnot cycle Exergy Second law efficiency 2 Problem 1 A heat engine receives reversibly 420 kj/cycle of heat from a source at 327 o C and rejects
More informationR13. II B. Tech I Semester Regular Examinations, Jan THERMODYNAMICS (Com. to ME, AE, AME) PART A
SET  1 II B. Tech I Semester Regular Examinations, Jan  2015 THERMODYNAMICS (Com. to ME, AE, AME) Time: 3 hours Max. Marks: 70 Note 1. Question Paper consists of two parts (PartA and PartB) 2. Answer
More information1. (10) Calorically perfect ideal air at 300 K, 100 kpa, 1000 m/s, is brought to rest isentropically. Determine its final temperature.
AME 5053 Intermediate Thermodynamics Examination Prof J M Powers 30 September 0 0 Calorically perfect ideal air at 300 K, 00 kpa, 000 m/s, is brought to rest isentropically Determine its final temperature
More informationCONCEPTS AND DEFINITIONS. Prepared by Engr. John Paul Timola
CONCEPTS AND DEFINITIONS Prepared by Engr. John Paul Timola ENGINEERING THERMODYNAMICS Science that involves design and analysis of devices and systems for energy conversion Deals with heat and work and
More informationIsentropic Efficiency in Engineering Thermodynamics
June 21, 2010 Isentropic Efficiency in Engineering Thermodynamics Introduction This article is a summary of selected parts of chapters 4, 5 and 6 in the textbook by Moran and Shapiro (2008. The intent
More informationFind: a) Mass of the air, in kg, b) final temperature of the air, in K, and c) amount of entropy produced, in kj/k.
PROBLEM 6.25 Three m 3 of air in a rigid, insulated container fitted with a paddle wheel is initially at 295 K, 200 kpa. The air receives 1546 kj of work from the paddle wheel. Assuming the ideal gas model,
More information374 Exergy Analysis. sys (u u 0 ) + P 0 (v v 0 ) T 0 (s s 0 ) where. e sys = u + ν 2 /2 + gz.
374 Exergy Analysis The value of the exergy of the system depends only on its initial and final state, which is set by the conditions of the environment The term T 0 P S is always positive, and it does
More informationUnit B1: List of Subjects
ES31 Energy Transfer Fundamentals Unit B: The First Law of Thermodynamics ROAD MAP... B1: The Concept of Energy B: Work Interactions B3: First Law of Thermodynamics B4: Heat Transfer Fundamentals Unit
More information(1)5. Which of the following equations is always valid for a fixed mass system undergoing an irreversible or reversible process:
Last Name First Name ME 300 Engineering Thermodynamics Exam #2 Spring 2008 March 28, 2008 Form A Note : (i) (ii) (iii) (iv) Closed book, closed notes; one 8.5 x 11 sheet allowed. 60 points total; 60 minutes;
More informationMAHALAKSHMI ENGINEERING COLLEGE
MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAPALLI621213. Department: Mechanical Subject Code: ME2202 U N IT  1 Semester: III Subject Name: ENGG. THERMODYNAMICS 1. 1 kg of gas at 1.1 bar, 27 o C is compressed
More informationChemical Engineering Thermodynamics Spring 2002
10.213 Chemical Engineering Thermodynamics Spring 2002 Test 2 Solution Problem 1 (35 points) High pressure steam (stream 1) at a rate of 1000 kg/h initially at 3.5 MPa and 350 ºC is expanded in a turbine
More informationChapter 3 PROPERTIES OF PURE SUBSTANCES
Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGrawHill, 2011 Chapter 3 PROPERTIES OF PURE SUBSTANCES Copyright The McGrawHill Companies, Inc. Permission
More informationMAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous) (ISO/IEC Certified) SUMMER 17 EXAMINATION
Important Instructions to examiners: 1) The answers should be examined by key words and not as wordtoword as given in the model answer scheme. 2) The model answer and the answer written by candidate
More information221. for gage pressure, the high and low pressures are expressed as. Noting that 1 psi = kpa,
 58E The systolic and diastolic pressures of a healthy person are given in mmhg. These pressures are to be expressed in kpa, psi, and meter water column. Assumptions Both mercury and water are incompressible
More informationChapter 7. Dr Ali Jawarneh. Department of Mechanical Engineering Hashemite University
Chapter 7 ENTROPY Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University Objectives Apply the second law of thermodynamics to processes. Define a new property called entropy to quantify
More informationThermodynamics Introduction and Basic Concepts
Thermodynamics Introduction and Basic Concepts by Asst. Prof. Channarong Asavatesanupap Mechanical Engineering Department Faculty of Engineering Thammasat University 2 What is Thermodynamics? Thermodynamics
More informationPTT 277/3 APPLIED THERMODYNAMICS SEM 1 (2013/2014)
PTT 77/3 APPLIED THERMODYNAMICS SEM 1 (013/014) 1 Energy can exist in numerous forms: Thermal Mechanical Kinetic Potential Electric Magnetic Chemical Nuclear The total energy of a system on a unit mass:
More informationMASS, MOMENTUM, AND ENERGY EQUATIONS
MASS, MOMENTUM, AND ENERGY EQUATIONS This chapter deals with four equations commonly used in fluid mechanics: the mass, Bernoulli, Momentum and energy equations. The mass equation is an expression of the
More informationUBMCC11  THERMODYNAMICS. B.E (Marine Engineering) B 16 BASIC CONCEPTS AND FIRST LAW PART A
UBMCC11  THERMODYNAMICS B.E (Marine Engineering) B 16 UNIT I BASIC CONCEPTS AND FIRST LAW PART A 1. What do you understand by pure substance? 2. Define thermodynamic system. 3. Name the different types
More information20 m neon m propane. g 20. Problems with solutions:
Problems with solutions:. A m tank is filled with a gas at room temperature 0 C and pressure 00 Kpa. How much mass is there if the gas is a) Air b) Neon, or c) Propane? Given: T7K; P00KPa; M air 9; M
More informationECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION. Instructor: R. Culham. Name: Student ID Number:
ECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION June 19, 2015 2:30 pm  4:30 pm Instructor: R. Culham Name: Student ID Number: Instructions 1. This is a 2 hour, closedbook examination. 2. Permitted
More informationContent. Entropy and principle of increasing entropy. Change of entropy in an ideal gas.
Entropy Content Entropy and principle of increasing entropy. Change of entropy in an ideal gas. Entropy Entropy can be viewed as a measure of molecular disorder, or molecular randomness. As a system becomes
More informationThermodynamics II. Week 9
hermodynamics II Week 9 Example Oxygen gas in a piston cylinder at 300K, 00 kpa with volume o. m 3 is compressed in a reversible adiabatic process to a final temperature of 700K. Find the final pressure
More informationExisting Resources: Supplemental/reference for students with thermodynamics background and interests:
Existing Resources: Masters, G. (1991) Introduction to Environmental Engineering and Science (Prentice Hall: NJ), pages 15 29. [ Masters_1991_Energy.pdf] Supplemental/reference for students with thermodynamics
More informationIntroduction to Thermodynamic Cycles Part 1 1 st Law of Thermodynamics and Gas Power Cycles
Introduction to Thermodynamic Cycles Part 1 1 st Law of Thermodynamics and Gas Power Cycles by James Doane, PhD, PE Contents 1.0 Course Oeriew... 4.0 Basic Concepts of Thermodynamics... 4.1 Temperature
More informationME 200 Final Exam December 14, :00 a.m. to 10:00 a.m.
CIRCLE YOUR LECTURE BELOW: First Name Last Name 7:30 a.m. 8:30 a.m. 10:30 a.m. 11:30 a.m. Boregowda Boregowda Braun Bae 2:30 p.m. 3:30 p.m. 4:30 p.m. Meyer Naik Hess ME 200 Final Exam December 14, 2015
More informationObjectives. Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation
Objectives Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation Conservation of Mass Conservation of Mass Mass, like energy, is a conserved
More informationEnergy and Energy Balances
Energy and Energy Balances help us account for the total energy required for a process to run Minimizing wasted energy is crucial in Energy, like mass, is. This is the Components of Total Energy energy
More informationBrown Hills College of Engineering & Technology
UNIT 4 Flow Through Nozzles Velocity and heat drop, Mass discharge through a nozzle, Critical pressure ratio and its significance, Effect of friction, Nozzle efficiency, Supersaturated flow, Design pressure
More informationChapter 12 PROPERTY RELATIONS. Department of Mechanical Engineering
Chapter 12 THERMODYNAMIC PROPERTY RELATIONS Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University it Objectives Develop fundamental relations between commonly encountered thermodynamic
More informationBasic Thermodynamics Prof. S.K Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Basic Thermodynamics Prof. S.K Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture  17 Properties of Pure SubstancesI Good morning to all of you. We were discussing
More informationUnit Workbook 2  Level 5 ENG U64 Thermofluids 2018 UniCourse Ltd. All Rights Reserved. Sample
Pearson BTEC Level 5 Higher Nationals in Engineering (RQF) Unit 64: Thermofluids Unit Workbook 2 in a series of 4 for this unit Learning Outcome 2 Vapour Power Cycles Page 1 of 26 2.1 Power Cycles Unit
More informationME 2322 Thermodynamics I PRELECTURE Lesson 23 Complete the items below Name:
Lesson 23 1. (10 pt) Write the equation for the thermal efficiency of a Carnot heat engine below: 1 L H 2. (10 pt) Can the thermal efficiency of an actual engine ever exceed that of an equivalent Carnot
More informationSPC 407 Sheet 5  Solution Compressible Flow Rayleigh Flow
SPC 407 Sheet 5  Solution Compressible Flow Rayleigh Flow 1. Consider subsonic Rayleigh flow of air with a Mach number of 0.92. Heat is now transferred to the fluid and the Mach number increases to 0.95.
More informationSECOND ENGINEER REG. III/2 APPLIED HEAT
SECOND ENGINEER REG. III/2 APPLIED HEAT LIST OF TOPICS A B C D E F G H I J K Pressure, Temperature, Energy Heat Transfer Internal Energy, Thermodynamic systems. First Law of Thermodynamics Gas Laws, Displacement
More informationEngineering Thermodynamics Solutions Manual
Engineering Thermodynamics Solutions Manual Prof. T.T. AlShemmeri Download free books at Prof. T.T. AlShemmeri Engineering Thermodynamics Solutions Manual 2 2012 Prof. T.T. AlShemmeri & bookboon.com
More information