ME1 Thermodynamics Lecture Notes 2024-2025 PDF

ME1 THERMODYNAMICS 2024-25 Alex M K P Taylor and Ricardo Martinez-Botas London, September 2024 Mechanical Engineering Department Imperial College London...

ME1 THERMODYNAMICS 2024-25 Alex M K P Taylor and Ricardo Martinez-Botas London, September 2024 Mechanical Engineering Department Imperial College London Lectures, Tutorials Section Contents and Tests Section 1: History linked to power plant development Lecture: weeks 2 & 3 Introduction to Relevance in the context of energy reserves Tutorial no. 1 & 2 Thermodynamics and the environment Macroscopic versus microscopic descriptions Section 2: of matter Basic Concepts The macroscopic viewpoint Systems, system states, and changes of state via energetic interactions Systems and Control Volumes Properties and states Equilibrium Processes and cycles Section 3: Forms of energy: kinetic, potential, and Lectures: wks 4, 5 & 6 Energy, Heat, internal Tutorial no. 3 Work and the Interaction with the surroundings: Heat Structured Tutorial 3 First Law Interaction with the surroundings: Work Displacement and shaft work First Law for a system (cyclic and non-cyclic) Section 4: Pure substances Lectures: wks 7, 8 & 9 Properties of Two-property rule, state diagrams Tutorial no. 4 Substances Intensive and extensive properties Internal energy, enthalpy and specific heats Ideal and perfect gases Phase change, vapour and liquid properties  Lectures, Tutorials Section Contents and Tests Section 5: System analysis Lectures: The First Law for The derivation of the first law for a flow weeks 10, 11, 16 & 17 Flow Processes process: general Tutorial no. 5 First law applied to a steady flow process Structured Tutorial 5 Throttling process Nozzles, pumps, compressors, and turbines Progress Test: (both Heat exchangers and mixing chambers Thermodynamics and Unsteady energy processes Fluid Mech. in one test) Section 6: Heat engines: general approach Lectures: The 2nd Law of Reversibility in thermodynamics weeks 18, 19, 20 & 21 Thermodynamics Clausius statement of the Second Law Tutorial no. 6 Temperature scale Structured Tutorial 6 Efficiency of a heat engine Heat pumps and coefficient of performance Section 7: Clausius inequality Lectures: weeks 22, Consequences of Definition of entropy 23, 24 & 25 (Revision) the 2nd Law State diagrams using entropy Tutorial no. 7 Tds relationships Structured Tutorial 7 Isentropic processes for perfect gas and steam Isentropic efficiency: compressors and turbines Application to a cycle Example Classes Topics: Weeks 30, 31 & 32, Steady flow energy equation applied to a check ME1 board simple power plant Clausius inequality Succession of processes in a system Isentropic efficiency in a reheat turbines and in an intercooled compressor system Second law using heat engines Summer Clinic Weeks 30, 31 & 32, Tutorials check ME1 board Thermofluids Check ME1 board Exam Content 1 INTRODUCTION TO THERMODYNAMICS........................................................................ 1 1.1 A VERY BRIEF HISTORY................................................................................................... 1 1.2 THE RELEVANCE OF THERMODYNAMCS IN THE 21ST CENTURY................................. 1 2 BASIC CONCEPTS............................................................................................................ 2 2.1 MACROSCOPIC VIEWPOINT: "CLASSICAL THERMODYNAMICS".................................. 2 2.2 FRAMEWORKS FOR STUDYING ENERGY TRANSFERS: OPEN & CLOSED SYSTEMS 2 2.3 PROPERTIES AND STATES.............................................................................................. 3 2.4 EQUILIBRIUM..................................................................................................................... 5 2.5 PROCESSES...................................................................................................................... 6 2.6 EXERCISE 2..................................................................................................................... 10 3 ENERGY, HEAT, WORK AND THE 1ST LAW................................................................... 15 3.1 FORMS OF ENERGY........................................................................................................ 15 3.2 HEAT TRANSFER............................................................................................................. 16 3.3 WORK TRANSFER........................................................................................................... 16 3.4 THE 1ST LAW OF THERMODYNAMICS............................................................................ 20 3.5 EXERCISE 3..................................................................................................................... 24 3.6 STRUCTURED TUTORIAL 3............................................................................................ 30 4 PROPERTIES OF SUBSTANCES.................................................................................... 36 4.1 INTRODUCTION............................................................................................................... 36 4.2 PURE SUBSTANCES....................................................................................................... 36 4.3 DIAGRAMS OF STATE..................................................................................................... 36 4.4 MODELLING THE BEHAVIOUR OF GASES.................................................................... 38 4.5 ENTHALPY AND SPECIFIC HEATS................................................................................. 39 4.6 THE BEHAVIOUR OF VAPOURS AND LIQUIDS.............................................................. 45 4.7 EXERCISE 4..................................................................................................................... 51 5 THE 1st LAW FOR FLOW PROCESSES.......................................................................... 57 5.1 INTRODUCTION............................................................................................................... 57 5.2 THE 1ST LAW EQUATION FOR OPEN SYSTEMS.......................................................... 58 5.3 THE STEADY-FLOW ENERGY EQUATION (SFEE)......................................................... 61 5.4 USING THE SFEE: SOME EXAMPLES............................................................................ 65 5.5 MASS AND VOLUME FLOW RATE – A REMINDER........................................................ 70 5.6 EXERCISE 5..................................................................................................................... 71 5.7 STRUCTURED TUTORIAL 5............................................................................................ 74 6 THE 2ND LAW OF THERMODYNAMICS........................................................................... 76 6.1 HEAT ENGINES, THERMAL RESERVOIRS AND THERMAL EFFICIENCY..................... 77 Content 6.2 REVERSIBLE PROCESSES............................................................................................. 79 6.3 CONSEQUENCES OF THE 2ND LAW FOR HEAT ENGINES............................................ 81 6.4 THE EFFICIENCY OF REVERSIBLE HEAT ENGINES..................................................... 84 6.5 THE "QUALITY" OF ENERGY........................................................................................... 85 6.6 THE CARNOT ENGINE AND CYCLE............................................................................... 86 6.7 REFRIGERATORS AND HEAT PUMPS (REVERSED HEAT ENGINES)......................... 88 6.8 EXERCISE 6..................................................................................................................... 89 6.9 STRUCTURED TUTORIAL 6............................................................................................ 92 7 CONSEQUENCES OF THE 2ND LAW OF THERMODYNAMICS...................................... 93 7.1 THE CLAUSIUS INEQUALITY.......................................................................................... 93 7.2 THE PROPERTY ENTROPY............................................................................................. 95 7.3 TWO IMPORTANT RELATIONSHIPS: ‘The T ds Equations’........................................... 98 7.4 ENTROPY CHANGE IN PROCESSES WITH PERFECT GASES..................................... 99 7.5 T – s DIAGRAMS FOR STEAM AND FOR PERFECT GASES....................................... 101 7.6 ISENTROPIC PROCESSES AND ISENTROPIC EFFICIENCY....................................... 101 7.7 ENTHALPY-ENTROPY DIAGRAMS............................................................................... 103 7.8 EXERCISE 7................................................................................................................... 105 7.9 STRUCTURED TUTORIAL 7.......................................................................................... 109 Introduction Chapter 1 1 INTRODUCTION TO THERMODYNAMICS You may have some idea of the subject, perhaps during Chemistry courses at school. It would be wise to put that knowledge to one side for a while and to start again as an engineer, since the emphasis is likely to be rather different (especially when we reach the topic of entropy). 1.1 A VERY BRIEF HISTORY The beginnings of engineering thermodynamics are to be found in the attempts of 18th century pioneers to build improved steam engines, then used mainly for pumping water out of mines. The aim was to consume less steam, therefore, to burn less fuel, for a given quantity of work done in raising water. James Watt, Scottish engineer (and instrument maker to Glasgow University) is well known for his engine of 1765. Developments such as Watt's were essentially practical in nature, with little underlying science. The science really began with the 1824 theory of French engineer Sadi Carnot, who tried with reasonable success to draw an analogy between the performance of water wheels (in terms of the water reservoir levels) and of "heat engines" (in terms of the temperature of the heat source etc.). Notable contributors to thermodynamics theory in the 19th century include the physicist (and brewery owner) James Joule, in demonstrating the 1st Law, and Lord Kelvin and William Rankine, both of Glasgow University, from whose work the 2nd Law was developed. It was not until the 20th century that the last (we believe!) misconceptions were eradicated. 1.2 THE RELEVANCE OF THERMODYNAMCS IN THE 21ST CENTURY High standards of living have been synonymous with high per capita energy consumption and high emissions of pollutants (particularly those responsible for acid rain and photochemical smog and for increasing the greenhouse effect). Virtually all domestic activity involves the consumption of energy, directly or otherwise, while the success of industries is often critically dependent on energy supplies and their cost. The economy of an entire nation can be heavily influenced by its primary energy sources (the fuels available to it), the ways in which that energy is converted into useful forms (particularly electricity) and the ways in which it is finally used. Two major problems face society today: (i) How can we maintain adequate and affordable supplies of energy in useful form, for the developing and the developed world, from dwindling reserves of coal, oil and gas ("fossil" fuels) or from alternative sources (nuclear materials, the sun, wind, waves, tides,...)? (ii) How can we avoid, or learn to live with, the resulting adverse effects on our surroundings, locally and globally (like air pollution or global warming)? Some recent political leaders have argued that time needs to be taken to work out possible solutions to these problems, but your generation will not have that luxury! You will have to solve the problems! The answers to global-scale questions such as how to limit carbon dioxide emissions, or more local ones such as how the U.K. should replace North Sea gas when it runs out, will depend to a large extent on the knowledge and skills of engineers with a sound grasp of the basic principles of Thermodynamics. 1 Basic Concepts Chapter 2 2 BASIC CONCEPTS 2.1 MACROSCOPIC VIEWPOINT: "CLASSICAL THERMODYNAMICS" We know that the temperature of a substance is determined by the energy of the molecules, and that the pressure in a gas or liquid is related to the number of molecules in a given space, their mass and their mean-square speed. It would be impossible to use such "microscopic" detail to analyse the behaviour of a quantity of material of interest to engineers (e.g., the gas inside a cylinder of an internal combustion engine), although there is a body of knowledge called Statistical Thermodynamics which is based on "average" behaviour of sufficiently large numbers of molecules; it is of interest to chemists and physicists as well as to advanced courses in engineering science. We would prefer to regard temperature as something measurable by a calibrated instrument called a thermometer (or thermocouple) and pressure as the local force per unit area on a surface, measurable by a pressure gauge (or transducer). At the scales (in terms of lengths, areas, volumes) we are concerned with, the empty space between and inside molecules is irrelevant; we prefer to regard a liquid or a gas as a "continuum", i.e., as homogeneous matter. This "macroscopic" viewpoint forms the basis of the only branch of Thermodynamics we shall cover, known as “Classical-“or “Engineering-” Thermodynamics. The continuum idea is also the foundation for your study of Fluid Mechanics. 2.2 FRAMEWORKS FOR STUDYING ENERGY TRANSFERS: SYSTEMS & CONTROL VOLUMES Just as we focus on a "particle" or a solid body when we apply Newton's laws of motion and calculate changes in the energy, velocity, position, etc. of the body under the influence of forces, we need an equivalent "object" on which to use the laws of Thermodynamics to calculate energy changes, changes in temperature, pressure, etc. Likening energy transfers to flows of money, we need to define something analogous to a bank account, on which to draw up a balance. If we look at a steam turbine in a power plant, we might choose the turbine itself as our "object". Steam flows in at one place and (with altered properties) out at another. Energy "flows" out as work (to drive the electrical generator) and as heat to the atmosphere (the casing is hot). The turbine is a fixed region of space, which, for the purpose of our "energy accounting", we call a CONTROL VOLUME (C.V.). Its surface, across which both ENERGY (in the form of heat or work) and MATERIAL (here steam) may flow, is called a C.V. SURFACE. (It is possible to choose C.V. surfaces, which expand or contract as a process takes place, but we shall not do so in this course. Such choices are made and are widespread in computational fluid dynamics, “CFD”) You will also use C.V. when applying the laws of Fluid Mechanics. 2 Basic Concepts Chapter 2 In Thermodynamics, we are not often interested in changes in the properties of the solids from which items of engineering plant are constructed, so here we might equally well consider our system surface as the inside surface of the turbine casing. In some situations, it makes more sense to focus not on a fixed region of space but on a fixed, identifiable quantity of material. An example is the air + petrol vapour mixture in the cylinder of a reciprocating engine, while the inlet and exhaust valves are shut, and the mixture is being compressed prior to ignition. For energy balance purposes, we call such a fixed quantity of material a CONTROL SYSTEM or simply a SYSTEM. The SYSTEM BOUNDARY is the real or imaginary surface, which separates the system from its SURROUNDINGS. Material clearly cannot cross a system boundary (here the cylinder head and walls plus the piston face), but energy (heat or work) may do so. In this example, the piston is moving upwards, and the system (air + petrol vapour) is changing its shape and volume but not its material contents or its mass. Note that some authors use the term “open system” to mean “control volume”. From now until §5, we shall concentrate on the closed system as a framework for applying the laws of thermodynamics, at this stage just the law of conservation of energy. A convenient feature of a closed system is that the principle of conservation of mass is automatically satisfied; we are dealing with a fixed mass. Before considering energy and energy transfers between a closed system and its surroundings, we must think about precise descriptions of the system at any point in a thermodynamic process: the ideas of properties and state, and the equilibrium between the system and its surroundings. This is all part of the "language" of thermodynamics; if we proceed too quickly to the more interesting parts of the course without an adequate grasp of this language, we shall come unstuck! 2.3 PROPERTIES AND STATES 2.3.1 Describing a system and its state To describe a system adequately, we must know (a) relevant facts about its material nature, and (b) its STATE, i.e., its condition at a given instant. Characteristics, which determine the state, are called PROPERTIES. Examples: (i) Motion of a smooth sphere on a smooth horizontal plane (hardly Thermodynamics!): System: the sphere. Material nature: mass and shape are sufficient; colour, moment of inertia etc. are not relevant. State: two coordinates 3 Basic Concepts Chapter 2 and two components of velocity form a sufficient definition; temperature, angular velocity etc. are irrelevant. (ii) The air in the cylinder of a diesel engine after it has been compressed and just before the fuel is injected: System: the air (not any of the engine parts). Material nature: its chemical composition, "air" is good enough, and its mass, perhaps 0.001 kg. State: its volume, perhaps 50 x 10-6 m3, and its temperature, perhaps 1000 K, are sufficient. Pressure could be calculated from these, so does not also have to be specified; we shall see shortly that any two of the properties volume, temperature and pressure are sufficient to fix the state. Note that the properties of a system in a given state do not depend on the process by which the system arrived in that state. If you were climbing Mount Everest and had reached the summit, the only relevant property defining your state might be the altitude you had reached (or your potential energy relative to what it was at sea level, which depends only on altitude - unless you had lost weight in the attempt!). The time taken to reach the summit and the amount of food consumed on the way would depend on the route taken, the weather, etc.; they are not properties, but your altitude at the summit would be the same regardless of how you got there, so it is a property. 2.3.2 Classifying properties Thermodynamic properties (those relevant to studies of energy) can be classified into two types. Those whose magnitudes are proportional to the system's mass are called EXTENSIVE properties. Examples: - potential energy, calculated as mgz and measured in J, where m is the mass of the system, g is the gravitational acceleration and z is the height above some datum level; - volume, V, measured in m3 (at fixed pressure and temperature, the volume of a system comprising 2 kg of a certain gas is twice that of one comprising 1 kg of the same gas) Those whose magnitudes do not depend on the system's mass are called INTENSIVE properties. Examples: - pressure, P; temperature, T; density, ; the reciprocal of density, i.e., volume per unit mass, which is called specific volume, v, and measured in m3/kg; - potential energy per unit mass, i.e., mgz ÷ m = gz, measured in J/kg Extensive properties can be used in fixing the state of a system only if we know the system's mass. In some thermodynamic analysis, particularly when we base it on an C.V. rather than a system, it is not necessary to know masses. (For example, in an "ideal" diesel engine, where we can ignore friction and heat loss, a knowledge of the air temperatures and pressures at the beginning and at the end of the compression stroke is sufficient to calculate the work done on the air by the piston per unit mass of air; the actual work done is then the product of this quantity, called the specific work, and the actual mass of air in the cylinder, which depends on the size of the cylinder.) It is therefore safer 4 Basic Concepts Chapter 2 to say that the state of a system is determined by the values of its intensive properties. Two systems have the same state if the values of all the relevant intensive properties in one system are the same as the corresponding properties in the other; the two systems need not have the same mass. 2.3.3 The "Two-property" rule (called “state postulate” in Ç&B) Experience shows that only a limited number of properties are necessary to fix the state of a system. For some situations relevant to engineers, the effects of electric and magnetic fields, and of surface tension, are negligible. If the effects of gravity and of motion (of the system as a whole) can also be neglected, the system is described as a SIMPLE COMPRESSIBLE SYSTEM, and the "two-property" rule states that: The state of a simple compressible system is fixed completely by the values of any two of its intensive properties, provided those two properties are independent of each other. "Independent" means that the value of one of those two properties is not determined by the value of the other; so, for example, density and specific volume will not fix a state. A common case of two properties not being independent is when the system consists of a single substance, e.g., water, in both the liquid and vapour (steam) phases; pressure is then determined by temperature and vice versa. (If water is boiling at atmospheric pressure, you know its temperature is 100°C.) If you know the pressure of such a two- phase system, the other property needed to determine the state must be something other than temperature. The small number of properties needed to describe such a system is a feature of Classical Thermodynamics. In a gravitational field, pressure varies with height, but in most systems of interest to us in Thermodynamics the variation is small, and we can assign a single value of pressure to the entire system. If motion of the whole system (not molecular motion) is important, only a few extra properties are needed: velocity components. 2.4 EQUILIBRIUM 2.4.1 Internal equilibrium If a system is completely isolated from its surroundings, i.e., no energy can cross the system boundary, it cannot be influenced by any external effects and, if none of its properties changes with time, it is said to be in INTERNAL EQUILIBRIUM. Example: gas (0.2 kg of CO2, say) in one side of a rigid, thermally insulated box, separated from an evacuated part by a diaphragm through which nothing can pass. ("Rigid" implies that the gas cannot be compressed or expanded by movement of the walls, so no work can be done, in the absence of any stirring device inside the box. The insulation prevents any heat transfer across the system boundary. If the gas properties are measured at intervals, they will eventually become 5 Basic Concepts Chapter 2 invariant with time; the system will be in internal equilibrium. Now suppose that the diaphragm breaks. Immediately afterwards, temperature T and pressure P will vary in time and space throughout the box (a small number of property values will no longer suffice to describe the instantaneous state); the system is not now in internal equilibrium. However, a new equilibrium state (T and P uniform and constant) will be reached when the gas fills the entire box, and all flow-like motion has ceased. 2.4.2 Equilibrium between a system and its surroundings The system (gas in a cylinder) in the left-hand diagram below is at a pressure P0 and temperature T0 greater than those of its surroundings, P1 and T1. It is in equilibrium with the surroundings only because of the thermal insulation (the piston is assumed to be a non-conductor) and the peg preventing outward motion of the piston. If the insulation is removed to allow a transfer of heat and the peg is also withdrawn, allowing the gas to expand and do work on the surrounding atmosphere, the system is no longer in equilibrium with its surroundings. The interaction between the system and the surroundings will continue until the causes of the heat transfer (the temperature difference) and the work transfer (the pressure difference) have been removed. When the system is at P1 and T1, it is once again in equilibrium with the surroundings. The term "mechanical equilibrium" is used when there is no tendency for any part of the system boundary to move, because the pressure of the system balances the external pressure (and any forces caused by weights, springs etc.). The term "thermal equilibrium" is used when there is no temperature difference across the system boundary hence no reason for any heat transfer. A combination of mechanical and thermal equilibrium (and some other types of equilibrium which need not concern us here) is referred to as "thermodynamic equilibrium". 2.5 PROCESSES 2.5.1 Definition When a system changes from one state of equilibrium (internally and with its surroundings) to another, we say that it has undergone a PROCESS. The succession of states through which the system passes during the change from the INITIAL STATE to the FINAL STATE is called the process PATH. These intermediate states are not equilibrium states, otherwise there would be no reason for the process to continue; a process can only take place when the equilibrium between system and surroundings is removed. To define a process completely, the initial and final states, the path and the energy transfers between system and surroundings should be specified. 6 Basic Concepts Chapter 2 2.5.2 Quasi-equilibrium processes The Thermodynamics of systems, which are not in equilibrium, is a difficult subject, not appropriate to an undergraduate course. How, then, can we deal with processes in engineering equipment, e.g., the changes in state of air inside the cylinder of a reciprocating engine or of steam as it flows through a turbine? Consider what would happen, during the air compression process in a diesel engine cylinder, if the piston were suddenly stopped. Just before that happened the air would be in a non-equilibrium state, with temperature, pressure and velocity varying throughout the cylinder. How long would it take, from the instant the piston stopped, for the system (the air) to settle down to a state of equilibrium, with all properties uniform  and constant? If that "settling time" were small compared with the time taken for any property to change noticeably as a result of piston movement (before we stopped it), we could say that the system had been nearly in equilibrium, at that point in the process. How nearly would depend on the piston speed, i.e., on the nature of the process, not of the system. In a real diesel engine (even a high-speed one running at, say, 4000 rev/min) "information" about the pressure increase, caused by the piston advancing into the air, is transmitted to all parts of the air much faster than the speed of the piston itself, and the air would indeed be nearly in equilibrium throughout this compression process. A process, which takes place slowly enough for the system to deviate only infinitesimally from equilibrium (internally and with its surroundings), is called a QUASI-EQUILIBRIUM (or quasi-static) process. Fortunately, real processes in much engineering equipment of interest to us can be considered to be quasi-equilibrium processes. 2.5.3 Diagrams of process paths The Two-Property Rule allows us to plot process paths on a two-dimensional diagram, as a graph of one property against any other independent property. For a system, pressure and specific volume are easily-measured properties and are commonly used to illustrate quasi-equilibrium process paths. Example for the compression process in a diesel engine cylinder: The initial state, or state 1, is the point marked by the number 1 (circled to remind us that it does not represent a quantity, such as a pressure of 1 bar). The specific volume in state 1 is v1 and the pressure is P1. The final state is state 2. The left-hand P - v diagram is for a non-equilibrium process in which the intermediate states are undefined; only the initial and final states, where the system is in equilibrium, can be plotted. The right-hand P - v diagram shows a quasi-equilibrium process, in which all the intermediate states can be considered to be equilibrium states and can therefore be plotted as points; the continuous line joining these states represents the process path.  Definition: "uniform" means that a property does not vary in space (at a given time); "constant" and "steady" mean that it does not vary in time (at a given position). 7 Basic Concepts Chapter 2 2.5.4 Cyclic Processes, or Cycles The sequence of processes taking place in a steadily-operating reciprocating engine repeats itself hundreds or thousands of times a minute: it executes a mechanical cycle. An idealised model of the processes in a diesel engine cylinder, for example, is as follows, where we take the system as air alone (neglecting the small additional mass of the injected fuel). This is known as the “air-standard” model. Process 1 - 2 (meaning from state 1 to state 2): compression of the air as the piston moves from its outermost (or "bottom dead centre") position to its innermost (or "top dead centre") position. Process 2 - 3 expansion of the air while there is a heat transfer to it (modelling the effect of the fuel combustion in a real engine), for part of the piston's return stroke, while the pressure stays constant. (In practice, the rise in pressure which would occur if combustion took place in a fixed-volume vessel is approximately counterbalanced by the reduction in pressure which would occur due to piston movement in the absence of combustion.) Process 3 - 4 continued expansion of the air to the maximum volume (piston's outermost position) while the pressure falls, modelling that part of the return stroke after combustion has been completed Process 4 – 1 instantaneous* reduction of pressure to its value in state 1 (* no piston movement therefore no change in volume). This models the removal of the combustion products from the cylinder and their replacement by a fresh charge of air; from a thermodynamic point of view, it is irrelevant that this occurs in a four-stroke engine by an inward stroke of the piston while the exhaust valve is open, followed by an outward stroke while the inlet valve is open. 8 Basic Concepts Chapter 2 A sequence of processes like this, where the final state of the last process is identical to the initial state of the first process, is called a (thermodynamic) CYCLIC PROCESS or simply a CYCLE. P - v diagrams of cycles are particularly useful for evaluating the net power output of an engine and its efficiency, meaning the net power divided by the rate of energy supply via the fuel. (With the idealised models in this course, we consider not the energy available from burning the fuel, a 2nd-year topic, but the equivalent rate of heat transfer to our system, a fixed mass of air alone.) 9 Basic Concepts Exercise 2 2.6 EXERCISE 2 Introduction & Basic Concepts Star rating system: Each question is assigned a hardness rating using the star sign next to the question number: * CORE level ** INTERMEDIATE level *** ADVANCED/EXAM level. *1 In power plant terminology, "station efficiency" means the net electrical power output of a power station divided by the theoretical rate of input of energy in the fuel. This energy input rate is the product of the mass of fuel consumed in unit time and the calorific value, or heating value, of the fuel. Calorific value is the energy theoretically converted from one form, chemical energy, to another, internal energy (which will be defined in Sec. 2) of the gaseous products, when unit mass of the fuel is completely burned. (a) A large coal-fired power station with a maximum electrical output of 1000 MW may have a station efficiency of 35%. If the coal has a calorific value of 22000 kJ/kg, how many tonnes of coal are consumed per day when such a plant is running continuously (i.e., for 24 h) at its maximum rating? (b) Typical U.K. coal has a sulphur content of 1.6% (by mass). Assuming that all of this is converted to sulphur dioxide (SO2) during combustion, what mass of SO2 will be emitted to the atmosphere per day by the power station in part (a)? (Approximate relative atomic masses S: 32, O: 16) *2 If a balloon is inflated by connecting it to a bottle of compressed gas, what would be the most appropriate "framework", system or control volume, for thermodynamic calculations on (a) the bottle, (b) the bottle's valve, (c) the bottle plus the balloon? State the reason in each case. *3 Two mountaineers set out from base camp, A, to reach the same higher-altitude camp, B. (This is analogous to a substance undergoing processes.) One takes a short but steep route, the other a longer route with a more gradual slope. When either of them has reached B, their final state, which of the following can be regarded as properties defining that state? (a) the latitude and longitude (b) the altitude (c) the distance travelled (along the mountain surface) during the climb One climber now returns to A. What word could be used to describe the process of ascending to B then descending to A, (d) using the same route in each direction? (e) descending by the route which the other climber used to reach B? 10 Basic Concepts Exercise 2 *4 A tank, with a leak-proof partition dividing it into two sections, has one section initially filled with a liquid (the system), as shown on the left. (a) The partition is suddenly removed (so that a process takes place). After some time, the liquid settles down to a new level, as shown on the right. i. Is the system in equilibrium in its initial state and in its final state? ii. On the centre diagram, sketch (very roughly) the liquid surface in some intermediate state. iii. Can this process be described as a quasi-equilibrium process? (Give the reason.) (b) From the same initial state, a different process occurs if, instead of removing the partition, it is left in place and a small hole is drilled in it so that liquid can flow slowly through to the other section. Sketch the liquid surface at an intermediate state in this new process, and state, with a reason, whether this is a quasi- equilibrium process. *5 The four sketches below represent different ways in which a gas (the system, shown shaded) in a cylinder can undergo an expansion process from an initial equilibrium state. In a, a diaphragm, which initially separated the gas from an evacuated section of the cylinder, has burst, allowing the gas to expand rapidly to fill the entire cylinder. In b, the gas is separated from the evacuated section by a lightweight piston attached to the end of the cylinder by a compression spring; initially, the spring was not compressed, and the piston was held in place by a catch; the expansion was caused by releasing the catch. In c, a heavy weight was placed on the piston, whose upper surface is now open to the atmosphere; initially the upward force on the piston due to the gas pressure just balanced the downward force due to atmospheric pressure plus the weight, and the expansion was caused by heating the gas. In d, a connecting rod links the piston to a rotating crankshaft, as in a reciprocating engine, and the expansion was again caused by heating the gas. 11 Basic Concepts Exercise 2 For each case, state, with reasons, whether the process is, is not, or could be a quasi-equilibrium process. If the answer is "could be", say what conditions would have to be satisfied for it to be a quasi-equilibrium process. **6 The valve on a compressed air bottle, containing air at an initial pressure of 130 bar, is opened slightly to discharge air slowly to the atmosphere (where the pressure is approximately 1 bar) until the pressure in the bottle has fallen to 80 bar. (Numbers are for illustration; no calculations are required.) (a) Could any of the following usefully be defined as a closed system when attempting to describe this process, and if not, why? (i) the air initially in the bottle (ii) the air which escapes (iii) the air which remains in the bottle. (b) Could the air, which is still in the bottle at the end of the process, be said to have undergone a quasi-equilibrium process? Why, and under what conditions? 12 Basic Concepts Exercise 2 _____________________________________________________________________ ANSWERS (Complete solutions in some cases) 1. (a) 11200 tonnes (b) 359 tonnes (each to 3 significant figures) 2. (a) C.V., because gas crosses the boundary (or "system surface"), at one place (or "port") (b) C.V., because gas crosses the system surface, at two ports (one inflow, one outflow) (c) System, because no gas crosses the boundary 3. (a) are both properties (at camp B the climbers have only one longitude and one latitude), so is (b) (assuming they remain on the surface, their altitude is uniquely determined by their longitude and latitude, rather like the value of any property of a simple compressible substance being fixed by the values of any other two independent properties). (c) is not a property, since it depends on the route between A and B (analogous to energy transfers, heat or work, between a system and its surroundings). (d) and (e) Cycle, because the sequence of processes begins and ends at A in both cases, regardless of the process paths (the routes). 4. (a) (i) Yes, in both states. No property changes with time and only a single value of the most relevant property, the height of the liquid surface, is needed to describe the system. (ii) No. In an intermediate state, surface height changes in a complicated way with position across the tank; a single value cannot be assigned to the system as a whole. Also, if the process were stopped by replacing the partition, the heights on each side would continue to oscillate for some finite time before reaching equilibrium. (b) Yes. At any instant during the process, only two properties would be needed to describe the state (the virtually uniform height on each side of the partition). Also, if the process were stopped by blocking the hole, the two heights would stop changing almost immediately; intermediate states would be almost equilibrium states). 5. (a) Not q-e. Unlike a sudden expansion of gas into the atmosphere, the system boundary can always be defined, as the cylinder's inside surface (since "adding a vacuum" to the system does not change its definition as a fixed, identifiable quantity of matter). However, in intermediate states the pressure (and other properties) varies throughout the gas. This is called an unresisted expansion. (b) Not q-e. Immediately after the catch is released to start the process, the spring exerts no force on the piston, so the expansion begins rapidly. As the expansion proceeds and the spring is compressed, the spring force increases from zero; this is a partially-resisted expansion. (c) Could be q-e if the heating is sufficiently slow, so that the gas pressure does not vary in space; it would then be a fully-resisted expansion. (In this particular 13 Basic Concepts Exercise 2 example, the pressure would also be constant throughout the process because the forces on the piston would balance at all times:- gas pressure x piston area = atmospheric pressure x piston area + weight of the "weight") (d) Could be q-e; similar to (c), with the effect of the weight replaced by the force exerted on the piston by the connecting rod. (Unlike (c), no reason why this should be a constant-pressure process.) 6. (a) (i) No. The boundary (except for that part formed by the bottle's inside surface) cannot be defined in the final state. (ii) No. No part of the boundary can be defined in the final state. (iii) Yes. The boundary is always the bottle's inside surface. (If unconvinced, take a microscopic view; at any time during the process, molecules of this gas can be found anywhere in the bottle; initially, molecules which will eventually escape can also be found anywhere in the bottle, so in the initial state only, we might regard the gas as two systems which share the same volume.) (b) Yes, provided the process takes place slowly so that the air remaining in the bottle has time to redistribute itself, keeping the pressure (and other properties) uniform throughout the bottle. 14 Energy, Heat, Work and the 1st Law Chapter 3 3 ENERGY, HEAT, WORK AND THE 1ST LAW 3.1 FORMS OF ENERGY You should be familiar with the concepts of kinetic and potential energy for solid bodies. If an entire thermodynamic system of mass m has a velocity magnitude C in some frame of reference, it too has a property KINETIC ENERGY (K.E.) of mC2/2 ; e.g. if we could identify and keep track of a particular 4 kg of steam as it flowed out of a steam turbine at 300 m/s, that system would have a K.E. of 4 × 3002 / 2 = 180000 J = 180 kJ. (The kilojoule is the most conveniently-sized energy unit for most thermodynamics applications, but you are advised for the time being to carry out calculations using the fundamental unit, the joule.) A system has a POTENTIAL ENERGY (P.E.) of mgz if its centre of mass is at a height z above some datum level (see Sec. 2.3.2); e.g., 20 kg of feed water entering the boiler of a steam plant at 4 m above floor level has a P.E. of 20 × 9.81 × 4 = 785 J relative to that datum. (Identifying the systems in these examples and defining their system boundaries might be difficult; we shall see later that the control volume is a more appropriate framework for handling such cases.) The energy associated with the molecular motion and inter-molecular forces in a substance (i.e. on the microscopic scale) is important in thermodynamics, since large changes can occur when a system changes state. This form of energy is related more to temperature than to the other properties of a system, and is the property called INTERNAL ENERGY (I.E.), with symbol U. Neglecting the energies associated with magnetic, electric and surface tension effects, the TOTAL ENERGY E of a system is then given by E = K.E. + P.E. + I.E. = mC2/2 + mgz + U.... (3-1) The specific total energy e (an intensive property) is the sum of the specific K.E, P.E. and I.E.: e = E/m = C2/2 + gz + U/m = C2/2 + gz + u. Note that we use small (lower-case) letters as the symbols for specific properties, expressed per unit mass, and capital (upper-case) letters for the extensive properties, proportional to the mass of the system. (Specific K.E. and P.E. do not have single symbols to represent them.) You must make sure that your handwriting distinguishes between u and U, for example! You will soon see that thermodynamics is only concerned with changes in the various forms of energy, not with their absolute values. The datum for any form of energy, the state at which we define the energy to be zero, is arbitrary. For situations where the system is an appropriate framework for handling problems, e.g. gas in a piston-cylinder arrangement, changes in K.E. and P.E. of the system are very often zero or insignificant compared with changes in I.E. In such cases, where a system undergoes a process from state 1 to state 2, we can write: 15 Energy, Heat, Work and the 1st Law Chapter 3 change in total energy = E = E2 - E1  U2 - U1 = m (u2 - u1). We next have to consider how a system's energy can change, i.e., how energy can cross the system boundary. Such energy transfers are classified in Sec. 3.2. 3.2 HEAT TRANSFER If a system changes state as a result of a temperature difference between system and surroundings, the transfer of energy across the system boundary is called a heat transfer. Note that in thermodynamic usage, "heat" is not a property of a system. A system has a certain energy E1 at the start of a process, the energy Q which then crosses the boundary may be called heat during the process, and at the end of the process the system has a new value, E2, of the property energy. The sign convention is that an energy transfer from the surroundings to the system is positive, and vice versa. A process in which no heat transfer occurs is called an adiabatic process. ("Adiabatic" has no other meaning.) In the 1st year course, we shall not concern ourselves with the detailed mechanisms of heat transfer (conduction, convection and radiation); a 2nd year course is devoted to these. 3.3 WORK TRANSFER In Mechanics, work is usually defined as the product of a force and the distance moved in the direction of the force. In Thermodynamics, we need a wider definition; two roughly equivalent versions are as follows. (a) A work transfer is any transfer of energy between a system and its surroundings which is not the result of a temperature difference. (b) An energy transfer from a system to its surroundings is a work transfer if the only effect on the surroundings could have been the rise of a weight. Example: a process where a gas expands in a vertical cylinder — 16 Energy, Heat, Work and the 1st Law Chapter 3 The word "only" is used because although a heat transfer can also cause the rise of a weight, this can never be the only effect on the surroundings; this will become clear when we deal with the 2nd Law of Thermodynamics. The words "could have been" are necessary to cover a case such as a process taking place in which an electric current cross the system boundary; the current could have powered a 2 "perfect" electric motor (no "I R" heating from the windings) which rotated a drum which wound up a rope to which a weight was attached. The case of a rotating shaft crossing the system boundary is also covered by this wording. The symbol for a work transfer is W and we shall use the mechanical engineer's traditional sign convention that W is positive for a transfer from system to surroundings (or work done by the system on the surroundings). This is the opposite of the convention for heat transfer (Q), arising from the subject's origins in steam engines; the early engineers wanted work from their engines and paid for fuel to provide heat to the water/steam in the engine, so it was natural to regard both these quantities as positive. Joule's famous experiments in Manchester in the 1840s showed that work and heat are equivalent forms of energy transfer. We now have to consider two important types of work transfer: displacement and shaft work. 3.3.1 Displacement Work Suppose that a gas in a vertical cylinder is at a pressure P greater than the pressure Pa of the surrounding atmosphere and is maintained in equilibrium by a large number of small weights placed on top of the piston (assuming the piston itself to be weightless). Now remove one of the weights. The piston will rise by a small amount and the gas pressure will fall by a small amount, until system (the gas) and surroundings reach a new equilibrium state. During this process, the difference between the upward force on the piston, due to gas pressure, and the downward force, due to the weights, will only differ very slightly from zero; the expansion of the gas is then said to be fully-resisted. We can consider this to be a quasi-equilibrium process. Note that the work done by the system all occurs at that part of the system boundary which moves, hence the terms "boundary 17 Energy, Heat, Work and the 1st Law Chapter 3 displacement work" or "moving boundary work", as Ç&B call it, or just "displacement work". The amount of work done in this process is the product of the force exerted on the moving boundary (the piston face) by the gas, P A, and the distance moved by the boundary, z. In the limit of infinitesimally small weights, z becomes an infinitesimally small distance dz and the work done is dW = P A dz. Now A dz is the volume dV through which the boundary moves, so dW = P dV.... (3-2) If we now continue to remove the small weights one by one, the gas will undergo a finite quasi-equilibrium process, and the work done will be obtained by integrating eqn 3-2: V2 W =  P dV for a quasi-equilibrium process... (3-3) V1 If the expansion were not fully resisted, e.g. if we removed several of the small weights at once, the process would not be a quasi-equilibrium (q-e) one, and W   P dV. Interpreting eqn 3-3 graphically, displacement work W is the area under the process path on a P — V diagram. To evaluate it numerically, we have to know the relationship between P and V. The left-hand diagram might represent the case above, where we remove weights from the piston. The right-hand diagram might represent the air compression process in a diesel engine. 3.3.2 Shaft work Displacement work is concerned with part of a system boundary moving in the direction of a normal (perpendicular) force, the pressure force. Work can also be done if a boundary moves in the direction of a tangential or shearing force; this is called shear work. Simple example: a vessel is filled with liquid, and its lid, in contact with the liquid, is pushed so that it slides across the top of the vessel; shear work is done on the liquid, related to the shear stress in the liquid at the boundary (the lid), the contact area between lid and liquid, 18 Energy, Heat, Work and the 1st Law Chapter 3 and the distance moved by the lid. This idea will be covered in more depth in Fluid Mechanics. Shear work transfer to a fluid commonly occurs during stirring processes, as in the left- hand diagram above. If the system is defined as the fluid alone, with boundary A (centre diagram), the paddle surfaces form part of the boundary, where both normal and shear forces are exerted on the fluid; displacement and shear work transfer occur together, in a way which is difficult to evaluate. However, if we redefine the system to include the paddle and the immersed part of its driving shaft, with boundary B (right-hand diagram), the only part of this boundary which is moving relative to the fluid is the shaft cross- section. Only shear forces are now involved. Work transfer as a result of shear forces in a rotating shaft is known as shaft work, with the symbol Wsh. This is the most common means of specifying work transfers in engineering plant. In “combined-cycle” plant, shaft work occurs to the air in the gas turbine engine's compressor, and from the combustion products or steam in the two turbines. Shaft power, i.e. rate of shaft work transfer, can be measured as W sh = torque x angular velocity e.g. in the case of a single shaft crossing a boundary, transmitting a torque of 2.1 x 10 6 N m while rotating at a steady 3000 rev/min (note that "r.p.m." is not an approved abbreviation), the shaft power = 2.1 x 106 N m x 2 x 3000/60 rad/s = 660 x 106 Nm/s (→ J/s → W) = 660 MW, the output of a typical steam turbine in a large power station. We usually have to consider shaft work in control volume analyses (Sec 5), rather than closed system analyses. 3.3.3 Work done in a cycle We can evaluate the net work done by a system in a cycle by summing (with the correct signs) the values of W for each of the processes making up the cycle, i.e. Wnet = W = W12 + W23 + W34 +.... where W23 (pronounced "W two three", not "W twenty-three") means the work done in process 2 → 3 (i.e. from state 2 to state 3), etc. Because a system returns to its initial state on completion of a cycle, and therefore to its initial volume V1 (and specific volume v1), the volume must decrease during some of the processes in the cycle if it increases 19 Energy, Heat, Work and the 1st Law Chapter 3 during others. W in some processes is therefore negative and in others positive, so the net work in the cycle may be negative, zero or positive. In the simple 2-process cycle sketched above, W12 is positive (V increases) and W21 is negative and of smaller magnitude. Wnet = W12 + W21 is therefore positive, and we can see that the magnitude of Wnet is the difference between the shaded areas in the left-hand and the centre P — V diagrams, i.e.  Wnet  for a cycle is the area enclosed in the P — V diagram. Considering just 1 kg of the substance forming a system of total mass m, the net specific work Wnet /m is the area enclosed in the P — v diagram. (Remember specific volume v = V/m.) It should be obvious (by thinking of the areas) that if the process paths proceed in a clockwise direction round a cycle, Wnet will be positive. This is true for the idealised model of a diesel engine cycle, the P — V diagram sketched in Sec. 2.5.4. (It would not be an engine if we did not get positive net work from it !) Cycles where the process paths go anti-clockwise involve net work transfer to the system; they occur in refrigerators and heat pumps, devices intended to produce heat transfers; we shall not consider them until Sec. 6, and then only briefly. 3.4 THE 1ST LAW OF THERMODYNAMICS On the macroscopic scale, it is a fundamental law of nature (proposed by Joule in the 1840s) that energy is conserved; it cannot be created or destroyed. If this is applied to a thermodynamic system undergoing a process from state 1 to state 2, we can write Q12 − W12 = E2 − E1 for process 1 → 2... (3-4) Q12 represents the algebraic sum of all the heat transfers during the process (+ve to the system and -ve from it) and W12 represents the algebraic sum of all the work transfers during the process (-ve to the system and +ve from it). 20 Energy, Heat, Work and the 1st Law Chapter 3 Q12 is the net energy added to the system by heat transfer(s) and –W12 is the net energy added by work transfer(s). These together increase the system's total energy from E1 to E2. Usually, we only have to deal with a single Q and/or a single W. For a system defined only as a quantity of fluid (excluding any structure or machinery), it does not matter where on the boundary these energy transfers occur, or when they occur. It also makes no difference whether or not it is a q-e process. In a cycle, the system's state, and therefore its total energy E, is the same at the start and end of the cycle. The change of energy of the system must then be zero, i.e. Q − W = 0 for a cycle... (3-5) where Q and W mean the algebraic sums (correct signs required!) of all the heat transfers and work transfers, respectively, round the complete cycle. In other words, (Q12 + Q23 +...) − (W12 + W23 +...) = 0. Later, in Sec. 5, we shall see how the First Law applies to control volumes, and derive an equation to use in place of eqn 3-4. Some very simple examples: (a) Concerning a single process: Gas (the system) in a cylinder is compressed by a piston, the magnitude of the work transfer being 30 kJ. If the internal energy of the gas increases by 20 kJ during this process, what heat transfer takes place? Eqn 3-4 applies. We are not concerned with changes of kinetic or potential energy of the system as a whole, so eqn 3-4 simplifies to Q12 – W12 = U2 – U1 where U is the system's internal energy. Compression means reduction of volume, so the displacement work (= PdV) is negative, i.e. W = -30 kJ.  Q12 = U2 – U1 + W = 20 + (–30) = –10 kJ, the minus sign indicating heat transfer from the gas to the surroundings. (b) Concerning a cycle: Water, filling a closed, rigid, insulated tank, is stirred for 100 s by a paddle wheel which is driven by a 1 kW motor. Then the insulation is removed, allowing a heat 21 Energy, Heat, Work and the 1st Law Chapter 3 transfer from the water until it returns to its initial state. What is the magnitude of the heat transfer Q21 in the second of the two processes? Eqn 3-5 applies. In the first process, there is no heat transfer (assuming the insulation to be perfect), i.e. Q12 = 0, while W12 = – 1 x 100 = –100 kJ. In the second process, there is no work transfer (no shaft work because the paddle is not operating, and no displacement work because there is no change of volume), i.e. W21 = 0. For a cycle of two processes, eqn 3-5 becomes (Q12 + Q21 ) - (W12 + W21 ) = 0,  Q21 = W12 + W21 – Q12 = –100 + 0 – 0 = –100 kJ, i.e. the magnitude of Q21 is 100 kJ. These two problems could be done in your head, using simple ideas of energy conservation, but for more complex problems it is wise to write down the appropriate version of the First Law equation, using the correct symbols for heat, work and internal energy with subscripts which identify the particular process. 22 Energy, Heat, Work and the 1st Law Chapter 3 Example Diesel engine performance calculation using idealised model of the cycle The model: System: taken as air throughout Processes: 1–2 Compression of air 2–3 Heat addition to air at const. pressure (modelling fuel injection and combustion) 3–4 Expansion of air 4–1 Heat rejection from air at const. volume (modelling expulsion of combustion gases then intake of fresh air) Given data: Air inlet conditions: pressure P1 = 1 bar, temperature T1 = 15°C (From the two- property rule, state 1 is fixed by these property values) Compression ratio V1 / V2 = v1 / v2 = 17 Specific heat addition q23 = 1613 kJ/kg Assumptions: Air behaves as a perfect gas, meaning that P v = R T and u = Cv T where R and Cv are constants with the following values: R = 287 J/kg K, Cv = 717 J/kg K (This will be covered in detail in Sec. 4) The compression and expansion processes are adiabatic and are described by the process-path relation P v1.4 = constant (The reason for this choice of process- path relation will become clear in Sec. 7.4) Find: the air properties P, v and T in each of the states 1, 2, 3 and 4 the heat transfer Q and displacement work transfer W in each of the processes 1– 2, 2–3, 3–4 and 4–1 the net work in the cycle the thermal efficiency of the cycle, defined as (net work) / (heat input) (Note carefully that net work means the sum of all the work transfers, each with its correct sign, but heat input means the sum of those heat transfers which have positive values) 23 Energy, Heat, Work and the 1st Law Exercise 3 3.5 EXERCISE 3 Energy, Heat, Work and the 1st Law Star rating system: Each question is assigned a hardness rating using the star sign next to the question number: * CORE level; ** INTERMEDIATE level; *** ADVANCED/EXAM level. *1. Consider a system comprising 2 kg of water in a lake above a waterfall; in this state, state 1, it is flowing sufficiently slowly that its kinetic energy can be neglected. It falls down the waterfall, a vertical distance of 50 m; immediately before it enters the pool at the bottom of the waterfall, it is in state 2. A little way downstream from the bottom of the waterfall, after all the turbulent motion has died down (and assuming we can still identify it), it is in state 3. Assume that there are no energy transfers (heat or work) between the system and its surroundings. (a) Evaluate the changes in potential energy PE, kinetic energy KE, internal energy U and total energy E between state 1 and state 2. (Remember that "change" always means final value minus initial value, not the other way round.) (b) Repeat part (a) for the changes between state 1 and state 3. (c) Assume that the specific internal energy u of water is a function only of its absolute temperature T and is given by u = 4190 T where u is the change in u, in J/kg, and T is the change in T, in K. (This is quite a good approximation for temperatures in the relevant range.) Evaluate the change in temperature of the system between state 1 and state 3. *2. (a) A gas inside a rigid, insulated vessel is stirred by a paddle wheel (or fan). The paddle wheel is driven by allowing a weight to fall, unwinding a rope from a drum attached to the shaft. State what form of energy transfer, to or from the system, takes place (i) if the system is defined as the gas plus the paddle wheel and that part of the shaft which is inside the vessel. (ii) if the system is defined as the gas plus the paddle wheel, shaft, drum, rope and weight. (b) A gas inside a rigid, insulated vessel is heated by an electric resistance heater. Ignoring the wires, state what form of energy transfer, to or from the system, takes place 24 Energy, Heat, Work and the 1st Law Exercise 3 (i) if the system is defined as the gas. (ii) if the system is defined as the gas plus the resistance heater. (iii) if the system is defined as the gas plus the resistance heater plus the battery. *3. A balloon made from light, flexible material, initially collapsed, is connected via a valve, initially closed, to a rigid storage bottle containing a gas at pressure 60 bar. The valve is partly opened and gas flows slowly into the balloon until its volume is 0.25 m3. Atmospheric pressure is 1.013 bar (1 bar = 105 N/m2 ). For a system defined as the gas initially in the bottle, and taking note of the words in italics above: (a) sketch the position of the system boundary at the start and end of this process; (b) calculate the work transfer across that part of the system boundary consisting of the inner surface of the balloon (N.B. not all the data above are relevant to this calculation); *4. A gas in a cylinder, with a leak-proof, frictionless piston, expands slowly from state 1 to state 2. (Leak-proof means that we can always define the system; frictionless and slowly mean a quasi-equilibrium process.) The cause of the process does not need to be known for this question. For each of the following types of process, (a) to (f), sketch the process path on a P - V diagram, shade the area representing the magnitude of the work transfer, then derive an expression (simplified as far as possible) for the displacement work done by the gas. (a) A constant-pressure process. (b) A constant-volume process. (c) A process where gas pressure varies linearly with piston displacement. (d) A process obeying the relationship PV n = constant, known as a "polytropic" process, where n is a constant (determined by the nature of the process) greater than 1, say 1.2. (e) Another process as in part (d), but where n = 1. (f) An adiabatic process in which the temperature changes from T1 to T2 , the mass of gas is m and its specific internal energy is related to its temperature by u = Cv T, where Cv is a constant and x denotes any change in property x. (Is it possible in this case to evaluate W in the same way as in parts (a) to (e)? If not, try using the 1st Law.) 25 Energy, Heat, Work and the 1st Law Exercise 3 **5. An idealised model of the cycle on which a spark-ignition (petrol or gasoline) four- stroke reciprocating engine works is as follows, based on a system comprising air only. Process Real SI engine Idealised model 1—2 Compression of Adiabatic compression of air Compression air + petrol vapour mixture P1V1n = P2V2n stroke 2—3 Rapid pressure rise after Heat transfer to air in a spark ignites mixture constant-volume process 3—4 Expansion of Adiabatic expansion of air Power stroke combustion products P3V3n = P4V4n 4—1 Exhaust stroke : Heat transfer from air in a exhaust valve opens, gases constant-volume process expelled at P = approx. Patmos.followed by (If the exhaust and Induction stroke :- induction strokes are exhaust valve closes, inlet modelled as constant- valve opens, cylinder fills pressure processes, at with new air + petrol vapour atmos. pressure, they mixture at P=approx. have no net effect; Patmos. one is the reverse of the other.) The diagrams above represent the real four-stroke spark-ignition engine, the arrows on the left-hand sketch showing the directions of piston motion and crankshaft rotation. (The P —V diagram could be plotted on an oscilloscope if the engine was equipped with a pressure transducer in one cylinder and a displacement transducer giving a signal proportional to instantaneous volume in the cylinder; it is easier in practice to provide a signal proportional to angle of rotation of the crankshaft. Such diagrams are known as "indicator diagrams".) 26 Energy, Heat, Work and the 1st Law Exercise 3 (a) Sketch the idealised model of the cycle on a P —V diagram. (b) Derive expressions for the work transfer in each of the four processes making up the idealised cycle, i.e. W12, W23, W34 and W41, and hence for the net work Wnet. (c) The relevant properties of air can be assumed to be related by the "equation of state" PV = mRT, where m is the mass of air in the cylinder, T is its absolute temperature and R, the "gas constant" for air, equals 287 J/kg K. (More about this in Sec. 4.) For both the compression and the expansion process, n may be taken as 1.4. To model a particular engine, take the maximum volume V1 as 1.6 litre, the compression ratio V1/V2 as 9 and the maximum temperature T3 as 3000°C. P1 may be taken as 1 bar and T1 as 15°C. For the idealised model of the cycle, calculate (i) the mass of air in the cylinder, (ii) the values of V, then P, then T, in each of states 2, 3 and 4, (iii) the work transfer in each of the four processes and the net work in one cycle. (d) If the net work per thermodynamic cycle, calculated in part (c)(iii), applied to a real four-stroke engine running at a speed of 5000 rev/min, what would be the engine's power output (in kW)? Why would a real engine with the same cylinder capacity, compression ratio and maximum temperature as in part (c) have a power output rather less than this? (e) A simple, commonly-used measure of engine performance is the "mean effective pressure" Pme , defined as the net work per cycle divided by the displaced or swept volume, i.e. Wnet / Vs or Wnet /(V1 - V2) (see the P – V diagram above). Interpret Pme graphically and evaluate it for the conditions in part (c). **6. A vertical cylinder of cross-sectional area 0.1 m2, with its lower end closed and upper end open, is fitted with a heavy piston, of mass m p = 203.9 kg. Initially, the piston is supported (e.g. on a ring projecting inwards a small distance from the cylindrical wall) such that there is a volume V1 of 0.05 m3 between its lower face and the closed end of the cylinder. This volume is occupied by a mass m of a gas, initially at a pressure P1 of 1 bar and temperature T1 of 26.8°C. Atmospheric pressure Pat , acting on the piston's upper face, is also 1 bar. The gas can be modelled as a "perfect gas". You will see in Sec. 4 of the course that this means its properties are related by Pv = RT (or PV = mRT ) (equation of state) and u = Cv T (or U = m Cv T ) (internal energy vs temperature is a straight line). For this gas, R = 0.290 kJ/kg K and Cv = 0.725 kJ/kg K. (Note that this R is not the same as the universal gas constant, R. If the system mass m represents n kilomoles of the substance, then nR is the same as mR. You are probably familiar with the equation of state in the form PV = nR T, but engineers prefer PV = mRT, since mass is usually a much more useful quantity than number of kilomoles.) 27 Energy, Heat, Work and the 1st Law Exercise 3 A heat transfer to the gas takes place, until the piston has risen through a distance of 200 mm.The process occurs sufficiently slowly and the effect of friction is sufficiently small that the system comprising the gas undergoes a quasi-equilibrium process. (a) Do three sketches of the cylinder and piston:- at the beginning of the process (gas in state 1), at the point where the piston is just about to rise off its support (gas in state 2), and at the end of the process (gas in state 3). Write on, or beside, each sketch the known values of the gas properties. (This may seem like a waste of time, but it is good practice and can later save time in searching through the wording of a question to look for given data.) (b) Sketch the process path on a P — V (or P — v ) diagram. (c) Use the equation of state in state 1 to show that m = 0.0575 kg (to 3 sig. fig.) (d) Considering the forces on the piston, show that P2 = 1.20 bar. What is P3? (e) Write down (in symbols) the 1st Law equation for the gas system in process 1—2, with the aim of finding Q12, the heat transfer necessary to begin lifting the piston. Satisfy yourself that the unknowns are W12 and T2, and that W12 = 0. Use the equation of state in states 1 and 2 (or just in state 2) to show that T2 = 360 K, and hence that Q12 = 2.50 kJ. (f) Write down (in symbols) the 1st Law equation for the gas system in process 2—3, with the aim of finding Q23 , the heat transfer while the piston is moving. Satisfy yourself that the unknowns are W23 and T3. Show that the displacement work W23 = 2.40 kJ. Use the equation of state in states 2 and 3 (or just in state 3) to show that T3 = 504 K, and hence that Q23 = 8.50 kJ. **7. A system undergoes the following cycle of quasi-equilibrium processes. Process 1 — 2: Constant volume, with heat transfer of 170 kJ to the system Process 2 — 3: Constant pressure, with heat transfer of 100 kJ from the system and work transfer of 42 kJ to the system Process 3 — 1: Adiabatic (a) Sketch the cycle on a P — V diagram. (If you cannot decide on the direction of process path 1 — 2 because the properties of the substance forming the system are not given, it is safe to assume that internal energy and pressure increase with temperature at constant volume.) (b) Evaluate the heat and work transfers in process 3 — 1. 28 Energy, Heat, Work and the 1st Law Exercise 3 _____________________________________________________________________ ANSWERS 1. (a) PE2 – PE1 = -981 J; KE2 – KE1 = +981 J; U2 – U1 = 0; E2 – E1 = 0 (b) PE3 – PE1 = -981 J; KE3 – KE1 = 0; U3 – U1 = +981 J; E3 – E1 = 0 2. (a) (i) Shaft work, negative (ii) None (Within the system, the weight loses potential energy and the gas gains internal energy, but no energy crosses the system boundary.) (b) (i) Heat, positive (ii) Electrical work (see notes, Sec. 3.3), negative (iii) None (Within the system, the battery loses chemical energy and the gas gains internal energy, but again no energy crosses the system boundary.) 3. Light, flexible material implies that the gas pressure inside the expanded balloon needs to be only very slightly greater than atmospheric pressure in order to stretch the material and support its weight. Hence the process can be approximated as a constant-pressure process with gas pressure equal to atmospheric pressure, the word slowly implying quasi-equilibrium. The initial pressure of the gas in the bottle is irrelevant here; it falls almost to atmospheric pressure as it flows through the valve. So the displacement work W = P dV which in this case becomes P  dV = P (V2 - V1) = 1.013 x 105 x 0.25 J = 25.3 kJ (positive; work done by gas on surrounding atmosphere). 4. (a) P1(V2 - V1) (b) 0 (No piston movement, therefore no displacement work) (c) ½ (P1 + P2) (V2 - V1) (d) (P2V2 - P1V1) / (1 - n ) (e) P1V1 ln (V2 / V1) 5. (c) (i) m = 1.94 x 10-3 kg (ii) V2 = 1.778 x 10-4 m3, P2 = 21.7 bar, T2 = 693.6 K V3 = 1.778 x 10-4 m3, P3 = 102.3 bar, T3 = 3273.2 K V4 = 1.600 x 10-3 m3, P4 = 4.72 bar, T4 = 1359.2 K (iii) W12 = -563.2 J, W23 = 0, W34 = +2659.2 J, W41 = 0. Wnet = W = 2096.0 J (d) 87.3 kW (e) 14.74 bar 6. (b) 1 — 2 is a constant volume process, 2 — 3 is a constant-pressure process (d) P3 = 1.20 bar 7. Q31 = 0 (given as adiabatic), W31 = 112 kJ 29 Energy, Heat, Work and the 1st Law Structured Tutorial 3 3.6 STRUCTURED TUTORIAL 3 Energy, heat, displacement work and the 1st Law of Thermodynamics This problem does not rely on material from Sec. 4 (Properties of Substances), although some basic ideas on gas properties are introduced. Consider a rigid cylinder of cross-sectional area A = 0.05 m2, open to the atmosphere at one end and fitted with a lightweight (i.e. negligible mass), leakproof, frictionless piston. The piston is linked by a spring to a fixed point beyond the open end of the cylinder. The space between the piston and the closed end is occupied by a gas, initially at a pressure P1 of 8 bar (absolute). Atmospheric pressure Patm can be taken as 1 bar. The spring, which has a stiffness or spring constant k of 25 kN/m, is initially in compression such that gas is in equilibrium with its surroundings. Suppose the gas is now heated slowly so that it expands, doing displacement work on the surroundings, until the piston's displacement x is 0.2 m. What must be the spring force F1 in the initial equilibrium state? Force balance on piston: so F1 = Does the system defined as the gas undergo a quasi-equilibrium (q-e) process? Can we calculate the displacement work transfer from the gas using W =  P dV ? To do this integral, we need the relationship between P and gas volume V. What is it? Consider the equilibrium of forces acting on the piston at any instant: PA (on left face) = (on right face) So P = 30 Energy, Heat, Work and the 1st Law Structured Tutorial 3 Now V = Suppose V1 = 0.01 m3. What are B, C, and the final volume V2 of the gas? B = C = V2 = What is the final pressure, P2? P2 = Now sketch the P — V diagram for this process. Now calculate W. 31 Energy, Heat, Work and the 1st Law Structured Tutorial 3 How can we find the heat transfer Q in this process? Use (Warning: don't confuse U and u, etc.; take care with handwriting) For simple compressible substances in general, u could be found if we knew the values of any two other independent intensive properties, such as P and v (v = V/m). Suppose that u for this gas depends on the value of only one other property, temperature T. This is a very good approximation for gases under a wide range of conditions. Let us assume that u = Cv T where Cv is a constant, still a good approximation provided that T is not too large. (More of this in lectures, Sec. 4; Cv is called the "specific heat at constant volume"; its use is not restricted to processes where the volume of the gas is constant.) So Q = W + m Cv (T2 – T1) (for this substance in this process). Suppose that Cv for this gas is 625 J/kgK, and that T1 is 400 K. All we need now are values for m and T2. For these, we need to know a relationship between the properties, which applies in any given state, called an "equation of state". A very good approximation for gases under a wide range of conditions is P v = R T, where R is another constant for the particular gas (not what what you may know as the universal gas constant; more of this in lectures, Sec. 4.) Suppose R is 250 J/kgK for this gas. Writing v = V/m, the equation of state becomes P V = m R T. What is the mass of gas, m ? Use m = What is the final temperature of the gas, T2 ? Use T2 = (or, So what is Q ? 32 Energy, Heat, Work and the 1st Law Structured Tutorial 3 Q = Check that the sign of Q is correct: the calculation gave a positive result, which is OK because the heat transfer was from the surroundings to the gas. Of the energy which was transferred to the gas by heating, about three quarters has gone to increase the internal energy of the gas and about one quarter has been transferred from the gas to become energy stored in the compressed spring. 33 Energy, Heat, Work and the 1st Law Structured Tutorial 3 A supplementary problem, which you will probably not have time to go through in detail during the tutorial, but should read through afterwards: Suppose the cylinder of the previous problem is closed at both ends, with the gas again initially at P1 = 8 bar (absolute) and the space on the other side of the piston completely evacuated. The piston is initially prevented from moving by a peg, and a spring is chosen such that there is initially no tension or compression. The peg is now removed. Suppose the piston moves through a distance of 0.2 m before it is stopped and held, e.g. by a ratchet mechanism. The gas then settles to a new equilibrium state. Is this a quasi-equilibrium (q-e) process, and if not, why? No. As soon as the peg is removed, a finite pressure difference exists across the piston, so the expansion is rapid; it is likely that a single value of pressure (or other property) is inadequate to describe the state of the gas at any instant. The expansion is initially unresisted. Would it still be a non-q-e process if the spring were initially compressed so that it exerted a force of 10 kN on the piston? In state 1, gas would exert force P1A = 8 x 105 x 0.05 N = 40 kN on piston, so piston would still have a large force imbalance; expansion would initially be partly- resisted. Answer is therefore Yes. What about an initial spring force of 40 kN? Would now be a q-e process, because fully - resisted (but no process would occur without some cause other than removing the peg.) What about a case with no spring but a piston having significant mass ? Resistance would now be caused by the inertia of the piston; how closely the process approached q-e would depend on the piston's mass. Going back to the case of the initially uncompressed spring and the lightweight piston, can we calculate the work transfer to or from the gas using W =  P dV ? No, because it is not a q-e process; at any instant during the process, P is not uniform throughout the gas. Is there a work transfer? Yes. Defining our system as the gas, there is energy in the compressed spring, part of the system's surroundings, at the end of the process. Would there have been a work transfer in the absence of the spring? No, because there would never have been any resistance. (You don't have to do work to "push on a vacuum"!) 34 Energy, Heat, Work and the 1st Law Structured Tutorial 3 How can W be found? By concentrating on the surroundings; need to find the work done on the spring. The spring force F increases linearly with piston displacement x. Here, F is initially zero, so it reaches 5 kN (i.e. 0 + kx ) after 0.2 m where the piston is stopped. To find the work Wspring done on the spring: Sketch, or imagine, the graph of F against x. |Wspring | =  F dx = ½( 0 + 5000 ) x 0.2 = 500 J or 0.5 kJ So Wspring = – 0.5 kJ (note sign convention; work transfer to spring, therefore – ve value) So what is W for the gas? |W | = |Wspring | = 0.5 kJ, so W = 0.5 kJ (intuitive way; work transfer from gas, must be +ve) or W = – Wspring = – ( – 0.5 kJ ) = 0.5 kJ (strict algebraic way; + sign appears automatically) (Choose which way suits you, but don't mix them.) 35 Properties of Substances Chapter 4 4 PROPERTIES OF SUBSTANCES 4.1 INTRODUCTION To predict the heat and work transfers in processes and cycles, we need to know the relevant properties of the system in any equilibrium state, or, for properties like internal energy which have an arbitrary datum, the change during the process. For example, when applying the 1st Law to the air in a diesel engine cylinder during the compression stroke, the change in specific internal energy u2 – u1 needs to expressed in terms of other properties which are known in advance (P1 , v1 and v2 ) or found from a knowledge of the process path (P2 or T2 ). Detailed knowledge of properties is also needed in order to select a suitable working fluid for a plant based on a particular thermodynamic cycle (although considerations of availability, cost, safety etc. are also very important). For example, what makes water/steam the best choice of working fluid for a conventional power plant consisting of boiler (heat in), turbine (work out), condenser (heat out) and pump (work in)? Why do refrigerators use substances such as ammonia or hydrofluorocarbon compounds as their working fluid, rather than water? In 1M, attention will be confined to pure substances. By the end of Section 4, we shall be able to relate the properties of gases such as air with sufficient accuracy to model their behaviour in some important types of process occurring in power plant and in other engineering applications. For liquids/vapours such as water/steam, tables and charts are needed in general. 4.2 PURE SUBSTANCES A substance whose chemical composition is the same throughout, and does not change with time, is called a pure substance. Examples: nitrogen gas liquid nitrogen air (several gases, but uniformly mixed) air/petrol vapour mixture products from burning (provided chemical reaction has ceased) water/steam mixture (chemically the same throughout, whatever the physical condition, e.g. water droplets in steam, vapour bubbles in water, water at the bottom of a vessel with steam above it,...) The following examples are not pure substances: mixtures of gaseous and liquid air (constituents have different boiling points; liquid is richer in nitrogen) air/petrol droplets (distinct regions with different chemical composition) 4.3 DIAGRAMS OF STATE A phase of a substance is identified by a particular configuration of the molecules: solid (molecules held in fixed positions by attractive forces), liquid (molecules still quite closely 36 Properties of Substances Chapter 4 spaced but no longer fixed), vapour or gas (widely spaced molecules in random motion). Processes where a substance changes phase are of great importance in engineering thermodynamics, e.g. the refrigerant evaporating to extract "latent heat" from the food compartment of a freezer, the steam partially condensing in the low-pressure turbine of a steam power plant. Phase changes are conveniently depicted on property diagrams such as the following pressure-temperature diagram (for a substance like water which expands on freezing). The triple point (0.00612 bar, 0.01°C for ice/water/steam) is the only state where solid, liquid and vapour phases can co- exist in equilibrium. Below the triple-point pressure, solids change phase directly to vapour (sublimation) when heated, e.g. solid carbon dioxide, "dry ice", for which the triple-point pressure is above atmospheric. The critical point is a state where liquid which is just about to evaporate and vapour which is just about to condense become identical, and supercritical fluid is the high-pressure continuation of the liquid and vapour phases where the two phases are indistinguishable; they will be discussed further in Sec. 4.6. From now on, diagrams will be simplified since we shall not be concerned with the thermodynamics of the solid phase or with solid-liquid or solid-vapour phase changes. All equilibrium states of a simple compressible system comprising a pure substance are represented by a point on a surface plotted in pressure – specific volume –temperature coordinates. P–v–T equilibrium surface for a pure substance (excluding solid region) Below the critical pressure (P at the critical point), there is a region where liquid and vapour are in equilibrium. For water/steam, the state of the contents of a kettle when it is 37 Properties of Substances Chapter 4 boiling lies somewhere in this region. Sec. 4.6 will introduce more terminology to describe this region, with important applications to the boiler and condenser of a steam power plant, for example. Some substances, e.g. air, are normally encountered in the vapour phase in states far away from the liquid+vapour (or two-phase) region, at temperatures well above the critical temperature. They are then normally called gases (or "permanent" gases) rather than vapours, and relationships between their properties can be greatly simplified. We shall next consider the properties of gases, and return to vapours and liquids in Sec. 4.6. 4.4 MODELLING THE BEHAVIOUR OF GASES There is a variety of empirically-based equations which relate P, v and T in a given state. They are known as equations of state. The simplest of these, the ideal-gas equation of state, can be derived theoretically (without recourse to measurements) if the following assumptions are made about the behaviour of the molecules of a gas: no momentum loss during collisions with container walls (molecules ~ rigid, elastic spheres) (reasonable since pressure in a sealed container does not fall with time) molecules have negligible volume (reasonable since gases are highly compressible) attractive force between adjacent molecules is negligible (reasonable since gases expand to fill the entire volume available) The kinetic theory of gases then yields the equation with which you are probably familiar: P V = n R T where n is the number of moles occupying volume V at pressure P and temperature T, and R is a constant, the same for all gases, called the universal gas constant and equal to 8.314 kJ/kmol K. (A kilomole, kmol, is the molar mass expressed in kg, not grams – SI, not cgs, units.) This is in accordance with the relationships derived experimentally in England by Robert Boyle in 1662 (PV = const. for fixed T) and in France by J. Charles in the early 1800s (T/V = const. for fixed P ). Engineers are normally more interested in the mass of a system than in the number of moles, so we will now derive two equivalent but more useful forms of this equation of state. Multiplying and dividing by the molar mass M (the mass in kg of one kmol of the gas, which equals the mass in g of one gram-mole), P V = (M n) ( R / M) T = m R T since system mass m = M n. R = R / M is known as the gas constant for the particular gas. Don’t confuse it with R. e.g. R for nitrogen is 8.314 / 28.01 = 0.297 kJ/kg K. Dividing both sides by m to remove the extensive properties so that the equation applies to one kg rather than the whole system, P (V / m) = P v = R T. 38 Properties of Substances Chapter 4 The assumptions and theory leading to this equation also imply that specific internal energy u depends on only one other property, temperature, as Joule demonstrated experimentally in 1843. It is sufficient to specify this model of gas properties as follows (remember it !) : IDEAL GAS MODEL P v = R T or P V = m R T... (4-1) and u = some function of T only... (4-2) It can be applied with good accuracy to model the behaviour of air and other common gases over the ranges of pressure and temperature relevant to most engineering equipment of interest to us. In 1M at least, we shall not need any greater sophistication, although Ç&B describe several more-complex equations of state starting with that of van der Waals:  a   P + 2  (v − b ) = R T  v  where non-zero a accounts for inter-molecular attraction and non-zero b for molecular volume. Expressions for u as a function of P as well as T would also be needed. If the processes of interest to us cover a sufficiently small range of temperature that we can approximate the ideal-gas u(T) relationship by a linear equation in T, i.e. a straight line fit, we have the perfect gas version of the ideal gas model. The now-constant slope of the u(T) curve allows us to write u2 – u1 = const. x (T2 – T1 ) for the change in specific internal energy in process 1–2, needed to obtain Q – W from the 1st Law equation. We shall use this simplification for almost all calculations on gas systems in 1M. More details of the perfect gas model will be given after defining specific heats in Sec. 4.5. Note that a perfect gas is an ideal gas; u is still a function of T only, but a particular function. 4.5 ENTHALPY AND SPECIFIC HEATS Consider a process in which there is heat transfer to a gas (the system) at constant pressure, e.g. in a vertical cylinder closed by a piston, with a weight on the piston to maintain the pressure at the initial value P1. 39 Properties of Substances Chapter 4 The 1st Law (for any process with negligible K.E. and P.E. change)