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Engineering Thermodynamics - A Graphical Approach

by Israel Urieli (latest update: 6/22/2015) https://www.ohio.edu/mechanical/thermo/ http://www.thermofluids.net/

http://www.ou.edu/theobjectlab/book.html http://www.academia.edu/11767578/THERMODYNA MICS_AN_ENGINEERING_APPROACH_8th_EDITION_2 015 This web resource is intended to be a totally self-contained learning resource in Engineering Thermodynamics, independent of any textbook. It is designed to be suitable for a two course sequence for Mechanical Engineering majors. It may, however, be used in any format and for any purpose, including self-study. The various unique pedagogical features of this web resource are discussed in Paper AC 2010-47, which was presented at the 2010 ASEE Annual Conference. It is licensed under a Creative CommonsAttribution-Noncommercial-Share Alike 3.0 United States license and as such is freely available. Comments and constructive criticism are welcomed by the author. In Part 1 we introduce the First and Second Laws of Thermodynamics. Rather than applying these laws in terms of components and processes we have chosen a more interesting approach of applying them to complete cycles or systems. The ideal Stirling cycle machine is developed as a prime example of both Laws (refer to a paper: A Meeting between Robert Stirling and Sadi Carnot in 1824 presented at the 2014 ISEC.), and complete ideal heat engines, steam power plants and refrigeration systems are evaluated in Chapters 3 and 4. Where appropriate, we introduce graphical two-dimensional plots to evaluate the performance of these systems rather than relying on equations and tables. This enables intuitive visualization of the solutions to a high degree of accuracy.

Part 1 - Introduction to the First and Second Laws of Thermodynamics Chapter 1: Introductory Concepts, Units, and Definitions Chapter 2: Properties of Pure Substances a) Phase Change, Property Tables and Diagrams b) The Ideal Gas Equation of State Thermodynamic Properties Tables and Charts Chapter 3: The First Law of Thermodynamics for Closed Systems a) The Energy Equation for Closed Systems b) Ideal Stirling Cycle Machines (Engines / Coolers) c) The Air Standard Diesel Cycle (Compression-Ignition) Engine d) The Air Standard Otto Cycle (Spark-Ignition) Engine Chapter 4: The First Law of Thermodynamics for Control Volumes a) The Energy Equation for Control Volumes b) Steam Power Plants c) Refrigeration Systems Chapter 5: The Second Law of Thermodynamics Chapter 6: Entropy - A New Property a) Defining and Evaluating Entropy We present an Entropy Summary Sheet, Isentropic Processes Summary Sheet,

and an Adiabatic Efficiency Summary Sheet of all the relevant equations relating to this Section. b) Aircraft Gas Turbine Engines In Part 2 we introduce the concept of Exergy to determine theoretical limits of performance of various thermodynamic components and systems, followed by advanced application of steam power plants. The chapter on Carbon Dioxide as a refrigerant does not appear in any textbook that I am aware of. Because of the Global Warming crisis, the currently used refrigerant, R134a, will be banned from usage in automobile air conditioning systems in Europe within a few years. Among the alternatives being developed we prefer to return to Carbon Dioxide as the refrigerant of choice. Finally we introduce mixtures of water vapor and air and their application in air-conditioning and cooling tower systems, and conclude with an introduction to combustion processes.

Part 2 - Applied Engineering Thermodynamics Chapter 7: Exergy - Maximum Available Work Potential a) Reversible Work, Irreversibility, Second Law Efficiency b) Examples of Adiabatic Control Volumes c) Heat Transfer from a Thermal Source Chapter 8: Steam Power Cycles a) Ideal Rankine and Reheat Cycles b) Regenerative Cycles - Open and Closed Feedwater Heaters Case Study - The General James M. Gavin Steam Power Plant Chapter 9: Carbon Dioxide (R744) The New Refrigerant

Chapter 10: Air - Water Vapor Mixtures a) Humidity and the Adiabatic Saturation Process b) The Psychrometric Chart and Air Conditioning Processes c) Cooling Towers for Steam Power Plants Chapter 11: Combustion Combustion Molar Enthalpy Tables

Chapter 1: Introductory Concepts, Units, and Definitions In this chapter various relevant concepts and definitions are introduced, and will be used throughout the course.

Thermodynamics and Energy Thermodynamics is the science of energy, including energy storage and energy in transit. The Conservation of Energy Principle states that energy cannot be created or destroyed, but can only change its form. The three forms of energy storage of greatest interest to us are Potential Energy (PE), Kinetic Energy (KE), and Internal Energy (U), which we introduce below. The two forms of energy in transit that we consider are Work (W) and Heat (Q), and the interactions between these various forms of energy are defined in terms of the First Law of Thermodynamics, which we introduce in Chapter 3. A Word on Units In this course we use the International System (SI) units exclusively, with occasional lapses. In the US this is an ongoing battle causing much confusion in the global technical environment (refer to wikipedia on this subject). Even the National Council of Examiners for Engineering and Surveying (NCEES) seems to be confused at this point - the Fundamentals of Engineering (FE) Reference Handbook and exam contain

exclusively SI units and then, when you reach maturity and are ready to take the Professional Engineering (PE) exam, you find that the English system of units (USCS) is acceptable, and in some cases used exclusively. This confusion reached a new climax when in the Fall of 1999, NASA's $125 million Mars Climate Orbiter broke up in the Martian atmosphere, because scientists in one of NASA's subcontractors failed to convert critical data from the English sytem to the SI system of units. Force and Work We begin with Newton's Second Law, as follows:

The Weight of a body is the force acting on that body due to the acceleration due to gravity (g = 9.807 [m/s2]), in accordance with the Universal Theory of Gravitation developed by Isaac Newton. Legend has it that Newton was inspired by an apple falling on his head, as is shown in a delightful website by Mike Guidry of the University of Tennessee on Sir Isaac Newton, in which we see a cartoon showing the apple falling on Newton's head. Well, this legend is extremely relevant, since the weight of a small apple is approximately one Newton. Furthermore, the mass of a plastic bottle containing one liter of water is approximately one kilogram. Quick Quiz - can you estimate how many Newtons (or apples) a liter of water weighs? At this point we note that the major confusion of the English system of units came about because of the decision to define mass and force independently as 1 lb (pound), when in fact they are related through Newton's Second Law. In order to justify this one has to separately define a pound mass (lbm) and a pound force (lbf), thus since the acceleraton due to gravity g = 32.2 [ft/s2] we have:

One attempt to solve this paradox has been the introduction of a new unit of mass, the "slug", thus: 1 slug = 32.2 lbm however I challenge anyone to go to the grocery store and request a slug of potatos. We now consider the work done (W), the energy in transit requiring both the applied force (F) and movement (x). If the force (F) is constant over the distance moved (x) then the work done is given by:

However, in general the force (F) is not constant over the distance x, thus we need to sum all the incremental work processes taking into consideration the variation of the force (F). This leads to the equivalent integral form for determining work done (W) as follows:

A Units Survival Kit for US Students Over the years we have developed a basic Units Survival Kit (for the SI challenged) in order to help convert between the USCS (English) system and the SI (International) system of units, as well as to develop a feel for the magnitudes of the various units.

Quick Quiz - we all know (from reading our speedometers) that 50 mph is equivalent to 80 km/hr. 1. What is the accuracy of this conversion? 2. Use this information to show that 9 mph is equivalent to 4 m/s. We find that with the above survival kit we can determine many unit conversions between SI & English units, typically as demonstrated in the following block:

As we progress and learn new concepts we will add to this Survival Kit.

Forms of Energy We introduce the various forms of energy of interest to us in terms of a solid body having a mass m [kg]. These include potential, kinetic and internal energy. Potential energy (PE) is associated with the elevation of the body, and can be evaluated in terms of the work done to lift the body from one datum level to another under a constant acceleration due to gravity g [m/s2], as follows:

Kinetic energy (KE) of a body is associated with its velocity [m/s] and can be evaluated in terms of the work required to change the velocity of the body, as follows:

Internal energy (U) of a body is that associated with the molecular activity of the body as indicated by its temperature T [°C], and can be evaluated in terms of the heat required to change the temperature of the body having a specific heat capacity C [J/kg.°C], as follows:

Cooking with Potential Energy In order to gain an intuitive appreciation for the relative magnitudes of the different forms of energy we consider the (tongue-in-cheek) example of an attempt to cook a turkey by potential energy. The turkey is brought to the top of a 100 m building (about 30 stories) and then dropped from the ledge. The potential energy is thus converted into kinetic energy, and finally on impact the kinetic energy is converted into internal energy. The increase in internal energy is represented by an increase in temperature, and hopefully, if this experiment is repeated enough times the temperature increase will allow the turkey to cook. This remarkable experiment was first reported by R.C.Gimmi and Gloria J Browne - "Cooking with Potential Energy", published in the Journal of Irreproducible Results (Vol 33, 1987, pp 21-22).

What a disappointment! At 0.33°C per fall it will require repeating the experiment 600 times just to reach the cooking temperature of 200°C.

Thermodynamic Systems For purposes of analysis we consider two types of Thermodynamic Systems:  Closed System - usually referred to as a System or a Control Mass. This type of system is separated from its surroundings by a physical boundary. Energy in transit in the form of Work or Heat can flow across the system boundary, however there can be no mass flow across the boundary. One typical example of a system is a piston / cylinder device in which the system is defined as the fixed

mass of fluid contained within the cylinder.

 Open System - usually referred to as a Control Volume. In this case, in addition to work or heat, we have mass flow of the working fluid across the system boundaries through inlet and outlet ports. In this course we will be exclusively concerned with steady flow control volumes, in that the net mass of working fluid within the system boundaries remains constant (ie mass flow in [kg/s] = mass flow out [kg/s]). The following sections refer mainly to systems we will consider control volumes in more detail starting with Chapter 4a.

Properties of a System The closed system shown above can be defined by its various Properties, such as its pressure (P), temperature (T), volume (V) and mass (m). We will introduce and define the various properties of thermodynamic interest as needed in context. Furthermore the properties can be either Extensive or Intensive (or Specific). An extensive property is one whose value depends on the mass of the system, as opposed to an intensive property (such as pressure or temperature) which is independent of the system mass. A specific property is an intensive property which has been obtained by dividing the extensive property by the mass of the system. Two examples follow - notice that specific properties will always have kilograms (kg) in the units denominater.

One often used exception to the above definitions is the concept of Specific Weight, defined as the weight per unit volume. We will not be using this concept throughout this text. State and Equilibrium The State of a system is defined by the values of the various intensive properties of the system. The State Postulate states that if two independent intensive property values are defined, then all the other intensive property values (and thus the state of the system) are also defined. This can significantly simplify the graphical representation of a system, since only two-dimensional plots are required. Note that pressure and temperature are not necessarily independent properties, thus a boiling liquid will change its state from liquid to vapor at a constant temperature and pressure. We assume that throughout the system Equilibrium conditions prevail, thus there are no temperature or pressure gradients or transient effects. At any instant the entire system is under chemical and phase equilibrium. Process and Cycle A Process is a change of state of a system from an initial to a final state due to an energy interaction (work or heat) with its surroundings. For example in the following diagram the system has undergone a compression process in the piston-cylinder device.

The Process Path defines the type of process undergone. Typical process paths are:  Isothermal (constant temperature process)

 Isochoric or Isometric (constant volume process)  Isobaric (constant pressure process)  Adiabatic (no heat flow to or from the system during the process) We assume that all processes are Quasi-Static in that equilibrium is attained after each incremental step of the process. A system undergoes a Cycle when it goes through a sequence of processes that leads the system back to its original state.

Pressure The basic unit of pressure is the Pascal [Pa], however practical units are kiloPascal [kPa], bar [100 kPa] or atm (atmosphere) [101.32 kPa]. The Gage (or Vacuum) pressure is related to the Absolute pressure as shown in the diagram below:

The basic method of measuring pressure is by means of a Manometer, as shown below:

The atmospheric pressure is measured by means of a Mercury Barometer as follows:

Solved Problem 1.1 - Using a Barometer to determine the Height of a Building This solved problem allows us to determine the height of a building in terms of the difference in atmospheric pressure between the top and bottom of the building. Note that according to legend, when this problem was put to Nobel Prize Winner Niels Bohr (while still a student), he came up with an interesting response, as shown on the following Australian website of the Virtual Teacher.

Problem 1.2 - Using two manometers to measure pressure drop and downstream pressure of compressed air flowing in a pipe

Temperature

Temperature is a measure of molecular activity, and a temperature difference between two bodies in contact (for example the immediate surroundings and the system) is the driving force leading to heat transfer between them. Both the Fahrenheit and the Celsius scales are in common usage in the US, hence it is important to be able to convert between them. Furthermore we will find that in some cases we require the Absolute (Rankine and Kelvin) temperature scales (for example when using the Ideal Gas Equation of State), thus we find it convenient to plot all four scales as follows:

Notice from the plot that -40°C equals -40°F, leading to convenient formulas for converting between the two scales as follows:

Quick Quiz - The temperature in Chicago in winter can be as low as 14°F. What is the temperature in °C, K, and °R. [-10°C, 263 K, 474°R] Note that by convention 263 K is read "263 Kelvin," and not "263 degrees Kelvin". _____________________________________________________________________ _________________

Engineering Thermodynamics by Israel Urieli is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License

_____________________________________________________-

Solved Problem 1.1 - Using a Barometer to determine the Height of a Building In this problem we use the basic mercury barometer to determine the height of a building. Consider the case that the barometer reading at the top of the building is 751 mm Hg, and at the bottom of the building is 760 mm Hg. Assume the density of mercury ρHg= 13,600 kg/m3, and that the average density of the column of air ρ air = 1.18 kg/m3. The approach to solution is illustrated in the following diagram. A free-body force diagram on the column of air allows us to determine the height as a function of the pressure difference ΔP from top to bottom.

Well, exactly 100m? - obviously this is a contrived example. When we first evaluated the height we came up with the result:

In Engineering Thermodynamics we normally present results to within 3 or 4 significant digits. The question that one really should ask is "Is this a reasonable method to measure the height of a building?" and the answer is a resounding NO! In

the following we do an uncertainty analysis and find that unless we also measure the air temperature during this experiment (why? temperature doesn't even appear in the above equations!) then this method has an accuracy of: height = 100 ± 42 m Unacceptable by any standards.

-________________________________________________________

Problem 1.2 - Using two manometers to measure pressure drop and downstream pressure of compressed air flowing in a pipe A Throttling Valve is often used to control the downstream pressure of a high pressure fluid (such as steam or air) flowing in a pipe. In the following diagram we have a water manometer to measure the pressure drop ΔP caused by the throttling valve as well as a mercury manometer to measure the downstream pressure of the air.

If the height difference in the water manometer h w is 150 cm, and that in the mercury manometer hHg is 225 cm, determine a) the pressure difference ΔP [14.7 kPa] and b) the downstream absolute pressure of the air P2 [400 kPa]. Assume that the density of

water is 1000 kg/m3, the density of mercury is 13,600 kg/m3, and that the atmospheric pressure is 100 kPa.

_____________________________________________________________

Chapter 2: Pure Substances a) Phase Change, Property Tables and Diagrams In this chapter we consider the property values and relationships of a pure substance (such as water) which can exist in three phases - solid, liquid and gas. We will not consider the solid phase in this course. In order to introduce the rather complex phase change interactions that occur in pure substances we consider an experiment in which we have liquid water in a pistoncylinder device at 20°C and 100kPa pressure.. Heat is added to the cylinder while the pressure is maintained constant until the temperature reaches 300°C, as shown in the following T-v diagram (temperature vs specific volume):

From State (1) to State (2) the water maintains its liquid phase and the specific volume increases very slightly until the temperature reaches close to 100°C (State (2) - Saturated Liquid). As more heat is added the water progressively changes phase from liquid to water vapor (steam) while maintaining the temperature at 100°C (Saturation Temperature - Tsat) until there is no liquid remaining in the cylinder (State (4) - Saturated Vapor). If heating continues then the water vapor temperature increases (T > Tsat) and is said to be in the Superheated (State (5)). Notice that during this entire process the specific volume of the water increased by more than three orders of magnitude, which made it necessary to use a logarithmic scale for the specific volume axis. We now consider repeating this experiment at various pressures, as shown in the following T-v diagram:

Notice that as we increase the applied pressure, the region between the saturated liquid and saturated vapor decreases until we reach the Critical Point, above which there is no clear distinction between the liquid and vapor states. It is common practice to join the loci of saturated liquid and saturated vapor points as shown in the T-v diagram below.

The saturation lines define the regions of interest as shown in the diagram, being the Compressed Liquid region, the Quality region enclosed by the saturation lines, and the Superheat region (which also includes the Transcritical region) to the right of the saturated vapor line and above the critical point. We will use Property Tables associated with the regions in order to evaluate the various properties. Notice that we have provided property tables of steam, Refrigerant R134a, and Carbon Dioxide, which we believe is destined to become the future refrigerant of common usage.

The Quality Region The Quality Region (also referred to as the Saturated Liquid-Vapor Mixture Region) is enclosed between the saturated liquid line and the saturated vapor line, and at any point within this region the quality of the mixture (also referred to as the dryness factor) is defined as the mass of vapor divided by the total mass of the fluid, as shown in the following diagram:

Notice that properties relating to the saturated liquid have the subscript f, and those relating to the saturated vapor have the subscript g. In order to evaluate the quality consider a volume V containing a mass m of a saturated liquid-vapor mixture.

Notice from the steam property tables that we have also included three new properties: internal energy u [kJ/kg], enthalpy h [kJ/kg], and entropy s [kJ/kg.K] all of which will be defined as needed in future sections. At this stage we note that the 3

equations relating quality and specific volume can also be evaluated in terms of these three additional properties.

The P-v Diagram for Water The above discussion was done in terms of the T-v diagram, however recall from Chapter 1 when we defined the State Postulate that any two independent intensive properties can be used to completely define all other intensive state properties. It is often advantageous to use the P-v diagram with temperature as the parameter as in the following diagram:

Notice that because of the extremely large range of pressure and specific volume values of interest, this can only be done on a log-log plot. This is extremely inconvenient, so both the T-v and the P-v diagrams are normally not drawn to scale, however are sketched only in order to help define the problem, which is then solved in terms of the steam tables. This approach is illustrated in the following solved problems.

Solved Problem 2.1 - Two kilograms of water at 25°C are placed in a piston cylinder device under 100 kPa pressure as shown in the diagram (State (1)). Heat is added to the water at constant pressure until the piston reaches the stops at a total volume of 0.4 m3 (State (2)). More heat is then added at constant volume until the temperature of the water reaches 300°C (State (3)). Determine (a) the quality of the fluid and the mass of the vapor at state (2), and (b) the pressure of the fluid at state (3).

Step 1: Always draw a complete diagram of the states and processes of the problem and include all the relevant information on the diagram. In this case there are three states and two processes (constant pressure and constant volume). Step 2: In the case of a closed system with a phase change fluid, always sketch a T_v or P_v diagram indicating all the relevant states and processes on the diagram. As mentioned above this diagram will not be drawn to scale, however it will help to define the problem and the approach to solution. In the case of steam, as we determine various values from the steam tables we add these values to the diagram, typically as shown below:

Notice that the T_v diagram is based exclusively on intensive properties, hence mass is not indicated on the diagram. Thus we indicate on the diagram that in order to determine the quality at state (2) we need to first evaluate the specific volume v 2, which can then be compared to the saturation values v f and vg at the pressure of 100 kPa. Thus v2 = V / m = 0.4 [m3] / 2 [kg] = 0.2 [m3 / kg]

Concerning state (3), the problem statement did not specify that it is in the superheat region. We needed to first determine the saturated vapor specific volume v g at 300°C. This value is 0.0216 m3 / kg, which is much less than the specific volume v 3 of 0.2 m3 / kg, thus placing state (3) well into the superheated region. Thus the two intensive properties which we use to determine the pressure at state (3) are T 3 = 300°C, and v3 = 0.2 m3 / kg. On scanning the superheat tables we find that the closest values lie somewhere between 1.2 MPa and 1.4 MPa, thus we use linear interpolation techniqes to determine the actual pressure P3 as shown below:

Solved Problem 2.2 - Two kilograms of water at 25°C are placed in a piston cylinder device under 3.2 MPa pressure as shown in the diagram (State (1)). Heat is added to the water at constant pressure until the temperature of the fluid reaches 350°C (State (2)). Determine the final volume of the fluid at state (2).

In this example since the pressure is known (3.2 MPa) and remains constant throughout the process, we find it convenient to draw a P-v diagram indicating the process (1) - (2) as follows.

As in the previous example, on scanning the superheat tables we find that we need to interpolate between pressure P = 3.0 MPa and P = 3.5 MPa in order to determine the specific volume at the required pressure of 3.2 MPa as follows:

Problem 2.3 - A piston-cylinder device contains a saturated mixture of steam and water having a total mass of 0.5 kg at a pressure of 160 kPa and an initial volume of 100 liters. Heat is then added and the fluid expands at constant pressure until it reaches a saturated vapor state.  a) Draw a diagram representing the process showing the initial and final states of the system.  b) Sketch this process on a P-v diagram with respect to the saturation lines, critical point, and relevant constant temperature lines, clearly indicating the initial and final states.  c) Determine the initial quality and temperature of the fluid mixture prior to heating. [quality x1 = 0.182, T1 = 113.3°C]  d) Determine the final volume of the steam after heating. [0.546 m3 (546 liters)] Note: 1000 liters - 1 m3.

Problem 2.4 - A pressure cooker allows much faster (and more tender) cooking by maintaining a higher boiling temperature of the water inside. It is well sealed, and steam can only escape through an opening on the lid, on which sits a metal petcock. When the pressure overcomes the weight of the petcock, the steam escapes, maintaining a constant high pressure while the water boils.

Assuming that the opening under the petcock has an area of 8 mm2, determine  a) the mass of the petcock required in order to maintain an operating pressure of 99 kPa gage. [80.7gm]  b) the corresponding temperature of the boiling water. [120.2°C] Note: Assume that the atmospheric pressure is 101 kPa. Draw a free body diagram of the petcock. On to Chapter 2b) of Pure Substances _____________________________________________________________________ _________________

Engineering Thermodynamics by Israel Urieli is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License

Chapter 2: Pure Substances b) The Ideal Gas Equation of State We continue with our discussion on Pure Substances. We find that for a pure substance in the superheated region, at specific volumes much higher than that at the critical point, the P-v-T relation can be conveniently expressed by the Ideal Gas Equation of State to a high degree of accuracy, as follows: Pv=RT where: R is constant for a particular substance and is called the Gas Constant Note that for the ideal gas equation both the pressure P and the temperature T must be expressed in absolute quantities. Consider for example the T-v duagram for water as shown below:

The shaded zone in the diagram indicates the region that can be represented by the Ideal Gas equation to an error of less than 1%. Note that at the critical point the error is 330%. The gas constant R can be expressed as follows:

The three commonly used formats to express the Ideal Gas Equation of State are:

Solved Problem 2.5 - A piston-cylinder device contains 0.5 kg saturated liquid water at a pressure of 200 kPa. Heat is added and the steam expands at constant pressure until it reaches 300°C.  a) Draw a diagram representing the process showing the initial and final states of the system.  b) Sketch this process on a T-v (temperature-specific volume) diagram with respect to the saturation lines, critical point, and relevant constant pressure lines, clearly indicating the initial and final states.  c) Using steam tables determine the initial temperature of the steam prior to heating.  d) Using steam tables determine the final volume of the steam after heating  e) Using the ideal gas equation of state determine the final volume of the steam after heating. Determine the percentage error of using this method compared to that of using the steam tables. Note: The critical point data and the ideal gas constant for steam can be found on the first page of the steam tables. Solution Approach: Even if questions a) and b) were not required, this should always be the first priority item in solving a thermodynamic problem.

c) Since state (1) is specified as saturated liquid at 200 kPa, we use the saturated pressure steam tables to determine that T1 = Tsat@ 200kPa = 120.2°C. d) From the T-v diagram we determine that state (2) is in the superheated region, thus we use the superheated steam tables to determine that v2 = v200kPa,300°C = 1.3162 m3/kg. Thus V2 = m,v2 = (0,5kg).(1.3162 m3/kg) = 0.658 m3 (658 liters)

Note that in doing a units check we find that the following conversion appears so often that we feel it should be added to our Units Conversion Survival Kit (recall Chapter 1):

Finally we determine the percentage error of using the ideal gas equation at state (2)

Problem 2.6 - Consider a rigid container having a volume of 100 liters, filled with steam at an initial state of 400 kPa and 300°C. The steam is then cooled until it reaches a temperature of 90°C.  a) Draw a diagram representing the process showing the initial and final states of the system.  b) Using steam tables determine the mass of steam in the container. [0.153 kg]  c) Using the ideal gas equation of state determine the mass of steam in the container. [0.151 kg] Determine the percentage error of using this method compared to that of using the steam tables. [1%]  d) Sketch this process on a T-v (temperature-specific volume) diagram with respect to the saturation lines, critical point, and relevant constant pressure lines, clearly indicating the initial and final states.

 e) Using steam tables determine the final pressure and quality of the fluid mixture after cooling. [70.2 kPa, X = 0.277] Note: The critical point data and ideal gas constant for steam can be found on the first page of the steam tables.

Solved Problem 2.7 - An automobile tire with a volume of 100 liters is inflated to a gage pressure of 210 kPa. Determine a) the mass of air in the tire if the temperature is 20°C, and b) the increase in gage pressure if the temperature in the tire reaches 50°C. Assume that atmospheric pressure is 100 kPa. Solution Approach: We always begin a thermodynamic problem with a sketch, indicating all the relevant information on the sketch, thus:

For part b) the temperature in the tire increases to 50°C (323K), however the volume and mass of air in the tire remains constant, thus:

(Note for the SI challenged - initially the pressure was 30 psig, and then rose to 35 psig. Try to validate these values)

Solved Problem 2.8 - In aircraft design it is common practice to specify a standard temperature distribution in the atmosphere near the surface of the earth (up to an elevation z of 10000m) as T(z) = (T0 + a.z)°C, where T0 at the earth's surface is 15°C, and a is the Temperature Lapse Rate (= ­0.00651°C / m). Using the Ideal Gas Equation of State, determine the pressure at an elevation of 3000m if at z = 0, P = 101 kPa. Solution Approach: We first develop the solution in terms of the Hydrostatic Equation on an elemental height of the column of air, the Ideal Gas Equation of State, and the Temperature Lapse Rate equation, as follows:

Solved Problem 2.9 - You may wonder why we would be interested in knowing the value of air pressure at 3000m altitude. In the following example we continue with the above development in order to evaluate the payload that can be lifted to an altitude of 3000m using a small hot air balloon (Volume =1000 m3) having an empty mass of 100kg. Assume that the temperature of the air in the balloon is 100°C. Solution Approach: In this case we develop the solution in terms of a force balance between the bouyancy force (weight of the displaced air) and the gravity force including the weight of the hot air, the balloon empty mass, and the payload mass. The maximum altitude occurs when those two forces are equal, as follows:

Finally - with 154 kg payload at least 2 persons can share and enjoy this wonderful experience. Unfortunately they will not be able to enjoy a decent cup of English tea. At a pressure of 69.9 kPa water will boil at (heavens forbid) < 90°C! (Saturation

temperature Tsat from the Steam Tables). Quick quiz: determine the temperature of a cup of tea in Denver, Colorado (elevation 5000 ft), or on the peaks of the Rocky Mountains (elevation 14000 ft. Hint: use the Units Survival Kit presented in Chapter1 to first convert from feet to meters) ________________________________________________________________

Non-Ideal Gas Behavior We noticed in the above T-v diagram for water that the gasses can deviate significantly from the ideal gas equation of state in regions nearby the critical point and there have been many equations of state recommended for use to account for this non-ideal behaviour. However, this non-ideal behaviour can be accounted for by a correction factor called the Compressibility Factor Z defined as follows:

thus when the compressibility factor Z approaches 1 the gas behaves as an ideal gas. Note that under the same conditions of temperature and pressure, the compressibility factor can be expressed as:

Different fluids have different values of critical point pressure and temperature data PCR and TCR, and these can be determined from the Table of Critical Point Data of Various Substances. Fortunately the Principle of Corresponding States states that we can normalize the pressure and temperature values with the critical values as follows:

All fluids normalized in this manner exhibit similar non-ideal gas behaviour within a few percent, thus they can all be plotted on a Generalised Compressibility Chart. A number of these charts are available, however we prefer to use the Lee-Kesler (logarithmic) Compressibilty Chart, The use of the compressibility chart is shown in the following example.

Solved Problem 2.10 - Carbon Dioxide gas is stored in a 100 liter tank at 6 MPa and 30°C. Determine the mass of CO 2 in the tank based on (a) values obtained from the CO2 tables of data, (b) the ideal gas equation of state, and (c) the generalized compressibility chart. Compare (b) and (c) to (a) and determine the percentage error in each case. Solution Approach: We first determine the Critical Point data for CO 2 from the Table of Critical Point Data of Various Substances

After evaluating the Reduced Pressure and Reduced Temperature we plot them on the Generalized Compressibility Chart in order to determine the Compressibility Factor, as shown below

The actual value of specific volume va is obtained from the CO2 Superheat Tables

The general rule is that if P > TCR then you are probably dealing with an ideal gas. If in doubt always check the Compressibility Factor Z on the Compressibility Chart.

Chapter 3: The First Law of Thermodynamics for Closed Systems Chapter 3: The First Law of Thermodynamics for Closed Systems a) The Energy Equation for Closed Systems We consider the First Law of Thermodynamics applied to stationary closed systems as a conservation of energy principle. Thus energy is transferred between the system and the surroundings in the form of heat and work, resulting in a change of internal energy of the system. Internal energy change can be considered as a measure of molecular activity associated with change of phase or temperature of the system and the energy equation is represented as follows:

Heat (Q) Energy transferred across the boundary of a system in the form of heat always results from a difference in temperature between the system and its immediate surroundings. We will not consider the mode of heat transfer, whether by conduction, convection or radiation, thus the quantity of heat transferred during any process will either be specified or evaluated as the unknown of the energy equation. By convention, positive heat is that transferred from the surroundings to the system, resulting in an increase in internal energy of the system

Work (W)

In this course we consider three modes of work transfer across the boundary of a system, as shown in the following diagram:

In this course we are primarily concerned with Boundary Work due to compression or expansion of a system in a piston-cylinder device as shown above. In all cases we assume a perfect seal (no mass flow in or out of the system), no loss due to friction, and quasi-equilibrium processes in that for each incremental movement of the piston equilibrium conditions are maintained. By convention positive work is that done by the system on the surroundings, and negative work is that done by the surroundings on the system, Thus since negative work results in an increase in internal energy of the system, this explains the negative sign in the above energy equation. Boundary work is evaluated by integrating the force F multiplied by the incremental distance moved dx between an initial state (1) to a final state (2). We normally deal with a piston-cylinder device, thus the force can be replaced by the piston area A multiplied by the pressure P, allowing us to replace A.dx by the change in volume dV, as follows:

This is shown in the following schematic diagram, where we recall that integration can be represented by the area under the curve.

Note that work done is a Path Function and not a property, thus it is dependent on the process path between the initial and final states. Recall in Chapter 1 that we introduced some typical process paths of interest:  Isothermal (constant temperature process)  Isochoric or Isometric (constant volume process)  Isobaric (constant pressure process)  Adiabatic (no heat flow to or from the system during the process) It is sometimes convenient to evaluate the specific work done which can be represented by a P-v diagram thus if the mass of the system is m [kg] we have finally:

We note that work done by the system on the surroundings (expansion process) is positive, and that done on the system by the surroundings (compression process) is negative.

Finally for a closed system Shaft Work (due to a paddle wheel) and Electrical Work (due to a voltage applied to an electrical resistor or motor driving a paddle wheel) will always be negative (work done on the system). Positive forms of shaft work, such as that due to a turbine, will be considered in Chapter 4 when we discuss open systems.

Internal Energy (u) The third component of our Closed System Energy Equation is the change of internal energy resulting from the transfer of heat or work. Since specific internal energy is a property of the system, it is usually presented in the Property Tables such as in the Steam Tables. Consider for example the following solved problem.

Solved Problem 3.1 - Recall the Solved Problem 2.2 in Chapter 2a in which we presented a constant pressure process. We wish to extend the problem to include the energy interactions of the process, hence we restate it as follows: Two kilograms of water at 25°C are placed in a piston cylinder device under 3.2 MPa pressure as shown in the diagram (State (1)). Heat is added to the water at constant pressure until the temperature of the steam reaches 350°C (State (2)). Determine the work done by the fluid (W) and heat transferred to the fluid (Q) during this process. Solution Approach: We first draw the diagram of the process including all the relevant data as follows:

Notice the four questions to the right of the diagram, which we should always ask before attempting to solve any thermodynamic problem. What are we dealing with liquid? pure fluid, such as steam or refrigerant? ideal gas? In this case it is steam, thus we will use the steam tables to determine the various properties at the various states. Is the mass or volume given? If so we will specify and evaluate the energy equation in

kiloJoules rather than specific quantities (kJ/kg). What about entropy? Not so fast - we have not yet considered enthalpy (below) - wait patiently until Chapter 6. Since work involves the integral of P.dv we find it convenient to sketch the Pv diagram of the problem as follows:

Notice on the P-v diagram how we determine the specific work done as the area under the process curve. We also notice that in the Compressed Liquid region the constant temperature line is essentially vertical. Thus all the property values at State (1) (compressed liquid at 25°C) can be determined from the saturated liquid table values at 25°C.

Enthalpy (h) - a New Property In the case studies that follow we find that one of the major applications of the closed system energy equation is in heat engine processes in which the system is approximated by an ideal gas, thus we will develop relations to determine the internal energy for an ideal gas. We will find also that a new property called Enthalpy will be useful both for Closed Systems and in particular for Open Systems, such as the components of steam power plants or refrigeration systems. Enthalpy is not a fundamental property, however is a combination of properties and is defined as follows:

As an example of its usage in closed systems, consider the following constant pressure process:

Applying the energy equation we obtain: However, since the pressure is constant throughout the process:

Substituting in the energy equation and simplifying:

Values for specific internal energy (u) and specific enthalpy (h) are available from the Steam Tables, however for ideal gasses it is necessary to develop equations for Δu and Δh in terms of Specific Heat Capacities. We develop these equations in terms of the differential form of the energy equation in the following web page: Specific Heat Capacities of an Ideal Gas We have provided property values for various ideal gases, including the gas constant and specific heat capacities in the following web page: Properties of Various Ideal Gases (at 300 K) __________________________________________________________________ On to Part b) of The First Law - Stirling Cycle Engines On to Part c) of The First Law - Diesel Cycle Engines On to Part d) of The First Law - Otto Cycle Engines

_____________________________________________________________________ _________________

Engineering Thermodynamics by Israel Urieli is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License

_______________________________________ Specific Heat Capacities of an Ideal Gas

For a simple system, internal energy (u) is a function of two independant variables, thus we assume it to be a function of temperature T and specific volume v, hence:

Substituting equation (2) in the energy equation (1) and simplifying, we obtain:

Now for a constant volume process (dv = 0):

That is, the specific constant volume heat capacity of a system is a function only of its internal energy and temperature. Now in his classic experiment of 1843 Joule showed

that the internal energy of an ideal gas is a function of temperature only, and not of pressure or specific volume. Thus for an ideal gas the partial derivatives can be replaced by ordinary derivatives, and the change in internal energy can be expressed as:

Consider now the enthalpy. By definition h = u + P v, thus differentiating we obtain:

Again for a simple system, enthalpy (h) is a function of two independant variables, thus we assume it to be a function of temperature T and pressure P, hence:

Substituting equation (6) in the energy equation (5), and simplifying:

Hence for a constant pressure process, since dP = 0:

That is, the specific constant pressure heat capacity of a system is a function only of its enthalpy and temperature. Now by definition

Now since for an ideal gas Joule showed that internal energy is a function of temperature only, it follows from the above equation that enthalpy is a function of temperature only. Thus for an ideal gas the partial derivatives can be replaced by ordinary derivatives, and the differential changes in enthalpy can be expressed as

Finally, from the definition of enthalpy for an ideal gas we have:

Values of R, CP, Cv and k for ideal gases are presented (at 300K) in the table on Properties of Various Ideal Gases. Note that the values of CP, Cv and k are constant with temperature only for mon-atomic gases such as helium and argon. For all other gases their temperature dependence can be considerable and needs to be considered. We find it convenient to express this dependence in tabular form and have provided a table of Specific Heat Capacities of Air.

Chapter 3: The First Law of Thermodynamics for Closed Systems b) Ideal Stirling Cycle Machines (Engines / Coolers) 1. The Stirling Cycle Engine Conceptually the Stirling engine is the simplest of all heat engines. It has no valves, and includes an externally heated space and an externally cooled space. It was invented by Robert Stirling, and an interesting website by Bob Sier includes a photograph of Robert Stirling, his original patent drawing of 1816, and an animated model of Stirling's original engine.

In its original single cylinder form the working gas (typically air or helium) is sealed within its cylinders by the piston and shuttled between the hot and cold spaces by a displacer. The linkage driving the piston and displacer will move them such that the gas will compress while it is mainly in the cool compression space and expand while in the hot expansion space. This is clearly illustrated in the adjacent animation which was produced by Richard Wheeler (Zephyris) of Wikipedia. Refer also to the animation produced by Matt Keveney in his Stirling engine animation website. Since the gas is at a higher temperature, and therefore pressure, during its expansion than during its compression, more power is produced during expansion than is reabsorbed during compression, and this net excess power is the useful output of the engine. Note that there are no valves or intermittent combustion, which is the major source of noise in an internal combustion engine. The same working gas is used over and over again, making the Stirling engine a sealed, closed cycle system. All that is added to the system is steady high temperature heat, and all that is removed from the system is low temperature (waste) heat and mechanical power. Athens, Ohio, is a hotbed of Stirling cycle machine activity, both engines and coolers, and includes R&D and manufacturing companies as well as internationally recognized consultants in the area of Stirling cycle computer analysis. The parent company of this activity is Sunpower, Inc. It was formed by William Beale in the early 1970's, mainly based on his invention of the free-piston Stirling engine which we describe below. Update (Jan. 2013): Sunpower was recently acquired by AMETEK, Inc in Pensylvania, however continues doing Stirling cycle machine development in Athens, Ohio. Update (Nov. 2013): Sunpower has recently introduced a 1 kW Stirling Developers Kit based on a free piston Stirling engine fired by Propane or natural gas. Some examples of single cylinder Stirling engines: Stirling Technology Inc. is a spinoff of Sunpower, and was formed in order to continue the development and manufacture of the 5 kW ST-5 Air engine. This large single cylinder engine burns biomass fuel (such as sawdust pellets or rice husks) and can function as a cogeneration unit in rural areas. It is not a free-piston engine, and uses a bell crank mechanism to obtain the correct displacer phasing. Another important early Stirling engine is Lehmann's machine on which Gusav Schmidt did the first reasonable analysis of

Stirling engines in 1871. Andy Ross of Columbus, Ohio built a small working replica of the Lehmann machine, as well as a model air engine. Solar Heat and Power Cogeneration: With the current energy and global warming crises, there is renewed interest in renewable energy systems, such as wind and solar energy, and distributed heat and power cogeneration systems. Cool Energy, Inc of Boulder, Colorado, is currently in advanced stages of developing a complete solar heat and power cogeneration system for home usage incorporating Stirling engine technology for electricity generation. This unique application includes evacuated tube solar thermal collectors, thermal storage, hot water and space heaters, and a Stirling engine/generator. Ideal Analysis: Please note that the following analysis of Stirling cycle engines is ideal, and is intended only as an example of First Law Analysis of closed systems. In the real world we cannot expect actual machines to perform any better than 40 - 50% of the ideal machine. The analysis of actual Stirling cycle machines is extremely complex and requires sophisticated computer analysis (see for example the course notes on: Stirling Cycle Machine Analysis.) The free-piston Stirling engine developed by Sunpower, Inc is unique in that there is no mechanical connection between the piston and the displacer, thus the correct phasing between them occurs by use of gas pressure and spring forces. Electrical power is removed from the engine by permanent magnets attached to the piston driving a linear alternator. Basically the ideal Stirling engine undergoes 4 distinct processes, each one of which can be separately analysed, as shown in the P-V diagram below. We consider first the work done during all four processes.  Process 1-2 is the compression process in which the gas is compressed by the piston while the displacer is at the top of the cylinder. Thus during this process the gas is cooled in order to maintain a constant temperature T C. Work W12 required to compress the gas is shown as the area under the P-V curve, and is evaluated as follows.

 Process 2-3 is a constant volume displacement process in which the gas is displaced from the cold space to the hot expansion space. No work is done, however as we shall see below, a significant amount of heat Q R is absorbed by the gas from the regenerator matrix.  Process 3-4 is the isothermal expansion process. Work W3-4 is done by the system and is shown as the area under the P-V diagram, while heat Q3-4 is added to the system from the heat source, maintaining the gas at a constant temperature TH.

 Finally, process 4-1 is a constant volume displacement process which completes the cycle. Once again we will see below that heat Q R is rejected by the working gas to the regenerator matrix. The net work Wnet done over the cycle is given by: Wnet = (W3-4 + W1-2), where the compression work W1-2 is negative (work done on the system). We now consider the heat transferred during all four processes, which will allow us to evaluate the thermal efficiency of the ideal Stirling engine. Recall from the previous section that in order to do a First Law analysis of an ideal gas to determine the heat transferred we needed to develop equations to determine the internal energy change Δu in terms of the Specific Heat Capacities of an Ideal Gas The two constant volume processes are formed by holding the piston in a fixed position, and shuttling the gas between the hot and cold spaces by means of the displacer. During process 4-1 the hot gas gives up its heat Q R by passing through a regenerator matrix, which is subsequently completely recovered during the process 23.

We will find in Chapter 5 that this is the maximum theoretical efficiency that is achievable from a heat engine, and usually referred to as the Carnot efficiency. For more information on this subject, refer to a paper: A Meeting between Robert Stirling and Sadi Carnot in 1824 presented at the 2014 ISEC.

Note that if no regenerator is present the heat QR must be supplied by the heater. Thus the efficiency will be significantly reduced to η th = Wnet / (Qin + QR). Furthermore the cooler will then have to reject the heat that is normally absorbed by the regenerator, thus the cooling load will be increased to Q out + QR. Recall that Q2-3 = QR = ­Q4-1. Note that the practical Stirling cycle has many losses associated with it and does not really involve isothermal processes, nor ideal regeneration. Furthermore since the Free-Piston Stirling cycle machines involve sinusoidal motion, the P-V diagram has an oval shape, rather than the sharp edges defined in the above diagrams. Nevertheless we use the ideal Stirling cycle to get an initial understanding and appreciation of the cycle performance. _____________________________________________________________________ _________________

Problem 3.2 - The Sunpower EG-1000 Stirling Engine/Generator _____________________________________________________________________ _____

2. The Stirling Cycle Cooler One important aspect of Stirling cycle machines that we need to consider is that the cycle can be reversed - if we put net work into the cycle then it can be used to pump heat from a low temperature source to a high temperature sink. Sunpower, Inc. has been actively involved in the deveplopment of Stirling cycle refrigeration systems and produces Stirling cycle croygenic coolers for liquifying oxygen. In 1984 Sunpower developed a free piston Duplex Stirling Machine having only three moving parts including one piston and two displacers, in which a gas fired Stirling cycle engine powered a Stirling cycle cooler. Global Cooling, Inc was established in 1995 as a spinoff of Sunpower, and was formed mainly in order to develop free-piston Stirling cycle coolers for home refrigerator applications. These systems, apart from being significantly more efficient than regular vapor-compression refrigerators, have the added advantage of being compact, portable units using helium as the working fluid (and not the HFC refrigerants such as R134a, having a Global Warming Potential of 1,300). More recently Global Cooling decided to concentrate their development efforts on systems in which there are virtually no competitive systems - cooling between -40°C and -80°C, and they established a new company name: Stirling Ultracold. We are fortunate to have obtained two original M100B coolers from Global Cooling. The one is used as a demonstrator unit, and is shown in operation in the following

photograph. The second unit is set up as a ME Senior Lab project in which we evaluate the actual performance of the machine under various specified loads and temperatures.

A schematic diagram followed by an animated schematic of the cooler (both courtesy of Global Cooling) are shown below

Conceptually the cooler is an extremely simple device, consisting essentially of only two moving parts - a piston and a displacer. The displacer shuttles the working gas (helium) between the compression and expansion spaces. The phasing between the piston and displacer is such that when the most of the gas is in the ambient compression space then the piston compresses the gas while rejecting heat to the ambient. The displacer then displaces the gas through the regenerator to the cold expansion space, and then both displacer and piston allow the gas to expand in this space while absorbing heat at a low temperature. _____________________________________________________________________ _________________

Problem 3.3 - Stirling Cycle Cooler M100B - Ideal Analysis Unfortunately the analysis of actual Stirling cycle machines is extremely complex and requires sophisticated computer analysis. We consider the idealised model of this cooler defined in terms of the P-V diagram shown below in order to determine the ideal performance of the M100B under typical operating conditions as described below. (Note that the values presented here are not actual values of the M100B, however were devised by your instructor for purposes of this exercise only).

Process (1)-(2) is the isothermal compression process at temperature T C = 30°C, during which heat QC is rejected to the ambient. Process (2)-(3) is the constant volume displacement process during which heat QR is rejected to the regenerator matrix. Process (3)-(4) is the isothermal expansion process at temperature T E = ­20°C, during which heat QE is absorbed from the freezer, and finally process (4)-(1) is the constant volume displacement process during which heat Q R is absorbed from the regenerator matrix. Thus the ideal Stirling cycle consists of four distinct processes, each one of which can be separately analysed. State (1) is defined at a maximum volume of 35 cm3 and a pressure of 1.9 MPa, and State (2) is defined at a minimum volume of 30 cm3. The energy transferred during both the compression and expansion processes is indicated on P-V diagrams as follows:

Since the working fluid is helium which is an ideal gas, we use the ideal gas equation of state throughout. Thus P V = m R T, where R = 2.077 kJ/kg K, and Δu = Cv ΔT, where Cv = 3.116 kJ/kg K. (refer: Ideal Gas Properties)

1. Determine the heat absorbed in the expansion space Q E during the expansion process (3) - (4) (Joules). Determine also the heat power absorbed (Watts). Note that the cycle frequency is the line frequency (f = 60 Hz). [QE = 8.56J (power = 513.6W)] 2. Determine the net work done per cycle (Joules): W net = WE + WC (Note that the compression work WC is always negative). Determine also the power supplied to the linear electric motor (Watts). [Wnet = -1.69J (power = -101W)] 3. Evaluate the Coefficient of Performance of the refrigerator defined as: COP R = QE / Wnet. (heat absorbed in the expansion space divided by the net work done). [COPR = 5.07] 4. Determine the amount of heat rejected by the working fluid Q R as it passes through the regenerator matrix during process (2) - (3). [QR = -16.46J (power = -988 W)] If there were no regenerator present then this heat would need to be removed from the gas by the expansion process in order to reduce the temperature to the cold temperature of the freezer. How would this affect the performance of the cooler? Discuss the importance of an effective regenerator in the Stirling cycle cooler.

Chapter 3: The First Law of Thermodynamics for Closed Systems b) Ideal Stirling Cycle Machines (Engines / Coolers) 1. The Stirling Cycle Engine Conceptually the Stirling engine is the simplest of all heat engines. It has no valves, and includes an externally heated space and an externally cooled space. It was invented by Robert Stirling, and an interesting website by Bob Sier includes a photograph of Robert Stirling, his original patent drawing of 1816, and an animated model of Stirling's original engine.

In its original single cylinder form the working gas (typically air or helium) is sealed within its cylinders by the piston and shuttled between the hot and cold spaces by a displacer. The linkage driving the piston and displacer will move them such that the gas will compress while it is mainly in the cool compression space and expand while in the hot expansion space. This is clearly illustrated in the adjacent animation which was produced by Richard Wheeler (Zephyris) of Wikipedia. Refer also to the animation produced by Matt Keveney in his Stirling engine animation website. Since the gas is at a higher temperature, and therefore pressure, during its expansion than during its compression, more power is produced during expansion than is reabsorbed during compression, and this net excess power is the useful output of the engine. Note that there are no valves or intermittent combustion, which is the major source of noise in an internal combustion engine. The same working gas is used over and over again, making the Stirling engine a sealed, closed cycle system. All that is added to the system is steady high temperature heat, and all that is removed from the system is low temperature (waste) heat and mechanical power. Athens, Ohio, is a hotbed of Stirling cycle machine activity, both engines and coolers, and includes R&D and manufacturing companies as well as internationally recognized consultants in the area of Stirling cycle computer analysis. The parent company of this activity is Sunpower, Inc. It was formed by William Beale in the early 1970's, mainly based on his invention of the free-piston Stirling engine which we describe below. Update (Jan. 2013): Sunpower was recently acquired by AMETEK, Inc in Pensylvania, however continues doing Stirling cycle machine development in Athens, Ohio. Update (Nov. 2013): Sunpower has recently introduced a 1 kW Stirling Developers Kit based on a free piston Stirling engine fired by Propane or natural gas. Some examples of single cylinder Stirling engines: Stirling Technology Inc. is a spinoff of Sunpower, and was formed in order to continue the development and manufacture of the 5 kW ST-5 Air engine. This large single cylinder engine burns biomass fuel (such as sawdust pellets or rice husks) and can function as a cogeneration unit in rural areas. It is not a free-piston engine, and uses a bell crank mechanism to obtain the correct displacer phasing. Another important early Stirling engine is Lehmann's machine on which Gusav Schmidt did the first reasonable analysis of

Stirling engines in 1871. Andy Ross of Columbus, Ohio built a small working replica of the Lehmann machine, as well as a model air engine. Solar Heat and Power Cogeneration: With the current energy and global warming crises, there is renewed interest in renewable energy systems, such as wind and solar energy, and distributed heat and power cogeneration systems. Cool Energy, Inc of Boulder, Colorado, is currently in advanced stages of developing a complete solar heat and power cogeneration system for home usage incorporating Stirling engine technology for electricity generation. This unique application includes evacuated tube solar thermal collectors, thermal storage, hot water and space heaters, and a Stirling engine/generator. Ideal Analysis: Please note that the following analysis of Stirling cycle engines is ideal, and is intended only as an example of First Law Analysis of closed systems. In the real world we cannot expect actual machines to perform any better than 40 - 50% of the ideal machine. The analysis of actual Stirling cycle machines is extremely complex and requires sophisticated computer analysis (see for example the course notes on: Stirling Cycle Machine Analysis.) The free-piston Stirling engine developed by Sunpower, Inc is unique in that there is no mechanical connection between the piston and the displacer, thus the correct phasing between them occurs by use of gas pressure and spring forces. Electrical power is removed from the engine by permanent magnets attached to the piston driving a linear alternator. Basically the ideal Stirling engine undergoes 4 distinct processes, each one of which can be separately analysed, as shown in the P-V diagram below. We consider first the work done during all four processes.  Process 1-2 is the compression process in which the gas is compressed by the piston while the displacer is at the top of the cylinder. Thus during this process the gas is cooled in order to maintain a constant temperature T C. Work W12 required to compress the gas is shown as the area under the P-V curve, and is evaluated as follows.

 Process 2-3 is a constant volume displacement process in which the gas is displaced from the cold space to the hot expansion space. No work is done, however as we shall see below, a significant amount of heat Q R is absorbed by the gas from the regenerator matrix.  Process 3-4 is the isothermal expansion process. Work W3-4 is done by the system and is shown as the area under the P-V diagram, while heat Q3-4 is added to the system from the heat source, maintaining the gas at a constant temperature TH.

 Finally, process 4-1 is a constant volume displacement process which completes the cycle. Once again we will see below that heat Q R is rejected by the working gas to the regenerator matrix. The net work Wnet done over the cycle is given by: Wnet = (W3-4 + W1-2), where the compression work W1-2 is negative (work done on the system). We now consider the heat transferred during all four processes, which will allow us to evaluate the thermal efficiency of the ideal Stirling engine. Recall from the previous section that in order to do a First Law analysis of an ideal gas to determine the heat transferred we needed to develop equations to determine the internal energy change Δu in terms of the Specific Heat Capacities of an Ideal Gas The two constant volume processes are formed by holding the piston in a fixed position, and shuttling the gas between the hot and cold spaces by means of the displacer. During process 4-1 the hot gas gives up its heat Q R by passing through a regenerator matrix, which is subsequently completely recovered during the process 23.

We will find in Chapter 5 that this is the maximum theoretical efficiency that is achievable from a heat engine, and usually referred to as the Carnot efficiency. For more information on this subject, refer to a paper: A Meeting between Robert Stirling and Sadi Carnot in 1824 presented at the 2014 ISEC.

Note that if no regenerator is present the heat QR must be supplied by the heater. Thus the efficiency will be significantly reduced to η th = Wnet / (Qin + QR). Furthermore the cooler will then have to reject the heat that is normally absorbed by the regenerator, thus the cooling load will be increased to Q out + QR. Recall that Q2-3 = QR = ­Q4-1. Note that the practical Stirling cycle has many losses associated with it and does not really involve isothermal processes, nor ideal regeneration. Furthermore since the Free-Piston Stirling cycle machines involve sinusoidal motion, the P-V diagram has an oval shape, rather than the sharp edges defined in the above diagrams. Nevertheless we use the ideal Stirling cycle to get an initial understanding and appreciation of the cycle performance. _____________________________________________________________________ _________________

Problem 3.2 - The Sunpower EG-1000 Stirling Engine/Generator _____________________________________________________________________ _____

2. The Stirling Cycle Cooler One important aspect of Stirling cycle machines that we need to consider is that the cycle can be reversed - if we put net work into the cycle then it can be used to pump heat from a low temperature source to a high temperature sink. Sunpower, Inc. has been actively involved in the deveplopment of Stirling cycle refrigeration systems and produces Stirling cycle croygenic coolers for liquifying oxygen. In 1984 Sunpower developed a free piston Duplex Stirling Machine having only three moving parts including one piston and two displacers, in which a gas fired Stirling cycle engine powered a Stirling cycle cooler. Global Cooling, Inc was established in 1995 as a spinoff of Sunpower, and was formed mainly in order to develop free-piston Stirling cycle coolers for home refrigerator applications. These systems, apart from being significantly more efficient than regular vapor-compression refrigerators, have the added advantage of being compact, portable units using helium as the working fluid (and not the HFC refrigerants such as R134a, having a Global Warming Potential of 1,300). More recently Global Cooling decided to concentrate their development efforts on systems in which there are virtually no competitive systems - cooling between -40°C and -80°C, and they established a new company name: Stirling Ultracold. We are fortunate to have obtained two original M100B coolers from Global Cooling. The one is used as a demonstrator unit, and is shown in operation in the following

photograph. The second unit is set up as a ME Senior Lab project in which we evaluate the actual performance of the machine under various specified loads and temperatures.

A schematic diagram followed by an animated schematic of the cooler (both courtesy of Global Cooling) are shown below

Conceptually the cooler is an extremely simple device, consisting essentially of only two moving parts - a piston and a displacer. The displacer shuttles the working gas (helium) between the compression and expansion spaces. The phasing between the piston and displacer is such that when the most of the gas is in the ambient compression space then the piston compresses the gas while rejecting heat to the ambient. The displacer then displaces the gas through the regenerator to the cold expansion space, and then both displacer and piston allow the gas to expand in this space while absorbing heat at a low temperature. _____________________________________________________________________ _________________

Problem 3.3 - Stirling Cycle Cooler M100B - Ideal Analysis Unfortunately the analysis of actual Stirling cycle machines is extremely complex and requires sophisticated computer analysis. We consider the idealised model of this cooler defined in terms of the P-V diagram shown below in order to determine the ideal performance of the M100B under typical operating conditions as described below. (Note that the values presented here are not actual values of the M100B, however were devised by your instructor for purposes of this exercise only).

Process (1)-(2) is the isothermal compression process at temperature T C = 30°C, during which heat QC is rejected to the ambient. Process (2)-(3) is the constant volume displacement process during which heat QR is rejected to the regenerator matrix. Process (3)-(4) is the isothermal expansion process at temperature T E = ­20°C, during which heat QE is absorbed from the freezer, and finally process (4)-(1) is the constant volume displacement process during which heat Q R is absorbed from the regenerator matrix. Thus the ideal Stirling cycle consists of four distinct processes, each one of which can be separately analysed. State (1) is defined at a maximum volume of 35 cm3 and a pressure of 1.9 MPa, and State (2) is defined at a minimum volume of 30 cm3. The energy transferred during both the compression and expansion processes is indicated on P-V diagrams as follows:

Since the working fluid is helium which is an ideal gas, we use the ideal gas equation of state throughout. Thus P V = m R T, where R = 2.077 kJ/kg K, and Δu = Cv ΔT, where Cv = 3.116 kJ/kg K. (refer: Ideal Gas Properties)

1. Determine the heat absorbed in the expansion space Q E during the expansion process (3) - (4) (Joules). Determine also the heat power absorbed (Watts). Note that the cycle frequency is the line frequency (f = 60 Hz). [QE = 8.56J (power = 513.6W)] 2. Determine the net work done per cycle (Joules): W net = WE + WC (Note that the compression work WC is always negative). Determine also the power supplied to the linear electric motor (Watts). [Wnet = -1.69J (power = -101W)] 3. Evaluate the Coefficient of Performance of the refrigerator defined as: COP R = QE / Wnet. (heat absorbed in the expansion space divided by the net work done). [COPR = 5.07] 4. Determine the amount of heat rejected by the working fluid Q R as it passes through the regenerator matrix during process (2) - (3). [QR = -16.46J (power = -988 W)] If there were no regenerator present then this heat would need to be removed from the gas by the expansion process in order to reduce the temperature to the cold temperature of the freezer. How would this affect the performance of the cooler? Discuss the importance of an effective regenerator in the Stirling cycle cooler.

Chapter 3: The First Law of Thermodynamics for Closed Systems c) The Air-Standard Diesel Cycle (Compression-Ignition) Engine The Air Standard Diesel cycle is the ideal cycle for Compression-Ignition (CI) reciprocating engines, first proposed by Rudolph Diesel over 100 years ago. The following link by the Kruse Technology Partnership describes the four-stroke diesel cycleoperation including a short history of Rudolf Diesel. The four-stroke diesel engine is usually used in motor vehicle systems, whereas larger marine systems usually use the two-stroke diesel cycle. Once again we have an excellent animation produced by Matt Keveney presenting the operation of the four-stroke diesel cycle.

The actual CI cycle is extremely complex, thus in initial analysis we use an ideal "airstandard" assumption, in which the working fluid is a fixed mass of air undergoing the complete cycle which is treated throughout as an ideal gas. All processes are ideal, combustion is replaced by heat addition to the air, and exhaust is replaced by a heat rejection process which restores the air to the initial state. The ideal air-standard diesel engine undergoes 4 distinct processes, each one of which can be separately analysed, as shown in the P-V diagrams below. Two of the four processes of the cycle are adiabatic processes (adiabatic = no transfer of heat), thus before we can continue we need to develop equations for an ideal gas adiabatic process as follows: The Adiabatic Process of an Ideal Gas (Q = 0) The analysis results in the following three general forms representing an adiabatic process:

where k is the ratio of heat capacities and has a nominal value of 1.4 at 300K for air. Process 1-2 is the adiabatic compression process. Thus the temperature of the air increases during the compression process, and with a large compression ratio (usually > 16:1) it will reach the ignition temperature of the injected fuel. Thus given the conditions at state 1 and the compression ratio of the engine, in order to determine the pressure and temperature at state 2 (at the end of the adiabatic compression process) we have:

Work W1-2 required to compress the gas is shown as the area under the P-V curve, and is evaluated as follows.

An alternative approach using the energy equation takes advantage of the adiabatic process (Q1-2 = 0) results in a much simpler process:

(thanks to student Nichole Blackmore for making me aware of this alternative approach) During process 2-3 the fuel is injected and combusted and this is represented by a constant pressure expansion process. At state 3 ("fuel cutoff") the expansion process continues adiabatically with the temperature decreasing until the expansion is complete. Process 3-4 is thus the adiabatic expansion process. The total expansion work is W exp = (W2-3 + W3-4) and is shown as the area under the P-V diagram and is analysed as follows:

Finally, process 4-1 represents the constant volume heat rejection process. In an actual Diesel engine the gas is simply exhausted from the cylinder and a fresh charge of air is introduced. The net work Wnet done over the cycle is given by: Wnet = (Wexp + W1-2), where as before the compression work W1-2 is negative (work done on the system). In the Air-Standard Diesel cycle engine the heat input Q in occurs by combusting the fuel which is injected in a controlled manner, ideally resulting in a constant pressure expansion process 2-3 as shown below. At maximum volume (bottom dead center) the burnt gasses are simply exhausted and replaced by a fresh charge of air. This is represented by the equivalent constant volume heat rejection process Q out = ­Q4-1. Both processes are analyzed as follows:

At this stage we can conveniently determine the engine efficiency in terms of the heat flow as follows:

_____________________________________________________________________ _____

The following problems summarize this section:

Problem 3.4 - A frictionless piston-cylinder device contains 0.2 kg of air at 100 kPa and 27°C. The air is now compressed slowly according to the relation P Vk = constant, where k = 1.4, until it reaches a final temperature of 77°C.  a) Sketch the P-V diagram of the process with respect to the relevant constant temperature lines, and indicate the work done on this diagram.  b) Using the basic definition of boundary work done determine the boundary work done during the process [-7.18 kJ].  c) Using the energy equation determine the heat transferred during the process [0 kJ], and verify that the process is in fact adiabatic. Derive all equations used starting with the basic energy equation for a non-flow system, the equation for internal energy change for an ideal gas (Δu), the basic equation for boundary work done, and the ideal gas equation of state [P.V = m.R.T]. Use values of specific heat capacity defined at 300K for the entire process.

Problem 3.5 - Consider the expansion stroke only of a typical Air Standard Diesel cycle engine which has a compression ratio of 20 and a cutoff ratio of 2. At the beginning of the process (fuel injection) the initial temperature is 627°C, and the air expands at a constant pressure of 6.2 MPa until cutoff (volume ratio 2:1). Subsequently the air expands adiabatically (no heat transfer) until it reaches the maximum volume.  a) Sketch this process on a P-v diagram showing clearly all three states. Indicate on the diagram the total work done during the entire expansion process.  b) Determine the temperatures reached at the end of the constant pressure (fuel injection) process [1800K], as well as at the end of the expansion process [830K], and draw the three relevant constant temperature lines on the P-v diagram.  c) Determine the total work done during the expansion stroke [1087 kJ/kg].  d) Determine the total heat supplied to the air during the expansion stroke [1028 kJ/kg].

Derive all equations used starting from the ideal gas equation of state and adiabatic process relations, the basic energy equation for a closed system, the internal energy and enthalpy change relations for an ideal gas, and the basic definition of boundary work done by a system (if required). Use the specific heat values defined at 1000K for the entire expansion process, obtained from the table of Specific Heat Capacities of Air.

Solved Problem 3.6 - An ideal air-standard Diesel cycle engine has a compression ratio of 18 and a cutoff ratio of 2. At the beginning of the compression process, the working fluid is at 100 kPa, 27°C (300 K). Determine the temperature and pressure of the air at the end of each process, the net work output per cycle [kJ/kg], and the thermal efficiency. Note that the nominal specific heat capacity values used for air at 300K are C P = 1.00 kJ/kg.K, Cv = 0.717 kJ/kg.K,, and k = 1.4. However they are all functions of temperature, and with the extremely high temperature range experienced in Diesel engines one can obtain significant errors. One approach (that we will adopt in this example) is to use a typical average temperature throughout the cycle. Solution Approach: The first step is to draw a diagram representing the problem, including all the relevant information. We notice that neither volume nor mass is given, hence the diagram and solution will be in terms of specific quantities. The most useful diagram for a heat engine is the P-v diagram of the complete cycle:

The next step is to define the working fluid and decide on the basic equations or tables to use. In this case the working fluid is air, and we have decided to use an average

temperature of 900K throughout the cycle to define the specific heat capacity values as presented in the table of Specific Heat Capacities of Air.

We now go through all four processes in order to determine the temperature and pressure at the end of each process.

Note that an alternative method of evaluating pressure P 2 is to simply use the ideal gas equation of state, as follows:

Either approach is satisfactory - choose whichever you are more comfortable with. We now continue with the fuel injection constant pressure process:

Notice that even though the problem requests "net work output per cycle" we have only calculated the heat in and heat out. In the case of a Diesel engine it is much simpler to evaluate the heat values, and we can easily obtain the net work from the energy balance over a complete cycle, as follows:

You may wonder at the unrealistically high thermal efficiency obtained. In this idealized analysis we have ignored many loss effects that exist in practical heat engines. We will begin to understand some of these loss mechanisms when we study the Second Law in Chapter 5. _____________________________________________________________________ _________ On to Part d) of The First Law - Otto Cycle Engines _____________________________________________________________________ _________________

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Chapter 3: The First Law of Thermodynamics for Closed Systems d) The Air-Standard Otto Cycle (Spark-Ignition) Engine The Air Standard Otto cycle is the ideal cycle for Spark-Ignition (SI) internal combustion engines, first proposed by Nikolaus Otto over 130 years ago, and which is currently used most motor vehicles. The following link by the Kruse Technology

Partnership presents a description of the four-stroke Otto cycle operation including a short history of Nikolaus Otto. Once again we have excellent animations produced by Matt Keveney presenting both the four-stroke and the two-stroke spark-ignition internal combustion engine engine operation The analysis of the Otto cycle is very similar to that of the Diesel cycle which we analysed in the previous section. We will use the ideal "air-standard" assumption in our analysis. Thus the working fluid is a fixed mass of air undergoing the complete cycle which is treated throughout as an ideal gas. All processes are ideal, combustion is replaced by heat addition to the air, and exhaust is replaced by a heat rejection process which restores the air to the initial state. The most significant difference between the ideal Otto cycle and the ideal Diesel cycle is the method of igniting the fuel-air mixture. Recall that in the ideal Diesel cycle the extremely high compression ratio (around 18:1) allows the air to reach the ignition temperature of the fuel. The fuel is then injected such that the ignition process occurs at a constant pressure. In the ideal Otto cycle the fuel-air mixture is introduced during the induction stroke and compressed to a much lower compression ratio (around 8:1) and is then ignited by a spark. The combustion results in a sudden jump in pressure while the volume remains essentially constant. The continuation of the cycle including the expansion and exhaust processes are essentially identical to that of the ideal Diesel cycle. We find it convenient to develop the analysis approach of the ideal Otto cycle through the following solved problem:

Solved Problem 3.7 - An ideal air-standard Otto cycle engine has a compression ratio of 8. At the beginning of the compression process, the working fluid is at 100 kPa, 27°C (300 K), and 800 kJ/kg heat is supplied during the constant volume heat addition process. Neatly sketch the pressure-volume [P-v] diagram for this cycle, and using the specific heat values for air at a typical average cycle temperaure of 900K determine:  a) the temperature and pressure of the air at the end of each process  b) the net work output/cycle [kj/kg], and  c) the thermal efficiency [ηth] of this engine cycle. Solution Approach:

The first step is to draw the P-v diagram of the complete cycle, including all the relevant information. We notice that neither volume nor mass have been provided, hence the diagram and solution will be in terms of specific quantities.

We assume that the fuel-air mixture is represented by pure air. The relevant equations of state, internal energy and adiabatic process for air follow:

We recall from the previous section that the nominal specific heat capacity values used for air at 300K are Cv = 0.717 kJ/kg.K,, and k = 1.4. However they are all functions of temperature, and with the extremely high temperature range experienced in internal combustion engines one can obtain significant errors. In this problem we use a typical average cycle temperature of 900K taken from the table of Specific Heat Capacities of Air.

We now go through all four processes in order to determine the temperature and pressure at the end of each process, as well as the work done and heat transferred during each process.

Note that the pressure P4 (as well as P2 above) could also be evaluated from the adiabatic process equation. We do so below as a vailidity check, however we find it more convenient to use the ideal gas equation of state wherever possible. Either method is satisfactory.

We continue with the final process to determine the heat rejected:

Notice that we have applied the energy equation to all four processes allowing us two alternative means of evaluating the "net work output per cycle" and the thermal efficiency, as follows:

Note that using constant specific heat values over the cycle we can determine the thermal efficiency directly from the ratio of specific heat capacities k with the following formula:

where r is the compression ratio

Quick Quiz: Using the heat and work energy equations derived above, derive this relation

Problem 3.8 - This is an extension of Solved Problem 3.7, in which we wish to use throughout all four processes the nominal standard specific heat capacity values for air at 300K. Using the values Cv = 0.717 kJ/kg.K, and k = 1.4, determine:  a) the temperature and pressure of the air at the end of each process [P2 = 1838 kPa, T2 = 689K, T3 = 1805K, P3 = 4815 kPa, P4 = 262 kPa, T4 = 786K]  b) the net work output/cycle [451.5 kJ/kg], and  c) the thermal efficiency of this engine cycle. [ηth = 56%] _____________________________________________________________________ _________________

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Chapter 4: The First Law of Thermodynamics for Control Volumes Chapter 4: The First Law of Thermodynamics for Control Volumes a) The Energy Equation for Control Volumes

In this course we consider three types of Control Volume Systems - Steam Power Plants, Refrigeration Systems, and Aircraft Jet Engines. Fortunately we will be able to separately analyse each component of the system independent of the entire system, which is typically represented as follows:

In addition to the energy flow across the control volume boundary in the form of heat and work, we will also have mass flowing into and out of the control volume. We will only consider Steady Flow conditions throughout, in which there is no energy or mass accumulation in the control volume, thus we will find it convenient to derive the energy equation in terms of power [kW] rather than energy [kJ]. Furthermore the term Control Volume indicates that there is no boundary work done by the system, and typically we have shaft work, such as with a turbine, compressor or pump. Mass Flow

Consider an elemental mass dm flowing through an inlet or outlet port of a control volume, having an area A, volume dV, length dx, and an average steady velocity , as follows.

Thus finally the mass flow rate

can be determined as follows:

Flow Energy

The fluid mass flows through the inlet and exit ports of the control volume accompanied by its energy. These include four types of energy - internal energy (u), kinetic enegy (ke), potential energy (pe), and flow work (w flow). In order to evaluate the flow work consider the following exit port schematic showing the fluid doing work against the surroundings through an imaginary piston:

It is of interest that the specific flow work is simply defined by the pressure P multiplied by the specific volume v. In the following section we can now develop the complete energy equation for a control volume. The Complete Energy Equation for a Control Volume

Consider the control volume shown in the following figure. Under steady flow conditions there is no mass or energy accumulation in the control volume thus the mass flow rate applies both to the inlet and outlet ports. Furthermore with a constant mass flow rate, it is more convenient to develop the energy equation in terms of power [kW] rather than energy [kJ] as was done previously.

The total power in due to heat and mass flow through the inlet port (1) must equal the total power out due to work and mass flow through the outlet port (2), thus:

The specific energy e can include kinetic and potential energy, however will always include the combination of internal energy (u) and flow work (Pv), thus we conveniently combine these properties in terms of the property enthalpy (as was done in Chapter 3a), as follows:

Note that z is the height of the port above some datum level [m] and g is the acceleration due to gravity [9.81 m/s2]. Substituting for energy e in the above energy equation and simplifying, we obtain the final form of the energy equation for a singleinlet single-outlet steady flow control volume as follows:

Notice that enthalpy h is fundamental to the energy equation for a control volume. The Pressure-Enthalpy (P-h) Diagram

When dealing with closed systems we found that sketching T-v or P-v diagrams was a significant aid in describing and understanding the various processes. In steady flow systems we find that the Pressure-Enthalpy (P-h) diagrams serve a similar purpose, and we will use them extensively. In this course we consider three pure fluids - water, refrigerant R134a, and carbon dioxide, and we have provided P-h diagrams for all three in the Property Tables section. We will illustrate their use in the following examples. TheP-h diagram for water is shown below. Study it carefully and try to understand the significance of the distinctive shapes of the constant temperature curves in the compressed liquid, saturated mixture (quality region) and superheated vapor regions.

_____________________________________________________________________ _________ On to Part b) of Control Volumes - Steam Power Plants On to Part c) of Control Volumes - Refrigerators and Heat Pumps

_____________________________________________________________________ _________________

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Chapter 4: The First Law of Thermodynamics for Control Volumes a) The Energy Equation for Control Volumes

In this course we consider three types of Control Volume Systems - Steam Power Plants, Refrigeration Systems, and Aircraft Jet Engines. Fortunately we will be able to separately analyse each component of the system independent of the entire system, which is typically represented as follows:

In addition to the energy flow across the control volume boundary in the form of heat and work, we will also have mass flowing into and out of the control volume. We will only consider Steady Flow conditions throughout, in which there is no energy or mass accumulation in the control volume, thus we will find it convenient to derive the energy equation in terms of power [kW] rather than energy [kJ]. Furthermore the term Control Volume indicates that there is no boundary work done by the system, and typically we have shaft work, such as with a turbine, compressor or pump. Mass Flow

Consider an elemental mass dm flowing through an inlet or outlet port of a control volume, having an area A, volume dV, length dx, and an average steady velocity , as follows.

Thus finally the mass flow rate

can be determined as follows:

Flow Energy

The fluid mass flows through the inlet and exit ports of the control volume accompanied by its energy. These include four types of energy - internal energy (u), kinetic enegy (ke), potential energy (pe), and flow work (w flow). In order to evaluate the flow work consider the following exit port schematic showing the fluid doing work against the surroundings through an imaginary piston:

It is of interest that the specific flow work is simply defined by the pressure P multiplied by the specific volume v. In the following section we can now develop the complete energy equation for a control volume. The Complete Energy Equation for a Control Volume

Consider the control volume shown in the following figure. Under steady flow conditions there is no mass or energy accumulation in the control volume thus the mass flow rate applies both to the inlet and outlet ports. Furthermore with a constant mass flow rate, it is more convenient to develop the energy equation in terms of power [kW] rather than energy [kJ] as was done previously.

The total power in due to heat and mass flow through the inlet port (1) must equal the total power out due to work and mass flow through the outlet port (2), thus:

The specific energy e can include kinetic and potential energy, however will always include the combination of internal energy (u) and flow work (Pv), thus we conveniently combine these properties in terms of the property enthalpy (as was done in Chapter 3a), as follows:

Note that z is the height of the port above some datum level [m] and g is the acceleration due to gravity [9.81 m/s2]. Substituting for energy e in the above energy equation and simplifying, we obtain the final form of the energy equation for a singleinlet single-outlet steady flow control volume as follows:

Notice that enthalpy h is fundamental to the energy equation for a control volume. The Pressure-Enthalpy (P-h) Diagram

When dealing with closed systems we found that sketching T-v or P-v diagrams was a significant aid in describing and understanding the various processes. In steady flow systems we find that the Pressure-Enthalpy (P-h) diagrams serve a similar purpose, and we will use them extensively. In this course we consider three pure fluids - water, refrigerant R134a, and carbon dioxide, and we have provided P-h diagrams for all three in the Property Tables section. We will illustrate their use in the following examples. TheP-h diagram for water is shown below. Study it carefully and try to understand the significance of the distinctive shapes of the constant temperature curves in the compressed liquid, saturated mixture (quality region) and superheated vapor regions.

_____________________________________________________________________ _________ On to Part b) of Control Volumes - Steam Power Plants On to Part c) of Control Volumes - Refrigerators and Heat Pumps

_____________________________________________________________________ _________________

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Chapter 4: The First Law of Thermodynamics for Control Volumes c) Refrigerators and Heat Pumps Introduction and Discussion In the early days of refrigeration the two refrigerants in common use were ammonia and carbon dioxide. Both were problematic - ammonia is toxic and carbon dioxide requires extremely high presures (from around 30 to 200 atmospheres!) to operate in a refrigeration cycle, and since it operates on a transcritical cycle the compressor outlet temperature is extremely high (around 160°C). When Freon 12 (dichloro-diflouromethane) was discovered it totally took over as the refrigerant of choice. It is an extremely stable, non toxic fluid, which does not interact with the compressor lubricant, and operates at pressures always somewhat higher than atmospheric, so that if any leakage occured, air would not leak into the system, thus one could recharge without having to apply vacuum. Unfortunately when the refrigerant does ultimately leak and make its way up to the ozone layer the ultraviolet radiation breaks up the molecule releasing the highly active chlorine radicals, which help to deplete the ozone layer. Freon 12 has since been banned from usage on a global scale, and has been essentially replaced by chlorine free R134a (tetraflouro-ethane) - not as stable as Freon 12, however it does not have ozone depletion characteristics. Recently, however, the international scientific consensus is that Global Warming is caused by human energy related activity, and various man made substances are defined on the basis of a Global Warming Potential (GWP) with reference to carbon dioxide (GWP=1). R134a has been found to have a GWP of 1300 and in Europe, within a few years, automobile air conditioning systems will be barred from using R134a as a refrigerant. The new hot topic is a return to carbon dioxide (R744) as a refrigerant (refer for example to the website: R744.com). The previous two major problems of high pressure and high compressor temperature are found in fact to be advantageous. The very high cycle pressure results in a high fluid density throughout the cycle, allowing miniturization of the systems for the same heat pumping power requirements. Furthermore the high outlet temperature will allow instant defrosting of automobile windshields (we don't have to wait until the car engine warms up) and can be used for

combined space heating and hot water heating in home usage (refer for example: Norwegian IEA Heatpump Program Annex28. In this chapter we cover the vapor-compression refrigeration cycle using refrigerant R134a, and will defer coverage of the carbon dioxide cycle to Chapter 9.

A Basic R134a Vapor-Compression Refrigeration System Unlike the situation with steam power plants it is common practice to begin the design and analysis of refrigeration and heat pump systems by first plotting the cycle on the P-h diagram. The following schematic shows a basic refrigeration or heat pump system with typical property values. Since no mass flow rate of the refrigerant has been provided, the entire analysis is done in terms of specific energy values. Notice that the same system can be used either for a refrigerator or air conditioner, in which the heat absorbed in the evaporator (qevap) is the desired output, or for a heat pump, in which the heat rejected in the condenser (qcond) is the desired output.

In this example we wish to evaluate the following:  Heat absorbed by the evaporator (qevap) [kJ/kg]  Heat rejected by the condenser (qcond) [kJ/kg]  Work done to drive the compressor (wcomp) [kJ/kg]  Coefficient of Performance (COP) of the system, either as a refrigerator or as a heat pump.

As with the Steam Power Plant, we find that we can solve each component of this system separately and independently of all the other components, always using the same approach and the same basic equations. We first use the information given in the above schematic to plot the four processes (1)-(2)-(3)-(4)-(1) on the P-h diagram. Notice that the fluid entering and exiting the condenser (State (2) to State (3)) is at the high pressure 1 MPa. The fluid enters the evaporator at State (4) as a saturated mixture at -20°C and exits the evaporator at State (1) as a saturated vapor. State (2) is given by the intersection of 1 MPa and 70°C in the superheated region. State (3) is seen to be in the subcooled liquid region at 30°C, since the saturation temperature at 1 MPa is about 40°C. The process (3)-(4) is a vertical line (h 3 = h4) as is discussed below. In the following section we develop the methods of evaluating the solution of this example using the R134a refrigerant tables. Notice that the refrigerant tables do not include the subcooled region, however since the constant temperature line in this region is essentially vertical, we use the saturated liquid value of enthalpy at that temperature.

Notice from the P-h diagram plot how we can get an instant visual appreciation of the system performance, in particular the Coefficient of Performance of the system by comparing the enthalpy difference of the compressor (1)-(2) to that of the evaporator (4)-(1) in the case of a refrigerator, or to that of the condenser (2)-(3) in the case of a heat pump. We now consider each component as a separate control volume and apply the energy equation, starting with the compressor. Notice that we have assumed that the kinetic and potential energy change of the fluid is negligeable, and that the compressor is adiabatic. The required values of enthalpy for the inlet and outlet ports are determined from the R134a refrigerant tables.

The high pressure superheated refrigerant at port (2) is now directed to a condenser in which heat is extracted from the refrigerant, allowing it to reach the subcooled liquid region at port (3). This is shown on the following diagram of the condenser:

The throttle is simply an expansion valve which is adiabatic and does no work, however enables a significant reduction in temperature of the refrigerant as shown in the following diagram:

The final component is the evaporator, which extracts heat from the surroundings at the low temperature allowing the refrigerant liquid and vapor mixture to reach the saturated vapor state at station (1).

In determining the Coefficient of Performance - for a refrigerator or air-conditioner the desired output is the evaporator heat absorbed, and for a heat pump the desired output is the heat rejected by the condenser which is used to heat the home. The required input in both cases is the work done on the compressor (ie the electricity bill). Thus COPR = qevap / wcomp = 145 / 65.5 = 2.2 COPHP = qcond / wcomp = 210 / 65.5 = 3.2 Notice that for the same system we always find that COP HP = COPR + 1. Notice also that the COP values are usually greater than 1, which is the reason why they are never referred to as "Efficiency" values, which always have a maximum of 100%. Thus the P-h diagram is a widely used and very useful tool for doing an approximate evaluation of a refrigerator or heat pump system. In fact, in the official Reference Handbook supplied by the NCEES to be used in the Fundamentals of Engineering

exam, only the P-h diagram is presented for R134a. You are expected to answer all the questions on this subject based on plotting the cycle on this diagram as shown above. _____________________________________________________________________ __

Problems 4.7 - Home Refrigerator & Home Refrigerator with an Internal Heat Exchanger

Problem 4.8 - Home Heat Pump System for Space Heating Problem 4.9 - Home Air Conditioner and Hot Water Heating System Problem 4.10 - A Novel Water/Steam Air Conditioner Problem 4.11 - A Home Geothermal Heat-Pump Problem 4.12 - Home Air Conditioner and Hot Water System with an Internal Heat Exchanger

Problem 4.13 - The BSU Geothermal Heat Pump System (Summertime) Problem 4.14 - The BSU Geothermal Heat Pump System (Wintertime) _____________________________________________________________________ _________________

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Chapter 5: The Second Law of Thermodynamics In this chapter we consider a more abstract approach to heat engine, refrigerator and heat pump cycles, in an attempt to determine if they are feasible, and to obtain the limiting maximum performance available for these cycles. The concept of mechanical and thermal reversibility is central to the analysis, leading to the ideal Carnot cycles. (Refer to Wikipedia: Sadi Carnot a French physicist, mathematician and engineer who gave the first successful account of heat engines, the Carnot cycle, and laid the foundations of the second law of thermodynamics). For more information on this subject, refer to a paper: A Meeting between Robert Stirling and Sadi Carnot in 1824 presented at the 2014 ISEC. We represent a heat engine and a heat pump cycle in a minimalist abstract format as in the following diagrams. In both cases there are two temperature reservoirs TH and TL, with TH > TL.

In the case of a heat engine heat QH is extracted from the high temperature source TH, part of that heat is converted to work W done on the surroundings, and the rest is rejected to the low temperature sink TL. The opposite occurs for a heat pump, in which work W is done on the system in order to extract heat QL from the low temperature source TL and "pump" it to the high temperature sink TH. Notice that the thickness of the line represents the amount of heat or work energy transferred. We now present two statements of the Second Law of Thermodynamics, the first regarding a heat engine, and the second regarding a heat pump. Neither of these statements can be proved, however have never been observed to be violated. The Kelvin-Planck Statement: It is impossible to construct a device which operates on a cycle and produces no other effect than the transfer of heat from a single body in order to produce work.

We prefer a less formal description of this statement in terms of a boat extracting heat from the ocean in order to produce its required propulsion work:

The Clausius Statement: It is impossible to construct a device which operates on a cycle and produces no other effect than the transfer of heat from a cooler body to a hotter body.

Equivalence of the Clausius and Kelvin-Planck Statements It is remarkable that the two above statements of the Second Law are in fact equivalent. In order to demonstrate their equivalence consider the following diagram. On the left we see a heat pump which violates the Clausius statement by pumping heat QL from the low temperature reservoir to the high temperature reservoir without any work input. On the right we see a heat engine rejecting heat QL to the low temperature reservoir.

If we now connect the two devices as shown below such that the heat rejected by the heat engine QL is simply pumped back to the high temperature reservoir then there will be no need for a low temperature reservoir, resulting in a heat engine which violates the Kelvin-Planck statement by extracting heat from a single heat source and converting it directly into work.

Mechanical and Thermal Reversibility Notice that the statements on the Second Law are negative statements in that they only describe what is impossible to achieve. In order to determine the maximum performance available from a heat engine or a heat pump we need to introduce the concept ofReversibilty, including both mechanical and thermal reversibility. We will attempt to clarify these concepts in terms of the following example of a reversible piston cylinder device in thermal equilibrium with the surroundings at temperature T0, and undergoing a cyclic compression/expansion process.

For mechanical reversibility we assume that the process is frictionless, however we also require that the process is a quasi-equilibrium one. In the diagram we notice that during compression the gas particles closest to the piston will be at a higher pressure than those farther away, thus the piston will be doing more compression work than it would do if we had waited for equilibrium conditions to occur after each incremental step. Similarly, thermal reversibility requires that all heat transfer is isothermal. Thus if there is an incremental rise in temperature due to compression then we need to wait until thermal equilibrium is established. During expansion the incremental fall in temperature will result in heat being transferred from the surroundings to the system until equilibrium is established. In summary, there are three conditions required for reversible operation:  All mechanical processes are frictionless.  At each incremental step in the process thermal and pressure equilibrium conditions are established.  All heat transfer processes are isothermal. Carnot's Theorem Carnot's theorem, also known as Carnot's rule, or the Carnot principle, can be stated as follows: No heat engine operating between two heat reservoirs can be more efficient than a reversible heat engine operating between the same two reservoirs. The simplest way to prove this theorem is to consider the scenario shown below, in which we have an irreversible engine as well as a reversible engine operating between the reservoirs TH and TL, however the irreversible heat engine has a higher efficiency than the reversible one. They both draw the same amount of heat QH from the high

temperature reservoir, however the irreversible engine produces more work WI than that of the reversible engine WR.

Note that the reversible engine by its nature can operate in reverse, ie if we use some of the work output (WR) from the irreversible engine in order to drive the reversible engine then it will operate as a heat pump, transferring heat QH to the high temperature reservoir, as shown in the following diagram:

Notice that the high temperature reservoir becomes redundent, and we end up drawing a net amount of heat (QLR - QLI) from the low temperature reservoir in order to produce a net amount of work (WI - WR) - a Kelvin-Planck violator - thus proving Carnot's Theorem. Corollary 1 of Carnot's Theorem: The first Corollary of Carnot's theorem can be stated as follows: All reversible heat engines operating between the same two heat reservoirs must have the same efficiency.

Thus regardless of the type of heat engine, the working fluid, or any other factor if the heat engine is reversible, then it must have the same maximum efficiency. If this is not the case then we can drive the reversible engine with the lower efficiency as a heat pump and produce a Kelvin-Planck violater as above. Corollary 2 of Carnot's Theorem: The second Corollary of Carnot's theorem can be stated as follows: The efficiency of a reversible heat engine is a function only of the respective temperatures of the hot and cold reservoirs. It can be evaluated by replacing the ratio of heat transfers QL and QH by the ratio of temperatures TL and TH of the respective heat reservoirs. Thus using this corollary we can evaluate the thermal efficiency of a reversible heat engine as follows:

Notice that we always go into "meditation mode" before replacing the ratio of heats with the ratio of absolute temperatures, which is only valid for reversible machines. The simplest conceptual example of a reversible heat engine is the Carnot cycle engine, as described in the following diagram:

Obviously a totally impractical engine which cannot be realized in practice, since for each of the four processes in the cycle the surrounding environment needs to be

changed from isothermal to adiabatic. A more practical example is the ideal Stirling cycle engine as described in the following diagram:

This engine has a piston for compression and expansion work as well as a displacer in order to shuttle the working gas between the hot and cold spaces, and was described previously in Chapter 3b. Note that under the same conditions of temperatures and compression ratio the ideal Carnot engine has the same efficiency however a significantly lower net work output per cycle than the Ideal Stirling cycle engine, as can be easily seen in the following diagram:

When the reversible engine is operated in reverse it becomes a heat pump or a refrigerator. The coefficient of Performance of these machines is developed as follows:

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Solved Problem 5.1 - Reversible Home Air Conditioner and Hot Water Heater _____________________________________________________________________ ___

Problem 5.2 - A heat pump is used to meet the heating requirements of a house and maintain it at 20°C. On a day when the outdoor air temperature drops to -10°C it is estimated that the house looses heat at the rate of 10 kW. Under these conditions the actual Coefficient of Performance (COPHP) of the heat pump is 2.5.  a) Draw a diagram representing the heat pump system showing the flow of energy and the temperatures, and determine:  b) the actual power consumed by the heat pump [4 kW]  c) the power that would be consumed by a reversible heat pump under these conditions [1.02 kW]  d) the power that would be consumed by an electric resistance heater under these conditions [10 kW]  e) Comparing the actual heat pump to the reversible heat pump determine if the performance of the actual heat pump is feasible,

Derive all equations used starting from the basic definition of COPHP. _____________________________________________________________________ ___

Problem 5.3 - During an experiment conducted in senior lab at 25°C, a student measured that a Stirling cycle refrigerator that draws 250W of power has removed 1000kJ of heat from the refrigerated space maintained at -30°C. The running time of the refrigerator during the experiment was 20min. Draw a diagram representing the refrigerator system showing the flow of energy and the temperatures, and determine if these measurements are reasonable [COPR = 3.33, COPR,rev = 4.42, ratio COPR/COPR,rev = 75% > 60% - not feasible]. State the reasons for your conclusions. Derive all equations used starting from the basic definition of the Coefficient of Performance of a refrigerator (COPR). _____________________________________________________________________ ___ On to Chapter 6: Entropy - A New Property _____________________________________________________________________ _________________

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Chapter 6: Entropy - a New Property a) Defining and Evaluating Entropy Following on the Second Law developed in Chapter 5 we consider the Clausius Inequality which leads to the definition of a new property Entropy (S - kJ / K) as follows:

A very strange definition indeed, and difficult to comprehend. It is defined in differential format as the reversible heat transfer divided by the temperature. In an attempt to try and understand it we rewrite the definition as follows:

It is advantageous to compare this definition with the equivalent definition of work as follows:

Thus it begins to make sense. Work done requires both a driving force (pressure P) and movement (volume change dV). We implicitly evaluated the work done for reversible processes - always neglecting friction or any other irreversibility. Similarly we can state that heat transfer requires both a driving force (temperature T) and some equivalent form of "movement" (entropy change dS). Since temperature can be considered as represented by the vibration of the molecules, it is this transfer of vibrational energy that we define as entropy. We now continue with the Increase in Entropy Principle which is also derived from the Clausius Inequality, and states that for any process, the total change in entropy of a system or control volume together with its enclosing adiabatic surroundings is always greater than or equal to zero. This total change of entropy is denoted the Entropy Generated during the process (Sgen [kJ/K] or sgen [kJ/kg.K]. For reversible processes the entropy generated will always be zero. We use the differential form of the energy equation to derive the T.ds relations which can be used to evaluate the change of entropy (Δs) for processes involving 2-phase fluids (Steam, R134a, CO2), solids or liquids, or ideal gasses. Finally we present a convenient Entropy Equation Summary Sheet which summarises the relevant relations concerning entropy generation and evaluation of entropy change Δs. The Isentropic Processes Summary Sheet extends the relations of entropy change to enable the evaluation of isentropic processes.

One of the important applications of isentropic processes is in determining the efficiency of various adiabatic components. These include turbines, compressors and aircraft jet nozzles. Thus we have made the statement that steam turbines are designed to be adiabatic, and that any heat loss from the turbine will result in a reduction in output power, however only now can we make the statement that the ideal turbine is isentropic. This enables us to evaluate the Adiabatic Efficiency (sometimes referred to as isentropic efficiency) of these components, and we extend the isentropic process sheet with an Adiabatic Efficiency Summary Sheet. There are two property diagrams involving entropy in common usage, the temperature-entropy (T-s) and enthalpy-entropy (h-s) "Mollier" diagrams. We will find that the h-s diagram is extremely useful for evaluating adiabatic turbines and compressors, and complements the P-h diagram which we used in Chapter 4 to evaluate entire steam power plants or refrigerator systems. The h-s diagram for steam is presented below:

The important characteristic of the h_s diagram is that the ideal adiabatic turbine can be conveniently plotted as a vertical line, allowing an intuitive visual appreciation of the turbine performance. We define the turbine adiabatic efficiency as follows:

Notice that for the for the actual turbine there will always be an increase in entropy, which means that the turbine adiabatic efficiency will always be less than 100%.

An Adiabatic Steam Turbine Example Consider an adiabatic steam turbine having a turbine adiabatic efficiency η T = 80%, operating under the conditions shown in the following diagram:

 Using steam tables, determine the enthalpy and entropy values at station (1) and station (2s) assuming that the turbine is isentropic. [h1 = 3479 kJ/kg, s1 = 7.764 kJ/kg.K; h2s = 2461 kJ/kg, s2s = s1]

 From the definition of turbine adiabatic efficiency (shown on the diagram), and given that ηT = 80%, determine the actual enthalpy and entropy values as well as the temperature at station (2a). [h2a = 2665 kJ/kg, s2a = 8.38 kJ/kg.K, T2a = 88°C]  Plot the actual and isentropic turbine processes (Stations (1)-(2a) and (1)-(2s)) on the enthalpy-entropy h-s "Mollier" diagram, and indicate the actual turbine specific work (wa) as well as the isentropic turbine specific work (w s) on the diagram.  Determine the actual power output of the turbine (kW). [1629 kW] The h-s diagram plot follows. Notice that we have indicated all the enthalpy and entropy values (which we determined from the steam tables) on the plot. This allows a check on the feasibility of our results.



Solved Problem 6.1 - Supercritical Steam Power Plant with Reheat for Athens, Ohio



Problem 6.2 - Turbine Adiabatic Efficiency

We take a somewhat simpler approach for incompressible liquids, as in the following problem. 

Problem 6.3 - Adiabatic Efficiency of a Power Plant Feedwater Pump

Unfortunately we find that at supercritical pressures and temperatures > 100°C, liquid water no longer behaves as an incompressible liquid, as in the following: 

Solved Problem 6.4 - Adiabatic Efficiency of a Supercritical Power Plant Feedwater Pump

Refrigerant R134a Compressors One of the interesting aspects of compressors is that they can be made more efficient by cooling. The reason why we still consider the adiabatic efficiency of compressors that are normally found in refrigeration, air-condition and heat pump systems is that it is considered to be impractical to cool them. Thus the ideal compressor (absorbing a minimum of power) is considered to be isentropic, and we define compressor adiabatic efficiency as follows:

We have also provided an R134a enthalpy-entropy (h-s) diagram which we find useful for evaluating adiabatic compressors that are normally found in refrigeration, air-condition and heat pump systems.



Solved Problem 6.5 - Adiabatic Efficiency of a R134a Compressor



Problem 6.6 - Adiabatic Evaluation of a R134a Compressor



Problem 6.7 - Recall in Chapter 4c that we provided Problem 4.7 concerning a home refrigerator, and examining it's performance before and after adding an internal heat exchanger. o a) Plot the actual and the isentropic compressor processes on the enthalpy-entropy (h-s) diagram provided, for both cases - with and without the internal heat exchanger. o b) Using the R134a tables determine the actual compressor adiabatic efficiency (ηC) for both cases [75%, 76%]



Problem 6.8 - Recall in Chapter 4c that we provided Problem 4.9 concerning an innovative home air conditioner and hot water heating system, in which we determined the COP for both the air conditioning and the water heating systems. Do sections a) and b) specified in Problem 6.7 above on the compressor of this home air conditioning system. [ηC = 76%]



Problem 6.9 - Adiabatic Efficiency of a High Pressure R134a Compressor

We will also extend the h-s diagram into the ideal gas region and use it to advantage when we consider gas turbine and jet engine systems in Part b) that follows. On to Part b) of Entropy - Aircraft Gas Turbune Engines _____________________________________________________________________ _________________

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Entropy Equation Summary Sheet

This summary sheet covers the relevant relations concerning entropy generation (sometimes referred to as entropy production), entropy evaluation and isentropic processes.

Isentropic Processes Summary Sheet Adiabatic Efficiencies of Turbines, Compressors, Nozzles

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Isentropic Process Summary Sheet

This summary sheet is an extension of the Entropy Equation Summary Sheet and covers the relevant relations concerning isentropic processes.

Note that these relations are identical to those that we developed previously for the Adiabatic Process of an Ideal Gas, which we used in Chapter 3c to analyze the adiabatic compression and expansion processes of an air-standard Diesel cycle engine. Adiabatic Efficiencies of Turbines, Compressors, Nozzles

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Adiabatic Efficiency Summary Sheet

This summary sheet is an extension of the Isentropic Process Summary Sheet and defines the adiabatic (isentropic) efficiency of turbines, compressors, and nozzles.

Back to the Engineering Thermodynamics Home Page

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Part 2 - Applied Engineering Thermodynamics Chapter 7: Exergy - Maximum Available Work Potential

Chapter 7: Exergy - Maximum Available Work Potential a) Reversible Work, Irreversibility and Second Law Efficiency In this Section we combine the First and Second Laws of thermodynamics in an attempt to determine theoretical limits of performance of various thermodynamic components and systems. Thus we introduce the concept of Exergy (aka Availability) - defined as the maximum work potential of a system or component at a given state in a specified environment. The environment is crucial in this definition since once the system or component has reached total thermodynamic equilibrium with its environment, and has used up all of its potential and kinetic energy relative to that environment, it is said to be in the Dead State. The environment is usually specified in terms of pressure and temperature as P 0 = 1 atmosphere. T0 = 25°C (77°F). In the following we attempt to introduce the concepts in terms of various examples.

Maximum Available Power Generation by a Wind Turbine This very intuitive first example defines the theoretical maximum available power from a wind generator as that which occurs when the kinetic energy of the air passing through the turbine rotor is reduced to zero. Clearly this is impractical, and in an interesting discussion of wind power on Wikipedia we find that Betz's Law imposes a theoretical limit of 59.3% of this maximum available power when the wind velocity is reduced by 1/3 while passing through the turbine rotor, and in fact the actual energy usage is much less.

An interesting application of wind power generation for home usage is the project of Dr Greg Kremer of the ME department at Ohio University. He has combined wind and solar power with battery backup connected to the electrical grid in his home. Using the conditions defining Dr. Kremer's wind turbine system (rotor diameter 3.53m) we determine the availability of his system as follows:

where:

Notice the dependence on the cube of the wind velocity. The average annual wind velocity in Athens, Ohio is 7mph (3.11m/s) giving a maximum available power of only 174W. However during the winter months (when the solar energy is lower) the velocity reaches 22.5mph (10m/s) giving a maximum available power of 5.79kW! Thus the wind/solar combination system seems like a compatible match, and so far Dr. Kremer has found that his net electrical power usage from the grid is negative! (His system feeds energy into the grid).

Hydroelectric Power Generation Our second example is that of hydroelectric power generation due to potential energy. Unlike wind power as described above, all of the available potential energy can be converted directly into work. Our favorite example is that of the Shoshone Hydro power plant in Glenwood Canyon, Colorado. A delightful description of this power plant is presented in Glenwood Canyon: An I-70 Odyssey by Matthew E Salek. The unique aspect of this plant is that unlike traditional plants which have the dam located at the same location, the Shoshone dam is located two miles upstream, and the water flows through a tunnel in the wall of the canyon to the power plant. At the power plant the water exits the Canyon wall and drops to the hydroelectric turbines to generate power.

where:

The Shoshone plant can provide up to 15MW power, which is enough power for about 15,000 households.

Exergy Analysis of a Control Volume In our third example we do an exergy analysis of a single-inlet single-outlet steadyflow control volume and define and evaluate the various concepts used. We have ignored kinetic and potential energy terms which simply directly contribute to the exergy as needed. We find it convenient to do the development in terms of specific quantities (by dividing throughout by the mass flow).

Energy (First Law):

Entropy Generation (Second Law):

Exergy Analysis: we first eliminate q from equations (1) and (2) as follows:

Notice in equation (3) that we have defined reversible work (w rev) as that in which no entropy is generated. We thus define a new term Irreversibility (irrev) as follows:

Thus from equation (3), when the irreversibility irrev = 0, the resulting Reversible Work is given by:

We now define the Second Law Efficiency (ηII) for either a work producing or a work absorbing device as follows:

The Exergy ψ (or Availability) of the working fluid at either the inlet or the outlet port is defined as the maximum available work when the state of that working fluid is reduced to the Dead State 0, thus:

Notice that on referring to the Control Volume diagram above, the reversible work equation (5) can be written in terms of the inlet (i) and outlet (e) states as follows:

Thus the Reversible Work of the control volume can also be defined in terms of the difference in exergy between the inlet and exit ports, thus:

In order to get an inuitive understanding of this analysis, consider the following equivalent system in which we use the heat transfer between the system and the surroundings in order to obtain reversible work.

However this reversible work wHE is a function of the temperature T of the control volume, which can vary significantly between the inlet state (i) and the outlet state (e). Thus we will need to sum the work output of an infinite number of elemental reversible heat engines, as shown in the equivalent diagram which follows:

This analysis was first presented to me by the late Gary Graham (of Ohio University) in 1995. Thus:

Solved Problem 7.1 - Exergy Analysis of a Geothermal Well Note: the following two problems are extensions of Solved Problem 6.10 in which we did an ideal analysis of the T700 helicopter gas turbine. In solving these problems you should derive all equations used starting from the basic energy equation for a flow system, the enthalpy difference (Δh) for an ideal gas, the equation for entropy generated (sgen) and entropy change (Δs) for an ideal gas, and the exergy equations for reversible work (wrev) and Second Law efficiency (ηII). Use values of Specific Heat Capacity (CP) at the average temperature of each process, which are obtained from the table of Specific Heat Capacities of Air.

Problem 7.2 - Recall Solved Problem 6.10 in which we did an ideal analysis of the T700 helicopter gas turbine. In this problem we wish to do a Second Law analysis of the compressor only of this gas turbine. Assume that air enters the compressor at 100kPa and 27°C and exits at 1500kPa and 427°C, with 40 kJ/kg of heat cooling transferred to the surroundings at 25°C. Determine the actual work (from the energy equation) [-451.6 kJ/kg], the reversible work (from an exergy analysis) [-383.4 kJ/kg], and the Second Law Efficiency (ηII) [85%] of this compressor.

Problem 7.3 - Recall again Solved Problem 6.10 in which we did an ideal analysis of the T700 helicopter gas turbine. In this problem we wish to do a Second Law analysis of the turbine which is required to drive the compressor of a gas turbine engine. Assume that the gas leaving the combustion chamber enters the turbine at 1500 kPa and and 927°C and exits at 400 kPa and 627°C, with 50 kJ/kg of heat loss transferred to the surroundings at 25°C. Assuming that the gas is pure air, determine  a) the actual work output (from the energy equation) [292.6 kJ/kg],  b) the entropy generated by this process [0.22 kJ/kg.K],  c) the available reversible work output (from an exergy analysis) [357.5 kJ/kg], and

 d) the Second Law efficiency (ηII) of this turbine [82%]. Exergy Analysis of an Air Nozzle On to Section b) of Exergy Analysis - Adiabatic Control Volumes _____________________________________________________________________ _________________

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Exergy Analysis of an Air Nozzle

In this example we consider the nozzle of an aircraft jet engine.

One may wonder at the purpose of determining the maximum available work potential of a nozzle, whose entire purpose is to maximise the kinetic energy at the exit state (2). Indeed, a perfect nozzle is isentropic, and has no work potential, since any work done would decrease the kinetic energy. In a practical nozzle there could be an increase in entropy from state (1) to state (2) as well as possible heat loss to the surroundings which will decrease the kinetic energy output and thus the exergy analysis is mainly in order to evaluate this lost kinetic energy potential, or the irreversibility. We recall that kinetic energy is entirely equivalent to exergy.

Recall the previous section in which we determined the irreversibility (irrev) as follows:

Thus referring to the figure above the irreversibility is given by:

However, since there is no actual work done in the nozzle, we have from the energy equation:

Thus by substituting q we obtain the irreversibility as:

Once again, in order to get an intuitive understanding of this analysis we consider the following equivalent nozzle in which we wish to capture the heat loss over the entire nozzle by means of summing the work output of an infinite number of elemental reversible heat engines.

The reason for the many elemental reversible heat engines is that the temperature T of the nozzle varies continuously between the inlet state (1) and the outlet state (2), thus:

However, since there is no actual work done in the nozzle, we have from the energy equation:

Thus by substitution we obtain the final available work potential form:

Since there is no actual work done, the irreversibility can be determined as follows:

Note that this equation is identical to that for the irreversibility given above.

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Chapter 7: Exergy - Maximum Available Work Potential b) Adiabatic Control Volumes Examples of Determining the Reversible Work, Irreversibility, and Second Law Efficiency for Adiabatic Control Volumes

This section is mainly concerned with an attempt to develop an intuitive understanding of the exergy equations developed in the previous section, by considering reversible equivalent circuits of some common adiabatic components. We repeat equations (3), (4) and (5) developed in Chapter 7a on the exergy analysis of a control volume.

Note that the reversible work wrev will either be the maximum available output work for work producing devices, or the minimum possible input work (negative value) for work absorbing devices. We now consider the Energy and Entropy Generation equations for adiabatic components (q = 0):

Applying all the above analysis to evaluating the Second Law Efficiencies (η II) of adiabatic work absorbing and work producing components we obtain:

Second Law Analysis of an Adiabatic Refrigeration Compressor We now apply the above Second Law analysis to an adiabatic refrigeration compressor. We wish to determine the minimum work wC rev required to drive the compressor between the inlet state (1) and the exit state (2). Note that the isentropic compression that we evaluated in Chapter 6 will not provide the answer, since state (2s) is not the same as the actual state (2).

The above equations are in fact correct however we have difficulty in understanding their significance. In examining the adiabatic compressor above we cannot understand why the environment (dead space) temperature T0 features so prominently in the equations, when in fact there seems to be no obvious interaction between the adiabatic compressor and the environment. Note that as far as the adiabatic compressor is concerned we will assume that the surroundings temperature T 0 is equal to the exit temperature. In an attempt to find some intuitive meaning to these equations we consider a reversible system having the same inlet and exit states as our actual compressor. This comprises a three component system consisting of an inlet heat exchanger, a reversible heat engine and an isentropic compressor as shown below:

A typical h-s diagram for this system is shown below, in which we have used typical inlet conditions of 140kPa, -10°C and exit conditions of 700kPa, 60°C. The reversible heat engine will provide extra work to drive the compressor, absorbing its heat from the environment temperature T0 while rejecting heat to the heat exchanger. The exit state (2) from the heat exchanger has been chosen such that the compression process (2) - (3) will be isentropic.

We now derive the exergy equations for the three component system above, and consider first the heat engine. Since the temperature T of the heat exchanger varies from the inlet temperature T1 to the outlet temperature T2, we use the differential energy equation form for the reversible heat engine.

Since T0 is constant, this equation can be integrated from the inlet state (1) to the outlet state (2), leading to:

This familiar final form was to be expected. The net minimum work required to drive the compressor is thus:determined as follows:

Notice that this result is identical to that shown above for the actual adiabatic compressor, since we added the heat exchanger, thus state (3) is in fact equivalent to the original state (2).

Problem 7.4 - Recall Solved Problem 6.5 in which we evaluated the Adiabatic Efficiency (ηC) of a refrigeration compressor, and determined it to be 92%. Redo this problem and determine the Second Law Efficiency (η II) of this compressor. [92%]

Second Law Analysis of an Adiabatic Steam Turbine We now apply the above Second Law Analysis to an adiabatic steam turbine. We wish to determine the maximum available turbine work output w T rev between the inlet state (1) and the exit state (2). We will then be able to determine the second law eficiency by comparing the actual work output to the reversible (maximum available) work output as follows:

Once again, in an attempt to find some intuitive meaning to these equations we consider a reversible system having the same inlet and exit states as the actual turbine, comprising an isentropic turbine, a heat pump pumping heat from the surroundings to the heat exchanger in the exit stream. Note that as far as the adiabatic turbine is concerned we will assume that the surroundings temperature T 0 is equal to the exit temperature.

The h-s diagram for this system is shown below, in which we have chosen as an example a steam turbine having inlet conditions 6MPa, 600°C, and outlet conditions 50kPa, 100°C.

Notice from the h-s diagram that the heat exchanger temperature varies from the saturation temperature at 50 kPa (81°C) to 100°C at state (3). In order to accommodate that change we develop the differential form of the heat pump as follows:

Since T0 is constant, this differential equation can be integrated from state (2) to state (3).

We can then subtract the work provided to the heat pump from the output work of the turbine leading to the final form of the maximum available work, as follows:

Notice that this result is identical to that shown in the box above for the actual adiabatic turbine, since state (3) is in fact equivalent to the original state (2).

Problem 7.5 - Recall Solved Problem 6.1 in which we evaluated the Adiabatic Efficiencies (ηT) of both the High Pressure and the Low Pressure steam turbines of the supercritical steam power plantand found them to be 77% and 90% respectively. Redo this problem and determine the respective Second Law Efficiencies (η II) of both turbines. [77%, 90%] On to Section c) - Exergy by Heat Transfer from a Thermal Source to a Control Volume

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Chapter 7: Exergy - Maximum Available Work Potential c) Heat Transfer to a Control Volume Consider the Control Volume as developed in Section a), however with the heat being transferred from a thermal source at temperature TH which is higher than the surroundings temperature T0.

Energy (First Law):

Entropy Generated (Second Law): Since heat is exchanged with a temperature source different from the dead space temperature T0 the entropy generated is:

We now consider a reversible process between the same inlet (i) and exit (e) states and heat transfer q. This requires that reversible heat transfer q 0,rev will occur from the surroundings to the control volume such that the entropy generated s gen = 0.

Thus:

Adding this new reversible heat source to the energy equation (1) above, and substituting equation (9) we obtain:

You may be confused as to how we can justify transferring heat q 0 reversibly from the surroundings at T0 to the control volume at a higher temperature. This unique approach due to Sonntag and Borgnakke (Introduction to Engineering Thermodynamics, Wiley, 2001) is done for convenience in order to validate the derivation of equation (10). In order to justify this, consider the equivalent system shown below, in which the hot source at TH is used as the heat source of a reversible heat engine, which in turn drives a heat pump to deliver the required heat q to the control volume. We will show that this system gives rise to the identical equation (10) as above.

From the energy equation for the reversible heat engine (recall Chapter 5) we have:

From the diagram we see that some of this work (w HP) is used to drive the reversible heat pump. Since the temperature T of the control volume varies from the inlet to the outlet, we consider the differential energy equation for the work into the heat pump:

Integrating across the entire control volume (inlet to outlet) we obtain:

From the diagram above we see that the total reversible work available from this system is given by:

Substituting from equations (11), (12) and the initial energy equation (1) above, we have:

Simplifying equation (13) leads to equation (10) as above - QED. (Quad Erat Demonstratum - Latin for "which was to be proved" used smugly by math gurus whenever they successfully conclude a proof - usually accompanied by a condescending smile) _____________________________________________________________________ _________________

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Chapter 8: Steam Power Cycles a) Ideal Rankine and Reheat Cycles We introduced the basic Steam Power Plant in Chapter 4b, however we could only evaluate the turbine and feedwater pump after we introduced the concept of entropy in Chapter 6a. In this Chapter we will consider ideal steam power cycles including isentropic turbines and pumps. This will allow us to determine the maximum performance of these systems, and evaluate the influence of various components on this performance. The pressure-enthalpy (P-h) diagram The common method of describing steam power plant systems is by plotting them on a T-s (temperature-entropy) diagram for steam. This is done exclusively in all thermodynamic text books that I have evaluated over many years. We find this approach cumbersome, non-intuitive and even incorrect in the description of feedwater pumps. We will make exclusive use of P-h (pressure-enthalpy) diagrams (which we inroduced in Chapter 4b) to describe the various steam power systems and we will require that all problems to be solved should first be drawn on the P-h diagram. A typical P-h diagram for steam is shown below. One aspect of the diagram that we have ignored until now are the various constant entropy lines drawn in the superheated and high quality section of the diagram. We will use these lines to indicate the path of an isentropic turbine for the ideal cycles described below. We do not require this for the feedwater pumps since, as we determined in Chapter 6a, the isentropic pump follows the constant temperature line.

An Ideal Rankine High Pressure (15MPa) Steam Power Cycle This is shown below as an Ideal Rankine cycle, which is the simplest of the steam power cycles. We have specifically split the turbine into a High Pressure (HP) turbine and a Low Pressure (LP) turbine since it is impractical for a single turbine to expand from 15MPa to 10kPa.

As stated above, our approach is that prior to doing any analysis we will always first sketch the complete cycle on a P-h diagram based exclusively on the pressure and temperature data presented, as follows:

Notice that since both turbines are considered ideal, they follow the isentropic curve (1)-(2). From the P-h diagram we see that the LP turbine output (station (2)) has a quality of 80%. This is unacceptable. The condensed water will cause erosion of the turbine blades, and we should always try to maintain a quality of above 90%. One example of the effects of this erosion can be seen on the blade tips of the final stage of the Gavin LP turbine. During 2000, all four LP turbines needed to be replaced because of the reduced performance resulting from this erosion. Thus once we have determined from the P-h diagram that this cycle is impractical we have no reason to continue with the analysis, and prefer to extend this into an Ideal Reheat steam power cycle as follows.

A High Pressure Ideal Reheat Steam Power Cycle We extend the above example to the more practical Ideal Reheat cycle as shown below. In this example the HP turbine expands the steam from 15 MPa to 1 MPa, and the steam is subsequently reheated back to 600°C before being expanded in the LP turbine to 10 kPa.

Again we plot this cycle on the P-h diagram and compare it to the previous situation of no reheat, as shown below:

We notice that reheating the output of the HP turbine back to 600°C (process (2)-(3)) allows both significantly more power output as well as increasing the quality at the LP turbine output (4) to 98%. The method of analysis is similar to that developed in Chapter 4b, in which we consider each component as a separate control volume.  1) Assuming both turbines are isentropic (adiabatic and reversible), and neglecting kinetic energy effects we use the steam tables to determine the combined work output of both turbines.

 2) In this ideal cycle we assume that the feedwater pump is isentropic. Furthermore, since the suction temperature of the water is 46°C, we assume that it behaves as an incompressible liquid, even at 15 MPa. Recall from Chapter 4b where we derived theenergy equation for an incompressible liquid.

 3) The total heat transfer to the boiler, including the reheat system:

 4) The thermal efficiency of the ideal Rankine reheat system, defined as the net work done (turbines, pump) divided by the total heat supplied to the boiler:

Notice that we could have also determined the thermal efficiency by the simpler method of evaluating the condenser heat transferred to the cooling water, thus:

This is an excellent check and should always be done to validate your answer. The reason why we have preferred to evaluate the output power directly is that it is the primary purpose of a steam power plant, thus we are always interested in the various components (turbines, pumps) that directly influence its value. Thermal efficiency is important in its own right, however only on condition that we can satisfy the output power requirements of the steam power plant. On to Section b) of Steam Power Cycles _____________________________________________________________________ _________________

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Chapter 8: Steam Power Cycles b) Regenerative Cycles - Open and Closed Feedwater Heaters

One approach to increasing the efficiency of steam power cycle is by extracting some of the steam from various stages of the turbine and using it to preheat the compressed liquid before it enters the boiler. This is done either by direct mixing of the fluids (Open Feedwater Heater) or through a heat exchanger (Closed Feedwater Heater), In many practical steam power plants various combinations of open and closed feedwater heaters are used, and systems using them are generally referred to as Regenerative Cycles. We continue with the Reheat cycle developed in Chapter 8a and examine the performance effects of adding open and closed feedwater heaters.

The Ideal Regenerative Reheat Cycle using an Open Feedwater Heater We continue with the Reheat cycle developed in Chapter 8a, and examine the effect of adding a regenerative heat exchanger in the form of an Open Feedwater Heater, as shown below. We will find that this system does result in an increase in thermal efficiency by preheating the water before it enters the boiler, however at the expense of a reduced power output. In the schematic diagram we notice that many of the state and enthalpy values (indicated in red) have already been evaluated in Chapter 8a. The mass fraction of the steam bled at the outlet of the HP turbine (2) as well as the state and enthalpy values at stations (6) through (9) will be evaluated below.

For this example we have chosen the mixture pressure of the open feedwater heater as 200kPa. Notice that a portion of the steam is bled off the outlet of the HP turbine at a pressure of 1MPa, passed through a throttling valve reducing its pressure to 200kPa, and then mixed with the compressed liquid (also at 200kPa), ultimately resulting in saturated liquid at station (8). We first need to determine the mass flow fraction y of the bled steam required to bring the output of the open feedwater heater (8) to a saturated liquid state.

Notice that the work output is reduced by having bled off a fraction y of the steam, and the boiler heat input is reduced by the increased temperature of the compressed liquid entering the boiler T9. Thus:

We always confirm our results by the alternate evaluation of efficiency using the heat flow out from the condenser to the cooling water:

Thus we see a slight increase in efficiency and reduction in power compared to the reheat system that we solved in Chapter 8a (45%, 1910 kJ/kg). Can we justify this added complexity for such a small gain in efficiency? This was discussed in Solved Problem 4.2 in which we noted that a de-aerator is a necessary vital component of a steam power plant with a condenser pressure of 10 kPa - air leaking into the system causes corrosion as well as a reduction in performance. The open feedwater heater naturally includes a de-aerator. On a previous visit to the Gavin power plant we were informed that the open feedwater heater can also conveniently include a liquid water storage tank which enables the feedwater pump to be the main power control of the system by varying the mass flow rate of the steam. We were also informed that using a single feedwater pump to increase the water pressure from 10 kPa to 15 MPa is impractical, and in the Gavin power plant, in addition to the condensate pump there is a booster pump to bring the pressure from 10 kPa to the de-aerator pressure.

Problem 8.1 - A 10 MPa Steam Power Plant with an Open Feedwater Heater Problem 8.2 - A Cogeneration Steam Power Plant with an Open Feedwater Heater _____________________________________________________________________ ___

The Ideal Regenerative Reheat Cycle using a Closed Feedwater Heater We again extend the Reheat cycle developed in Chapter 8a, and examine the effect of adding a regenerative heat exchanger in the form of a Closed Feedwater Heater, as shown below. We have not included an open feedwater heater even though we learned in the previous example that it is a necessary component of a high pressure steam cycle, since we are using this example to develop the techniques for analyzing closed feedwater heaters. The various state values shown on the schematic (in red) have been evaluated in previous sections.

From the schematic diagram we see that a portion of the steam is bled off from the output of the HP turbine at state (2), and then used to heat the high pressure liquid, ultimately condensing to a subcooled liquid at state (8). A finite temperature difference is a necessary condition for heat transfer to occur, and from the temperature distribution plot shown above we see that the closed feedwater heater behaves as a counterflow heat exhanger, in which the compressed liquid water entering at state (6) is heated to the saturation temperature of the bled steam (T sat@1MPa = 180°C). In typical closed feedwater heaters (such as those used at the Gavin Power Plant) there are three distinct zones of heat transfer as shown in the simplified schematic diagram below:

The bled steam first enters the desuperheating zone enclosure and is cooled while raising the temperature of the feedwater leaving the heater to a level approaching or equal to the steam saturation temperature. The condensing zone is the largest heat transfer region within the heater shell. The major portion of heat transfer takes place here as the steam condenses and gives up its latent heat. The subcooling zone, which is enclosed in a separate shrouded area within the shell, further cools the condensed steam while heating the incoming feedwater. The following photograph shows one of the seven sets of closed feedwater heaters used in the Gavin Power Plant Typically we find that the subcooled liquid is reduced to within around 4°C to 6°C above the incoming feedwater temperature.

The P-h diagram plot of this system follows, and we notice on the diagram that the high pressure feedwater is heated from state (6) to state (7) before entering the boiler, while a mass fraction y of the bled steam is cooled from the HP turbine outlet state (2) to a subcooled liquid at state (8). The subcooled liquid is then passed through a throttling valve before being returned to the condenser at state (9). As we have learned from our studies of refrigerators, a throttling valve is simply represented on the Ph diagram by a vertical line, since from the energy equation we find that h 8 equals h9.

We consider first the evaluation of the mass flow fraction y, which is bled from the output of the HP turbine on order to heat the compressed water at state (8) to the saturation temperature of the steam at station (2):

Once again we notice that the work output is reduced by having bled off a fraction y of the steam, and the boiler heat input is reduced by the increased temperature of the compressed liquid entering the boiler T7. Thus:

Thus the system thermal efficiency becomes:

The alternative evaluation of thermal efficiency using the heat transfer from the condenser to the cooling water:

Notice that the thermal efficiency is slightly more than that of the previous example with an open feedwater heater, however with even less power output. In practice one normally finds a combination of open and closed feedwater heaters, and in the Gavin power plant there are seven closed feedwater heaters and one de-aerator/open

feedwater heater. In the following section we develop the analysis technique of this type of system.

Problem 8.3 - A Reheat Steam Power Plant with a Closed Feedwater Heater _____________________________________________________________________ ___

The Ideal Regenerative Reheat Cycle using both an Open and a Closed Feedwater Heater In a practical power plant one may find various combinations of closed and open feedwater heaters. For example the Gavin Power Plant has one open- and 7 closedfeedwater heaters in each of the two sections of the plant (refer to the Case Study at the end of this section). In the following example we have chosen one open and one closed feedwater heaters in order to illustrate the method of analysis, however the same approach will apply to any combination of feedwater heaters.

Once again we evaluate the required mass flow fractions y 1 and y2 of the bled steam in order to bring the compressed water at the entrance to the boiler (state (10)) to the correct state. An enthalpy inventory and energy balance on both the closed and open feedwater heaters leads to the following:

The resultant net work output and heat input now become:

And finally we obtain the thermal efficiency of the overall system as follows:

As always, we check this result against the equivalent method of considering only heat in and heat rejected in the condenser to the cooling water

.

Case Study - The General James M. Gavin Steam Power Plant _____________________________________________________________________ _________________

Engineering Thermodynamics by Israel Urieli is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License

Case Study - The General James M. Gavin Steam Power Plant Background - The General James M. Gavin Plant's units 1 and 2 are identical, each with a generating capacity of 1300 MW. Unit 1 was completed in 1974 and Unit 2 was completed the following year. With a total generating capacity of 2,600 MW, Gavin Plant ranks as the largest generating station in the state of Ohio. It is located along the Ohio River at Cheshire, Ohio, and has an average daily coal consumption of 25,000 tons at full capacity. The coal arrives by barge and is stored in the plant's coal yard. Conveyer belts carry the coal from the yard into the plant where pulverizers grind the coal into a fine, talcum powder-like consistency. The powdered coal is injected into the steam generator where it is burned at high temperature providing the heat power which drives the power plant.

Schematic Diagram for Analysis - The formal schematic diagram of the Gavin Plant is extremely complex. There are six turbines on two separate parallel shafts, each driving a hydrogen cooled electrical generator producing 26,000 volts. Transformers outside the plant building step up this voltage to 765,000 volts so that it can be transmitted efficiently over a long distance. The high pressure (HP) turbine drives one shaft together with low pressure (LP) turbines A and B, and the

intermediate pressure (Reheat) turbine drives the second shaft together with LP turbines C and D. The following represents a much simplified schematic diagram for purposes of doing an initial analysis of the system. Some of the state values shown were not available and represent estimates on the part of your instructor in order to enable a complete analysis. Notice that the feedwater pump is driven by a separate 65,000HP turbine (FPT) which taps some of the steam from the outlet of the reheat turbine, returning the steam to the condenser hotwell. The feedwater pump pressurizes the water to 30 MPa, however the pressure at the HP turbine inlet drops to 25 MPa since the steam has had to pass through 350 miles of piping in the steam generator. The flow control valve together with the speed control of the feedwater pump enables control of output power matching it to the demand. The system has four low pressure closed feedwater heaters, one open feedwater heater / de-aerator, and three high pressure closed feedwater heaters.

As always, prior to doing any analysis we always first sketch the complete cycle on a P-h diagram based on the data provided in the system diagram. This leads to the following diagram:

Notice from the P-h diagram how the three high pressure closed feedwater heaters progressively heat the steam from state (10) to state (11), thus the steam generator is only required to heat the steam from state (11) to state (1) leading to an increase in thermal efficiency. Similarly the four low pressure closed feedwater heaters progressively raise the temperature of the liquid from state (7) to state (8), thus reducing the fractional amount of steam required (y5) in order to raise the temperature of the liquid from state (8) to state (9). It is true that as we draw off steam from the turbines for all the heaters, we reduce the output power accordingly, however the net effect of this process is to increase the overall thermal efficiency of the system. One important consideration is the choice of the state (5) at the outlet of the low pressure turbines. The quality (x = 0.93) shown on the flow diagram is not a measurable quantity, and the identical pressure and temperature conditions exist throughout the quality region. The only guide that we have is the knowledge that steam turbine adiabatic efficiencies vary between 85% and 90%, thus in order to ensure that we are choosing reasonable state values we plot all three turbines on the companion h-s diagram indicating both the isentropic as well as the actual processes on the diagram as follows:

Thus from the diagram we determined that the choice of quality x = 0.93 brought us into the correct efficiency range. This is an extremely critical choice, since by choosing a quality that is too low can lead to erosion of the turbine blades and a reduction of performance. One example of the effects of this erosion can be seen on the blade tips of the final stage of the Gavin LP turbine. During 2000, all four LP turbines needed to be replaced because of the reduced performance resulting from this erosion. (Refer:Tour of the Gavin Power Plant - Feb. 2000) We now do an enthalpy inventory of the known state points on the cycle using either the Steam Tables or more conveniently directly from the NIST Chemistry WebBook (avoiding the need for interpolation), leading to the following table: State 1 2 3 4

Position HP turbine inlet HP turbine outlet Reheat turbine inlet LP turbine inlet

Enthalpy h [kJ/kg] h1 = h25MPa, 550°C = 3330 [kJ/kg] h2 = h5MPa, 300°C = 2926 [kJ/kg] h3 = h4.5MPa, 550°C = 3556 [kJ/kg] h4 = h800kPa, 350°C = 3162 [kJ/kg]

5 6 7 9 10

LP turbine outlet (quality region) Hotwell outlet (subcooled liquid) Condensate Pump outlet Open Feedwater Heater (saturated liquid) Feedwater Pump outlet (compressed liquid)

h5 = h10kPa, quality X=0.93 = hf+X.(hg­hf) hf = 192 [kJ/kg], hg = 2584 [kJ/kg] => h5 = 2417 [kJ/kg] h6 = hf@40°C = 168 [kJ/kg] h7 = h6 = 168 [kJ/kg] T9 = Tsat@800kPa = 170°C h9 = hf@800kPa = 721 [kJ/kg] T10 =T9+5°C = 175°C h10 = h30MPa, 175°C = 756 [kJ/kg] (Compressed liquid)

Note: State points (8) and (11) result respectively from the low- and high-pressure closed feedwater heaters and are evaluated below. Notice that the temperature T 10 is 5°C higher than the temperature T9. Normally we consider liquid water to be incompressible, thus pumping it to a higher pressure does not result in an increase of its temperature. However on a recent visit to the Gavin Power Plant we discovered that at 30MPa pressure and more than 100°C, water is no longer incompressible, and compression will always result in a temperature increase of up to 7°C. We cannot use the simple incompressible liquid formula to determine pump work, however need to evaluate the difference in enthalpy from the Compressed Liquid Water tables, leading to the enthalpy h10 shown in the table. Finally, do not forget that all values of enthalpy obtained should be checked for validity against the above P-h and h-s diagrams.

Analysis - We need to determine the mass fractions of all the feedwater heaters yi as well as that drawn off for the feedwater pump turbine, in order to evaluate the heat input and the total power output of the system. We find it convenient to separate the system into a high pressure section including the HP and Reheat turbines, and a low pressure section including the two LP turbine sets. Using the techniques of enthalpy balance on the open and closed feedwater heaters developed in Chapter 8b, we obtain the mass fraction equations of the high pressure section as summarized in the following diagram.

In order to enable evaluation of the enthalpies at the various state points in the diagram we estimated the various intermediate temperature values at the turbine taps from the above P-h and h-s diagrams. The closed feedwater heaters are all of type counterflow heat exchangers, and we make the assumption that the outlet temperature equals the saturation temperature of the respective turbine tap, and that the drain temperature is 5°C above the inlet temperature value. The resulting enthalpy inventory of the intermediate state points follows:

State

Position

Enthalpy h [kJ/kg]

t8 HP Turbine tap ht8 = h8MPa, 350°C = 2988 [kJ/kg] Closed Feedwater 11 Heater #8 outlet Closed Feedwater f7 Heater #7 outlet Closed d8 Feedwater Heater #8 drain Closed Feedwater f6 Heater #6 outlet Closed d7 Feedwater Heater #7 drain Reheat Turbine t6 tap Closed d6 Feedwater Heater #6 drain

T11 = Tsat@8MPa = 295°C h11 = h30MPa, 295°C = 1304 [kJ/kg]

Tf7 = Tsat@5MPa = 264°C hf7 = h30MPa, 264°C = 1154 [kJ/kg] Td8 = Tf7+5°C = 269°C hd8 = h8MPa, 269°C = 1179 [kJ/kg] Tf6 = Tsat@2MPa = 212°C hf6 = h30MPa, 212°C = 918 [kJ/kg] Td7 = Tf6+5°C = 217°C hd7 = h5MPa, 217°C = 931 [kJ/kg] ht6 = h2MPa, 450°C = 3358 [kJ/kg] Td6 = T10+5°C = 180°C hd6 = h2MPa, 180°C = 764 [kJ/kg]

The resultant fractional mass flow rates to the high pressure heat exchanger section follows: Mass flow path HP Turbine tap t8 to Closed Feedwater Heater #8 HP Turbine outlet 2 to Closed Feedwater Heater #7 Reheat Turbine tap t6 to Closed Feedwater Heater #6 Reheat Turbine outlet 4 to Open Feedwater

State conditions

Fractional mass flow

8MPa, 350°C

y8 = 0.083

5MPa, 300°C

y7 = 0.108

2MPa, 450°C

y6 = 0.050

800kPa, 350°C y5 = 0.025

Heater #5 Similar to the high pressure section above we obtain the mass fraction equations for the low pressure section as summarized in the following diagram:

The enthalpy inventory of the intermediate state points indicated on the above diagram follows: State t4 8 t3 f3 d4 t2 f2 d3 t1 f1 d2 d1

Position LP A&C Turbine tap Closed Feedwater Heater #4 outlet LP B&D Turbine tap Closed Feedwater Heater #3 outlet Closed Feedwater Heater #4 drain LP A&C Turbine tap Closed Feedwater Heater #2 outlet Closed Feedwater Heater #3 drain LP B&D Turbine tap Closed Feedwater Heater #1 outlet Closed Feedwater Heater #2 drain Closed Feedwater Heater #1 drain

Enthalpy h [kJ/kg] ht4 = h450kPa, 280°C = 3025 [kJ/kg] T8 = Tsat@450kPa = 148°C h8 = h800kPa, 148°C = 624 [kJ/kg] ht3 = h250kPa, 220°C = 2909 [kJ/kg] Tf3 = Tsat@250kPa = 127°C hf3 = h800kPa, 127°C = 534 [kJ/kg] Td4 = Tf3+5°C = 132°C hd4 = h450kPa, 132°C = 555 [kJ/kg] ht2 = h100kPa, 120°C = 2717 [kJ/kg] Tf2 = Tsat@100kPa = 100°C hf2 = h800kPa, 100°C = 420 [kJ/kg] Td3 = Tf2+5°C = 105°C hd7 = h250kPa, 105°C = 440 [kJ/kg] ht1 = h40kPa, quality X=0.98 = hf+X.(hfg) hf = 318 [kJ/kg], hfg = 2319 [kJ/kg] => ht1 = 2590 [kJ/kg] Tf1 = Tsat@40kPa = 76°C hf1 = h800kPa, 76°C = 319 [kJ/kg] Td2 = Tf1+5°C = 81°C hd2 = h100kPa, 81°C = 339 [kJ/kg] Td1 = T6+5°C = 45°C hd1 = h40kPa, 45°C = hf@45°C = 188 [kJ/kg]

The resulting fractional mass flow rates to the low pressure heat exchanger section follows: Mass flow path Reheat Turbine outlet 4 to Feedwater Pump Turbine (mass fraction 65,4[kg/s]/1234[kg/s])

State conditions

Fractional mass flow

800kPa, 350°C yFPT = 0.053

LP TurbineA&C tap t4 to Heater #4 LP TurbineB&D tap t3 to Heater #3 LP TurbineA&C tap t2 to Heater #2 LP TurbineB&D tap t1 to Heater #1

450kPa, 280°C 250kPa, 220°C 100kPa, 120°C 40kPa, quality X=0.98

y4 = 0.027 y3 = 0.033 y2 = 0.029 y1 = 0.041

From the above diagrams, an energy equation balance on the various components of the system leads to the following equations for the total turbine work output (wT kJ/kg), the total heat input to the steam generator (q in kJ/kg) and the thermal efficiency th.

Performance Results - Finally we have all the data and equations required to determine the performance with the following results:  The work done by the HP, Reheat, and LP turbine set

 The total heat input to the steam generator including the reheat section:

 The thermal efficiency of the system. Up until now we have not considered the boiler efficiency. This is dependent on many factors, including the grade of coal used, the heat transfer and heat loss mechanisms in the boiler, and so on. A typical design value of boiler efficiency for a large power plant is 88%.

 The Feedwater pump and turbine performance

 The power output of the turbines, and heat power to the steam generator:

Note: It is always a good idea to validate ones calculations by evaluating the thermal efficiency using only the heat supplied to the steam generator and that rejected by the condenser.

This is the same efficiency value as obtained by the direct method, thus validating the method.

Discussion - We were extremely satisfied that a system as complex as the Gavin Power Plant is amenable to this simplified analysis. Notice that no matter how complex the system is, we can easily plot the entire system on a P-h diagram in order to obtain an immediate intuitive understanding and evaluation of the system performance. The diagram also serves as a usefull validity check by comparing each value of enthalpy evaluated to the values on the enthalpy axis of the P-h diagram. The analytical power output (1455 MW) is higher than the actual power output of 1300 MW mainly because of the significant electrical power required to run the power plant and the heat and pressure drop losses inherent in a large complex system. In order to justify the complexity of the seven closed feedwater heaters we analysed two simpler systems for comparison. In all cases we used the same steam mass flow rate of 1234 kg/s and the same feedwater pump turbine system as above. Note that the open feedwater heater also acts as a de-aerator and storage tank, and is thus a necessary component of the system.  No closed feedwater heaters in the system. This allows all of the steam to be directed to the turbines resulting in a much higher power output of 1652 MW, however with a reduction in thermal efficiency from 46% to 41%.  Using only the three high pressure closed feedwater heaters and not the four low pressure closed feedwater heaters. This requires a significant increase in the

steam tapped from the outlet of the reheat turbine to be directed to the open feedwater heater resulting in a lower power output of 1397 MW with a thermal efficiency of 45%. Thus use of the seven closed feedwater heaters is justified, resulting in the maximum thermal efficiency together with a satisfactory power output, _____________________________________________________________________ __

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Chapter 9: Carbon Dioxide (R744) - The New Refrigerant Introduction and Discussion In the early days of refrigeration the two refrigerants in common use were ammonia and carbon dioxide. Both were problematic - ammonia is toxic and carbon dioxide requires extremely high presures (from around 30 to 200 atmospheres!) to operate in a refrigeration cycle, and since it operates on a transcritical cycle the compressor outlet temperature is extremely high (around 160°C). When Freon 12 (dichloro-diflouromethane) was discovered it totally took over as the refrigerant of choice. It is an extremely stable, non toxic fluid, which does not interact with the compressor lubricant, and operates at pressures always somewhat higher than atmospheric, so that if any leakage occured, air would not leak into the system, thus one could recharge without having to apply vacuum. Unfortunately when the refrigerant does ultimately leak and make its way up to the ozone layer the ultraviolet radiation breaks up the molecule releasing the highly active chlorine radicals, which help to deplete the ozone layer. Freon 12 has since been banned from usage on a global scale, and has been essentially replaced by chlorine free R134a (tetraflouro-ethane) - not as stable as Freon 12, however it does not have ozone depletion characteristics.

Recently, however, the international scientific consensus is that Global Warming is caused by human energy related activity, and various man made substances are defined on the basis of a Global Warming Potential (GWP) with reference to carbon dioxide (GWP = 1). R134a has been found to have a GWP of 1300 and in Europe, within a few years, automobile air conditioning systems will be barred from using R134a as a refrigerant. The new hot topic is a return to carbon dioxide as a refrigerant. The previous two major problems of high pressure and high compressor temperature are found in fact to be advantageous. The very high cycle pressure results in a high fluid density throughout the cycle, allowing miniturization of the systems for the same heat pumping power requirements. Furthermore the high outlet temperature will allow instant defrosting of automobile windshields (we don't have to wait until the car engine warms up) and can be used for combined space heating and hot water heating in home usage (refer for example: Norwegian IEA Heatpump Program Annex28). Recent update March 2013 - Volkswagen, Daimler, Audi, BMW and Porsche have announced plans to develop CO2 MAC (Mobile Air Conditioning) systems (refer: Environmental Leader) Property Tables for Carbon Dioxide (R744) We were not able to find any published tables for Carbon Dioxide (R744) refrigerant, hence decided to create our own. The following set of tables was developed using software from the NIST (National Institute for Standards and Technology) and has been organized in a format suitable for evaluating refrigeration and heat pump systems Thermodynamic Properties of Carbon Dioxide R744 The P-h diagram for Carbon Dioxide (R744)

The h-s Diagram for Carbon Dioxide (R744)

A Heat Pump System using Carbon Dioxide Refrigerant (R744) In addition to being a environmentally benign fluid, there can be significant advantages to using carbon dioxide in a home air-conditioning/heat-pump system environment. Consider the following system diagram:

Notice that in addition to serving as an air conditioner/space heater, the high compressor outlet temperature can be used to provide hot water at a significant economy over the regular gas or electric hot water heater. Thus the heat flow to the hot water heater cools the gas from 160°C to 70°C, and the heat flow to the space heater further reduces the gas temperature to 45°C. In order to determine the enthalpy at outlet station (4) we need to consider the energy equation applied to the internal heat exchanger. Since we assume that it is externally adiabatic, all the heat transfer is internal, as shown in the following:

Drawing all the processes of the above scheme on a P-h diagram we obtain the following diagram. Notice the sketches on the diagram of the various components as well as the internal heat exchanger indicating the heat flow from the gas cooler outlet (3) - (4) to the compressor inlet (6) - (1).

Problem 9.1 - Use the R744 refrigerant property tables in order to evaluate the following:  Determine the work done on the compressor [97 kJ/kg].  Determine the heat rejected to the hot water heater [164 kJ/kg], and that rejected to the space heater [97 kJ/kg].  Determine the Coefficient of Performance of the hot water heater [COPhw=1.7] and that of the space heater [COPspace=1] (Recall that COP is defined as the desired heat transferred divided by the work done on the compressor).  Determine the Coefficient of Performance of the air conditioner [COPa/c=1.7]. (Notice from the P-h diagram that the internal heat exchanger significantly increases the capacity of the air conditioner.)

Problem 9.2 - For the following additional questions we can assume that the compressor power is 1kW. (Note - we can use the COP values to answer these questions - we do not need to evaluate the mass flow rate of the refrigerant ):  Determine how long it will take to heat 100 liters of water in the tank from 30°C to 60°C [2 hours]  During the summer months when the air conditioner is operating, determine the volumetric flow rate of the air [5.1 m3/min] flowing through the evaporator cooling duct in order to reduce the air temperature from 30°C to 13°C. (Note assume a pressure of 100kPa and temperature of 25°C to evaluate the specific volume of the air)  During the winter months when the heat pump is operating, and using the same fan as above determine increase in temperature of the air [10°C] flowing through the space heating duct.

Problem 9.3 - Notice that the compressor does not follow an isentropic process. This is a practical system with data adapted from a Visteon Corp. automobile a/c design. In this exercise you should do the following:  Plot the compressor process (1)-(2) on the enthalpy-entropy (h-s) diagram provided above. Plot also the isentropic compressor process and (using the Property Tables) determine the compressor isentropic efficiency η C. Recall that the compressor isentropic efficiency η C is defined as the isentropic work done on the compressor divided by the actual work done. Indicate both the isentropic work done and the actual work done on the h-s diagram. [ηC=74%]

Problem 9.4 - A R744 (CO2) Home Geothermal Heat-Pump - It is a well known fact that there is a year round constant temperature only a few meters below the earths surface. In this problem we wish to evaluate a system which is designed to use this underground thermal source to advantage. _____________________________________________________________________ _________________

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Chapter 10: Air - Water Vapor Mixtures a) Humidity and the Adiabatic Saturation Process

Atmospheric air includes dry air and water vapor. Recall that for an ideal gas, enthalpy (h) is a function of temperature only (Δh = C P.ΔT). Notice also from the hs diagram for steam that at relatively low temperatures (