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MIT Course 16 Fall 2002 Thermal Energy 16.050 Prof. Z. S. Spakovszky Notes by E.M. Greitzer Z. S. Spakovszky Table of Contents PART 0 - PRELUDE: REVIEW OF “UNIFIED ENGINEERING THERMODYNAMICS” 0.1 What it's all about 0.2 Definitions and fundamentals ideas of thermodynamics 0.3 Review of thermodynamics concepts 0-1 0-1 0-2 PART 1 - THE SECOND LAW OF THERMODYNAMICS 1.A- Background to the Second Law of Thermodynamics 1.A.1 Some Properties of Engineering Cycles: Work and Efficiency 1.A.2 Carnot Cycles 1.A.3 Brayton Cycles (or Joule Cycles): The Power Cycle for a Gas Turbine Jet Engine 1.A.4 Gas Turbine Technology and Thermodynamics 1.A.5 Refrigerators and heat pumps 1.A.6 Reversibility and Irreversibility in Natural Processes 1.A.7 Difference between Free Expansion of a Gas and Reversible Isothermal Expansion 1.A.8 Features of reversible Processes 1.B The Second Law of Thermodynamics 1.B.1 Concept and Statements of the Second Law 1.B.2 Axiomatic statements of the Laws of Thermodynamics 1.B.3 Combined First and Second Law Expressions 1.B.4 Entropy Changes in an Ideal Gas 1.B.5 Calculation of Entropy Change in Some Basic Processes 1.C Applications of the Second Law 1.C.1 Limitations on the Work that Can be Supplied by a Heat Engine 1.C.2 The thermodynamic Temperature Scale 1.C.3 Representation of Thermodynamic Processes in T-s coordinates 1.C.4 Brayton Cycle in T-s coordinates 1.C.5 Irreversibility, Entropy Changes, and Lost Work 1.C.6 Entropy and Unavailable Energy 1.C.7 Examples of Lost Work in Engineering Processes 1.C.8 Some Overall Comments on Entropy, Reversible and Irreversible Processes 1A-1 1A-3 1A-5 1A-8 1A-11 1A-12 1A-14 1A-16 1B-1 1B-3 1B-5 1B-6 1B-7 1C-1 1C-3 1C-4 1C-5 1C-8 1C-11 1C-14 1C-23 1.D Interpretation of Entropy on the Microscopic Scale - The Connection between Randomness and Entropy 1.D.1 Entropy Change in Mixing of Two ideal Gases 1D-1 1.D.2 Microscopic and Macroscopic Descriptions of a System 1D-2 1.D.3 A Statistical Definition of Entropy 1D-3 1.D.4 Connection between the Statistical Definition of Entropy and Randomness 1D-5 1.D.5 Numerical Example of the Approach to the Equilibrium Distribution 1.D.6 Summary and Conclusions 1D-6 1D-11 PART 2 - POWER AND PROPULSION CYCLES 2.A Gas Power and Propulsion Cycles 2.A.1 The Internal Combustion Engine (Otto Cycle) 2.A.2 Diesel Cycle 2.A.3 Brayton Cycle 2.A.4 Brayton Cycle for Jet Propulsion: the Ideal Ramjet 2.A.5 The Breguet Range Equation 2.A.6 Performance of the Ideal Ramjet 2.A.7 Effect of departures from Ideal Behavior 2A-1 2A-4 2A-5 2A-6 2A-8 2A-11 2A-14 2.B Power Cycles with two-Phase media 2.B.1Behavior of Two-Phase Systems 2.B.2 Work and Heat Transfer with Two-Phase Media 2.B.3 The Carnot Cycle as a Two-Phase Power Cycle 2.B.4 Rankine Power Cycles 2.B.5 Enhancements of, and Effect of Design Parameters on Rankine Cycles 2.B.6 Combined Cycles in Stationary Gas Turbine for Power Production 2.B.7 Some Overall Comments on Thermodynamic Cycles 2B-1 2B-5 2B-8 2B-13 2B-15 2B-19 2B-21 2.C Introduction to Thermochemistry 2.C.1 Fuels 2.C.2 Fuel-Air Ratio 2.C.3 Enthalpy of Formation 2.C.4 First Law analysis of Reacting systems 2.C.5 Adiabatic Flame Temperature 2C-1 2C-2 2C-2 2C-4 2C-7 PART 3 - INTRODUCTION TO ENGINEERING HEAT TRANSFER 1.0 2.0 2.1 2.2 2.3 3.0 3.1 3.2 3.3 4.0 5.0 6.0 7.0 8.0 Heat Transfer Modes.......................................................................................................... HT-5 Conduction Heat Transfer .................................................................................................. HT-5 Steady-State One-Dimensional Conduction.................................................................... HT-8 Thermal Resistance Circuits ..........................................................................................HT-10 Steady Quasi-One-Dimensional Heat Flow in Non-Planar Geometry ........................... HT-14 Convective Heat Transfer..................................................................................................HT-18 The Reynolds Analogy..................................................................................................HT-19 Combined Conduction and Convection .........................................................................HT-24 Dimensionless Numbers and Analysis of Results ..........................................................HT-29 Temperature Distributions in the Presence of Heat Sources...............................................HT-32 Heat Transfer From a Fin ..................................................................................................HT-35 Transient Heat Transfer (Convective Cooling or Heating) .................................................HT-40 Some Considerations in Modeling Complex Physical Processes........................................HT-42 Heat Exchangers ...............................................................................................................HT-43 8.1 Efficiency of a Counterflow Heat Exchanger.................................................................HT-50 9.0 Radiation Heat Transfer (Heat transfer by thermal radiation).............................................HT-52 9.1 Ideal Radiators ..............................................................................................................HT-53 9.2 Kirchhoff's Law and "Real Bodies" ...............................................................................HT-55 9.3 Radiation Heat Transfer Between Planar Surfaces .........................................................HT-55 9.4 Radiation Heat Transfer Between Black Surfaces of Arbitrary Geometry ......................HT-60 ACKNOWLEDGEMENT Preparation of these notes has benefited greatly from the expertise of a number of individuals, and we are pleased to acknowledge this help. Jessica Townsend, Vincent Blateau, Isabel Pauwels, and David Milanes, the successive Teaching Assistants in this core department course, provided ideas, corrected errors, inserted “Muddy Points”, supplied the index, and in general, created a much more readable document. Any errors that remain, or lack of readability, are thus the sole responsibility of the authors. We also appreciate the work of Diana Park and Robin Palazzolo, who contributed greatly to the editing and graphics. Finally, we are grateful to have had the opportunity to discuss some of the material with Professor Frank Marble of Caltech, whose understanding, insight, and ability to describe thermofluids concepts provide a model of how to address important technical problems. PART 0 PRELUDE: REVIEW OF "UNIFIED ENGINEERING THERMODYNAMICS" PART 0 - PRELUDE: REVIEW OF “UNIFIED ENGINEERING THERMODYNAMICS” [IAW pp 2-22, 32-41 (see IAW for detailed SB&VW references); VN Chapter 1] 0.1 What it’s All About The focus of thermodynamics in 16.050 is on the production of work, often in the form of kinetic energy (for example in the exhaust of a jet engine) or shaft power, from different sources of heat. For the most part the heat will be the result of combustion processes, but this is not always the case. The course content can be viewed in terms of a “propulsion chain” as shown below, where we see a progression from an energy source to useful propulsive work (thrust power of a jet engine). In terms of the different blocks, the thermodynamics in Unified Engineering and in this course are mainly about how to progress from the second block to the third, but there is some examination of the processes represented by the other arrows as well. The course content, objectives, and lecture outline are described in detail in Handout #1. Energy source chemical nuclear, etc. Heat (combustion process) Mechanical work, electric power... Useful propulsive work (thrust power) 0.2 Definitions and Fundamental Ideas of Thermodynamics As with all sciences, thermodynamics is concerned with the mathematical modeling of the real world. In order that the mathematical deductions are consistent, we need some precise definitions of the basic concepts. A continuum is a smoothed-out model of matter, neglecting the fact that real substances are composed of discrete molecules. Classical thermodynamics is concerned only with continua. If we wish to describe the properties of matter at a molecular level, we must use the techniques of statistical mechanics and kinetic theory. A closed system is a fixed quantity of matter around which we can draw a boundary. Everything outside the boundary is the surroundings. Matter cannot cross the boundary of a closed system and hence the principle of the conservation of mass is automatically satisfied whenever we employ a closed system analysis. The thermodynamic state of a system is defined by the value of certain properties of that system. For fluid systems, typical properties are pressure, volume and temperature. More complex systems may require the specification of more unusual properties. As an example, the state of an electric battery requires the specification of the amount of electric charge it contains. Properties may be extensive or intensive. Extensive properties are additive. Thus, if the system is divided into a number of sub-systems, the value of the property for the whole system is equal to the sum of the values for the parts. Volume is an extensive property. Intensive properties do not depend on the quantity of matter present. Temperature and pressure are intensive properties. Specific properties are extensive properties per unit mass and are denoted by lower case letters. For example: specific volume = V/m = v. 0-1 Specific properties are intensive because they do not depend on the mass of the system, A simple system is a system having uniform properties throughout. In general, however, properties can vary from point to point in a system. We can usually analyze a general system by sub-dividing it (either conceptually or in practice) into a number of simple systems in each of which the properties are assumed to be uniform. If the state of a system changes, then it is undergoing a process. The succession of states through which the system passes defines the path of the process. If, at the end of the process, the properties have returned to their original values, the system has undergone a cyclic process. Note that although the system has returned to its original state, the state of the surroundings may have changed. Muddy points Specific properties (MP 0.1) What is the difference between extensive and intensive properties? (MP 0.2) 0.3 Review of Thermodynamic Concepts The following is a brief discussion of some of the concepts introduced in Unified Engineering, which we will need in 16.050. Several of these will be further amplified in the lectures and in other handouts. If you need additional information or examples concerning these topics, they are described clearly and in-depth in the Unified Notes of Professor Waitz, where detailed references to the relevant sections of the text (SB&VW) are given. They are also covered, although in a less detailed manner, in Chapters 1 and 2 of the book by Van Ness. 1) Thermodynamics can be regarded as a generalization of an enormous body of empirical evidence. It is extremely general, and there are no hypotheses made concerning the structure and type of matter that we deal with. ;; ; 2) Thermodynamic system: A quantity of matter of fixed identity. Work or heat (see below) can be transferred across the system boundary, but mass cannot. Gas, Fluid System Boundary 3) Thermodynamic properties: For engineering purposes, we want "averaged" information, i.e., macroscopic not microscopic (molecular) description. (Knowing the position and velocity of each of 1020+ molecules that we meet in typical engineering applications is generally not useful.) 4) State of a system: The thermodynamic state is defined by specifying values of a (small) set of measured properties which are sufficient to determine all the remaining properties. 0-2 5) Equilibrium: The state of a system in which properties have definite (unchanged) values as long as external conditions are unchanged is called an equilibrium state. Properties (P, pressure, T, temperature, ρ, density) describe states only when the system is in equilibrium. Mechanical Equilibrium ; ;;; Thermal Equilibrium Po Mass Mg + PoA = PA Gas T1 Gas at Pressure, P Gas T2 Insulation Copper Partition Over time, T1 → T2 6) Equations of state: For a simple compressible substance (e.g., air, water) we need to know two properties to set the state. Thus: P = P(v,T), or v = v(P, T), or T = T(P,v) where v is the volume per unit mass, 1/ρ. Any of these is equivalent to an equation f(P,v,T) = 0 which is known as an equation of state. The equation of state for an ideal gas, which is a very good approximation to real gases at conditions that are typically of interest for aerospace applications is: – = RT, Pv where –v is the volume per mol of gas and R is the "Universal Gas Constant", 8.31 kJ/kmol-K. A form of this equation which is more useful in fluid flow problems is obtained if we divide by the molecular weight, M: Pv = RT, or P = ρ RT where R is R/M, which has a different value for different gases. For air at room conditions, R is 0.287 kJ/kg-K. 7) Quasi-equilibrium processes: A system in thermodynamic equilibrium satisfies: a) mechanical equilibrium (no unbalanced forces) b) thermal equilibrium (no temperature differences) c) chemical equilibrium. For a finite, unbalanced force, the system can pass through non-equilibrium states. We wish to describe processes using thermodynamic coordinates, so we cannot treat situations in which such imbalances exist. An extremely useful idealization, however, is that only "infinitesimal" unbalanced forces exist, so that the process can be viewed as taking place in a series of "quasiequilibrium" states. (The term quasi can be taken to mean "as if"; you will see it used in a number of contexts such as quasi-one-dimensional, quasi-steady, etc.) For this to be true the process must be slow in relation to the time needed for the system to come to equilibrium internally. For a gas 0-3 10 at conditions of interest to us, a given molecule can undergo roughly 10 molecular collisions per second, so that, if ten collisions are needed to come to equilibrium, the equilibration time is on the order of 10-9 seconds. This is generally much shorter than the time scales associated with the bulk properties of the flow (say the time needed for a fluid particle to move some significant fraction of the lighten of the device of interest). Over a large range of parameters, therefore, it is a very good approximation to view the thermodynamic processes as consisting of such a succession of equilibrium states. ; ;;; 8) Reversible process For a simple compressible substance, Work = ∫PdV. If we look at a simple system, for example a cylinder of gas and a piston, we see that there can be two pressures, Ps, the system pressure and Px, the external pressure. Ps Px The work done by the system on the environment is Work = ∫PxdV. This can only be related to the system properties if Px ≈ Ps. For this to occur, there cannot be any friction, and the process must also be slow enough so that pressure differences due to accelerations are not significant. Px with friction P ➀ ➁ Ps (V) ➀ ∫ Px dV ≠ 0 ➁ but ➀ ∫ Ps dV = 0 Vs Work during an irreversible process ≠ ∫ Ps dV Under these conditions, we say that the process is reversible. The conditions for reversibility are that: a) If the process is reversed, the system and the surroundings will be returned to the original states. b) To reverse the process we need to apply only an infinitesimal dP. A reversible process can be altered in direction by infinitesimal changes in the external conditions (see Van Ness, Chapter 2). 9) Work: For simple compressible substances in reversible processes , the work done by the system on the environment is ∫PdV. This can be represented as the area under a curve in a Pressurevolume diagram: 0-4 I P P II V1 V2 ≠ W1-2 W1-2 I II V Volume Work depends on the path Work is area under curve of P(V) a) b) c) d) e) Work is path dependent; Properties only depend on states; Work is not a property, not a state variable; When we say W1-2, the work between states 1 and 2, we need to specify the path; For irreversible (non-reversible) processes, we cannot use ∫PdV; either the work must be given or it must be found by another method. Muddy points How do we know when work is done? (MP 0.3) 10) Heat Heat is energy transferred due to temperature differences. a) b) c) d) e) Heat transfer can alter system states; Bodies don't "contain" heat; heat is identified as it comes across system boundaries; The amount of heat needed to go from one state to another is path dependent; Heat and work are different modes of energy transfer; Adiabatic processes are ones in which no heat is transferred. 11) First Law of Thermodynamics For a system, ∆E = Q − W E is the energy of the system, Q is the heat input to the system, and W is the work done by the system. E = U (thermal energy) + Ekinetic + Epotential + .... If changes in kinetic and potential energy are not important, ∆ U = Q −W . a) U arises from molecular motion. b) U is a function of state, and thus ∆U is a function of state (as is ∆E ). c) Q and W are not functions of state. Comparing (b) and (c) we have the striking result that: d) ∆U is independent of path even though Q and W are not! 0-5 Muddy points What are the conventions for work and heat in the first law? (MP 0.4) When does E->U? (MP 0.5) 12) Enthalpy: A useful thermodynamic property, especially for flow processes, is the enthalpy. Enthalpy is usually denoted by H, or h for enthalpy per unit mass, and is defined by: H = U + PV. In terms of the specific quantities, the enthalpy per unit mass is h = u + Pv = u + P / ρ . 13) Specific heats - relation between temperature change and heat input For a change in state between two temperatures, the “specific heat” is: Specific heat = Q/(Tfinal - Tinitial) We must, however, specify the process, i.e., the path, for the heat transfer. Two useful processes are constant pressure and constant volume. The specific heat at constant pressure is denoted as Cp and that at constant volume as Cv, or cp and cv per unit mass. ∂h ∂u c p = and cv = ∂T p ∂T v For an ideal gas and du = cvdT. dh = cpdT The ratio of specific heats, cp/cv is denoted by γ. This ratio is 1.4 for air at room conditions. The specific heats cv. and cp have a basic definition as derivatives of the energy and enthalpy. Suppose we view the internal energy per unit mass, u, as being fixed by specification of T, the temperature and v, the specific volume, i.e., the volume per unit mass. (For a simple compressible substance, these two variables specify the state of the system.) Thus, u = u(T,v). The difference in energy between any two states separated by small temperature and specific volume differences, dT and dv is du = ∂u dT + ∂u dv ∂T v ∂v T The derivative ∂u ∂T v represents the slope of a line of constant v on a u-T plane. The derivative is also a function of state, i. e., a thermodynamic property, and is called the specific heat at constant volume, cv. The name specific heat is perhaps unfortunate in that only for special circumstances is the derivative related to energy transfer as heat. If a process is carried out slowly at constant volume, no work will be done and any energy increase will be due only to energy transfer as heat. For such a 0-6 process, cv does represent the energy increase per unit of temperature (per unit of mass) and consequently has been called the "specific heat at constant volume". However, it is more useful to think of cv in terms of its definition as a certain partial derivative, which is a thermodynamic property, rather than a quantity related to energy transfer as heat in the special constant volume process. The enthalpy is also a function of state. For a simple compressible substance we can regard the enthalpy as a function of T and P, that is view the temperature and pressure as the two variables that define the state. Thus, h = h(T,P). Taking the differential, dh = ∂h dT + ∂h dP ∂T P ∂P T The derivative ∂h ∂T P is called the specific heat at constant pressure, denoted by cp. The derivatives cv and cp constitute two of the most important thermodynamic derivative functions. Values of these properties have been experimentally determined as a function of the thermodynamic state for an enormous number of simple compressible substances. 14) Ideal Gases The equation of state for an ideal gas is PV = NRT, where N is the number of moles of gas in the volume V. Ideal gas behavior furnishes an extremely good approximation to the behavior of real gases for a wide variety of aerospace applications. It should be remembered, however, that describing a substance as an ideal gas constitutes a model of the actual physical situation, and the limits of model validity must always be kept in mind. One of the other important features of an ideal gas is that its internal energy depends only upon its temperature. (For now, this can be regarded as another aspect of the model of actual systems that the perfect gas represents, but it can be shown that this is a consequence of the form of the equation of state.) Since u depends only on T, du = cv (T)dT In the above equation we have indicated that cv can depend on T. Like the internal energy, the enthalpy is also only dependent on T for an ideal gas. (If u is a function of T, then, using the perfect gas equation of state, u + Pv is also.) Therefore, dh = cP(T)dT. Further, dh = du + d(Pv) = cv dT + R dT. Hence, for an ideal gas, cv = cP - R. In general, for other substances, u and h depend on pressure as well as on temperature. In this respect, the ideal gas is a very special model. 0-7 The specific heats do not vary greatly over wide ranges in temperature, as shown in VWB&S Figure 5.11. It is thus often useful to treat them as constant. If so u2 - u1 = cv (T2 - T1) h2 - h1 = cp (T2 - T1) These equations are useful in calculating internal energy or enthalpy differences, but it should be remembered that they hold only for an ideal gas with constant specific heats. In summary, the specific heats are thermodynamic properties and can be used even if the processes are not constant pressure or constant volume. The simple relations between changes in energy (or enthalpy) and temperature are a consequence of the behavior of an ideal gas, specifically the dependence of the energy and enthalpy on temperature only, and are not true for more complex substances. Adapted from "Engineering Thermodynamics", Reynolds, W. C and Perkins, H. C, McGraw-Hill Publishers 15) Specific Heats of an Ideal Gas 1. All ideal gases: (a) (b) (c) The specific heat at constant volume (cv for a unit mass or CV for one kmol) is a function of T only. The specific heat at constant pressure (cp for a unit mass or CP for one kmol) is a function of T only. A relation that connects the specific heats cp, cv, and the gas constant is cp - cv = R where the units depend on the mass considered. For a unit mass of gas, e. g., a kilogram, cp and cv would be the specific heats for one kilogram of gas and R is as defined above. For one kmol of gas, the expression takes the form: CP - CV = R, (d) where CP and CV have been used to denote the specific heats for one kmol of gas and R is the universal gas constant. The specific heat ratio, γ, = cp/cv (or CP/CV), is a function of T only and is greater than unity. 2. Monatomic gases, such as He, Ne, Ar, and most metallic vapors: (a) (b) (c) cv (or CV) is constant over a wide temperature range and is very nearly equal to (3/2)R [or (3/2)R , for one kmol]. cp (or CP) is constant over a wide temperature range and is very nearly equal to (5/2)R [or (5/2)R , for one kmol]. γ is constant over a wide temperature range and is very nearly equal to 5/3 [γ = 1.67]. 0-8 3. So-called permanent diatomic gases, namely H2, O2, N2, Air, NO, and CO: (a) (b) (c) cv (or CV ) is nearly constant at ordinary temperatures, being approximately (5/2)R [(5/2)R , for one kmol], and increases slowly at higher temperatures. cp (or CP ) is nearly constant at ordinary temperatures, being approximately (7/2)R [(7/2)R , for one kmol], and increases slowly at higher temperatures. γ is constant over a temperature range of roughly 150 to 600K and is very nearly equal to 7/5 [γ = 1.4]. It decreases with temperature above this. 4. Polyatomic gases and gases that are chemically active, such as CO2, NH3, CH4, and Freons: The specific heats, cv and cp, and γ vary with the temperature, the variation being different for each gas. The general trend is that heavy molecular weight gases (i.e., more complex gas molecules than those listed in 2 or 3), have values of γ closer to unity than diatomic gases, which, as can be seen above, are closer to unity than monatomic gases. For example, values of γ below 1.2 are typical of Freons which have molecular weights of over one hundred. Adapted from Zemansky, M. W. and Dittman, R. H., "Heat and Thermodynamics", Sixth Edition, McGraw-Hill book company, 1981 16) Reversible adiabatic processes for an ideal gas From the first law, with Q = 0, du = cvdT, and Work = Pdv du + Pdv = 0 Also, using the definition of enthalpy dh = du + Pdv + vdP. (i) (ii) The underlined terms are zero for an adiabatic process. Re-writing (i) and (ii), γ cvdT = - γPdv cpdT = vdP. Combining the above two equations we obtain -γ Pdv = vdP or -γ dv/v = dP/P (iii) Equation (iii) can be integrated between states 1 and 2 to give γln(v2/v1) = ln(P2/P1), or, equivalently, (P2v2γ )( / P1v1γ )= 1 For an ideal gas undergoing a reversible, adiabatic process, the relation between pressure and volume is thus: Pv γ = constant, or P = constant × ρ γ . 17) Examples of flow problems and the use of enthalpy a) Adiabatic, steady, throttling of a gas (flow through a valve or other restriction) Figure 0-1 shows the configuration of interest. We wish to know the relation between properties upstream of the valve, denoted by “1” and those downstream, denoted by “2”. 0-9 P1, v1, u1 ... P2, v2, u2 ... Valve Figure 0-1: Adiabatic flow through a valve, a generic throttling process To analyze this situation, we can define the system (choosing the appropriate system is often a critical element in effective problem solving) as a unit mass of gas in the following two states. Initially the gas is upstream of the valve and just through the valve as indicated. In the final state the gas is downstream of the valve plus just through the valve. The figures on the left show the actual configuration just described. In terms of the system behavior, however, we could replace the fluid external to the system by pistons which exert the same pressure that the external fluid exerts, as indicated schematically on the right side of Figure 0-2 below. Initial state = system pistons Final state = pistons system Figure 0-2: Equivalence of actual system and piston model The process is adiabatic, with changes in potential energy and kinetic energy assumed to be negligible. The first law for the system is therefore ∆ U = −W . The work done by the system is W = P2V 2 − P1V1 . Use of the first law leads to U 2 + P2 V2 = U1 + P1V1 . In words, the initial and final states of the system have the same value of the quantity U+PV. For the case examined, since we are dealing with a unit mass, the initial final states of the system have the same value of u+Pv. 0-10 Muddy points When is enthalpy the same in initial and final states? (MP 0.6) b) Another example of a flow process, this time for an unsteady flow, is the transient process of filling a tank, initially evacuated, from a surrounding atmosphere, which is at a pressure P0 and a temperature T0 . The configuration is shown in Figure 0-3. P0 ,T0 Vacuum System (all the gas that goes into the tank) P0 Valve V0 Figure 0-3: A transient problem—filling of a tank from the atmosphere At a given time, the valve at the tank inlet is opened and the outside air rushes in. The inflow stops when the pressure inside is equal to the pressure outside. The tank is insulated, so there is no heat transfer to the atmosphere. What is the final temperature of the gas in the tank? This time we take the system to be all the gas that enters the tank. The initial state has the system completely outside the tank, and the final state has the system completely inside the tank. The kinetic energy initially and in the final state is negligible, as is the change in potential energy so the first law again takes the form, ∆ U = −W . Work is done on the system, of magnitude P0V0 , where V0 is the initial volume of the system, so ∆ U = P0V 0 . In terms of quantities per unit mass (∆ U = m∆u ,V 0 = mv 0 , where m is the mass of the system), ∆ u = u final − u i = P0 v 0 . The final value of the internal energy is u final = ui + P0 v0 = hi = h0 . For a perfect gas with constant specific heats, u = c vT ; h = c p T , c vT final = c p T0 , 0-11 cp T0 = γT0 . cv The final temperature is thus roughly 200oF hotter than the outside air! Tfinal = It may be helpful to recap what we used to solve this problem. There were basically four steps: 1 Definition of the system 2 Use of the first law 3 Equating the work to a “PdV” term 4 Assuming the fluid to be a perfect gas with constant specific heats. A message that can be taken from both of these examples (as well as from a large number of other more complex situations, is that the quantity h = u + Pv occurs naturally in problems of fluid flow. Because the combination appears so frequently, it is not only defined but also tabulated as a function of temperature and pressure for a number of working fluids. Muddy points In the filling of a tank, why (physically) is the final temperature in the tank higher than the initial temperature? (MP 0.7) 18) Control volume form of the system laws (Waitz pp 32-34, VWB&S, 6.1, 6.2) The thermodynamic laws (as well as Newton’s laws) are for a system, a specific quantity of matter. More often, in propulsion and power problems, we are interested in what happens in a fixed volume, for example a rocket motor or a jet engine through which mass is flowing. For this reason, the control volume form of the system laws is of great importance. A schematic of the difference is shown below. Rather than focus on a particle of mass which moves through the engine, it is more convenient to focus on the volume occupied by the engine. This requires us to use the control volume form of the thermodynamic laws. Engine System at ti System at time tf Figure 0-4: Control volume and system for flow through a propulsion device The first of these is conservation of mass. For the control volume shown, the rate of change of mass inside the volume is given by the difference between the mass flow rate in and the mass flow rate out. For a single flow coming in and a single flow coming out this is • • dmCV = min − mout . dt If the mass inside the control volume changes with time it is because some mass is added or some is taken out. The first law of thermodynamics can be written as a rate equation: dE • • = Q− W . dt To derive the first law as a rate equation for a control volume we proceed as with the mass conservation equation. The physical idea is that any rate of change of energy in the control 0-12 volume must be caused by the rates of energy flow into or out of the volume. The heat transfer and the work are already included and the only other contribution must be associated with the mass flow in and out, which carries energy with it. The figure below shows a schematic of this idea. m⋅ i Pi Ti vi ei ⋅ ⋅ Wboundary Wshaft dEcv dt ⋅ Q m⋅ e Pe Te ve ee Figure 0-5: Schematic diagram illustrating terms in the energy equation for a control volume The fluid that enters or leaves has an amount of energy per unit mass given by e = u + c 2 / 2 + gz , where c is the fluid velocity. In addition, whenever fluid enters or leaves a control volume there is a work term associated with the entry or exit. We saw this in example 16a, and the present derivation is essentially an application of the ideas presented there. Flow exiting at station “e” must push back the surrounding fluid, doing work on it. Flow entering the volume at station “i” is pushed on by, and receives work from the surrounding air. The rate of flow work at exit is given by the product of the pressure times the exit area times the rate at which the external flow is “pushed back”. The last of these, however, is equal to the volume per unit mass times the rate of mass flow. Put another way, in a time dt, the work done on the surroundings by the flow at the exit station is dW flow = Pvdme . The net rate of flow work is W flow = Pe v e m& e − Pi vi m& i . Including all possible energy flows (heat, shaft work, shear work, piston work etc.), the first law can then be written as: d ECV = ∑ Q& + ∑W& shaft + ∑W& shear + ∑ W& piston + ∑W& flow + ∑ m& (u + 12 c 2 + gz) ∑ dt 0-13 where Σ includes the sign associated with the energy flow. If heat is added or work is done on the system then the sign is positive, if work or heat are extracted from the system then the sign is negative. NOTE: this is consistent with ∆E = Q – W, where W is the work done by the system on the environment, thus work is flowing out of the system. We can then collect the specific energy term e included in Ecv and the specific flow term Pv to make the enthalpy appear: 2 2 Total energy associated with mass flow: e + Pv = u + c / 2 + gz + Pv = h + c / 2 + gz = ht , where ht is the stagnation enthalpy (IAW, p.36). Thus, the first law can be written as: d dt ∑E CV = ∑ Q& + ∑ W& shaft + ∑ W& shear + ∑W& piston + ∑ m& ( h + 12 c 2 + gz) . For most of the applications done in this course, there will be no shear work and no piston work. Hence, the first law for a control volume will be most often used as: ( ) ( ) • • • • dE CV 2 2 = Q CV − W shaft + m i hi + ci / 2 + gzi − m e he + ce / 2 + gze . dt The rate of work term is the sum of the shaft work and the flow work. In writing the control volume form of the equation we have assumed only one entering and one leaving stream, but this could be generalized to any number of inlet and exit streams. Muddy points What distinguishes shaft work from other works? (MP 0.8) For problems of interest in aerospace applications the velocities are high and the term that is associated with changes in the elevation is small. From now on, we will neglect this term unless explicitly stated. The control volume form of the first law is thus ( ) ( • • • • dE CV 2 2 = Q CV − W shaft + mi hi + ci / 2 − m e he + c e / 2 dt • • • ) • = QCV − W shaft + m i hti − me hte . For steady flow (d/dt = 0) the inlet and exit mass flow rates are the same and the control volume form of the first law becomes the “Steady Flow Energy Equation” (SFEE) • Steady Flow Energy Equation: ( ) • • m hte − hti = QCV − W shaft . The steady flow energy equation finds much use in the analysis of power and propulsion devices and other fluid machinery. Note the prominent role of enthalpy. 0-14 Using what we have just learned we can attack the tank filling problem solved in (16b) from an alternate point of view using the control volume form of the first law. In this problem the shaft work is zero, and the heat transfer, kinetic energy changes, and potential energy changes are neglected. In addition there is no exit mass flow. control volume • control surface m (mass flow) Figure 0-6: A control volume approach to the tank filling problem The control volume form of the first law is therefore • dU = mi hi . dt The equation of mass conservation is dm • = mi . dt Combining we have dU dm hi . = dt dt Integrating from the initial time to the final time (the incoming enthalpy is constant) and using U = mu gives the result u final = hi = h0 as before. Muddy points Definition of a control volume (MP 0.9) What is the difference between enthalpy and stagnation enthalpy? (MP 0.10) 0-15 Muddiest Points on Part 0 0.1 Specific properties Energy, volume, enthalpy are all extensive properties. Their value depends not only on the temperature and pressure but also on “how much”, i.e., what the mass of the system is. The internal energy of two kilograms of air is twice as much as the internal energy of one kilogram of air. It is very often useful to work in terms of properties that do not depend on the mass of the system, and for this purpose we use the specific volume, specific energy, specific enthalpy, etc., which are the values of volume, energy, and enthalpy for a unit mass (kilogram) of the substance. For a system of mass m, the relations between the two quantities are: V = mv ; U = mu ; H = mh 0.2 What is the difference between extensive and intensive properties? Intensive properties are properties that do not depend on the quantity of matter. For example, pressure and temperature are intensive properties. Energy, volume and enthalpy are all extensive properties. Their value depends on the mass of the system. For example, the enthalpy of a certain mass of a gas is doubled if the mass is doubled; the enthalpy of a system that consists of several parts is equal to the sum of the enthalpies of the parts. 0.3 How do we know when work is done? A rigorous test for whether work is done or not is whether a weight could have been raised in the process under consideration. I will hand out some additional material to supplement the notes on this point, which seems simple, but can be quite subtle to unravel in some situations. 0.4 What are the conventions for work and heat in the first law? Heat is positive if it is given to the system. Work is positive if it is done by the system. 0.5 When does E->U? We deal with changes in energy. When the changes in the other types of energy (kinetic, potential, strain, etc) can be neglected compared to the changes in thermal energy, then it is a good approximation to use ∆U as representing the total energy change. 0.6 When is enthalpy the same in initial and final states? Initial and final stagnation enthalpy is the same if the flow is steady and if there is no net shaft work plus heat transfer. If the change in kinetic energy is negligible, the initial and final enthalpy is the same. The “tank problem” is unsteady so the initial and final enthalpies are not the same. See the discussion of steady flow energy equation in notes [(17) in Section 0]. 0.7 In the filling of a tank, why (physically) is the final temperature in the tank higher than the initial temperature? Work is done on the system, which in this problem is the mass of gas that is pushed into the tank. 0.8 What distinguishes shaft work from other works? The term shaft work arises in using a control volume approach. As we have defined it, “shaft work” is all work over and above work associated with the “flow work” (the work done by pressure forces). Generally this means work done by rotating machinery, which is carried by a shaft from the control volume to the outside world. There could also be work over and above the pressure force work done by shear stresses at the boundaries of the control volume, but this is seldom important if the control boundary is normal to the flow direction. If we consider a system (a mass of fixed identity, say a blob of gas) flowing through some device, neglecting the effects of raising or lowering the blob the only mode of work would be the work to compress the blob. This would be true even if the blob were flowing through a turbine or compressor. (In doing this we are focusing on the same material as it undergoes the unsteady compression or expansion processes in the device, rather than looking at a control volume, through which mass passes.) The question about shaft work and non shaft work has been asked several times. I am not sure how best to answer, but it appears that the difficulty people are having might be associated with being able to know when one can say that shaft work occurs. There are several features of a process that produces ( or absorbs) shaft work. First of all, the view taken of the process is one of control volume, rather than control mass (see the discussion of control volumes in section 0 or in IAW). Second, there need to be a shaft or equivalent device ( a moving belt, a row of blades) that can be identified as the work carrier. Third, the shaft work is work over and above the flow work that is done by (or received by) the streams that exit and enter the control volume. 0.9 Definition of a control volume. A control volume is an enclosure that separates a quantity of matter from the surroundings or environment. The enclosure does not necessarily have to consist of a solid boundary like the walls of a vessel. It is only necessary that the enclosure forms a closed surface and that its properties are defined everywhere. An enclosure may transmit heat or be a heat insulator. It may be deformable and thus capable of transmitting work to the system. It may also be capable of transmitting mass. PART 1 THE SECOND LAW OF THERMODYNAMICS PART 1 - THE SECOND LAW OF THERMODYNAMICS 1.A. Background to the Second Law of Thermodynamics [IAW 23-31 (see IAW for detailed VWB&S references); VN Chapters 2, 3, 4] 1.A.1 Some Properties of Engineering Cycles; Work and Efficiency As motivation for the development of the second law, we examine two types of processes that concern interactions between heat and work. The first of these represents the conversion of work into heat. The second, which is much more useful, concerns the conversion of heat into work. The question we will pose is how efficient can this conversion be in the two cases. i F + R Block on rough surface Viscous liquid Resistive heating Figure A-1: Examples of the conversion of work into heat Three examples of the first process are given above. The first is the pulling of a block on a rough horizontal surface by a force which moves through some distance. Friction resists the pulling. After the force has moved through the distance, it is removed. The block then has no kinetic energy and the same potential energy it had when the force started to act. If we measured the temperature of the block and the surface we would find that it was higher than when we started. (High temperatures can be reached if the velocities of pulling are high; this is the basis of inertia welding.) The work done to move the block has been converted totally to heat. The second example concerns the stirring of a viscous liquid. There is work associated with the torque exerted on the shaft turning through an angle. When the stirring stops, the fluid comes to rest and there is (again) no change in kinetic or potential energy from the initial state. The fluid and the paddle wheels will be found to be hotter than when we started, however. The final example is the passage of a current through a resistance. This is a case of electrical work being converted to heat, indeed it models operation of an electrical heater. All the examples in Figure A-1 have 100% conversion of work into heat. This 100% conversion could go on without limit as long as work were supplied. Is this true for the conversion of heat into work? To answer the last question, we need to have some basis for judging whether work is done in a given process. One way to do this is to ask whether we can construct a way that the process could result in the raising of a weight in a gravitational field. If so, we can say “Work has been done”. It may sometimes be difficult to make the link between a complicated thermodynamic process and the simple raising of a weight, but this is a rigorous test for the existence of work. One example of a process in which heat is converted to work is the isothermal (constant temperature) expansion of an ideal gas, as sketched in the figure. The system is the gas inside the chamber. As the gas expands, the piston does work on some external device. For an ideal gas, the internal energy is a function of temperature only, so that if the temperature is constant for some process the internal energy change is zero. To keep the temperature constant during the expansion, 1A-1 heat must be supplied. Because ∆U = 0, the first law takes the form Q=W. This is a process that has 100% conversion of heat into work. P, T The work exerted by the system is given by Patm Work received, W 2 Work = ∫ PdV 1 where 2 and 1 denote the two states at the beginning and end of the process. The equation of state for an ideal gas is Q P = NRT/V, with N the number of moles of the gas contained in the chamber. Using the equation of state, the expression for work can be written as 2 V (A.1.1) Work during an isothermal expansion = NRT ∫ dV / V = NRT ln 2 . V1 1 For an isothermal process, PV = constant, so that P1 / P2 = V2 / V1 . The work can be written in terms of the pressures at the beginning and end as P Work during an isothermal expansion = NRT ln 1 . (A.1.2) P2 The lowest pressure to which we can expand and still receive work from the system is atmospheric pressure. Below this, we would have to do work on the system to pull the piston out further. There is thus a bound on the amount of work that can be obtained in the isothermal expansion; we cannot continue indefinitely. For a power or propulsion system, however, we would like a source of continuous power, in other words a device that would give power or propulsion as long as fuel was added to it. To do this, we need a series of processes where the system does not progress through a one-way transition from an initial state to a different final state, but rather cycles back to the initial state. What is looked for is in fact a thermodynamic cycle for the system. We define several quantities for a cycle: QA is the heat absorbed by the system QR is the heat rejected by the system W is the net work done by the system. The cycle returns to its initial state, so the overall energy change, ∆U , is zero. The net work done by the system is related to the magnitudes of the heat absorbed and the heat rejected by W = Net work = QA − QR . The thermal efficiency of the cycle is the ratio of the work done to the heat absorbed. (Efficiencies are often usefully portrayed as “What you get” versus “What you pay for”. Here what we get is work and what we pay for is heat, or rather the fuel that generates the heat.) In terms of the heat absorbed and rejected, the thermal efficiency is: η = thermal efficiency = = QA − QR Q = 1− R . QA QA Work done Heat absorbed (A.1.3) 1A-2 The thermal efficiency can only be 100% (complete conversion of heat into work) if QR = 0 , and a basic question is what is the maximum thermal efficiency for any arbitrary cycle? We examine this for two cases, the Carnot cycle and the Brayton (or Joule) cycle which is a model for the power cycle in a jet engine. 1.A.2 Carnot Cycles A Carnot cycle is shown below. It has four processes. There are two adiabatic reversible legs and two isothermal reversible legs. We can construct a Carnot cycle with many different systems, but the concepts can be shown using a familiar working fluid, the ideal gas. The system can be regarded as a chamber filled with this ideal gas and with a piston. a 3 Q2 2 1 P 4 b d Q1 T2 c T1 T2 Q2 Reservoir T1 Insulating stand Q1 Reservoir V Figure A-2: Carnot cycle – thermodynamic diagram on left and schematic of the different stages in the cycle for a system composed of an ideal gas on the right The four processes in the Carnot cycle are: 1) The system is at temperature T2 at state (a). It is brought in contact with a heat reservoir, which is just a liquid or solid mass of large enough extent such that its temperature does not change appreciably when some amount of heat is transferred to the system. In other words, the heat reservoir is a constant temperature source (or receiver) of heat. The system then undergoes an isothermal expansion from a to b, with heat absorbed Q2 . 2) At state b, the system is thermally insulated (removed from contact with the heat reservoir) and then let expand to c. During this expansion the temperature decreases to T1. The heat exchanged during this part of the cycle, Qbc = 0. 3) At state c the system is brought in contact with a heat reservoir at temperature T1. It is then compressed to state d, rejecting heat Q1 in the process. 4) Finally, the system is compressed adiabatically back to the initial state a. The heat exchange Qda = 0 . The thermal efficiency of the cycle is given by the definition η = 1− QR Q = 1+ 1 . QA Q2 (A.2.1) In this equation, there is a sign convention implied. The quantities Q A ,Q R as defined are the magnitudes of the heat absorbed and rejected. The quantities Q1,Q 2 on the other hand are defined with reference to heat received by the system. In this example, the former is negative and the latter is positive. The heat absorbed and rejected by the system takes place during isothermal processes and we already know what their values are from Eq. (A.1.1): 1A-3 [ ] Q1 = Wcd = NRT [ln(Vd / Vc )] = - [ln(Vc / Vd )]. Q2 = Wab = NRT2 ln(Vb / Va ) 1 ( Q1 is negative.) The efficiency can now be written in terms of the volumes at the different states as: η = 1+ T1 [ ln( Vd / Vc ) ] T2 [ ln( Vb / Va ) ] . (A.2.2) The path from states b to c and from a to d are both adiabatic and reversible. For a reversible adiabatic process we know that PV γ = constant. Using the ideal gas equation of state, we have TV γ −1 = constant. Along curve b-c, therefore T2 Vbγ −1 = T1Vcγ −1 . Along the curve d-a, T2 Vaγ −1 = T1Vdγ −1 . Thus, γ −1 γ −1 T2 / T1 ) Va Vd ( Vd Va = = , or Vd / Vc = Va / Vb . , which means that Vc Vb Vc (T2 / T1 ) Vb Comparing the expression for thermal efficiency Eq. (A.2.1) with Eq. (A.2.2) shows two consequences. First, the heats received and rejected are related to the temperatures of the isothermal parts of the cycle by Q1 Q2 + = 0. (A.2.3)) T1 T2 Second, the efficiency of a Carnot cycle is given compactly by T ηc = 1 − 1 . Carnot cycle efficiency (A.2.4) T2 The efficiency can be 100% only if the temperature at which the heat is rejected is zero. The heat and work transfers to and from the system are shown schematically in Figure A-3. Q2 T2 System W (net work) T1 Q1 Figure A-3: Work and heat transfers in a Carnot cycle between two heat reservoirs 1A-4 Muddy points T Since η = 1 − 1 , looking at the P-V graph, does that mean the farther apart the T1, T2 T2 isotherms are, the greater efficiency? And that if they were very close, it would be very inefficient? (MP 1A.1) In the Carnot cycle, why are we only dealing with volume changes and not pressure changes on the adiabats and isotherms? (MP 1A.2) Is there a physical application for the Carnot cycle? Can we design a Carnot engine for a propulsion device? (MP 1A.3) How do we know which cycles to use as models for real processes? (MP 1A.4) 1.A.3 Brayton Cycles (or Joule Cycles): The Power Cycle for a Gas Turbine Jet Engine For a Brayton cycle there are two adiabatic legs and two constant pressure legs. Sketches of an engine and the corresponding cycle are given in Figure A-4. Combustor Combustor q2 PCompressor exit Inlet Compressor b c Turbine and nozzle P Nozzle Patm Turbine Inlet and compressor a q1 V d Heat rejection to atmosphere Figure A-4: Sketch of the jet engine components and corresponding thermodynamic states Gas turbines are also used for power generation and for closed cycle operation (for example for space power generation). A depiction of the cycle in this case is shown in Figure A-5. 1A-5 ⋅ Q Equivalent heat transfer at constant pressure 2 3 ⋅ ⋅ Wcomp Compressor Wnet Turbine 1 4 ⋅ Q Equivalent heat transfer at constant pressure Figure A-5: Thermodynamic model of gas turbine engine cycle for power generation The objective now is to find the work done, the heat absorbed, and the thermal efficiency of the cycle. Tracing the path shown around the cycle from a-b-c-d and back to a, the first law gives (writing the equation in terms of a unit mass), ∆ua −b −c − d − a = 0 = q2 + q1 − w . The net work done is w = q2 + q1 , where q1 , q2 are defined as heat received by the system ( q1 is negative). We thus need to evaluate the heat transferred in processes b-c and d-a. For a constant pressure process the heat exchange per unit mass is dh = c p dT = dq , or [dq]cons tan t P = dh . The heat exchange can be expressed in terms of enthalpy differences between the relevant states. Treating the working fluid as an ideal gas, for the heat addition from the combustor, q2 = hc − hb = c p (Tc − Tb ) . The heat rejected is, similarly, q1 = ha − hd = c p (Ta − Td ) . The net work per unit mass is given by [ ] Net work per unit mass = q1 + q2 = c p (Tc − Tb ) + (Ta − Td ) . The thermal efficiency of the Brayton cycle can now be expressed in terms of the temperatures: 1A-6 [ ] Net work c p (Tc − Tb ) − (Td − Ta ) = Heat in c p [Tc − Tb ] (T − Ta ) = 1 − Ta (Td / Ta − 1) . = 1− d Tb (Tc / Tb − 1) (Tc − Tb ) η= (A.3.1) To proceed further, we need to examine the relationships between the different temperatures. We know that points a and d are on a constant pressure process as are points b and c, and Pa = Pd ; Pb = Pc . The other two legs of the cycle are adiabatic and reversible, so Pd Pa = Pc Pb == > Td Tc γ / (γ −1) T = a Tb γ / (γ −1) . Td Ta T T = , or, finally, d = c . Using this relation in the expression for thermal Tc Tb Ta Tb efficiency, Eq. (A.1.3) yields an expression for the thermal efficiency of a Brayton cycle: T (A.3.2) Ideal Brayton cycle efficiency: η B = 1 − a Tb Tatmospheric = 1− . Tcompressor exit Therefore The temperature ratio across the compressor, Tb / Ta = TR . In terms of compressor temperature ratio, and using the relation for an adiabatic reversible process we can write the efficiency in terms of the compressor (and cycle) pressure ratio, which is the parameter commonly used: ηB = 1 − 1 1 = 1− . TR ( PR)(γ −1) / γ (A.3.3) Figure A-6 shows pressures and temperatures through a gas turbine engine (the afterburning J57, which powers the F-8 and the F-101). Figure A-6: Gas turbine engine pressures and temperatures 1A-7 Overall Pressure Ratio (OPR), Sea Level, T-O Equation (A.3.3) says that for a high cycle efficiency, the pressure ratio of the cycle should be increased. Figure A-7 shows the history of aircraft engine pressure ratio versus entry into service, and it can be seen that there has been a large increase in cycle pressure ratio. The thermodynamic concepts apply to the behavior of real aerospace devices! Trent 890 GE90 40 30 20 Trent 775 CF6-80C2A8 CF6-80E1A4 CF6-80C2A8 CFM56-5C4 PW4084 PW4052 PW4168 RB211-524D4 CF6-50E CF6-50A RB211-22 JT9D-7R4G TF39-1 JT9D-70 CF6-6 CFM56-2 JT9D-3A Spey 512-14 Spey 512 JT8D-17 JT8D-1 Spey 505 Conway 508 JT3D 10 0 1960 CFM56-5B CFM56-3C JT8D-219 Tay 611 Spey 555 Tay 651 Conway 550 1970 1980 Year of Certification 1990 2000 Figure A-7: Gas turbine engine pressure ratio trends (Jane’s Aeroengines, 1998) Muddy points When flow is accelerated in a nozzle, doesn’t that reduce the internal energy of the flow and therefore the enthalpy? (MP 1A.5) Why do we say the combustion in a gas turbine engine is constant pressure? (MP 1A.6) Why is the Brayton cycle less efficient than the Carnot cycle? (MP 1A.7) If the gas undergoes constant pressure cooling in the exhaust outside the engine, is that still within the system boundary ? (MP 1A.8) Does it matter what labels we put on the corners of the cycle or not? (MP 1A.9) Is the work done in the compressor always equal to the work done in the turbine plus work out (for a Brayton cyle) ? (MP 1A.10) 1.A.4 Gas Turbine Technology and Thermodynamics The turbine entry temperature, Tc , is fixed by materials technology and cost. (If the temperature is too high, the blades fail.) Figures A-8 and A-9 show the progression of the turbine entry temperatures in aeroengines. Figure A-8 is from Rolls Royce and Figure A-9 is from Pratt&Whitney. Note the relation between the gas temperature coming into the turbine blades and the blade melting temperature. 1A-8 Rotor inlet gas temperature vs Cooling effectiveness. Figure A-8: Rolls-Royce high temperature technology Figure A-9: Turbine blade cooling technology [Pratt & Whitney] For a given level of turbine technology (in other words given maximum temperature) a design question is what should the compressor TR be? What criterion should be used to decide this? Maximum thermal efficiency? Maximum work? We examine this issue below. Tb2 Cycle with Tb → Tc Tb1 P Cycle with lower Tb Patm T = Tc T = Tatm = Ta V Figure A-10: Efficiency and work of two Brayton cycle engines The problem is posed in Figure A-10, which shows two Brayton cycles. For maximum efficiency we would like TR as high as possible. This means that the compressor exit temperature approaches the turbine entry temperature. The net work will be less than the heat received; as Tb → Tc the heat received approaches zero and so does the net work. The net work in the cycle can also be expressed as ∫ Pdv , evaluated in traversing the cycle. This is the area enclosed by the curves, which is seen to approach zero as Tb → Tc . 1A-9 The conclusion from either of these arguments is that a cycle designed for maximum thermal efficiency is not very useful in that the work (power) we get out of it is zero. A more useful criterion is that of maximum work per unit mass (maximum power per unit mass flow). This leads to compact propulsion devices. The work per unit mass is given by: [ ] Work / unit mass = c p (Tc − Tb ) − (Td − Ta ) Max. turbine temp. (Design constraint) Atmospheric temperature The design variable is the compressor exit temperature, Tb , and to find the maximum as this is varied, we differentiate the expression for work with respect to Tb : dT dT dT dWork = cp c −1 − d + a . dTb dTb dTb dTb The first and the fourth term on the right hand side of the above equation are both zero (the turbine entry temperature is fixed, as is the atmospheric temperature). The maximum work occurs where the derivative of work with respect to Tb is zero: dT dWork = 0 = −1 − d . (A.4.1) dTb dTb To use Eq. (A.4.1), we need to relate Td and Tb . We know that Td Tc TT = or Td = a c . Ta Tb Tb Hence, dTd − Ta Tc = . dTb Tb2 Plugging this expression for the derivative into Eq. (A.4.1) gives the compressor exit temperature for maximum work as Tb = Ta Tc . In terms of temperature ratio, Tb T = c . Ta Ta The condition for maximum work in a Brayton cycle is different than that for maximum efficiency. The role of the temperature ratio can be seen if we examine the work per unit mass which is delivered at this condition: TT Work / unit mass = c p Tc − Ta Tc − a c + Ta . Ta Tc Compressor temperature ratio for maximum work: Ratioing all temperatures to the engine inlet temperature, T T Work / unit mass = c p Ta c − 2 c + 1 . Ta Ta To find the power the engine can produce, we need to multiply the work per unit mass by the mass flow rate: 1A-10 • T T Power = m c p Ta c − 2 c + 1 ; Maximum power for an ideal Brayton cycle Ta Ta kg J J K = = Watts .) (The units are s kg - K s (A.4.2) Figures A-11a. and A-11b. available from: B.L. Koff Spanning the Globe with Jet Propulsion AIAA Paper 2987, AIAA Annual Meeting and Exhibit, 1991. C. E. Meece, Gas Turbine Technologies of the Future, International Symposium on Air Breathing Engines, 1995, paper 95-7006. a) Gas turbine engine core b) Core power vs. turbine entry temperature Figure A-11: Aeroengine core power [Koff/Meese, 1995] Figure A-11 shows the expression for power of an ideal cycle compared with data from actual jet engines. Figure 11a shows the gas turbine engine layout including the core (compressor, burner, and turbine). Figure 11b shows the core power for a number of different engines as a function of the turbine rotor entry temperature. The equation in the figure for horsepower (HP) is the same as that we just derived, except for the conversion factors. The analysis not only shows the qualitative trend very well but captures much of the quantitative behavior too. A final comment (for now) on Brayton cycles concerns the value of the thermal efficiency. The Brayton cycle thermal efficiency contains the ratio of the compressor exit temperature to atmospheric temperature, so that the ratio is not based on the highest temperature in the cycle, as the Carnot efficiency is. For a given maximum cycle temperature, the Brayton cycle is therefore less efficient than a Carnot cycle. Muddy points • What are the units of w in power = m w ? (MP 1A.11) Precision about the assumptions made in the Brayton cycle for maximum efficiency and maximum work (MP 1A.12) You said that for a gas turbine engine modeled as a Brayton cycle the work done is w=q1 +q2 , where q2 is the heat added and q1 is the heat rejected. Does this suggest that the work that you get out of the engine doesn't depend on how good your compressor and turbine are?…since the compression and expansion were modeled as adiabatic. (MP 1A.13) 1.A.5 Refrigerators and Heat Pumps The Carnot cycle has been used for power, but we can also run it in reverse. If so, there is now net work into the system and net heat out of the system. There will be a quantity of heat Q2 rejected at the higher temperature and a quantity of heat Q1 absorbed at the lower temperature. The former of 1A-11 these is negative according to our convention and the latter is positive. The result is that work is done on the system, heat is extracted from a low temperature source and rejected to a high temperature source. The words “low” and “high” are relative and the low temperature source might be a crowded classroom on a hot day, with the heat extraction being used to cool the room. The cycle and the heat and work transfers are indicated in Figure A-12. In this mode of operation Q2 P T2 a System b W d c T1 V Q1 Figure A-12: Operation of a Carnot refrigerator the cycle works as a refrigerator or heat pump. “What we pay for” is the work, and “what we get” is the amount of heat extracted. A metric for devices of this type is the coefficient of performance, defined as Coefficient of performance = Q1 Q1 = . W Q1 + Q2 For a Carnot cycle we know the ratios of heat in to heat out when the cycle is run forward and, since the cycle is reversible, these ratios are the same when the cycle is run in reverse. The coefficient of performance is thus given in terms of the absolute temperatures as T1 . Coefficient of performance = T2 − T1 This can be much larger than unity. The Carnot cycles that have been drawn are based on ideal gas behavior. For different working media, however, they will look different. We will see an example when we discuss two-phase situations. What is the same whatever the medium is the efficiency for all Carnot cycles operating between the same two temperatures. Muddy points Would it be practical to run a Brayton cycle in reverse and use it as rerigerator? (MP 1A.14) 1.A.6 Reversibility and Irreversibility in Natural Processes We wish to characterize the “direction” of natural processes; there is a basic “directionality” in nature. We start by examining a flywheel in a fluid filled insulated enclosure as shown in Figure A-13. 1A-12 State A: flywheel spinning, system cool State B: flywheel stationary Figure A-13: Flywheel in insulated enclosure at initial and final states A question to be asked is whether we could start with state B and then let events proceed to state A? Why or why not? The first law does not prohibit this. The characteristics of state A are that the energy is in an organized form, the molecules in the flywheel have some circular motion, and we could extract some work by using the flywheel kinetic energy to lift a weight. In state B, in contrast, the energy is associated with disorganized motion on a molecular scale. The temperature of the fluid and flywheel are higher than in state A, so we could probably get some work out by using a Carnot cycle, but it would be much less than the work we could extract in state A. There is a qualitative difference between these states, which we need to be able to describe more precisely. Muddy points Why is the ability to do work decreased in B? How do we know? (MP 1A.15) Another example is a system composed of many bricks, half at a high temperature TH and half at a low temperature TL (see IAW p. 42). With the bricks separated thermally, we have the ability to obtain work by running a cycle between the two temperatures. Suppose we put two bricks together. Using the first law we can write CTH + CTL = 2CTM . (TH + TL ) / 2 = TM where C is the “heat capacity” = ∆Q / ∆T . (For solids the heat capacities (specific heats) at constant pressure and constant volume are essentially the same.) We have lost the ability to get work out of these two bricks. Can we restore the system to the original state without contact with the outside? The answer is no. Can we restore the system to the original state with contact with the outside? The answer is yes. We could run a refrigerator to take heat out of one brick and put it into the other, but we would have to do work. We can think of the overall process involving the system (the two bricks in an insulated setting) and the surroundings (the rest of the universe) as: System is changed Surroundings are unchanged. The composite system (system and the surroundings) is changed by putting the bricks together. The process is not reversible—there is no way to undo the change and leave no mark on the surroundings. 1A-13 What is the measure of change in the surroundings? a) Energy? This is conserved. b) Ability to do work? This is decreased. The measurement and characterization of this type of changes is the subject of the second law of thermodynamics. 1.A.7 Difference between Free Expansion of a Gas and Reversible Isothermal Expansion The difference between reversible and irreversible processes is brought out through examination of the isothermal expansion of an ideal gas. The question to be asked is what is the difference between the “free expansion” of a gas and the isothermal expansion against a piston? To answer this, we address the steps that we would have to take to reverse, in other words, to undo the process. By free expansion, we mean the unrestrained expansion of a gas into a volume as shown at the right. Initially all the gas is in the volume designated as V1 with the rest of the insulated enclosure a vacuum. The total volume ( V1 plus the evacuated volume) is V2 . At a given time a hole is opened in the partition and the gas rushes through to fill the rest of the enclosure. V1, T1 Vacuum During the expansion there is no work exchanged with the surroundings because there is no motion of the boundaries. The enclosure is insulated so there is no heat exchange. The first law tells us therefore that the internal energy is constant ( ∆U = 0). For an ideal gas, the internal energy is a function of temperature only so that the temperature of the gas before the free expansion and after the expansion has been completed is the same. Characterizing the before and after states; Before: State 1, V = V1 , T = T1 After: State 2, V = V2 , T = T1 . Q=W=0, so there is no change in the surroundings. To restore the original state, i.e., to go back to the original volume at the same temperature (V2 → V1 at constant T = T1 ) we can compress the gas isothermally (using work from an external agency). We can do this in a quasi-equilibrium manner, with Psystem ≈ Pexternal . If so the work that 2 we need to do is W = ∫ PdV . We have evaluated the work in a reversible isothermal expansion 1 (Eq. A.1.1), and we can apply the arguments to the case of a reversible isothermal compression. The work done on the system to go from state “2” to state “1” is V W = Work done on system = NR T1 ln 2 . V1 From the first law, this amount of heat must also be rejected from the gas to the surroundings if the temperature of the gas is to remain constant. A schematic of the compression process, in terms of heat and work exchanged is shown in Figure A-14. 1A-14 System W (work in) Q (heat out) Figure A-14: Work and heat exchange in the reversible isothermal compression process At the end of the combined process (free expansion plus reversible compression): a) The system has been returned to its initial state (no change in system state). b) The surroundings (us!) did work on the system of magnitude W. c) The surroundings received an amount of heat, Q, which is equal to W. d) The sum of all of these events is that we have converted an amount of work, W, into an amount of heat, Q, with W and Q numerically equal in Joules. g The net effect is the same as if we let a weight fall and pull a block along a rough surface, as at right. There is 100% conversion of work into heat. Block Weight The results of the free expansion can be contrasted against a process of isothermal expansion against a pressure dP which is slightly different than that of the system, as shown below. System P, T Work received, W Q Q Figure A-15: Work and heat transfer in reversible isothermal expansion 1A-15 W During the expansion, work is done on the surroundings of magnitude W = ∫ PdV , where P can be taken as the system pressure. As evaluated in Eq. (A.1.1), the magnitude of the work done by the V system is W=NR T1 ln 2 . At the end of the isothermal expansion, therefore: V1 a) The surroundings have received work W b) The surroundings have given up heat, Q, numerically equal to W. We now wish to restore the system to its initial state, just as we did in the free expansion. To do this we need to do work on the system and extract heat from the system, just as in the free expansion. In fact, because we are doing a transition between the same states along the same path, the work and heat exchange are the same as those for the compression process examined just above. The overall result when we have restored the system to the initial state, however, is quite different for the reversible expansion and for the free expansion. For the reversible expansion, the work we need to do on the system to compress it has the same magnitude as the work we received during the expansion process. Indeed, we could raise a weight during the expansion and then allow it to be lowered during the compression process. Similarly the heat put into the system by us (the surroundings) during the expansion process has the same magnitude as the heat received by us during the compression process. The result is that when the system has been restored back to its initial state, so have the surroundings. There is no trace of the overall process on either the system or the surroundings. That is another meaning of the word “reversible”. Muddy points With the isothermal reversible expansion is Pexternal constant? If so, how can we have Psystem ≅ Pexternal ? (MP 1A.16) Why is the work done equal to zero in the free expansion? (MP 1A.17) Is irreversibility defined by whether or not a mark is left on the outside environment? (MP 1A.18) 1.A.8 Features of reversible processes Reversible processes are idealizations or models of real processes. One familiar, and widely used, example is Bernoulli’s equation, which you saw last year. They are extremely useful for defining limits to system or device behavior, for enabling identification of areas in which inefficiencies occur, and in giving targets for design. An important feature of a reversible process is that, depending on the process, it represents the maximum work that can be extracted in going from one state to another, or the minimum work that is needed to create the state change. This is shown clearly in the discussion on page 46 of the Waitz notes. Said differently: The work done by a system during a reversible process is the maximum work we can get. The work done on a system in a reversible process is the minimum work we need to do to achieve that state change. A process must be quasi-static (quasi-equilibrium) to be reversible. This means that the following effects must be absent or negligible. 1) Friction If Pexternal ≠ Psystem , as shown in (8) of Part 0, we would have to do work to bring the system from one volume to another and return it to the initial condition. [Review (8) of Part 0] 1A-16 2) Free (unrestrained) expansion 3) Heat transfer through a finite temperature difference. Q2 T2 Q 1 = Q2 System T1 Q1 Figure A-16: Heat transfer across a finite temperature difference Suppose we have heat transfer from a high temperature to a lower temperature as shown above. How do we restore the situation to the initial conditions? One thought would be to run a Carnot refrigerator to get an amount of heat, Q, from the lower temperature reservoir to the higher temperature reservoir. We could do this but the surroundings, again us, would need to provide some amount of work (which we could find using our analysis of the Carnot refrigerator). The net (and only) result at the end of the combined process would be a conversion of an amount of work into heat. For reversible heat transfer from a heat reservoir to a system, the temperatures of the system and the reservoir must be Theat reservoir = Tsystem ± dT . In other words the difference between the temperatures of the two entities involved in the heat transfer process can only differ by an infinitesimal amount, dT. All natural processes are irreversible to some extent, and reversible processes are idealized models. In natural processes, the conditions for mechanical, thermal, and chemical equilibrium may not be satisfied. In addition, dissipative effects (viscosity, friction) exist. Reversible processes are quasiequilibrium, with no dissipative effects. It cannot be emphasized too strongly that there are a number of engineering situations where the effect of irreversibility can be neglected and the reversible process furnishes an excellent approximation to reality. The second law, which is the next topic we address, allows us to make a quantitative statement concerning the irreversibility of a given physical process. Portrait of Nicolas Sadi Carnot, painted by Louis Leopold Boilly, 1813. Figure A-17: this is Nicolas Sadi Carnot (1796-1832), an engineer and an officer in the French army. Carnot’s work is all the more remarkable because it was made without benefit of the first law, which was not discovered until 30 years later. [Atkins, The Second Law]. 1A-17 Muddy points Is heat transfer across a finite temperature difference only irreversible if no device is present between the two to harvest the potential difference ? (MP 1A.19) 1A-18 Muddiest Points on Part 1A 1A.1 Since h = 1 – T1/T2, looking at the P-V graph, does that mean the farther apart the T1, T2 isotherms are, the greater the efficiency? And that if they were very close, it would be very inefficient? This is correct. However, there is a limit on the maximum achievable efficiency. We cannot convert the absorbed heat into 100% work, that is, we always must reject some amount of heat. The amount of heat we must reject is QR = - T1/T2*QA (see notes for derivation). Thus for given values of T2 and QA , QR depends only on the temperature of the cold reservoir T1, which is limited by the temperatures naturally available to us. These temperatures are all well above absolute zero, and there are no means to reduce QR to negligible values. The consequence of this is that the Carnot cycle efficiency cannot approach one (η = 1 only if QR = 0, which is not possible). 1A.2 In the Carnot cycle, why are we only dealing with volume changes and not pressure changes on the adiabats and isotherms? We are not neglecting the pressure terms and we are also dealing with pressure changes. On the adiabats we know that dq = 0 (adiabatic process), so that for reversible processes we can write the first law as du = -Pdv and, using enthalpy, also as dh = vdP. With dh = cp dT and du = cv dT for an ideal gas, we can write the ratio of dh/du as dh/du = cp/cv = γ = -(v dP) / ( Pdv). By arranging terms we obtain dP/P = -γ dv/v. For a process we can integrate from 1 to 2 and get P2v2γ = P1v1γ, or Pvγ = const. This relation shows how pressure and volume changes are related to one another during an adiabatic reversible process. During an isothermal process, the temperature stays constant. Using the equation of state for an ideal gas Pv = RT, we find that Pv = const on an isotherm. Again, this relation tells us how pressure changes are related to volume changes during an isothermal process. Note that in the P-V diagram, adiabats (Pvγ = constant) are steeper curves than isotherms (Pv = constant). 1A.3 Is there a physical application for the Carnot cycle? Can we design a Carnot engine for a propulsion device? We will see that Carnot cycles are the best we can do in terms of efficiency. A constant temperature heat transfer process is, however, difficult to attain in practice for devices in which high rates of power are required. The main role of the Carnot engine is therefore as a standard against which all other cycles are compared and which shows us the direction in which design of efficient cycles should go. 1A.4 How do we know which cycles to use as models for real processes? We have discussed this briefly for the Brayton cycle, in that we looked at the approximation that was made in saying heat addition occurred at constant pressure. You can also see that the Carnot cycle is not a good descriptor of a gas turbine engine! We will look further at this general point, not only for the Brayton cycle, but also for the Rankine cycle and for some internal combustion engine cycles. I will try to make clear what are the approximations and why the cycle under study is being used as a model. 1A.5 When flow is accelerated in a nozzle, doesn't that reduce the internal energy of the flow and therefore the enthalpy? Indeed both enthalpy and internal energy are reduced. The stagnation enthalpy is the quantity that is constant. 1A.6 Why do we say that the combustion in a gas turbine engine is at constant pressure? This is an approximation, and a key question is indeed how accurate it is and what the justification is. The pressure change in the combustor can be analyzed using the 1dimensional compressible flow equations. The momentum equation is dP = − ρcdc , where c is the velocity. If we divide both sides by P, we obtain: dP γc 2 dc γc 2 dc dc =− =− 2 = −γM 2 , P γP/ρ c a c c where a is the speed of sound and M is the Mach number. Changes in velocity are due to changes in density and in flow-through area, as given by the 1-dimensional continuity equation ρca = constant Hence ln ρ + ln c + ln a = constant Differentiating, dρ dc dA dc dA dρ + + = 0, or =− − ρ c A c A ρ Velocity changes are therefore related to area changes (geometry) and density changes (basically heat input). For a gas turbine combustion process the change in density is comparable with (a significant fraction of) the initial density and the area change is several times the initial area. This means that the change in velocity divided by the initial velocity is roughly of the order of magnitude of unity. The momentum equation thus tells us that for small Mach number (say 0.1) the ratio dP/P will be much less than one, so that the pressure can be approximated as constant. In reality the pressure does drop in the combustor, but the overall drop from inlet to exit is about 3-4%, small compared to the initial level of pressure, so that the approximation of constant pressure is a useful one. The rapidity of the combustion process does not really have anything to do with this approximation. We could have a process, such as a nozzle, in which there was combustion at the same time that the pressure was dropping. As seen from the momentum equation, the heat addition does not "directly" affect the pressure - changes in pressure are associated with changes in velocity. 1A.7 Why is the Brayton cycle less efficient than the Carnot cycle? Consider the Brayton cycle and the corresponding work done as being approximated by a number of elementary Carnot cycles, as shown by the dashed lines in Figure 1. All of these Carnot cycles have the same pressure ratio, thus the same temperature ratio, and thus the same efficiency. The temperature ratio that figures into the efficiency of the elementary Carnot cycles is the inlet temperature divided by the compressor exit temperature, not the maximum cycle temperature, which is at the combustor exit. The basic reason for the lower efficiency is that heat is absorbed at an average temperature that is lower than the maximum temperature and rejected at an average temperature higher than the minimum temperature. We will come back to this important point (which has implications for all cycles), but if you cannot wait, see Section 1-C of the notes. Brayton cycle (solid line) P Elementary Carnot Cycle (dashed lines) v Figure 1 - Brayton cycle considered as a number of elementary Carnot cycles, all having the same pressure ratio and therefore the same temperature ratio, which is lower than the overall cycle temperature ratio, Tmax / Tmin . 1A.8 If the gas undergoes constant pressure cooling in the exhaust outside the engine, is that still within the system boundary? When we analyze the state changes as we trace them around the cycle, we are viewing the changes in a system, a mass of fixed identity. Thus we follow the mass as it moves through the device and the cooling of the gas outside the engine is happening to our system. 1A.9 Does it matter what labels we put on the corners of the cycle or not? It does not matter what labels we use on the corners of the cycle. A cycle is a series of processes. Independent of where you start in the cycle, it always brings you back to the state where you started. 1A.10 Is the work done in the compressor always equal to the work done in the turbine plus work out (for a Brayton cyle)? NO. The work done in the compressor plus net work out equals the total turbine work. Using the 1st law, the net work we get out of the Brayton cycle is w = q2 + q1 = cp [(Ta – Td) + (Tc – Tb)] (see notes for details). Rearranging the temperatures we can also write w = cp [(Tc – Td) – (Tb – Ta)] = Dhturbine – Dhcompressor. Thus the net work is the difference between the enthalpy drop across the turbine (we get work from the turbine) and the enthalpy rise through the compressor (we have to supply work to the compressor). • 1A.11 What are the units of w in power = m w ? The units of power are J/s (kJ/s, MJ/s) or Watts (kW, MW). The mass flow is kg/s. The units of w, work per unit mass, are thus J/kg. For the aeroengine, we can think of a given diameter (frontal area) as implying a given mass flow (think of a given Mach number and hence a given ratio of flow to choked flow). If so, for a given fan diameter power scales directly as work per unit mass. 1A.12 Precision about the assumptions made in the Brayton cycle for maximum efficiency and maximum work We have first derived a general expression for the thermal efficiency of an ideal Brayton cycle (see Equation A.3.3 in your notes). The assumptions we made for the cycle were that both the compressor and turbine are ideal, such that they can be modeled adiabatic and reversible. We then looked at possible ideal Brayton cycles that would yield (A) maximum efficiency and (B) maximum work, keeping the assumptions of an ideal cycle (the assumptions of adiabatic and reversible compression and expansion stem from the choice of an ideal cycle). One way to construct an ideal Brayton cycle in the P-V diagram is to choose the inlet temperature Ta and inlet pressure Pa, the compressor pressure ratio Pb/Pa or temperature ratio Tb/Ta, and the turbine inlet temperature Tc. Apart from setting the inlet conditions (these mainly depend on the flight altitude and Mach number and the day), we decided to fix the turbine inlet temperature (fixed by material technology or cost). So the only two "floating" cycle parameters that remain to be defined are the compressor exit temperature Tb and the turbine exit temperature Td. Looking at equation A.3.3 we know that the higher the compressor temperature ratio Tb/Ta the higher the thermal efficiency. So, for (A) maximum efficiency we would chose the compressor exit temperature as high as possible, that is in the limit Tb = Tc. Constructing this cycle in the P-V diagram and letting Tb approach Tc shows that the area enclosed by the cycle, or in other words the net work, becomes zero. Thus a cycle constructed under the given inlet conditions and constraints on Tc is not very useful because we don't get any work out of it. For the derivation of Tb for maximum work (keeping Tc fixed as above), see notes for details. 1A.13 You said that for a gas turbine engine modeled as a Brayton cycle the work done is w=q1+q2, where q2 is the heat added and q1 is the heat rejected. Does this suggest that the work that you get out of the engine doesn't depend on how good your compressor and turbine are?…since the compression and expansion were modeled as adiabatic. Using the 1st law, the net work we get out of the cycle is w = q2 + q1 = cp [(Ta – Td) + (Tc – Tb)] (see notes for details). Rearranging the temperatures we can also write w = cp [(Tc – Td) – (Tb – Ta)] = ∆hturbine – ∆hcompressor. Thus the net work is the difference between the enthalpy drop across the turbine (we get work from the turbine) and the enthalpy rise through the compressor (we have to supply work to the compressor, this is done through the drive shaft that connects turbine and compressor). In class we analyzed an ideal Brayton cycle with the assumptions of adiabatic reversible compression and expansion processes, meaning that the work done by the turbine is the maximum work we can get from the given turbine (operating between Tc and Td), and the work needed to drive the given compressor is the minimum work required. In the assumptions the emphasis is put on reversible rather than adiabatic. For real engines the assumption of adiabatic flow through the compressor and turbine still holds. This is an approximation – the surface inside the compressor or turbine where heat can be transferred is much smaller than the mass flow of the fluid moving through the machine so that the heat transfer is negligible – we will discuss the different concepts of heat transfer later in class. However the compression and expansion processes in real engines are irreversible due to non-ideal behavior and loss mechanisms occurring in the turbomachinery flow. Thus the thermal efficiency and work for a real jet engine with losses depend on the component efficiencies of turbine and compressor and are less than for an ideal jet engine. We will discuss these component efficiencies in more detail in class. 1A.14 Would it be practical to run a Brayton cycle in reverse and use it as rerigerator? Yes. In fact people in Cryogenics use reversed Brayton cycles to cool down systems where very low temperatures are required (e.g. space applications, liquefaction of propellants). One major difference between a regular Brayton cycle (such as a jet-engine or a gas-turbine) and a reversed Brayton cycle is the working fluid. In order to make a reversed Brayton cycle practical we have to choose a working fluid that is appropriate for the application. Extremely low temperatures can be achieved when using a regenerator – a heat exchanger that preheats the fluid before it enters the compressor and cools the fluid further down before it enters the turbine. In this configuration the fluid is expanded to much lower temperatures, and more heat can be absorbed from the cooling compartment. 1A.15 Why is the ability to do work decreased in B? How do we know? In state A, the energy is in organized form and the molecules move along circular paths around the spinning flywheel. We could get work out this system by using all of the kinetic energy of the flywheel and for example lift a weight with it. The energy of the system in state B (flywheel not spinning) is associated with disorganized motion (on the molecular scale). The temperature in state B is higher than in state A. We could also extract work from state B by running for example an ideal Carnot cycle between TB and some heat reservoir at lower temperature. However the work we would get from this ideal Carnot cycle is less than the work we get from state A (all of the kinetic energy), because we must reject some heat when we convert heat into work (we cannot convert heat into 100% work). Although the energy of the system in state A is the same as in state B (we know this from 1st law) the "organization" of the energy is different, and thus the ability to do work is different. 1A.16 With the isothermal reversible expansion, is Pexternal constant? If so, how can we have Psystem≅Pexternal? For a reversible process, if the external pressure were constant, there would need to be a force that pushed on the piston so the process could be considered quasi-equilibrium. This force could be us, it could be a system of weights, or it could be any other work receiver. Under these conditions the system pressure would not necessarily be near the F external pressure but we would have Psystem ≅ Pexternal + work receiver . We can of course Apiston think of a situation in which the external pressure was varied so it was always close to the system pressure, but that is not necessary. 1A.17 Why is the work done equal to zero in the free expansion? In this problem, the system is everything inside the rigid container. There is no change in volume, no “dV”, so no work done on the surroundings. Pieces of the gas might be expanding, pushing on other parts of the gas, and doing work locally inside the container (and other pieces might be compressed and thus receive work) during the free expansion process, but we are considering the system as a whole, and there is no net work done. 1A.18 Is irreversibility defined by whether or not a mark is left on the outside environment? A process is irreversible when there is no way to undo the change without leaving a mark on the surroundings or "the rest of the universe". In the example with the bricks, we could undo the change by putting a Carnot refrigerator between the bricks (both at TM after putting them together) and cooling one brick down to TL and heating the other brick to TH to restore the initial state. To do this we have to supply work to the refrigerator and we will also reject some heat to the surroundings. Thus we leave a mark on the environment and the process is irreversible. 1A.19 Is heat transfer across a finite temperature difference only irreversible if no device is present between the two to harvest the potential difference? If we have two heat reservoirs at different temperatures, the irreversibility associated with the transfer of heat from one to the other is indeed dependent on what is between them. If there is a copper bar between them, all the heat that comes out of the high temperature reservoir goes into the low temperature reservoir, with the result given in Section 1.B.5. If there were a Carnot cycle between them, some (not all ) heat from the high temperature reservoir would be passed on to the low temperature reservoir, the process would be reversible, and work would be done. The extent to which the process is irreversible for any device can be assessed by computing the total entropy change (device plus surroundings) associated with the heat transfer. 1.B: The Second Law of Thermodynamics [IAW 42-50; VN Chapter 5; VWB&S-6.3, 6.4, Chapter 7] 1.B.1 Concept and Statements of the Second Law (Why do we need a second law?) The unrestrained expansion, or the temperature equilibration of the two bricks, are familiar processes. Suppose you are asked whether you have ever seen the reverse of these processes take place? Do two bricks at a medium temperature ever go to a state where one is hot and one is cold? Will the gas in the unrestrained expansion ever spontaneously return to occupying only the left side of the volume? Experience hints that the answer is no. However, both these processes, unfamiliar though they may be, are compatible with the first law. In other words the first law does not prohibit their occurrence. There thus must be some other “great principle” that describes the direction of natural processes, that tells us which first law compatible processes will not be observed. This is contained in the second law. Like the first law, it is a generalization from an enormous amount of observation. There are several ways in which the second law of thermodynamics can be stated. Listed below are three that are often encountered. As described in class (and as derived in almost every thermodynamics textbook), although the three may not appear to have much connection with each other, they are equivalent. 1) No process is possible whose sole result is the absorption of heat from a reservoir and the conversion of this heat into work. [Kelvin-Planck statement of the second law] Q T2 System W This is not possible T1 2) No process is possible whose sole result is the transfer of heat from a cooler to a hotter body. [Clausius statement of the second law] Q T2 For T1 < T2 , this is not possible T1 Q 1B-1 3) There exists a property called entropy, S, which is a thermodynamic property of a system. For a reversible process, changes in this property are given by dS = (dQreversible)/T The entropy change of any system and its surroundings, considered together, is positive and approaches zero for any process which approaches reversibility. ∆ Stotal > 0 For an isolated system, i.e., a system that has no interaction with the surroundings, changes in the system have no effect on the surroundings. In this case, we need to consider the system only, and the first and second laws become: ∆ E system = 0 ∆ S system > 0 For an isolated system the total energy (E = U + Kinetic Energy + Potential Energy + ....) is constant. The entropy can only increase or, in the limit of a reversible process, remain constant. All of these statements are equivalent, but (3) gives a direct, quantitative measure of the departure from reversibility. Entropy is not a familiar concept and it may be helpful to provide some additional rationale for its appearance. If we look at the first law, dU = dQ − dW the term on the left is a function of state, while the two terms on the right are not. For a simple compressible substance, however, we can write the work done in a reversible process as dW = PdV , so that dU = dQ − PdV ; First law for a simple compressible substance, reversible process. Two out of the three terms in this equation are expressed in terms of state variables. It seems plausible that we ought to be able to express the third term using state variables as well, but what are the appropriate variables? If so, the term dQ = ( ) [ ] should perhaps be viewed as analogous to dW = PdV where the parenthesis denotes an intensive state variable and the square bracket denotes an extensive state variable. The second law tells us that the intensive variable is the temperature, T, and the extensive state variable is the entropy, S. The first law for a simple compressible substance in terms of state variables is thus dU = TdS − PdV . (B.1.1) Because Eq. (B.1.1) includes the second law, it is referred to as the combined first and second law. Because it is written in terms of state variables, it is true for all processes, not just reversible ones. We list below some attributes of entropy: a) S is an extensive variable. The entropy per unit mass, or specific entropy, is s. 1B-2 b) The units of entropy are Joules per degree Kelvin (J/K). The units for specific entropy are J/K-kg. dQrev , where the numerator is the heat given to the system and the c) For a system, dS = T denominator is the temperature of the system at the location where the heat is received. d) dS = 0 for pure work transfer. Muddy points Why is dU = TdS − PdV always true? (MP 1B.1) What makes dQrev different than dQ? (MP 1B.2) 1.B.2 Axiomatic Statements of the Laws of Thermodynamics1 (i.) Introduction As a further aid in familiarization with the second law of thermodynamics and the idea of entropy, we draw an analogy with statements made previously concerning quantities that are closer to experience. In particular, we wish to (re-) present the Zeroth and First Laws of thermodynamics in the same framework as we have used for the Second Law. In this so-called "axiomatic formulation", the Zeroth, First and Second Laws are all introduced in a similar fashion. (ii.) Zeroth Law We start with a statement which is based on two observations: a) If two bodies are in contact through a thermally-conducting boundary for a sufficiently long time, no further observable changes take place; thermal equilibrium is said to prevail. b) Two systems which are individually in thermal equilibrium with a third are in thermal equilibrium with each other; all three systems have the same value of the property called temperature. The closely connected ideas of temperature and thermal equilibrium are formally expressed in the “Zeroth Law of Thermodynamics”: Zeroth Law There exists for every thermodynamic system in equilibrium a property called temperature. Equality of temperature is a necessary and sufficient condition for thermal equilibrium. The Zeroth law thus defines a property (temperature) and describes its behavior. (iii.) First Law Observations also show that for any system there is a property called the energy. The First Law asserts that one must associate such a property with every system. First Law There exists for every thermodynamic system a property called the energy. The change of energy of a system is equal to the mechanical work done on the system in an adiabatic process. In a non-adiabatic process, the change in energy is equal to the heat added to the system minus the mechanical work done by the system. 1 From notes of Professor F. E. C. Culick, California Institute of Technology (with minor changes) 1B-3 On the basis of experimental results, therefore, one is led to assert the existence of two new properties, the temperature and internal energy, which do not arise in ordinary mechanics. In a similar way, a further remarkable relationship between heat and temperature will be established, and a new property, the entropy, defined. Although this is a much less familiar property, it is to be stressed that the general approach is quite like that used to establish the Zeroth and First Laws. A general principle and a property associated with any system are extracted from experimental results. Viewed in this way, the entropy should appear no more mystical than the internal energy. The increase of entropy in a naturally occurring process is no less real than the conservation of energy. (iv.) Second Law Although all natural processes must take place in accordance with the First Law, the principle of conservation of energy is, by itself, inadequate for an unambiguous description of the behavior of a system. Specifically, there is no mention of the familiar observation that every natural process has in some sense a preferred direction of action. For example, the flow of heat occurs naturally from hotter to colder bodies, in the absence of other influences, but the reverse flow certainly is not in violation of the First Law. So far as that law is concerned, the initial and final states are symmetrical in a very important respect. The Second Law is essentially different from the First Law; the two principles are independent and cannot in any sense be deduced from one another. Thus, the concept of energy is not sufficient, and a new property must appear. This property can be developed, and the Second Law introduced, in much the same way as the Zeroth and First Laws were presented. By examination of certain observational results, one attempts to extract from experience a law which is supposed to be general; it is elevated to the position of a fundamental axiom to be proved or disproved by subsequent experiments. Within the structure of classical thermodynamics, there is no proof more fundamental than observations. A statement which can be adopted as the Second Law of thermodynamics is: Second Law There exists for every thermodynamic system in equilibrium an extensive scalar property called the entropy, S, such that in an infinitesimal reversible change of state of the system, dS = dQ/T, where T is the absolute temperature and dQ is the amount of heat received by the system. The entropy of a thermally insulated system cannot decrease and is constant if and only if all processes are reversible. As with the Zeroth and First Laws, the existence of a new property is asserted and its behavior is described. (v.) Reversible Processes In the course of this development, the idea of a completely reversible process is central, and we can recall the definition, “a process is called completely reversible if, after the process has occurred, both the system and its surroundings can be wholly restored by any means to their respective initial states”. Especially, it is to be noted that the definition does not, in this form, specify that the reverse path must be identical with the forward path. If the initial states can be restored by any means whatever, the process is by definition completely reversible. If the paths are identical, then one usually calls the process (of the system) reversible, or one may say that the state of the system follows a reversible path. In this path (between two equilibrium states 1 and 2), (i) the system passes through the path followed by the equilibrium states only, and (ii) the system will take the reversed path 2 to 1 by a simple reversal of the work done and heat added. Reversible processes are idealizations not actually encountered. However, they are clearly 1B-4 useful idealizations. For a process to be completely reversible, it is necessary that it be quasi-static and that there be no dissipative influences such as friction and diffusion. The precise (necessary and sufficient) condition to be satisfied if a process is to be reversible is the second part of the Second Law. The criterion as to whether a process is completely reversible must be based on the initial and final states. In the form presented above, the Second Law furnishes a relation between the properties defining the two states, and thereby shows whether a natural process connecting the states is possible. Muddy points What happens when all the energy in the universe is uniformly spread, ie, entropy at a maximum? (MP 1B.3) 1.B.3 Combined First and Second Law Expressions First Law: dU = dQ − dW - Always true Work and heat exchange in terms of state variables: dQ = TdS; dW = PdV - Only true for reversible processes. dU = dQ − PdV ; Simple compressible substance, reversible process dU = dQ − PdV − XdY ; Substance with other work modes (e.g., stress-strain), X is a pressure-like quantity, Y is a volume like quantity dU = TdS − dW ; Only true for a reversible process First law in terms of state variables: dU = TdS − PdV ; This is a relation between properties and is always true In terms of specific quantities (per unit mass): du = Tds − Pdv Combined first and second law (a) or Gibbs equation (a) The combined first and second law expressions are often more usefully written in terms of the enthalpy, or specific enthalpy, h = u + Pv: dh = du + Pdv + vdP = Tds − Pdv + Pdv + vdP , using the first law. dh = Tds + vdP Or, since v = 1/ ρ dh = Tds + dP . ρ Combined first and second law (b) or Gibbs equation (b) In terms of enthalpy (rather than specific enthalpy) the relation is dH = TdS + VdP . 1B-5 1.B.4 Entropy Changes in an Ideal Gas Many aerospace applications involve flow of gases (e.g., air) and we thus examine the entropy relations for ideal gas behavior. The starting point is form (a) of the combined first and second law, du = Tds − Pdv . For an ideal gas, du = cv dT . Thus dT P + dv. T T Using the equation of state for an ideal gas ( Pv = RT ), we can write the entropy change as an expression with only exact differentials: Tds = cv dT + Pdv or ds = cv ds = cv dT dv +R . T v (B.4.1) Integrating between two states “1” and “2”: T2 ∆s = s2 − s1 = ∫ cv T1 v 2 dv dT + R∫ . v1 v T For constant specific heat T v ∆s = s2 − s1 = cv ln 2 + Rln 2 . T1 v1 R In non-dimensional form (using = (γ − 1) ) cv T v ∆s = ln 2 + (γ − 1)ln 2 . cv T1 v1 Entropy change of an ideal gas (B.4.2) Equation (B.4.2) is in terms of specific quantities. For N moles of gas T V ∆S = N ln 2 + (γ − 1)ln 2 . Cv V1 T1 Rather than temperature and volume, we can develop an alternative form of the expression, in terms of pressure and volume, for entropy change, which allows us to examine an assumption we have used over the past year. The ideal gas equation of state can be written as lnP + lnv = lnR + lnT. Taking differentials of both sides yields dP dv dT + = P v T Using the above equation in Eq. (B.4.1), and making use of the relations c p = cv + R; c p / cv = γ , we find 1B-6 or dP dv dv ds = cv + +R , v v P ds dP dv . = +γ cv P v Integrating between two states 1 and 2 P v γ P2 v2 ∆s = ln + γln = ln 2 2 . cv P1 v1 P1 v1 (B.4.3) Using both sides of (B.4.3) as exponents we obtain P2 v2γ = Pvγ γ Pv 1 1 [ ] 2 1 = e ∆s / cv . (B.4.4) Equation (B.4.4) describes a general process. For the specific situation in which ∆s = 0 , i.e., the entropy is constant, we recover the expression Pvγ = constant. It was stated that this expression applied to a reversible, adiabatic process. We now see, through use of the second law, a deeper meaning to the expression, and to the concept of a reversible adiabatic process, in that both are characteristics of a constant entropy, or isentropic, process. Muddy points Why do you rewrite the entropy change in terms of Pv γ? (MP 1B.4) What is the difference between isentropic and adiabatic? (MP 1B.5) 1.B.5 Calculation of Entropy Change in Some Basic Processes a) Heat transfer from, or to, a heat reservoir. A heat reservoir is a constant temperature heat source or sink. Because the temperature is uniform, there is no heat transfer across a finite temperature difference and the heat exchange is reversible. From the definition of entropy ( dS = dQrev / T ) , Q ∆S = , T where Q is the heat into the reservoir (defined here as positive if heat flows into the reservoir) TH QH Heat transfer from/to a heat reservoir Q TH b) Heat transfer between two heat reservoirs The entropy changes of the two reservoirs are the sum of the entropy change of each. If the high temperature reservoir is at TH and the low temperature reservoir is at TL , the total entropy change is QH Device (block of copper) no work no change in state TL Heat transfer between two reservoirs 1B-7 −Q Q Q ∆S = (TH − TL ) + = TH TL TH TL The second law says that the entropy change must be equal to or greater than zero. This corresponds to the statement that heat must flow from the higher temperature source to the lower temperature source. This is one of the statements of the second law given in Section 1.B.1. Muddy points In the single reservoir example, why can the entropy decrease? (MP 1B.6)9 Why does the entropy of a heat reservoir change if the temperature stays the same? (MP9 1B.7)9 How can the heat transfer from or to a heat reservoir be reversible? (MP 1B.8)9 How can ∆S be less than zero in any process? Doesn't entropy always increase? (MP 1B.9)9 Q = ∆S for a reservoir, could you add Q to any size reservoir and still get the same ∆S? If T (MP 1B.10) c) Possibility of obtaining work from a single heat reservoir We can regard the proposed process as the absorption of heat, Q, by a device or system, operating in a cycle, rejecting no heat, and producing work. The total entropy change is the sum of the change in the reservoir, the system or device, and the surroundings. The entropy change of the reservoir is ∆S = −Q / TH . The entropy change of the device is zero, because we are considering a complete cycle (return to initial state) and entropy is a function of state. The surroundings receive work only so the entropy change of the surroundings is zero. Work from a single heat reservoir The total entropy change is ∆S total = ∆Sreservoir + ∆Sdevice + ∆Ssurroundings = −Q / TH + 0 + 0 The total entropy change in the proposed process is thus less than zero, ∆S total < 0 which is not possible. The second law thus tells us that we cannot get work from a single reservoir only. The “only” is important; it means without any other changes occurring. This is the other statement of the second law we saw in Section 1.B.1. Muddy points What is the difference between the isothermal expansion of a piston and the (forbidden) production of work using a single reservoir? (MP 1B.11) For the "work from a single heat reservoir" example, how do we know there is no ∆Ssurr? (MP 1B.12) 1B-8 How does a cycle produce zero ∆S? I thought that the whole thing about cycles was an entropy that the designers try to minimize. (MP 1B.13) d) Entropy changes in the “hot brick problem” We can examine in a more quantitative manner the changes that occurred when we put the two bricks together, as depicted on the left-hand side of the figure below. The process by which the two bricks come to the same temperature is not a reversible one, so we need to devise a reversible path. To do this imagine a large number of heat reservoirs at varying temperatures spanning the range TH − dT,............,TL + dT , as in the right hand side of the figures. The bricks are put in contact with them sequentially to raise the temperature of one and lower the temperature of the TH TM TL TH TM TH - dT Temperature equalization of two bricks ........ TL TL + dT Reservoirs used in reversible state transformation other in a reversible manner. The heat exchange at any of these steps is dQ = CdT . For the high temperature brick, the entropy change is: ∆Shot brick = ∫ T CdT = C ln M T TH TM TH where C is the heat capacity of the brick (J/kg). This quantity is less than zero. For the cold brick, ∆Scold brick = ∫ TM TL T CdT = C ln M . T TL The entropy change of the two bricks is ∆S bricks TM TM TM2 = C ln + ln = C ln > 0. TH TL TL TH The process is not reversible. e) Difference between the free expansion and the reversible isothermal expansion of an ideal gas The essential difference between the free expansion in an insulated enclosure and the reversible isothermal expansion of an ideal gas can also be captured clearly in terms of entropy changes. For a state change from initial volume and temperature V1 , T1 to final volume and (the same) temperature V2 , T1 the entropy change is 2 ∆S = ∫ dS = ∫ 1 2 1 2 PdV dU +∫ , 1 T T 1B-9 or, making use of the equation of state and the fact that dU = 0 for an isothermal process, V ∆S = NR ln 2 . V1 This is the entropy change that occurs for the free expansion as well as for the isothermal reversible expansion processes—entropy changes are state changes and the two system final and end states are the same for both processes. For the free expansion: V ∆S system = NR ln 2 ; ∆S surroundings = 0 V1 There is no change in the entropy of the surroundings because there is no interaction between the system and the surroundings. The total entropy change is therefore, V ∆S total = ∆S system + ∆S surroundings = NR ln 2 > 0. V1 There are several points to note from this result. i) ∆S total > 0 so the process is not reversible 2 dQ dQ = 0 ; the equality between ∆S and is only for a reversible ii) ∆S system > ∫ T 1 T process iii) There is a direct connection between the work needed to restore the system to the original state and the entropy change: V W = NRT ln 2 = T∆S2 −1 V1 The quantity T∆S has a physical meaning as “lost work” in the sense of work which we lost the opportunity to utilize. We will make this connection stronger in Section 1.C. For the reversible isothermal expansion: The entropy is a state variable so the entropy change of the system is the same as before. In this case, however, heat is transferred to the system from the surroundings ( Q surroundings < 0) so that Q surroundings surroundings ∆S = < 0. T The heat transferred from the surroundings, however, is equal to the heat received by the system: Q surroundings = Q system = W . V Q surroundings −W surroundings ∆S = = = - NR ln 2 . T T V1 The total change in entropy (system plus surroundings) is therefore Q Q − = 0. T T The reversible process has zero total change in entropy. ∆S total = ∆S system + ∆S surroundings = 1B-109 Muddy points On the example of free expansion versus isothermal expansion, how do we know that the9 pressure and volume ratios are the same? We know for each that P9 2 >P 1 and V2 >V1 . (MP 1B.14)9 V Where did ∆Ssystem = NRln 2 come from? (MP 1B.15) V1 1B-119 Muddiest Points on Part 1B 1B.1 Why is dU = TdS − PdV always true? This is a relation between state variables. As such it is not path dependent, only depends on the initial and final states, and thus must hold no matter how we transition from initial state to final state. What is not always true, and what holds only for reversible processes are the relations Tds = dq and Pdv = dw. One example of this is the free expansion where dq = dw = 0, but where the quantities Tds and Pdv (and the integrals of these quantities) are not zero. 1B.2 What makes dQrev different than dQ? The term dQrev denotes the heat exchange during a reversible process. We use the notation dQ to denote heat exchange during any process, not necessarily reversible. The distinction between the two is important for the reason given above in (3). 1B.3 What happens when all the energy in the universe is uniformly spread, ie, entropy at a maximum? I quote from The Refrigerator and the Universe, by Goldstein and Goldstein: “The entropy of the universe is not yet at its maximum possible value and it seems to be increasing all the time. Looking forward to the future, Kelvin and Clausius foresaw a time when the maximum possible entropy would be reached and the universe would be at equilibrium forever afterward; at this point, a state called the “heat death” of the universe, nothing would happen forever after”. The book also gives comments on the inevitability of this fate. 1B.4 Why do you rewrite the entropy change in terms of Pvγ? We have discussed the representation of thermodynamic changes in P-v coordinates a number of times and it is familiar, as is the idea of the “ Pv γ = constant ” process. I want to relate this to the more general expression involving the entropy change (Equation B.4.4) to show (i) when the simple form applied and (ii) how valid an approximation it was. Using the entropy change, we now have a quantitative metric for doing just that. 1B.5 What is the difference between isentropic and adiabatic? Isentropic means no change in entropy (dS = 0). An adiabatic process is a process with no heat transfer (dQ = 0). We defined for reversible processes TdS = dQ. So generally an adiabatic process is not necessarily isentropic – only if the process is reversible and adiabatic we can call it isentropic. For example a real compressor can be assumed adiabatic but is operating with losses. Due to the losses the compression is irreversible. Thus the compression is not isentropic. 1B.6 In the single reservoir example, why can the entropy decrease? When we looked at the single reservoir, our “system” was the reservoir itself. The example I did in class had heat leaving the reservoir, so that Q was negative. Thus the entropy change of the reservoir is also negative. The second law, however, guarantees that there is a positive change in entropy somewhere else in the surroundings that will be as large, or larger, than this decrease. 1B.7 Why does the entropy of a heat reservoir change if the temperature stays the same? A heat reservoir is an idealization (like an ideal gas, a rigid body, an inviscid fluid, a discrete element mass-spring-damper system). The basic idea is that the heat capacity of the heat reservoir is large enough so that the transfer of heat in whatever problem we address does not apprecibly alter the temperature of the reservoir. In grappling with approximations such as this it is useful to think about extreme cases. Therefore, suppose the thermal reservoir is the atmosphere. The mass of the atmosphere is roughly 1019 kg (give or take an order of magnitude). Let us calculate the temperature rise due to the heat dumped into the atmosphere by a jet engine during a transcontinental flight. A large gas turbine engine might produce on the order of 100 MW of heat, so that the rise in atmospheric temperature, δTatm , for the heat transfer Q associated with a 6 hour flight is given by M atm c pδTatm = 6 × 3600 × 10 8 J . Substituting for the atmospheric mass and the specific heat gives a value for temperature change of roughly 10-10 K. To a very good approximation, we can say that the temperature of this heat reservoir is constant and we can evaluate the entropy change of the reservoir as Q/T. 1B.8 How can the heat transfer from or to a heat reservoir be reversible? We made the assumption that the heat reservoir is very large, and therefore it is a constant temperature heat source or sink. Since the temperature is uniform there is no heat transfer across a finite temperature difference and this heat exchange is reversible. We discussed this in the second example "Heat transfer between two heat reservoirs". 1B.9 How can ∆S be less than zero in any process? Doesn't entropy always increase? The second law says that the total entropy (system plus surroundings) always increases. (See Section 1.B.1). This means that either the system or the surroundings can have it entropy decrease if there is heat transfer between the two, although the sum of all entropy changes must be positive. For an isolated system, with no heat transfer to the surroundings, the entropy must always increase. Q = ∆S for a reservoir, could you add Q to any size reservoir and still get the T same ∆S? 1B.10 If Yes, as long as the system you were adding heat to fulfilled the conditions for being a reservoir. 1B.11 What is the difference between the isothermal expansion of a piston and the (forbidden) production of work using a single reservoir? The difference is contained in the word sole in the Kelvin-Planck statement of the second law given in Section 1.B.1 of the notes. For the isothermal expansion the changes are: a) The reservoir loses heat Q b) The system does work W (equal in magnitude to Q) c) The system changes its volume and pressure. d) The system changes its entropy (the entropy increases by Q/T). For the “forbidden” process, a) The reservoir loses heat Q b) The system does work W (= Q) and that’s all the changes that there are. I leave it to you to calculate the total entropy changes (system plus surroundings) that occur in the two processes. 1B.12 For the "work from a single heat reservoir" example, how do we know there is no ∆Ssurr? Our system was the heat reservoir itself. In the example we had heat leaving the reservoir, thus Q was negative and the entropy change of the reservoir was also negative. Using the second law, it is guaranteed that somewhere else in the surroundings a positive entropy change will occur that is as large or larger than the decrease of the entropy of the reservoir. 1B 13 How does a cycle produce zero ∆S? I thought that the whole thing about cycles was an entropy that the designers try to minimize. The change in entropy during a cycle is zero because we are considering a complete cycle (returning to initial state) and entropy is a function of state (holds for ideal and real cycles!). The entropy you are referring to is entropy that is generated in the components of a nonideal cycle. For example in a real jet engine we have a non-ideal compressor, a non-ideal combustor and also a non-ideal turbine. All these components operate with some loss and generate entropy – this is the entropy that the designers try to minimize. Although the change in entropy during a non-ideal cycle is zero, the total entropy change (cycle and heat reservoirs!) is ∆Stotal > 0. Basically the entropy generated due to irreversibilities in the engine is additional heat rejected to the environment (to the lower heat reservoir). We will discuss this in detail in Section 1.C.1. 1B.14 On the example of free expansion versus isothermal expansion, how do we know that the pressure and volume ratios are the same? We know for each that P2>P1 and V2>V1. During the free-expansion no work is done and no heat is transferred (insulated system). Thus the internal energy stays constant and so does the temperature. This means that P1V1 = P2V2 holds also for the free-expansion and that the pressure and volume ratios are the same when comparing free-expansion to reversible isothermal expansion. V 1B.15 Where did ∆Ssystem = NRln 2 come from? V1 We were using the 1st and 2nd law combined (Gibbs) and in the example discussed there was no change in internal energy (dU=0). If we then integrate dS = P/TdV using P/T = NR/V (with N being the number of moles of gas in volume V and R is the universal gas constant) we obtain ∆Ssystem = NR ln(V2/V1). 1.C Applications of the Second Law [VN-Chapter 6; VWB&S-8.1, 8.2, 8.5, 8.6, 8.7, 8.8, 9.6] 1.C.1 Limitations on the Work that Can be Supplied by a Heat Engine The second law enables us to make powerful and general statements concerning the maximum work that can be QH derived from any heat engine which operates in TH a cycle. To illustrate these ideas, we use a Carnot cycle which is shown schematically at the right. The engine operates between two heat Carnot cycle reservoirs, exchanging heat We QH with the high temperature reservoir at TH and QL with the reservoir at TL. . The entropy changes of the two reservoirs are: Q TL ∆SH = H ; QH < 0 TH QL QL ∆SL = ; QL > 0 TL The same heat exchanges apply to the system, but with opposite signs; the heat received from the high temperature source is positive, and conversely. Denoting the heat transferred to the engines by subscript “e”, QHe = −QH ; QLe = −QL . The total entropy change during any operation of the engine is, ∆S total = ∆ SH + ∆ SL + ∆Se { { { Re servoir at TH Re servoir at TL Engine For a cyclic process, the third of these ( ∆Se ) is zero, and thus (remembering that QH < 0 ), ∆S total = ∆SH + ∆SL = QH QL + TH TL For the engine we can write the first law as ∆Ue = 0 (cyclic process) = QHe + QLe − We . Or, We = QHe + QLe = −QH − QL . Hence, using (C.1.1) T We = −QH − TL ∆S total + QH L TH 1C-1 (C.1.1) T = ( −QH )1 − L − TL ∆S total . TH The work of the engine can be expressed in terms of the heat received by the engine as T We = QHe 1 − L − TL ∆S total . TH ( ) The upper limit of work that can be done occurs during a reversible cycle, for which the total entropy change ( ∆S total ) is zero. In this situation: T Maximum work for an engine working between TH and TL : We = QHe 1 − L TH Also, for a reversible cycle of the engine, ( ) QH QL + = 0. TH TL These constraints apply to all reversible heat engines operating between fixed temperatures. The thermal efficiency of the engine is Work done W = Heat received QHe T = 1 − L = ηCarnot . TH η= The Carnot efficiency is thus the maximum efficiency that can occur in an engine working between two given temperatures. We can approach this last point in another way. The engine work is given by or, We = −QH − TL ∆S total + QH (TL / TH ) TL ∆S total = −QH + QH (TL / TH ) − We The total entropy change can be written in terms of the Carnot cycle efficiency and the ratio of the work done to the heat absorbed by the engine. The latter is the efficiency of any cycle we can devise: QH T W QH ∆S total = e 1 − L − e = e ηCarnot − η Any other . TL TH QHe TL cycle The second law says that the total entropy change is equal to or greater than zero. This means that the Carnot cycle efficiency is equal to or greater than the efficiency for any other cycle, with the equality only occurring if ∆S total = 0 . 1C-2 Muddy points So, do we lose the capability to do work when we have an irreversible process and entropy increases? (MP 1C.1) Why do we study cycles starting with the Carnot cycle? Is it because I is easier to work with? (MP 1C.2) 1.C.2 The Thermodynamic Temperature Scale The considerations of Carnot cycles in this section have not mentioned the working medium. They are thus not limited to an ideal gas and hold for Carnot cycles with any medium. Because we derived the Carnot efficiency with an ideal gas as a medium, the temperature definition used in the ideal gas equation is not essential to the thermodynamic arguments. More specifically, we can define a thermodynamic temperature scale that is independent of the working medium. To see this, consider the situation shown below in Figure C-1, which has three reversible cycles. There is a high temperature heat reservoir at T3 and a low temperature heat reservoir at T1. For any two temperatures T1 , T2 , the ratio of the magnitudes of the heat absorbed and rejected in a Carnot cycle has the same value for all systems. T1 Q1 WA Q1 A WC C Q2 T2 Q2 WB Q3 B Q3 T3 Figure C-1: Arrangement of heat engines to demonstrate the thermodynamic temperature scale We choose the cycles so Q1 is the same for A and C. Also Q 3 is the same for B and C. For a Carnot cycle Q η = 1 + L = F(TL , TH ) ; η is only a function of temperature. QH Also Q1 Q2 = F(T1 , T2 ) Q2 Q3 = F(T2 , T3 ) Q1 Q3 = F(T1 , T3 ) . But 1C-3 Q1 Q1 Q2 = . Q3 Q2 Q3 Hence F(T1 , T3 ) = F(T1 , T2 ) × F(T2 , T3 ) . 1 424 3 14442444 3 Cannot be a function of T2 Not a function of T2 We thus conclude that F(T1 , T2 ) has the form f (T1 ) / f (T2 ) , and similarly F(T2 , T3 ) = f (T2 ) / f (T3 ). The ratio of the heat exchanged is therefore f (T1 ) Q1 . = F(T1 , T3 ) = Q3 f (T3 ) In general, f (TH ) QH , = QL f (TL ) so that the ratio of the heat exchanged is a function of the temperature. We could choose any function that is monotonic, and one choice is the simplest: f (T ) = T . This is the thermodynamic scale of temperature, QH QL = TH TL . The temperature defined in this manner is the same as that for the ideal gas; the thermodynamic temperature scale and the ideal gas scale are equivalent 1.C.3 Representation of Thermodynamic Processes in T-s coordinates. It is often useful to plot the thermodynamic state transitions and the cycles in terms of temperature (or enthalpy) and entropy, T,S, rather than P,V. The maximum temperature is often the constraint on the process and the enthalpy changes show the work done or heat received directly, so that plotting in terms of these variables provides insight into the process. A Carnot cycle is shown below in these coordinates, in which it is a rectangle, with two horizontal, constant temperature legs. The other two legs are reversible and adiabatic, hence isentropic ( dS = dQrev / T = 0), and therefore vertical in T-s coordinates. Isothermal TH a b Adiabatic T TL d c s Carnot cycle in T,s coordinates If the cycle is traversed clockwise, the heat added is 1C-4 Heat added: QH = ∫abTdS = TH ( Sb − Sa ) = TH ∆S . The heat rejected (from c to d) has magnitude QL = TL ∆S . The work done by the cycle can be found using the first law for a reversible process: dU = dQ − dW . = TdS − dW (This form is only true for a reversible process). We can integrate this last expression around the closed path traced out by the cycle: ∫ dU = ∫ TdS − ∫ dW However dU is an exact differential and its integral around a closed contour is zero: 0 = ∫ TdS − ∫ dW . The work done by the cycle, which is represented by the term ∫ dW , is equal to ∫ Tds , the area enclosed by the closed contour in the T-S plane. This area represents the difference between the heat absorbed ( ∫ TdS at the high temperature) and the heat rejected ( ∫ TdS at the low temperature). Finding the work done through evaluation of ∫ TdS is an alternative to computation of the work in a reversible cycle from ∫ PdV . Finally, although we have carried out the discussion in terms of the entropy, S, all of the arguments carry over to the specific entropy, s; the work of the reversible cycle per unit mass is given by ∫ Tds. Muddy points How does one interpret h-s diagrams? (MP 1C.3) Is it always OK to "switch" T-s and h-s diagram? (MP 1C.4) What is the best way to become comfortable with T-s diagrams? (MP 1C.5) What is a reversible adiabat physically? (MP 1C.6) 1.C.4 Brayton Cycle in T-s Coordinates The Brayton cycle has two reversible adiabatic (i.e., isentropic) legs and two reversible, constant pressure heat exchange legs. The former are vertical, but we need to define the shape of the latter. For an ideal gas, changes in specific enthalpy are related to changes in temperature by dh = c p dT , so the shape of the cycle in an h-s plane is the same as in a T-s plane, with a scale factor of c p between the two. This suggests that a place to start is with the combined first and second law, which relates changes in enthalpy, entropy, and pressure: dh = Tds + dp . ρ On constant pressure curves dP=0 and dh = Tds . The quantity desired is the derivative of temperature, T, with respect to entropy, s, at constant pressure: (∂T ∂s) p . From the combined first and second law, and the relation between dh and dT, this is 1C-5 T ∂T (C.4.1) = ∂s p c p The derivative is the slope of the constant pressure legs of the Brayton cycle on a T-s plane. For a given ideal gas (specific c p ) the slope is positive and increases as T. We can also plot the Brayton cycle in an h-s plane. This has advantages because changes in enthalpy directly show the work of the compressor and turbine and the heat added and rejected. The slope of the constant pressure legs in the h-s plane is (∂h ∂s) p = T . Note that the similarity in the shapes of the cycles in T-s and h-s planes is true for ideal gases only. As we will see when we examine two-phase cycles, the shapes look quite different in these two planes when the medium is not an ideal gas. Plotting the cycle in T-s coordinates also allows another way to address the evaluation of the Brayton cycle efficiency which gives insight into the relations between Carnot cycle efficiency and efficiency of other cycles. As shown in Figure C-2, we can break up the Brayton cycle into many small Carnot cycles. The " i th " Carnot cycle has an efficiency of ηci = 1 − Tlowi Thighi , [ ( )] where the indicated lower temperature is the heat rejection temperature for that elementary cycle and the higher temperature is the heat absorption temperature for that cycle. The upper and lower curves of the Brayton cycle, however, have constant pressure. All of the elementary Carnot cycles therefore have the same pressure ratio: ( P Thigh ) = PR = constant (the same for all the cycles). P(Tlow ) From the isentropic relations for an ideal gas, we know that pressure ratio, PR, and temperature γ −1 / γ ratio, TR, are related by : PR( ) = TR . 1C-6 Figure C-2 available from: Kerrebrock, Aircraft Engines and Gas Turbines, 2nd Ed. MIT Press. Figure 1.3, p.8. Figure C-2: Ideal Brayton cycle as composed of many elementary Carnot cycles [Kerrebrock] ( ) The temperature ratios Tlowi Thighi of any elementary cycle “i” are therefore the same and each of the elementary cycles has the same thermal efficiency. We only need to find the temperature ratio across any one of the cycles to find what the efficiency is. We know that the temperature ratio of the first elementary cycle is the ratio of compressor exit temperature to engine entry (atmospheric for an aircraft engine) temperature, T2 / T0 in Figure C-2. If the efficiency of all the elementary cycles has this value, the efficiency of the overall Brayton cycle (which is composed of the elementary cycles) must also have this value. Thus, as previously, Tinlet η Brayton = 1 − . Tcompressor exit A benefit of this view of efficiency is that it allows us a way to comment on the efficiency of any thermodynamic cycle. Consider the cycle shown on the right, which operates between some maximum and minimum temperatures. We can break it up into small Carnot cycles Tmax and evaluate the efficiency of each. It can be seen that the efficiency of any of the small cycles drawn will be less than the T efficiency of a Carnot cycle between Tmax and Tmin . This Tmin graphical argument shows that the efficiency of any other thermodynamic cycle operating between these maximum and minimum temperatures has an efficiency less than that of a Carnot cycle. s Arbitrary cycle operating between Tmin, Tmax Muddy points If there is an ideal efficiency for all cycles, is there a maximum work or maximum power for all cycles? (MP 1C.7) 1C-7 1.C.5 Irreversibility, Entropy Changes, and “Lost Work” Consider a system in contact with a heat reservoir during a reversible process. If there is heat Q absorbed by the reservoir at temperature T, the change in entropy of the reservoir is ∆S = Q / T . In general, reversible processes are accompanied by heat exchanges that occur at different temperatures. To analyze these, we can visualize a sequence of heat reservoirs at different temperatures so that during any infinitesimal portion of the cycle there will not be any heat transferred over a finite temperature difference. During any infinitesimal portion, heat dQrev will be transferred between the system and one of the reservoirs which is at T. If dQrev is absorbed by the system, the entropy change of the system is dS system = dQrev . T The entropy change of the reservoir is dQrev . T The total entropy change of system plus surroundings is dS reservoir = − dS total = dS system + dS reservoir = 0 . This is also true if there is a quantity of heat rejected by the system. The conclusion is that for a reversible process, no change occurs in the total entropy produced, i.e., the entropy of the system plus the entropy of the surroundings: ∆S total = 0. We now carry out the same type of analysis for an irreversible process, which takes the system between the same specified states as in the reversible process. This is shown schematically at the right, with I and R denoting the irreversible and reversible processes. In the irreversible process, the system receives heat dQ and does work dW. The change in internal energy for the irreversible process is dU = dQ − dW (Always true - first law). For the reversible process dU = TdS − dWrev . Irreversible and reversible state changes Because the state change is the same in the two processes (we specified that it was), the change in internal energy is the same. Equating the changes in internal energy in the above two expressions yields dQactual − dWactual = TdS − dWrev . 1C-8 The subscript “actual” refers to the actual process (which is irreversible). The entropy change associated with the state change is dS = dQactual 1 + [dWrev − dWactual ] . T T (C.5.1) If the process is not reversible, we obtain less work (see IAW notes) than in a reversible process, dWactual < dWrev , so that for the irreversible process, dS > dQactual . T (C.5.2) There is no equality between the entropy change dS and the quantity dQ/T for an irreversible process. The equality is only applicable for a reversible process. The change in entropy for any process that leads to a transformation between an initial state “a” and a final state “b” is therefore ∆S = Sb − Sa ≥ ∫ab dQactual T where dQactual is the heat exchanged in the actual process. The equality only applies to a reversible process. The difference dWrev − dWactual represents work we could have obtained, but did not. It is referred to as lost work and denoted by Wlost . In terms of this quantity we can write, dS = dQactual dWlost . + T T (C.5.3) The content of Equation (C.5.3) is that the entropy of a system can be altered in two ways: (i) through heat exchange and (ii) through irreversibilities. The lost work ( dWlost in Equation C.5.3) is always greater than zero, so the only way to decrease the entropy of a system is through heat transfer. To apply the second law we consider the total entropy change (system plus surroundings). If the surroundings are a reservoir at temperature T, with which the system exchanges heat, ( ) dS reservoir = dS surroundings = − dQactual . T The total entropy change is dS total = dS system + dS surroundings = 1C-9 dQactual dWlost dQactual + − T T T dWlost ≥ 0. T The quantity ( dWlost / T ) is the entropy generated due to irreversibility. dS total = Yet another way to state the distinction we are making is dS system = dS from heat transfer + dSgenerated due to = dSheat transfer + dSGen . (C.5.4) irreversible processes The lost work is also called dissipation and noted dφ. Using this notation, the infinitesimal entropy change of the system becomes: dSsystem = dSheat transfer + or TdSsystem = dQ r + dφ dφ T Equation (C.5.4) can also be written as a rate equation, dS ˙ ˙ = S = Sheat transfer + S˙Gen . dt Either of equation (C.5.4) or (C.5.5) can be interpreted to mean that the entropy of the system, S, is affected by two factors: the flow of heat Q and the appearance of additional entropy, denoted by dSGen, due to irreversibility1. This additional entropy is zero when the process is reversible and always positive when the process is irreversible. Thus, one can say that the system develops sources which create entropy during an irreversible process. The second law asserts that sinks of entropy are impossible in nature, which is a more graphic way of saying that dSGen and ṠGen are positive definite, or zero, for reversible processes. 1 dQ Q˙ The term S˙heat transfer = , or , which is associated with heat transfer to T T dt the system, can be interpreted as a flux of entropy. The boundary is crossed by heat and the ratio of this heat flux to temperature can be defined as a flux of entropy. There are no restrictions on the sign of this quantity, and we can say that this flux either contributes towards, or drains away, the system's entropy. During a reversible process, only this flux can affect the entropy of the system. This terminology suggests that we interpret entropy as a kind of weightless fluid, whose quantity is conserved (like that of matter) during a reversible process. During an irreversible process, however, this fluid is not conserved; it cannot disappear, but rather is created by sources throughout the system. While this interpretation should not be taken too literally, it provides an easy mode of expression and is in the same category of concepts such as those associated with the phrases "flux of 1 This and the following paragraph are excerpted with minor modifications from A Course in Thermodynamics, Volume I, by J. Kestin, Hemisphere Press (1979) 1C-10 (C.5.5) energy" or "sources of heat". In fluid mechanics, for example, this graphic language is very effective and there should be no objections to copying it in thermodynamics. Muddy points Do we ever see an absolute variable for entropy? So far, we have worked with deltas only (MP 1C.8) dQrev dQrev I am confused as to dS = as opposed to dS ≥ .(MP 1C.9) T T dQ For irreversible processes, how can we calculate dS if not equal to (MP T 1C.10) 1.C.6 Entropy and Unavailable Energy (Lost Work by Another Name) Consider a system consisting of a heat reservoir at T2 in surroundings (the atmosphere) at T0 . The surroundings are equivalent to a second reservoir at T0 . For an amount of heat, Q, transferred from the reservoir, the maximum work we could derive is Q times the thermal efficiency of a Carnot cycle operated between these two temperatures: Maximum work we could obtain = Wmax = Q(1 − T0 / T2 ) . (C.6.1) Only part of the heat transferred can be turned into work, in other words only part of the heat energy is available to be used as work. Suppose we transferred the same amount of heat from the reservoir directly to another reservoir at a temperature T1 < T2. The maximum work available from the quantity of heat, Q , before the transfer to the reservoir at T1 is, Wmax = Q(1 − T0 / T2 ) ; [Maximum work between T2 , T0 ]. T2 ,T0 The maximum amount of work available after the transfer to the reservoir at T1 is, Wmax = Q(1 − T0 / T1 ); [Maximum work between T1 , T0 ]. T1 ,T0 There is an amount of energy that could have been converted to work prior to the irreversible heat transfer process of magnitude E ′ , T T T T E ′ = Q 1 − 0 − 1 − 0 = Q 0 − 0 , T1 T2 T2 T1 or Q Q E ′ = T0 − . T1 T2 1C-11 However, Q / T1 is the entropy gain of the reservoir at T1 and (- Q / T2 ) is the entropy decrease of the reservoir at T2 . The amount of energy, E ′ , that could have been converted to work (but now cannot be) can therefore be written in terms of entropy changes and the temperature of the surroundings as E ′ = T0 ∆Sreservoir + ∆Sreservoir at T1 at T2 = T0 ∆Sirreversible heat transfer process E′ = “Lost work”, or energy which is no longer available as work. The situation just described is a special case of an important principle concerning entropy changes, irreversibility and the loss of capability to do work. We thus now develop it in a more general fashion, considering an arbitrary system undergoing an irreversible state change, which transfers heat to the surroundings (for example the atmosphere), which can be assumed to be at constant temperature, T0 . The change in internal energy of the system during the state change is ∆U = Q − W . The change in entropy of the surroundings is (with Q the heat transfer to the system) Q . T0 Now consider restoring the system to the initial state by a reversible process. To do this we need to do work, Wrev on the system and extract from the system a quantity of heat Qrev . (We did this, for example, in “undoing” the free expansion process.) The change in internal energy is (with the quantities Qrev and Wrev both regarded, in this example, as positive for work done by the surroundings and heat given to the surroundings)2 ∆S surroundings = − ∆Urev = −Qrev + Wrev . In this reversible process, the entropy of the surroundings is changed by Q ∆S surroundings = rev . T For the combined changes (the irreversible state change and the reversible state change back to the initial state), the energy change is zero because the energy is a function of state, ∆Urev + ∆U = 0 = Q − W + ( −Qrev + Wrev ) . Thus, Qrev − Q = Wrev − W . In the above equation, and in the arguments that follow, the quantities Qrev and Wrev are both regarded as positive for work done by the surroundings and heat given to the surroundings. Although this is not in accord with the convention we have been using, it seems to me, after writing the notes in both ways, that doing this gives easier access to the ideas. I would be interested in your comments on whether this perception is correct. 2 1C-12 For the system, the overall entropy change for the combined process is zero, because the entropy is a function of state, ∆S system;combined process = ∆Sirreversible process + ∆Sreversible process = 0. The total entropy change is thus only reflected in the entropy change of the surroundings: ∆S total = ∆Ssurroundings . The surroundings can be considered a constant temperature heat reservoir and their entropy change is given by (Q − Q ) . ∆S total = rev T0 We also know that the total entropy change, for system plus surroundings is, 0 ∆S total = ∆Sirreversible + ∆Sreversible process system + surroundings process The total entropy change is associated only with the irreversible process and is related to the work in the two processes by ∆S total = (Wrev − W ) . T0 The quantity Wrev − W represents the extra work required to restore the system to the original state. If the process were reversible, we would not have needed any extra work to do this. It represents a quantity of work that is now unavailable because of the irreversibility. The quantity Wrev can also be interpreted as the work that the system would have done if the original process were reversible. From either of these perspectives we can identify ( Wrev − W ) as the quantity we denoted previously as E ′ , representing lost work. The lost work in any irreversible process can therefore be related to the total entropy change (system plus surroundings) and the temperature of the surroundings by Lost work = Wrev − W = T0 ∆S total . To summarize the results of the above arguments for processes where heat can be exchanged with the surroundings at T0 : 1) Wrev − W represents the difference between work we actually obtained and work that would be done during a reversible state change. It is the extra work that would be needed to restore the system to its initial state. 2) For a reversible process, Wrev = W ; ∆S total = 0 1C-13 3) For an irreversible process, Wrev > W ; ∆S total > 0 4) (Wrev − W ) = E ′ = T0 ∆S total is the energy that becomes unavailable for work during an irreversible process. Muddy points Is ∆S path dependent? (MP 1C.11) Are Q rev and Wrev the Q and W going from the final state back to the initial state? (MP 1C.12) 1.C.7 Examples of Lost Work in Engineering Processes a) Lost work in Adiabatic Throttling: Entropy and Stagnation Pressure Changes A process we have encountered before is adiabatic throttling of a gas, by a valve or other device as shown in the figure at the right. The 1 2 velocity is denoted by c. There is no shaft work and no heat transfer and the flow is steady. Under these conditions we can use the first law for a control volume (the Steady Flow Energy Equation) to make a statement about the c1 c2 conditions upstream and downstream of the valve: P1 P2 Adiabatic throttling T T2 1 h + c2 / 2 = h + c2 / 2 = h , 1 1 2 2 t where ht is the stagnation enthalpy, corresponding to a (possibly fictitious) state with zero velocity. The stagnation enthalpy is the same at stations 1 and 2 if Q=W=0, even if the flow processes are not reversible. For an ideal gas with constant specific heats, h = c p T and ht = c p Tt . The relation between the static and stagnation temperatures is: Tt (γ − 1)c 2 = 1 + (γ − 1)c 2 , c2 = 1+ = 1+ 2c p T 2γ{ T RT 2a 2 Tt γ − 1 2 = 1+ M , T 2 a2 where a is the speed of sound and M is the Mach number, M = c/a. In deriving this result, use has only been made of the first law, the equation of state, the speed of sound, and the definition of the Mach number. Nothing has yet been specified about whether the process of stagnating the fluid is reversible or irreversible. When we define the stagnation pressure, however, we do it with respect to isentropic deceleration to the zero velocity state. For an isentropic process 1C-14 P2 T2 = P1 T1 γ / (γ −1) . The relation between static and stagnation pressures is T γ /(γ −1) = t . P T Pt The stagnation state is defined by Pt , Tt . In addition, sstagnation state = sstatic state . The static and stagnation states are shown below in T-s coordinates. Pt T P Tt 2 c 2 T s Figure C-1: Static and stagnation pressures and temperatures Stagnation pressure is a key variable in propulsion and power systems. To see why, we examine the relation between stagnation pressure, stagnation temperature, and entropy. The form of the combined first and second law that uses enthalpy is Tds = dh − 1 dP . ρ (C.7.1) This holds for small changes between any thermodynamic states and we can apply it to a situation in which we consider differences between stagnation states, say one state having properties (Tt , Pt ) Bt and the other having properties (Tt + dTt , Pt + dPt ) (see at right). The corresponding static states are At also indicated. Because the entropy is the same at static and T stagnation conditions, ds needs no subscript. Writing (1.C.8) in terms of stagnation B c p dTt c p dTt R 1 − − dPt . dPt = conditions yields ds = ρ t Tt Tt Tt Pt A Both sides of the above are perfect differentials and can be integrated as s dsA-B Stagnation and static states 1C-15 Pt Tt ∆s γ = ln 2 − ln 2 . R γ − 1 Tt1 Pt1 For a process with Q = W = 0, the stagnation enthalpy, and hence the stagnation temperature, is constant. In this situation, the stagnation pressure is related directly to the entropy as, Pt ∆s = − ln 2 . R Pt1 (C.7.2) Pt1 The figure on the right shows this relation on a T-s diagram. Pt2 We have seen that the entropy is related to the loss, or Tt irreversibility. The stagnation pressure plays the role of an indicator of loss if the stagnation temperature is constant. T The utility is that it is the stagnation pressure (and ∆s1-2 temperature) which are directly measured, not the entropy. The throttling process is a representation of flow through s inlets, nozzles, stationary turbomachinery blades, and the use of stagnation pressure as a measure of loss is a practice that has widespread application. Eq. (C.7.2) can be put in several useful approximate forms. First, we note that for aerospace applications we are (hopefully!) concerned with low loss devices, so that the stagnation pressure ( ) change is small compared to the inlet level of stagnation pressure ∆Pt / Pt1 = Pt − Pt 2 / Pt 1 << 1. 1 Expanding the logarithm [using ln (1-x) ≅ -x + ….], P ∆P ∆P ln t2 = ln1 − t ≈ t , Pt1 Pt1 Pt1 or, ∆s ∆Pt ≈ . R Pt1 Another useful form is obtained by dividing both sides by c2/2 and taking the limiting forms of the expression for stagnation pressure in the limit of low Mach number (M<<1). Doing this, we find: ∆Pt T∆s ≅ 2 ρc 2 / 2 c /2 The quantity on the right can be interpreted as the change in the “Bernoulli constant” for incompressible (low Mach number) flow. The quantity on the left is a non-dimensional entropy change parameter, with the term T ∆s now representing the loss of mechanical energy associated with the change in stagnation pressure. ( ) ( ) To summarize: 1) for many applications the stagnation temperature is constant and the change in stagnation pressure is a direct measure of the entropy increase 2) stagnation pressure is the quantity that is actually measured so that linking it to entropy (which is not measured) is useful 1C-16 3) we can regard the throttling process as a “free expansion” at constant temperature Tt1 from the initial stagnation pressure to the final stagnation pressure. We thus know that, for the process, the work we need to do to bring the gas back to the initial state is Tt ∆s , which is the ”lost work” per unit mass. Muddy points Why do we find stagnation enthalpy if the velocity never equals zero in the flow? (MP 1C.13) Why does Tt remain constant for throttling? (MP 1C.14) b) Adiabatic Efficiency of a Propulsion System Component (Turbine) A schematic of a turbine and the accompanying thermodynamic diagram are given in Figure C-2. There is a pressure and temperature drop through the turbine and it produces work. P1 1 . 1 m Work h or T 2 2s P2 ∆h 2 s ∆s Figure C-2: Schematic of turbine and associated thermodynamic representation in h-s coordinates There is no heat transfer so the expressions that describe the overall shaft work and the shaft work per unit mass are: • ( ) • m ht2 − ht1 = W shaft (h t2 ) (C.7.3) − ht1 = wshaft If the difference in the kinetic energy at inlet and outlet can be neglected, Equation (C.7.3) reduces to (h2 − h1 ) = wshaft . The adiabatic efficiency of the turbine is defined as actual work . ηad = ideal work ( ∆s = 0) For a given pressure ratio The performance of the turbine can be represented in an h-s plane (similar to a T-s plane for an ideal gas) as shown in Figure C-2. From the figure the adiabatic efficiency is ηad = h1 − h2 h1 − h2 s − (h2 − h2 s ) = h1 − h2 s h1 − h2 s 1C-17 The adiabatic efficiency can therefore be written as ∆h . ηad = 1 − Ideal work The non-dimensional term ( ∆h /Ideal work) represents the departure from isentropic (reversible) processes and hence a loss. The quantity ∆h is the enthalpy difference for two states along a constant pressure line (see diagram). From the combined first and second laws, for a constant pressure process, small changes in enthalpy are related to the entropy change by Tds = dh., or approximately, T2 ∆s = ∆h . The adiabatic efficiency can thus be approximated as ηad = 1 − T2 ∆s Lost work = 1− . Ideal work h1 − h2 s The quantity T∆s represents a useful figure of merit for fluid machinery inefficiency due to irreversibility. Muddy points How do you tell the difference between shaft work/power and flow work in a turbine, both conceptually and mathematically? (MP 1C.15) c) Isothermal Expansion with Friction In a more general look at the isothermal expansion, we now drop the restriction to frictionless processes. As seen in the diagram at the right, work is done to overcome friction. If the kinetic energy of the piston is negligible, a balance of forces tells us that P, T Work receiver Friction Isothermal expansion with friction Wsystem on piston = Wdone by + Wreceived . friction During the expansion, the piston and the walls of the container will heat up because of the friction. The heat will be (eventually) transferred to the atmosphere; all frictional work ends up as heat transferred to the surrounding atmosphere. W friction = Q friction 1C-18 The amount of heat transferred to the atmosphere due to the frictional work only is thus, Q friction = Wsystem − Wreceived . 12 4 4 3 on piston Work 1 4 24 3 received Work produced The entropy change of the atmosphere (considered as a heat reservoir) due to the frictional work is (∆Satm ) due to frictional = Q friction Tatm work only = Wsystem − Wreceived Tatm The difference between the work that the system did (the work we could have received if there were no friction) and the work that we actually received can be put in a (by now familiar) form as Wsystem − Wreceived = Tatm ∆Satm = Lost or unavailable work Muddy points Is the entropy change in the equations two lines above the total entropy change? If so, why does it say ∆Satm? (MP 1C.16) d) Entropy Generation, Irreversibility, and Cycle Efficiency As another example, we show the links between entropy changes and cycle efficiency for an irreversible cycle. The conditions are: i) A source of heat at temperature, T Q ii) A sink of heat (rejection of heat) at T0 T iii) An engine operating in a cycle irreversibly During the cycle the engine extracts heat Q, rejects heat Q0 and produces work,W: W = Q − Q0 . Work ∆S = ∆Sengine + ∆Ssurroundings . The engine operates in a cycle and the entropy change for the complete cycle is zero. Therefore, ∆S = 0 + ∆Sheat + ∆Sheat . source sin k 144 244 3 ∆Ssurroundings The total entropy change is, ∆S total = ∆Sheat + ∆Sheat = source sin k −Q Q0 + . T T0 1C-19 To Qo Suppose we had an ideal reversible engine working between these same two temperatures, which extracted the same amount of heat, Q, from the high temperature reservoir, and rejected heat of magnitude Q0rev to the low temperature reservoir. The work done by this reversible engine is Wrev = Q − Q0 rev . For the reversible engine the total entropy change over a cycle is ∆S total = ∆Sheat + ∆Sheat = source sin k −Q Q0 rev + = 0. T T0 Combining the expressions for work and for the entropy changes, Q0 = Q0 rev + Wrev − W The entropy change for the irreversible cycle can therefore be written as ∆S total = −Q Q0 rev Wrev − W + + . T T0 T0 14243 =0 The difference in work that the two cycles produce is equal to the entropy that is generated during the cycle: T0 ∆S total = Wrev − W . The second law states that the total entropy generated is greater than zero for an irreversible process, so that the reversible work is greater than the actual work of the irreversible cycle. An “engine effectiveness”, Eengine , can be defined as the ratio of the actual work obtained divided by the work that would have been delivered by a reversible engine operating between the two temperatures T and T0 . Eengine = ηengine ηreversible engine Eengine = = Actual work obtained W = Wrev Work that would be delivered by a reversible cycle between T , T0 Wrev − T0 ∆S total T ∆S total =1− 0 Wrev Wrev The departure from a reversible process is directly reflected in the entropy change and the decrease in engine effectiveness. 1C-20 Muddy points Why does ∆Sirrev=∆Stotal in this example? (MP 1C.17) In discussing the terms "closed system" and "isolated system", can you assume that you are discussing a cycle or not? (MP 1C.18) Does a cycle process have to have ∆S=0? (MP 1C.19) In a real heat engine, with friction and losses, why is ∆S still 0 if TdS=dQ+dφ? (MP 1C.20) e) Propulsive Power and Entropy Flux The final example in this section combines a number of ideas presented in this subject and in Unified in the development of a relation between entropy generation and power needed to propel a vehicle. Figure C-3 shows an aerodynamic shape (airfoil) moving through the atmosphere at a constant velocity. A coordinate system fixed to the vehicle has been adopted so that we see the airfoil as fixed and the air far away from the airfoil moving at a velocity c0 . Streamlines of the flow have been sketched, as has the velocity distribution at station “0” far upstream and station “d” far downstream. The airfoil has a wake, which mixes with the surrounding air and grows in the ∆c c0 A2 A0 Wake Streamlines (control surface) A1 Actual wake profile Figure C-3: Airfoil with wake and control volume for analysis of propulsive power requirement downstream direction. The extent of the wake is also indicated. Because of the lower velocity in the wake the area between the stream surfaces is larger downstream than upstream. We use a control volume description and take the control surface to be defined by the two stream surfaces and two planes at station 0 and station d. This is useful in simplifying the analysis because there is no flow across the stream surfaces. The area of the downstream plane control surface is broken into A1, which is area outside the wake and A2 , which is the area occupied by wake fluid, i.e., fluid that has suffered viscous losses. The control surface is also taken far enough away from the vehicle so that the static pressure can be considered uniform. For fluid which is not in the wake (no viscous forces), the momentum equation is cdc = − dP / ρ . Uniform static pressure therefore implies uniform velocity, so that on A1 the velocity is equal to the upstream value, c0 . The downstream velocity profile is actually continuous, as indicated. It is approximated in the analysis as a step change to make the algebra a bit simpler. (The conclusions apply to the more 1C-21 general velocity profile as well and we would just need to use integrals over the wake instead of the algebraic expressions below.) The equation expressing mass conservation for the control volume is ρ 0 A0 c0 = ρ 0 A1c0 + ρ 2 A2 c2 . (C.7.5) The vertical face of the control surface is far downstream of the body. By this station, the wake fluid has had much time to mix and the velocity in the wake is close to the free stream value, c0 . We can thus write, wake velocity = c2 = (c0 − ∆c) ; ∆c / c0 << 1. (C.7.6) (We chose our control surface so the condition ∆c / c0 << 1 was upheld.) The integral momentum equation (control volume form of the momentum equation) can be used to find the drag on the vehicle. ρ0 A0 c02 = − Drag + ρ0 A1c02 + ρ2 A2 c22 . (C.7.7) There is no pressure contribution in Eq. (C.7.7) because the static pressure on the control surface is uniform. Using the form given for the wake velocity, and expanding the terms in the momentum equation out we obtain, [ ρ0 A0 c02 = − Drag + ρ0 A1c02 + ρ2 A2 c02 − 2c0 c2 + ( ∆c)2 ] (C.7.8) The last term in the right hand side of the momentum equation, ρ 2 A2 ( ∆c) 2 , is small by virtue of the choice of control surface and we can neglect it. Doing this and grouping the terms on the right hand side of Eq. (C.7.8) in a different manner, we have [ ] c0 [ ρ0 A1c0 ] = c0 ρ0 A1c0 + ρ2 A2 (c0 − ∆c) + {− Drag − ρ2 A2 c0 ∆c} The terms in the square brackets on both sides of this equation are the continuity equation multiplied by c0 . They thus sum to zero leaving the curly bracketed terms as Drag = − ρ2 A2 c0 ∆c . (C.7.9) The wake mass flow is ρ2 A2 c2 = ρ2 A2 (c0 − ∆ c). All this flow has a velocity defect (compared to the free stream) of ∆c , so that the defect in flux of momentum (the mass flow in the wake times the velocity defect) is, to first order in ∆c , Momentum defect in wake = − ρ2 A2 c0 ∆c , = Drag. 1C-22 The combined first and second law gives us a means of relating the entropy and velocity: Tds = dh − dP / ρ . The pressure is uniform (dP=0) at the downstream station. There is no net shaft work or heat transfer to the wake so that the mass flux of stagnation enthalpy is constant. We can also approximate that the condition of constant stagnation enthalpy holds locally on all streamlines. Applying both of these to the combined first and second law yields Tds = dht − cdc . For the present situation, dht = 0; cdc = c0 ∆c , so that T0 ∆s = − c0 ∆c (C.7.10) In Equation (C.7.10) the upstream temperature is used because differences between wake quantities and upstream quantities are small at the downstream control station. The entropy can be related to the drag as Drag = ρ 2 A2 T0 ∆s (C.7.11) The quantity ρ 2 A2 c0 ∆s is the entropy flux (mass flux times the entropy increase per unit mass; in the general case we would express this by an integral over the locally varying wake velocity and density). The power needed to propel the vehicle is the product of drag x flight speed, Drag × co . From Eq. (C.7.11), this can be related to the entropy flux in the wake to yield a compact expression for the propulsive power needed in terms of the wake entropy flux: Propulsive power needed = T0 ( ρ 2 A2 c0 ∆s) = T0 × Entropy flux in wake (C.7.12) This amount of work is dissipated per unit time in connection with sustaining the vehicle motion. Equation (C.7.12) is another demonstration of the relation between lost work and entropy generation, in this case manifested as power that needs to be supplied because of dissipation in the wake. Muddy points Is it safe to say that entropy is the tendency for a system to go into disorder? (MP 1C.21) 1.C.8 Some Overall Comments on Entropy, Reversible and Irreversible Processess [Mainly excerpted (with some alterations) from: Engineering Thermodynamics, William C. Reynolds and Henry C. Perkins, McGraw-Hill Book Company, 1977] 1C-23 Muddy points Isn't it possible for the mixing of two gases to go from the final state to the initial state? If you have two gases in a box, they should eventually separate by density, right? (MP 1C.22) Muddiest Points on Part 1C 1C.1 So, do we lose the capability to do work when we have an irreversible process and entropy increases? Absolutely. We will see this in a more general fashion very soon. The idea of lost work is one way to view what “entropy is all about”! 1C.2 Why do we study cycles starting with the Carnot cycle? Is it because it is easier to work with? Carnot cycles are the best we can do in terms of efficiency. We use the Carnot cycle as a standard against which all other cycles are compared. We will see in class that we can break down a general cycle into many small Carnot cycles. Doing this we can gain insight in which direction the design of efficient cycles should go. 1C.3 How does one interpret h-s diagrams? I find h-s diagrams useful, especially in dealing with propulsion systems, because the difference in stagnation h can be related (from the Steady Flow Energy Equation) to shaft work and heat input. For processes that just have shaft work (compressors or turbines) the change in stagnation enthalpy is the shaft work. For processes that just have heat addition or rejection at constant pressure, the change in stagnation enthalpy is the heat addition or rejection. 1C.4 Is it always OK to "switch" T-s and h-s diagram? No! This is only permissible for perfect gases with constant specific heats. We will see, when we examine cycles with liquid-vapor mixtures, that the h-s diagrams and the T-s diagrams look different. 1C.5 What is the best way to become comfortable with T-s diagrams? I think working with these diagrams may be the most useful way to achieve this objective. In doing this, the utility of using these coordinates (or h-s coordinates) should also become clearer. I find that I am more comfortable with T-s or h-s diagrams than with P-v diagrams, especially the latter because it conveys several aspects of interest to propulsion engineers: work produced or absorbed, heat produced or absorbed, and loss. 1C.6 What is a reversible adiabat physically? Let's pick an example process involving a chamber filled with a compressible gas and a piston. We assume that the chamber is insulated (so no heat-transfer to or from the chamber) and the process is thus adiabatic. Let us also assume that the piston is ideal, such that there is no friction between the walls of the chamber and the piston. The gas is at some Temperature T1. We now push the piston in and compress the gas. The internal energy of the system will then increase by the amount of work we put in and the gas will heat up and be at higher pressure. If we now let the piston expand again, it will return to its original position (no friction, ideal piston) and the work we took from the environment will be returned (we get the exact same amount of work back and leave no mark on the environment). Also, the temperature and the pressure of the gas return to the initial values. We thus have an adiabatic reversible process. For both compression and expansion we have no change in entropy of the system because there is no heat transfer and also no irreversibility. If we now draw this process in the h-s or T-s diagram we get a vertical line since the entropy stays constant: S = constant or ∆S = 0 and we can also call this process an isentropic process. 1C.7 If there is an ideal efficiency for all cycles, is there a maximum work or maximum power for all cycles? Yes. As with the Brayton cycle example, we could find the maximum as a function of the appropriate design parameters. 1C.8 Do we ever see an absolute variable for entropy? So far, we have worked with deltas only. It is probably too strong a statement to say that for “us” the changes in entropy are what matters, but this has been my experience for the type of problems aerospace engineers work on. Some values of absolute entropy are given in Table A.8 in SB&VW. We will also see, in the lectures on Rankine cycles, that the entropy of liquid water at a temperature of 0.01 C and a pressure of 0.6113 kPa has been specified as zero for problems involving two-phase (steam and water) behavior. 1C.9 I am confused as to dS = dQrev dQrev as opposed to dS ≥ . T T Both of these are true and both can always be used. The first is the definition of entropy. The second is a statement of how the entropy behaves. Section 1C.5 attempts to make dQ dWlost . the relationship clearer through the development of the equality dS = + T T dQ ? T We need to define a reversible process between the two states in order to calculate the entropy (see muddy point 3, above). See VN Chapter 5 (especially) for discussion of entropy or section 1C.5. If you are still in difficulty, come and see me. 1C.10 For irreversible processes, how can we calculate dS if not equal to 1C.11 Is ∆S path dependent? No. Entropy is a function of state (see Gibbs) and thus ∆S is path independent. For example we could have three different paths connecting the same two states and therefore have the same change in entropy ∆S path I = ∆S path II = ∆S path III . 1C.12 Are Q rev and Wrev the Q and W going from the final state back to the initial state? Yes. We have an irreversible process from state 1 to state 2. We then used a reversible process to restore the initial state 1 (we had to do work on the system and extract heat from the system). 1C.13 Why do we find stagnation enthalpy if the velocity never equals zero in the flow? The stagnation enthalpy (or temperature) is a useful reference quantity. Unlike the static temperature it does not vary along a streamline in an adiabatic flow, even if irreversible. It was thus the natural reference temperature to use in describing the throttling process. In addition, changes in stagnation pressure are direct representations of the shaft work or heat associated with a fluid component. The enthalpy is not, unless we assume that changes in KE are small. Measurement of stagnation temperature thus allows direct assessment of shaft work in a turbine or compressor, for example. 1C.14 Why does Tt remain constant for throttling? Because for a steady adiabatic flow with no shaft work done the Steady Flow Energy Equation yields constant stagnation enthalpy even though the flow processes might not be reversible (see notes). For a perfect gas h = cpT, thus the stagnation temperature remains constant. 1C.15 How do you tell the difference between shaft work/power and flow work in a turbine, both conceptually and mathematically? Let us look at the expansion of a flow through a turbine using both the control mass approach and the control volume approach. Using the control mass approach we can model the situation by tracking 1kg of air as follows: state 1 – before the expansion we have 1kg of air upstream of the turbine. We then push the gas into the turbine and expand it through the blade rows. After the expansion we take 1kg of air out of the turbine and the mass of air is downstream of the turbine – state 2. The work done by the gas is work done by the turbine (blades moved around by the gas) plus the work done by pressures (flow work). Using the first law we can then write for the change of internal energy of 1kg of air: u2 – u1 = - wshaft + p1v1 – p2v2 (adiabatic turbine: dq = 0) When entering the turbine, the fluid has to push the surroundings out of the way to make room for itself (it has volume v1 and is at p1) – the work to do this is +p1v1. When leaving the turbine the fluid is giving up room and the work to keep that volume v2 at pressure p2 is freed; thus –p2v2. We can then write for the shaft work wshaft = u1 + p1v1 – (u2 + p2v2). The right hand side of the above equation is the change in enthalpy (h1 - h2). This is another example to show how useful enthalpy is (enthalpy is the total energy of a fluid: the internal energy plus the extra energy that it is credited with by having a volume v at pressure p). The shaft work outputted by the turbine is equal to the change in enthalpy (enthalpy contains the flow work!). wshaft = h1 - h2. We can also solve this problem by using the 1st law in general form (control volume approach). . . . . . d/dt{Σ ECV} = Σ Q + Σ Wshaft + Σ Wshear + Σ Wpiston + Σ m ( h + _ c2 + gz). Note that in this form the flow work is buried in h already! For this turbine, we can drop the unsteady term on the left and neglect heat fluxes (adiabatic turbine), shear work and piston work (no pistons but blades, so we keep the shaft work). Further we assume that changes in potential energy and kinetic energy are negligible and we obtain for 1kg/s air 0 = - wshaft + h1 - h2. We obtain the same result as before: wshaft = h1 - h2. 1C.16 Is the entropy change in the equations two lines above the total entropy change? If so, why does it say ∆Satm? The entropy change in question is the entropy change due to the heat produced by friction only. Wsystem − Wreceived Q friction to frictional = = . (∆Satm )due work only Tatm Tatm ( ) 1C.17 Why does ∆Sirrev=∆Stotal in this example? When we wrote this equality, we were considering a system that was returned to its original state, so that there were no changes in any of the system properties. All evidence of irreversibility thus resides in the surroundings. 1C.18 In discussing the terms "closed system" and "isolated system", can you assume that you are discussing a cycle or not? The terms closed system and isolated system have no connection to whether we are discussing a cycle or not. They are attributes of a system (any system), whether undergoing cyclic behavior, one-way transitions, or just sitting there. 1C.19 Does a cycle process have to have ∆S=0? Entropy is a state function. If the process is cyclic, initial and final states are the same. So, for a cyclic process, ∆S = 0 . 1C.20 In a real heat engine, with friction and losses, why is ∆S still 0 if TdS=dQ+dφ? The change in entropy during a real cycle is zero because we are considering a complete cycle (returning to initial state) and entropy is a function of state (holds for both ideal and real cycles!). Thus if we integrate dS = dQ/T + dΦ/T around the real cycle we will obtain ∆Scycle = 0. What essentially happens is that all irreversibilities (dΦ's) are turned into additional heat that is rejected to the environment. The amount of heat rejected in the real cycle QRreal is going to be larger than the amount of heat rejected in an ideal cycle QRideal QRideal = QA TR/TA (from ∆Scycle = 0) QRreal = QA TR/TA + TR∆SΦ (from ∆Scycle = 0) We will see this better using the T-s diagram. The change of entropy of the surroundings (heat reservoirs) is ∆Ssurr = -QA/TA + QR real/T R = ∆SΦ > 0. So ∆Scycle = 0 even for real cycles, but ∆Stotal = ∆Scycle + ∆Ssurr = ∆SΦ > 0. 1C.21 Is it safe to say that entropy is the tendency for a system to go into disorder? Entropy can be given this interpretation from a statistical perspective, and this provides a different, and insightful view of this property. At the level in which we have engaged the concept, however, we focus on the macroscopic properties of systems, and there is no need to address the idea of order and disorder ; as we will see, entropy is connected to the loss of our ability to do work, and that is sufficient to make it a concept of great utility for the evaluation and design of engineering systems. We will look at this in a later lecture. If you are interested in pursuing this, places to start might be the book by Goldstein and Goldstein referred to above, Great Ideas in Physics by Lightman ( paperback book by an MIT professor), or Modern Thermodynamics, by Kondepudi and Prigogine. 1C.22 Isn't it possible for the mixing of two gases to go from the final state to the initial state? If you have two gases in a box, they should eventually separate by density, right? Let us assume that gas X is oxygen and gas Y is nitrogen. When the membrane breaks the entire volume will be filled with a mixture of oxygen and nitrogen. This may be considered as a special case of an unrestrained expansion, for each gas undergoes an unrestrained expansion as it fills the entire volume. It is impossible for these two uniformly mixed gases to separate without help from the surroundings or environment. A certain amount of work is necessary to separate the gases and to bring them back into the left and right chambers. 1.D: Interpretation of Entropy on the Microscopic Scale - The Connection between Randomness and Entropy 1.D.1 Entropy Change in Mixing of Two Ideal Gases Consider an insulated rigid container of gas separated into two halves by a heat conducting partition so the temperature of the gas in each part is the same. One side contains air, the other side another gas, say argon, both regarded as ideal gases. The mass of gas in each side is such that the pressure is also the same. The entropy of this system is the sum of the entropies of the two parts: Ssystem = Sair + Sarg on . Suppose the partition is taken away so the gases are free to diffuse throughout the volume. For an ideal gas, the energy is not a function of volume, and, for each gas, there is no change in temperature. (The energy of the overall system is unchanged, the two gases were at the same temperature initially, so the final temperature is the same as the initial temperature.) The entropy change of each gas is thus the same as that for a reversible isothermal expansion from the initial specific volume vi to the final specific volume, v f . For a mass m of ideal gas, the ( ) entropy change is ∆S = mR ln v f vi . The entropy change of the system is [ ] [ ] ∆Ssystem = ∆Sair + ∆Sarg on = mair Rair ln ( v f )air ( vi )air + marg on Rarg on ln ( v f )arg on ( vi )arg on . (D.1.1) Equation (D.1.1) states that there is an entropy increase due to the increased volume that each gas is able to access. Examining the mixing process on a molecular level gives additional insight. Suppose we were able to see the gas molecules in different colors, say the air molecules as white and the argon molecules as red. After we took the partition away, we would see white molecules start to move into the red region and, similarly, red molecules start to come into the white volume. As we watched, as the gases mixed, there would be more and more of the different color molecules in the regions that were initially all white and all red. If we moved further away so we could no longer pick out individual molecules, we would see the growth of pink regions spreading into the initially red and white areas. In the final state, we would expect a uniform pink gas to exist throughout the volume. There might be occasional small regions which were slightly more red or slightly more white, but these fluctuations would only last for a time on the order of several molecular collisions. In terms of the overall spatial distribution of the molecules, we would say this final state was more random, more mixed, than the initial state in which the red and white molecules were confined to specific regions. Another way to say this is in terms of “disorder”; there is more disorder in the final state than in the initial state. One view of entropy is thus that increases in entropy are connected with increases in randomness or disorder. This link can be made rigorous and is extremely useful in describing systems on a microscopic basis. While we do not have scope to examine this topic in depth, the purpose of Section 1.D is to make plausible the link between disorder and entropy through a statistical definition of entropy. 1D-1 1.D.2 Microscopic and Macroscopic Descriptions of a System The microscopic description of a system is the complete description of each particle in this system. In the above example, the microscopic description of the gas would be the list of the state of each molecule: position and velocity in this problem. It would require a great deal of data for this description; there are roughly 1019 molecules in a cube of air one centimeter on a side at room temperature and pressure. The macroscopic description, which is in terms of a few (two!) properties is thus far more accessible and useable for engineering applications, although it is restricted to equilibrium states. To address the description of entropy on a microscopic level, we need to state some results concerning microscopic systems. These, and the computations and arguments below are taken almost entirely from the excellent discussion in Chapter 6 of Engineering Thermodynamics by Reynolds and Perkins (1977)*. For a given macroscopic system, there are many microscopic states. A key idea from quantum mechanics is that the states of atoms, molecules, and entire systems are discretely quantized. This means that a system of particles under certain constraints, like being in a box of a specified size, or having a fixed total energy, can exist in a finite number of allowed microscopic states. This number can be very big, but it is finite. The microstates of the system keep changing with time from one quantum state to another as molecules move and collide with one another. The probability for the system to be in a particular quantum state is defined by its quantum-state probability pi . The set of the pi is called the distribution of probability. The sum of the probabilities of all the allowed quantum states must be unity, hence for any time t, ∑ pi = 1 (D.2.1) i When the system reaches equilibrium, the individual molecules still change from one quantum state to another. In equilibrium, however, the system state does not change with time; so the probabilities for the different quantum states are independent of time. This distribution is then called the equilibrium distribution, and the probability pi can be viewed as the fraction of time a system spends in the i th quantum state. In what follows, we limit consideration to equilibrium states. We can get back to macroscopic quantities from the microscopic description using the probability distribution. For instance, the macroscopic energy of the system would be the weighted average of the successive energies of the system (the energies of the quantum states); weighted by the relative time the system spends in the corresponding microstates. In terms of probabilities, the average energy, E , is E = ∑ pi ε i , where ε i is the energy of a quantum state. i * Reynolds, W.C., and Perkins, H.C., Engineering Thermodynamics, McGraw-Hill Book Co., 1977. 1D-2 (D.2.2) The probability distribution provides information on the randomness of the equilibrium quantum states. For example, suppose the system can only exist in three states (1, 2 and 3). If the distribution probability is p1=1, p2=0, p3=0 the system is in quantum state 1 and there is no randomness. If we were asked what quantum state the system is in, we would be able to say it is always in state 1. If the distribution were p1=0.3, p2=0.2, p3=0.5 p1=0.5, p2=0.2, p3=0.3 or the randomness would not be zero and would be equal in both cases. We would be more uncertain about the instantaneous quantum state than in the first situation. Maximum randomness corresponds to the case where the three states are equally probable: p1=1/3, p2=1/3, p3=1/3 In this case, we can only guess the instantaneous state with 33 per cent probability. 1.D.3 A Statistical Definition of Entropy The list of the pi is a precise description of the randomness in the system, but the number of quantum states in almost any industrial system is so high this list is not useable. We thus look for a single quantity, which is a function of the pi that gives an appropriate measure of the randomness of a system. As shown below, the entropy provides this measure. There are several attributes that the sought for function should have. The first is that the average of the function over all of the microstates should have an extensive behavior, in other words the microscopic description of the entropy of a system C, composed of parts A and B is given by SC = SA + SB . (D.3.1) Second is that entropy should increase with randomness and should be largest for a given energy when all the quantum states are equiprobable. The average of the function over all the microstates is defined by S = f = ∑ pi f ( pi ) (D.3.2) i where the function f ( pi ) is to be found. Suppose that system A has n microstates and system B has m microstates. The entropies of systems A, B, and C, are defined by 1D-3 n SA = ∑ pi f ( pi ) (D.3.3a) i =1 m ( ) = ∑ ∑ p f (p ) =∑ p f (p )∑ p f (p ) SB = ∑ p j f p j SC j =1 n m i =1 j =1 m n ij ij i =1 (D.3.3b) i i j =1 j (D.3.3c) j In Equations (D.3.2 and D.3.3), the term pij means the probability of a microstate in which system A is in state i and system B is in state j. For Equation (D.3.1) to hold, n m i =1 j =1 ( ) n m i =1 j =1 ( ) SC = ∑ pi f ( pi ) ∑ p j f p j = ∑ pi f ( pi ) + ∑ p j f p j = SA + SB . (D.3.4) The function f must be such that this is true regardless of the values of the probabilities pi and p j . This will occur if f ( ) = ln( ) because ln( a ⋅ b) = ln( a) + ln(b) . Making this substitution, the expression for Sc in Equation (D.3.4) becomes n m n m i =1 j =1 i =1 j =1 ( ) ∑ ∑ pi p j ln( pi ) + ∑ ∑ pi p j ln p j . (D.3.5a) Rearranging the sums, D.3.5a becomes n m m n ∑ pi ln( pi ) ∑ p j + ∑ p j ln p j ∑ pi . i =1 i =1 j =1 j =1 ( ) (D.3.5b) Because n m i =1 j =1 ∑ pi = ∑ p j = 1, (D.3.6) the left hand side of Equation (D.3.4) can be written as n m i =1 j =1 ( ) SC = ∑ pi ln( pi ) + ∑ p j ln p j . (D.3.7) This means that Equation (D.3.4) is satisfied for any pi , p j , n, m . Reynolds and Perkins show that the most general f ( pi ) is f = C ln( pi ) , where C is an arbitrary constant. Because the pi are less than unity, the constant is chosen to be negative to make the entropy positive. Based on the above a the statistical definition of entropy can be given as: S = − k ∑ pi ln pi . (D.3.8) i 1D-4 The constant k is known as the Boltzmann constant, k = 1.380 × 10 −23 J . K (D.3.9) The value of k is (another wonderful result!) given by k = R/ N Avogadro , where R is the universal gas constant, 8.3143 J/(mol-K) and N Avogadro is Avogadro's number, 6.02 × 10 23 molecules per mol. Sometimes k is called the gas constant per molecule. With this value for k, the statistical definition of entropy is identical with the macroscopic definition of entropy. 1.D.4 Connection between the Statistical Definition of Entropy and Randomness We need now to examine the behavior of the statistical definition of entropy as regards randomness. Because a uniform probability distribution reflects the largest randomness, a system with n allowed states will have the greatest entropy when each state is equally likely. In this situation, the probabilities become pi = p = 1 , Ω (D.4.1) where Ω is the total number of microstates. The entropy is thus 1 1 1 1 1 ln = − kΩ ln = − k ln Ω Ω Ω i =1 Ω Ω Ω S = −k ∑ S = k ln Ω . (D.4.2) Equation (D.4.1) states that the larger the number of possible states the larger the entropy. The behavior of the entropy stated in Equation (D.4.2) can be summarized as follows: a) S is maximum when Ω is maximum, which means many permitted quantum states, hence much randomness, b) S is minimum when Ω is minimum. In particular, for Ω=1, there is no randomness and S=0. These trends are in accord with our qualitative ideas concerning randomness. Equation (D.4.2) is carved on Boltzmann's tombstone (he died about a hundred years ago) in Vienna. We can also examine the additive property of entropy with respect to probabilities. If we have two systems, A and B, which are viewed as a combined system, C, the quantum states for the combined system are the combinations of the quantum states from A and B. The quantum state where A is in its state x and B in its state y would have a probability pAx ⋅ pBy because the two probabilities are independent. The number of probabilities for the combined system, ΩC , is thus defined by ΩC = Ω A ⋅ Ω B . The entropy of the combined system is 1D-5 SC = k ln(Ω A ⋅ Ω B ) = k ln Ω A + k ln Ω B = SA + SB (D.4.3) Equation (D.4.2) is sometimes taken as the basic definition of entropy, but it should be remembered that it is only appropriate when each quantum state is equally likely. Equation (D.3.8) is more general and applies equally for equilibrium and non-equilibrium situations. A simple numerical example shows trends in entropy changes and randomness for a system which can exist in three states. Consider the five probability distributions p2 = 0, p3 = 0 ; S = − k (1ln1 + 0 ln 0 + 0 ln 0) = 0 p2 = 0.2, p3 = 0 ; S = − k[0.8 ln(0.8) + 0.2 ln(0.2) + 0 ln(0)] = 0.500 k p2 = 0.1, p3 = 0.1; S = − k[0.8 ln(0.8) + 0.1ln(0.1) + 0.1ln(0.1)] = 0.639k p2 = 0.3, p3 = 0.2 ; S = − k[0.5 ln(0.5) + 0.3 ln(0.3) + 0.2 ln(0.2)] = 1.03k 1 1 v) p1 = 1 / 3, p2 = 1 / 3, p3 = 1 / 3 ; S = −3k ln = 1.099k . 3 3 i) ii) iii) iv) p1 = 1.0, p1 = 0.8, p1 = 0.8, p1 = 0.5, The first distribution has no randomness. For the second, we know that state 3 is never found. States (iii) and (iv) have progressively greater uncertainty about the distribution of states and thus higher randomness. State (v) has the greatest randomness and uncertainty and also the largest entropy. 1.D.5. Numerical Example of the Approach to the Equilibrium Distribution Reynolds and Perkins give a numerical example which illustrates the above concepts and also the tendency of a closed isolated system to tend to equilibrium. Reynolds and Perkins,Engineering Thermodynamics, McGraw-Hill, 1977. Sec. 6.7. pp.177-183. 1D-6 1.D.6 Summary and Conclusions a) Entropy as defined from a microscopic point of view is a measure of randomness in a system. b) The entropy is related to the probabilities pi of the individual quantum states of the system by S = − k ∑ pi ln pi i where k, the Boltzmann constant is given by R/ N Avogadro . c) For a system in which there are Ω quantum states, all of which are equally probable (for which the probability is pi = 1/ Ω ), the entropy is given by S = k ln Ω . The more quantum states, the more the randomness and uncertainty that a system is in a particular quantum state. d) From the statistical point of view there is a finite, but exceedingly small possibility that a system that is well mixed could suddenly "unmix" and that all the air molecules in the room could suddenly come to the front half of the room. The unlikelihood of this is well described by Denbigh [Principles of Chemical Equilibrium, 1981] in a discussion of the behavior of an isolated system: "In the case of systems containing an appreciable number of atoms, it becomes increasingly improbably that we shall ever observe the system in a non-uniform condition. For example, it is calculated that the probability of a relative change of density, ∆ρ ρ , of only 0.001% in 1 8 cm3 of air is smaller than 10 −10 and would not be observed in trillions of years. Thus, according to the statistical interpretation the discovery of an appreciable and spontaneous decrease in the entropy of an isolated system, if it is separated into two parts, is not impossible, but exceedingly improbable. We repeat , however, that it is an absolute impossibility to know when it will take place." 1D-7 e) The definition of entropy in the form S = − k ∑ pi ln pi arises in other aerospace fields, notably i that of information theory. In this context, the constant k is taken as unity and the entropy becomes a dimensionless measure of the uncertainty represented by a particular message. There is no underlying physical connection with thermodynamic entropy, but the underlying uncertainty concepts are the same. f) The presentation of entropy in this subject is focused on the connection to macroscopic variables and behavior. These involve the definition of entropy given in Section 1.B of the notes and the physical link with lost work, neither of which makes any mention of molecular (microscopic) behavior. The approach in other sections of the notes is only connected to these macroscopic processes and does not rely at all upon the microscopic viewpoint. Exposure to the statistical definition of entropy, however, is helpful as another way not only to answer the question of "What is entropy?" but also to see the depth of this fundamental concept and the connection with other areas of technology. 1D-8 PART 2 POWER AND PROPULSION CYCLES PART 2 – POWER AND PROPULSION CYCLES 2A – Gas Power and Propulsion Cycles [SB&VW - 11.8, 11.9, 11.10, 11.11, 11.12, 11.13, 11.14] In this section we analyze several gas cycles used in practical applications for propulsion and power generation, using the air standard cycle. The air standard cycle is an approximation to the actual cycle behavior, and the term specifically refers to analysis using the following assumptions: • Air is the working fluid (the presence of combustion products is neglected) • Combustion is represented by heat transfer from an external heat source • The cycle is ‘completed’ by heat transfer to the surroundings • All processes are internally reversible • Air is a perfect gas with constant specific heats 2.A.1 The Internal combustion engine (Otto Cycle) The different processes of an idealized Otto cycle (internal combustion engine) are shown in Figure 2A-1: 3 P Adiabatic reversible QH 2 4 P0 QL 5 1 V2 = V3 V1 = V4 V Figure 2A-1: Ideal Otto cycle i. ii. iii. iv. v. vi. vii. Intake stroke, gasoline vapor and air drawn into engine (5 -> 1) Compression stroke, P, T increase (1->2) Combustion (spark), short time, essentially constant volume (2->3) Model: heat absorbed from a series of reservoir at temperatures T2 to T3 Power stroke: expansion (3 ->4) Valve exhaust: valve opens, gas escapes (4->1) Model: rejection of heat to series of reservoirs at temperatures T4 to T1 Exhaust stroke, piston pushes remaining combustion products out of chamber 1->5 2A-1 The actual cycle does not have these sharp transitions between the different processes and might be as sketched in Figure 2A-2 Not P isentropic Exhaust valve opens Spark P0 Exhaust valve closes V Figure 2A-2: Sketch of actual Otto cycle Efficiency of an ideal Otto cycle The starting point is the general expression for the thermal efficiency of a cycle: η= Q + QL Q work = H = 1+ L . heat input QH QH The convention, as previously, is that heat exchange is positive if heat is flowing into the system or engine, so QL is negative. The heat absorbed occurs during combustion when the spark occurs, roughly at constant volume. The heat absorbed can be related to the temperature change from state 2 to state 3 as: QH = Q23 = ∆U23 (W23 = 0) = ∫TT23 Cv dT = Cv (T3 − T2 ) The heat rejected is given by (for a perfect gas with constant specific heats) QL = Q41 = ∆U 41 = Cv (T1 − T4 ) Substituting the expressions for the heat absorbed and rejected in the expression for thermal efficiency yields η = 1− T4 − T1 T3 − T2 2A-2 We can simplify the above expression using the fact that the processes from 1 to 2 and from 3 to 4 are isentropic: T4 V1γ −1 = T3V2γ −1 , T1V1γ −1 = T2 V2γ −1 (T4 − T1 )V1γ −1 = (T3 − T2 )V2γ −1 T4 − T1 V2 = T3 − T2 V1 γ −1 V1 = r is called the compression ratio. In terms of compression ratio, the V2 efficiency of an ideal Otto cycle is: The quantity ηOtto = 1 − V1 V2 γ −1 1 = 1 − γ −1 . r The ideal Otto cycle efficiency is shown at the right, as a function of the compression ratio. As the compression ratio, r, increases, ηOtto increases, but so does T2 . If T2 is too high, the mixture will ignite without a spark (at the wrong location in the cycle). Thermal efficiency, ηth 70 1 60 50 40 30 20 10 0 0 2 4 6 8 10 12 Compression ratio, rv 14 Ideal Otto cycle thermal efficiency Engine work, rate of work per unit enthalpy flux: The non-dimensional ratio of work done (the power) to the enthalpy flux through the engine is given by ˙ η W˙ Q Power = = 23 Otto ˙ pT1 ˙ pT1 mc Enthalpy flux mc There is often a desire to increase this quantity, because it means a smaller engine for the same power. The heat input is given by ˙ =m Q 23 ˙ fuel (∆h fuel ) , where • ∆h fuel is the heat of reaction, ie the chemical energy liberated per unit mass of fuel • ṁ fuel is the fuel mass flow rate . The non-dimensional power is ˙ fuel ∆h fuel m W˙ 1 = 1 − γ −1 ˙ pT1 ˙ mc m c pT1 r . 2A-3 16 The quantities in this equation, evaluated at stoichiometric conditions are: ˙ fuel m 1 ∆h fuel 4 × 107 ≈ , ≈ 3 ˙ m 15 c pT1 10 × 288 so, W˙ 1 ≈ 9 1 − γ −1 . ˙ pT1 mc r Muddy points How is ∆ h fuel calculated? (MP 2A.1) What are "stoichiometric conditions"? (MP 2A.2) 2.A.2. Diesel Cycle The Diesel cycle is a compression ignition (rather than spark ignition) engine. Fuel is sprayed into the cylinder at P2 (high pressure) when the compression is complete, and there is ignition without a spark. An idealized Diesel engine cycle is shown in Figure 2A-3. QH P 2 3 Adiabatic reversible 4 QL 1 V2 V3 V4 = V1 V Figure 2A-3 Ideal Diesel cycle The thermal efficiency is given by: η Diesel = 1 + η Diesel = 1 − C (T − T ) QL = 1+ v 1 4 QH C p (T3 − T2 ) (T4 T1 − 1) γT2 (T3 T2 − 1) T1 This cycle can operate with a higher compression ratio than Otto cycle because only air is compressed and there is no risk of auto-ignition of the fuel. Although for a given compression ratio the Otto cycle has higher efficiency, because the Diesel engine can be operated to higher compression ratio, the engine can actually have higher efficiency than an Otto cycle when both are operated at compression ratios that might be achieved in practice. 2A-4 Muddy points When and where do we use cv and cp? Some definitions use dU=cvdT. Is it ever dU=cpdT? (MP 2A.3) Explanation of the above comparison between Diesel and Otto. (MP 2A.4) 2.A.3 Brayton Cycle The Brayton cycle is the cycle that represents the operation of a gas turbine engine. The “simple gas turbine” can be operated in open cycle or closed cycle (recirculating working fluid) modes, as shown below. QH Fuel Heat exchanger Combustion chamber Compressor Turbine Air wnet Compressor Products Turbine wnet Heat exchanger QL (a) (b) Figure 2A-4: Gas turbine engine operating on the Brayton cycle – (a) open cycle operation, (b) closed cycle operation Efficiency of the Brayton cycle: We derived the ideal Brayton cycle efficiency in Section 1.A: Tinlet 1 = 1− η Brayton = 1 − (γ −1) / γ Tcompressor exit PR . Net work per unit mass flow in a Brayton cycle: The net mechanical work of the cycle is given by: Net mechanical work/unit mass = wturbine − wcompressor where , wcompressor = − ∆h12 = − ∆hcomp wturbine = − ∆h34 = − ∆hturb If kinetic energy changes across the compressor and turbine are neglected, the temperature ratio, TR, across the compressor and turbine is related to the enthalpy changes: ∆hcomp ∆hturb , TR − 1 = = h1 h4 2A-5 ∆hturb = − ∆hcomp h4 h1 The net work is thus h net work = ∆hcomp 4 − 1 h1 The turbine work is greater than the work needed to drive the compressor. The thermodynamic states in an enthalpy-entropy (h,s) diagram, and the work of the compressor and turbine, are shown below for an ideal Brayton cycle. h T4=Tmax 4 P3 qA wturb 3 5 wcomp qR P0 0 T0=Tinlet s Figure 2A-5: Brayton cycle in enthalpy-entropy (h-s) representation showing compressor and turbine work Muddy points What is shaft work? (MP 2A.5) 2.A.4 Brayton Cycle for Jet Propulsion: the Ideal Ramjet A schematic of a ramjet is given in Figure 2A-6 below. Station Numbers . 0 inlet 3 fuel, mf 1 exhaust 5 streamtube 4 streamtube cθ p0 T0 c3 p3 T 3 diffuser πd T4 burner τd πd c5 p5 T 5 nozzle Figure 2A-6: Ideal ramjet [adapted from J. L. Kerrebrock, Aircraft Engines and Gas Turbines] 2A-6 In the ramjet there are “no moving parts”. The processes that occur in this propulsion device are: 0->3 isentropic diffusion (slowing down) and compression, with a decrease in Mach number, M0 → M3 << 1 3->4 Constant pressure combustion, 4->5 Isentropic expansion through the nozzle. Thrust of an ideal engine ramjet The coordinate system and control volume are chosen to be fixed to the ramjet. The thrust, F, is given by: F = ṁ(c 5 − c0 ) , where c5 and co are the inlet and exit flow velocities. The thrust can be put in terms of nondimensional parameters as follows: F c = 5 ṁa0 a5 a5 c0 − , where a = γRT is the speed of sound. a0 a0 F a T = M 5 5 − M0 = M 5 5 − M0 a0 T0 ṁa0 γ Using M32 , M 42 P γ − 1 2 γ −1 << 1 in the expression for stagnation pressure, T = 1 + M 2 P , P3 ≈ PT3 = PT0 ; P4 ≈ PT4 = PT5 ; P4 ≈ P3 The ratios of stagnation pressure to static pressure at inlet and exit of the ramjet are: PT0 PT P P = 3 = 4 = e P0 P0 P0 Pe 14 24 3 1 424 3 determines M0 determines Me The ratios of stagnation to static pressure at exit and at inlet are the same, with the consequence that the inlet and exit Mach numbers are also the same. M 5 = M0 . To find the thrust we need to find the ratio of the temperature at exit and the temperature at inlet. This is given by: γ −1 2 1+ M0 TT , TT5 TT T5 2 = = 5 = 4 = τb γ −1 2 T0 T0 TT0 TT3 1+ M5 2 2A-7 where τ b is the stagnation temperature ratio across the combustor (burner). The thrust is thus: F = M0 ( τ b − 1) ṁa0 Cycle efficiency in terms of aerodynamic parameters: η Brayton = 1 − T0 Tcompressor exit = 1− T0 T T = 1 − 0 , and 0 = T3 TT0 TT0 1 , so: γ −1 2 1+ M0 2 γ −1 2 M0 2 : Ramjet thermodynamic cycle efficiency in terms of flight η Brayton = γ −1 2 1+ M0 2 Mach number, M0 . For propulsion engines, the figures of merit includes more than thrust and η Brayton . The specific impulse, I sp measures how effectively fuel is used: I sp = F F = ; Specific Impulse, ˙ fg f m ˙g m where m ˙f = f m ˙ is the fuel mass flow rate. To find the fuel-air ratio, f, we employ a control volume around the combustor and carry out an energy balance. Before doing this, however, it is useful to examine the way in which Isp appears in expressions for range. Muddy points What exactly is the specific impulse, Isp, a measure of? (MP 2A.6) How is Isp found for rockets in space where g ~ 0? (MP 2A.7) Why does industry use TSCP rather than Isp? Is there an advantage to this? (MP 2A.8) Why isn't mechanical efficiency an issue with ramjets? (MP 2A.9) How is thrust created in a ramjet? (MP 2A.10) Why don't we like the numbers 1 and 2 for the stations? Why do we go 0-3? (MP 2A.11) For the Brayton cycle efficiency, why does T3 =Tt0? (MP 2A.12) 2.A.5 The Breguet Range Equation [See Waitz Unified Propulsion Notes, No. IV (see the 16.050 Web site)] Consider an aircraft in level flight, with weight W. The rate of change of the gross weight of the vehicle is equal to the fuel weight flow: 2A-8 L D F W g ṁ f = − dW F =− dt Isp ( ) ( ) W = L= D LD = F LD The rate of change of aircraft gross weight is thus dW W =− L dt D I sp . Suppose L/D and Isp remain constant along flight path: ( ) dW dt =− L I W D sp . ( ) We can integrate this equation for the change in aircraft weight to yield a relation between the weight change and the time of flight: ln W t , where Wi is the initial weight. =− L I Wi D sp ( ) If Wf is the final weight of vehicle and tinitial=0, the relation between vehicle parameters and flight time, t f , is W L I sp ln i = t f D Wf . The range is the flight time multiplied by the flight speed, or, Range = c0 t f = L D { × c0 × I ) ({ sp propulsion system designer aircraft designer 2A-9 × ln Wi W f 123 structural designer The above equation is known as the Breguet range equation. It shows the influence of aircraft, propulsion system, and structural design parameters. Relation of overall efficiency, Isp, and thermal efficiency Suppose ∆h fuel is the heating value (‘heat of combustion’) of fuel (i.e., the energy per unit of fuel mass), in J/kg. The rate of energy release is ṁ f ∆h fuel , so F ∆h fuel Fc0 ∆h fuel c0 I sp = c0 = ˙ f g ∆h fuel m ˙ f ∆h fuel g m and Thrust power (usefulwork) Fc0 = = ηoverall (overall propulsion system efficiency) Ideal available energy ṁ f ∆h fuel g L Wi c0 I sp and Range = ∆h fuel η ln overall ∆h fuel g D Wf = ηthermalη propulsiveηcombustion ηoverall = ηoverall The Propulsion Energy Conversion Chain The above concepts can be depicted as parts of the propulsion energy conversion train mentioned in Part 0, which shows the process from chemical energy contained in the fuel to energy useful to the vehicle. Chemical 1 energy Heat ⋅ • m f ∆h fuel m ηcombustion ≅ 1 Useful work: Thrust power Mechanical work ηthermal c 2 exit 2 − c0 2 2 η propulsive = 2 1 + cexit / c0 Figure 2A-7: The propulsion energy conversion chain. The combustion efficiency is near unity unless conditions are far off design and we can regard the two main drivers as the thermal and propulsive1 efficiencies. The evolution of the overall efficiency of aircraft engines in terms of these quantities is shown below in Figure 2A-8. 1 The transmission efficiency represents the ratio between compressor and turbine power, which is less than unity due to parasitic frictional effects. As with the combustion efficiency, however, this is very close to one and the horizontal axis can thus be regarded essentially as propulsive efficiency. 2A-10 0.8 0.1 Overall Efficiency 0.3 0.4 0.2 Core Thermal Efficiency 0.8 0.7 0.6 0.5 0.5 0.6 0.4 Future trend 0.7 '777' Engines 0.6 0.3 SFC Advanced UDF CF6-80C2 Low BPR 0.5 Turbojets Current High BPR 0.4 UDF Engine 0.3 Whittle 0.2 0.3 0.4 0.5 0.6 0.7 Propulsive x Transmission Efficiency 0.8 Figure 2A-8: Overall efficiency of aircraft engines as functions of thermal and propulsive efficiencies [data from Koff]. Muddy points How can we idealize fuel addition as heat addition? (MP 2A.13) 2.A.6 Performance of the Ideal Ramjet We now return to finding the ramjet fuel-air ratio, f. Using a control volume around the burner, we get: ˙ =m Heat given to the fluid: Q ˙ f ∆h fuel = m ˙ f∆h fuel . Q 3 From the steady flow energy equation: 4 ˙ 4 ht4 − m ˙ 3 ht 3 = m ˙ 3 f∆h fuel m The exit mass flow is not greatly different from the inlet mass flow, m˙ 4 = m˙ 3 (1 + f ) ≈ m˙ 3 , because the fuel-air ratio is much less than unity (generally several percent). We thus neglect the difference between the mass flows and obtain ( ) ht4 − ht3 = c p Tt4 − Tt3 = f ∆h fuel ( ) γ −1 2 Tt3 c p (τ b − 1) = f ∆h fuel , with Tt3 = Tt0 = T0 1 + 2 M0 14243 Θ0 2A-11 Fuel-air ratio, f: f= τ b −1 , ∆h fuel c p T0Θ 0 The fuel-air ratio, f, depends on the fuel properties ( ∆h fuel ), the desired flight parameters ( Θ 0 ), the ramjet performance ( τ b ), and the temperature of the atmosphere ( T0 ). Specific impulse, Isp: The specific impulse for the ramjet is given by c0 ∆h fuel τ − ( b 1) c T F 1 p 0 I sp = = f ṁg g Θ0 (τ b − 1) The specific impulse can be written in terms of fuel properties and flight and vehicle characteristics as, a ∆h M0 . Isp = 0 fuel × gc pT0 Θ0 τ b + 1 1 424 3 14 4244 3 ( fuel properties ) flight characteristics, ramjet temp increase We wish to explore the parameter dependency of the above expression, which is a complicated formula. How can we do this? What are the important effects of the different parameters? How do we best capture the ramjet performance behavior? To make effective comparisons, we need to add some additional information concerning the operational behavior. An important case to examine is when f is such that all the fuel burns, i.e. when we have stoichiometric conditions. What happens in this situation as the flight Mach number, M0, increases? T0 is fixed so Tt3 increases, but the maximum temperature does not increase much because of dissociation: the reaction does not go to completion at high temperature. A useful approximation is therefore to take Tt4 constant for stoichiometric operation. In the stratosphere, from 10 to 30 km, T0 ≈ constant ≈ 212 K . The maximum temperature ratio is TT T τ max = max = 4 = const , T0 T0 τb = TT4 TT3 = TT4 TT3 T0 T0 = τ max Θ0 For the stoichiometric ramjet: I sp = F F a0 a0 = = M0 ( τ b − 1) ˙g m ˙ a0 f stoich g fm f stoich g 2A-12 Using the expression for τ b , the specific impulse is τ a0 I sp = M0 max − 1 Θ0 fstoich g Representative performance values: A plot of the performance of the stoichiometric ramjet is shown in Figure 2A-9. . m F qb = 10 τ = 10 f =max f stoic 2.5 F a0 2 . m a0 I g stoic f 1.5 a0 1 Isp gfstoich a0 0.5 f = fstoich 1 2 3 4 5 6 7 Figure 2A-9: Thrust per unit mass flow and specific impulse for ideal ramjet with stoichiometric combustion [adapted from Kerrebrock] M0 The figure shows that for the parameters used, the best operating range of a hydrocarbon-fueled ramjet is 2 ≤ M0 ≤ 4 . The parameters used are τ max = 10 , a0 ≈ 300 ms −1 in the stratosphere, a0 ≈ 450 s . fstoich = 0.067 for hydrocarbons g fstoich Recapitulation: In this section we have: Examined the Brayton Cycle for propulsion Found η Brayton as a function of M0 Found ηoverall and the relation between ηoverall and ηBrayton F Examined and I sp as a function of M0 for the ideal ramjet. ṁ a0 Muddy points What is the relation between h t4 − h t3 = f∆h f and the existence of the maximum value of Tt4? (MP 2A.14a) Why didn’t we have a 2s point for the Brayton cycle with non-ideal components ? (MP 2A.14b) What is the variable fstoich? (MP 2A.15) 2A-13 2.A.7 Effect of Departures from Ideal Behavior - Real Cycle behavior [See also charts 69-82 in 16.050: Gas Turbine Engine Cycles] What are the sources of non-ideal performance and departures from reversibility? - Losses (entropy production) in the compressor and the turbine - Stagnation pressure decrease in the combustor - Heat transfer We take into account here only irreversibility in the compressor and in the turbine. Because of these irreversibilities, we need more work, ∆hcomp (the changes in kinetic energy from inlet to exit of the compressor are neglected), to drive the compressor than in the ideal situation. We also get less work, ∆hturb , back from the turbine. The consequence, as can be inferred from Figure 2A-10 below, is that the net work from the engine is less than in the cycle with ideal components. P3 h T4 = Tmax 4 wturb qA P0 3 5 3s 5s wcomp qR 0 T0 = Tinlet s Figure 2A-10: Gas turbine engine (Brayton) cycle showing effect of departure from ideal behavior in compressor and turbine To develop a quantitative description of the effect of these departures from reversible behavior, consider a perfect gas with constant specific heats and neglect kinetic energy at the inlet and exit of the turbine and compressor. We define the turbine adiabatic efficiency as: ηturb = actual h − h5 wturb = 4 ideal h4 − h5S wturb actual ideal where wturb is specified to be at the same pressure ratio as wturb . (See charts 69-76 in 16.050 Gas Turbine Cycles.) There is a similar metric for the compressor, the compressor adiabatic efficiency: 2A-14 ηcomp = ideal wcomp actual wcomp = h3S − h0 h3 − h0 again for the same pressure ratio. Note that the ratio is the actual work delivered divided by the ideal work for the turbine, whereas the ratio is the ideal work needed divided by the actual work required for the compressor. These are not thermal efficiencies, but rather measures of the degree to which the compression and expansion approach the ideal processes. We now wish to find the net work done in the cycle and the efficiency. The net work is given either by the difference between the heat received and rejected or the work of the compressor and turbine, where the convention is that heat received is positive and heat rejected is negative and work done is positive and work absorbed is negative. qH + q L = (h4 − h3 ) − (h5 − h0 ) { { Net work = heat in heat out wturb + wcomp = (h4 − h5 ) − (h3 − h0 ) The thermal efficiency is: Net work ηthermal = Heat input We need to calculate T3 , T5 From the definition of ηcomp , T3 − T0 With ( T3S T0 ) (T = 3S − T0 ηc )=T ( T3S 0 γ −1 γ ) T0 −1 ηc γ −1 P γ = isentropic temperature ratio = exit = Π comp Pinlet comp γ −1 γ Π comp −1 + T0 T3 = T0 ηcomp Similarly, by the definition of η = turb ( T4 − T5 = ηturb T4 − T5S actual work received , we can find T5 : P ideal work for same ext Pinlet ) γ −1 T5S γ = ηturb T4 1 − = η 1 − Π T turb 4 Turb T4 γ −1 γ T5 = T4 − ηturb T4 1 − Π turb 2A-15 The thermal efficiency can now be found: ηthermal = 1 + T −T QL = 1− 5 0 QH T4 − T3 γ −1 1 with Π c = = Π , and τ S = Π γ the isentropic cycle temperature ratio, Πt 1 T4 1 − ηturb 1 − − T0 τS ηthermal = 1 − 1 τ S − 1) + 1 T4 − T0 ( ηcomp or, ηthermal T4 1 −τS 1 − ηcompηturb τ S T0 = T4 1 + ηc − 1 − τ S T0 There are several non-dimensional parameters that appear in this expression for thermal efficiency. We list these just below and show their effects in subsequent figures: Parameters reflecting design choices γ −1 γ τ S = Π : cycle pressure ratio T4 : maximum turbine inlet temperature T0 Parameters reflecting the ability to design and execute efficient components ηcomp : compressor efficiency ηturb : turbine efficiency In addition to efficiency, net rate of work is a quantity we need to examine, W˙ net = W˙ turbine − W˙ compressor Putting this in a non-dimensional form: W˙ net 1 1 T = − ηturb 4 1 − (τ S − 1) + ˙ c p T0 m ηcomp T τS 14402443 144 2443 work to drive compressor work extracted from flow by turbine η T4 T W˙ net 1 T 0 = (τ S − 1) − ˙ c p T0 τ η m S comp 2A-16 Trends in net power and efficiency are shown in Figure 2A-11 for parameters typical of advanced civil engines. Some points to note in the figure: • For any ηcomp , ηturb ≠ 1 , the optimum pressure ratio ( Π ) for maximum ηth is not the highest that can be achieved, as it is for the ideal Brayton cycle. The ideal analysis is too idealized in this regard. The highest efficiency also occurs closer to the pressure ratio for maximum power than in the case of an ideal cycle. Choosing this as a design criterion will therefore not lead to the efficiency penalty inferred from ideal cycle analysis. • There is a strong sensitivity to the component efficiencies. For example, for ηturb = ηcomp = 0.85 , the cycle efficiency is roughly two-thirds of the ideal value. • The maximum power occurs at a value of τ S or pressure ratio ( Π ) less than that for max η . (this trend is captured by ideal analysis). • The maximum power and maximum ηthermal are strongly dependent on the maximum temperature, T4 . T0 Muddy points T How can 4 be the maximum turbine inlet temperature? (MP 2A.16) T0 When there are losses in the turbine that shift the expansion in T-s diagram to the right, does this mean there is more work than ideal since the area is greater? (MP 2A.17) 2A-17 Figure 2A-11 available from: Cumpsty, Jet Propulsion. Cambridge University Press, February 1998. Figure 4.3 p.33. Figure 2A-11: Non-dimensional power and efficiency for a non-ideal gas turbine engine - (a) Non-dimensional work as a function of cycle pressure ratio for different values of turbine entry temperature divided by compressor entry temperature, (b) Overall cycle efficiency as a function of pressure ratio for different values of turbine entry temperature divided by compressor entry temperature, (c) Overall cycle efficiency as a function of cycle pressure ratio for different component efficiencies. [from Cumpsty, Jet Propulsion] 2A-18 A scale diagram of a Brayton cycle with non-ideal compressor and turbine behaviors, in terms of temperature-entropy (h-s) and pressure-volume (P-v) coordinates is given below as Figure 2A12. Figure 2A-12: Scale diagram of non-ideal gas turbine cycle. Nomenclature is shown in the figure. Pressure ratio 40, T0 = 288, T4 =1700, compressor and turbine efficiencies = 0.9 [adapted from Cumpsty, Jet Propulsion] Muddy points For an afterburning engine, why must the nozzle throat area increase if the temperature of the fluid is increased? (MP 2A.18a) Why doesn’t the pressure in the afterburner go up if heat is added? (MP 2A.18b) Why is the flow in the nozzle choked? (MP 2A.18c) What’s the point of having a throat if it creates a retarding force? (MP 2A.18d) Why isn’t the stagnation temperature conserved in this steady flow? (MP 2A.18e) 2A-19 Muddiest points on part 2A 2A.1 How is ∆hfuel calculated? For now, we rely on tabulated values. In the lectures accompanying Section 2.C of the notes, we will see how one can calculate the heat , ∆h fuel , liberated in a given reaction. 2A.2 What are "stoichiometric conditions"? Stoechiometric conditions are those in which the proportions of fuel and air are such that there is not an excess of each one--all the fuel is burned, and all the air (oxidizer) is used up doing it. See Notes Sections 2.C. 2A.3 When and where do we use cv and cp? Some definitions use dU=cvdT. Is it ever dU=cpdT? The answer is no. The definitions of cp and cv are derived in the notes on page 0-6. cp is the specific heat at constant pressure and for an ideal gas dh = cp dT always holds. Similarly cv is the specific heat at constant volume and for an ideal gas du = cv dT always holds. A discussion on this is also given in the notes on pages 0-6 and 0-7. If you think about how you would measure the specific heat c = q/(Tfinal – Tinitial) for a certain known change of state you could do the following experiments. For a process during which heat ∆q is transferred (reversibly) and the volume stays constant (e.g. a rigid, closed container filled with a substance, or the heat transfer in an Otto engine during combustion – the piston is near the top-dead-center and the volume is approximately constant for the heat transfer) the first law is du = dq since v = const. Using the definition du = cv dT we obtain for the specific heat at constant volume cv = ∆q/ ∆T , where both the heat transferred ∆q and the temperature difference ∆T can be measured. Similarly we can do an experiment involving a process where the pressure is kept constant during the reversible heat transfer ∆q (e.g. a rigid container filled with a substance that is closed by a lid with a certain weight, or the heat transfer in a jet engine combustor where the pressure is approximately constant during heat addition). The first law can be written in terms of enthalpy as dh – vdp = dq, and since p = const we obtain dh = dq. Using the definition dh = cp dT we obtain for the specific heat at constant pressure cp = ∆q/ ∆T . 2A.4 Explanation of the above comparison between Diesel and Otto. Basically we can operate the diesel cycle at much higher compression ratio than the Otto cycle because only air is compressed and we don't run into the auto-ignition problem (knocking problem). Because of the higher compression ratios in the diesel engine we get higher efficiencies. 2A.5 What is shaft work? I am not sure how best to answer, but it appears that the difficulty people are having might be associated with being able to know when one can say that shaft work occurs. There are several features of a process that produces (or absorbs) shaft work. First of all the view taken of the process is one of control volume, rather than control mass (see the discussion of control volumes in Section 0 or in IAW). Second, there needs to be a shaft or equivalent device (a moving belt, a row of blades) that can be identified as the work carrier. Third, the shaft work is work over and above the “flow work” that is done by (or received by) the streams that exit and enter the control volume. 2A.6 What exactly is the specific impulse, Isp, a measure of? The specific impulse is a measure of how well the fuel is used in creating thrust. For a rocket engine, the specific impulse is the effective exit velocity divided by the acceleration of gravity, g. In terms of relating the specific impulse to some characteristic time, we can write the definition of I sp as I sp m& g = F . From this, one can regard the specific impulse as the time that it would take to flow a quantity of fuel that has a weight equal to the thrust force 2A.7 How is Isp found for rockets in space where g ~ 0? The impulse I given to a rocket is the thrust force integrated over the burn time. Traditionally, for the case of constant exhaust velocity cex, the specific impulse has been used Isp = I /(mp g) = cex / g0, where mp is the propellant mass and g0 is the Earth's surface gravity. Thus Isp is measured in seconds and is a force per weight flow. Often today, however, specific impulse is measured in units meters/second [m/s], recognizing that force per mass flow is more logical. The specific impulse is then simply equal to the exhaust velocity Isp = cex. 2A.8 Why does industry use TSCP rather than Isp? Is there an advantage to this? I am not sure why Thrust*Specific Fuel Consumption was originally used. The gas turbine industry uses TSFC; the rocket propulsion industry uses essentially its inverse, Specific Impulse. Perhaps an advantage is that TSFC is a number of magnitude unity, whereas specific impulse is not. 2A.9 Why isn't mechanical efficiency an issue with ramjets? As defined, the mechanical efficiency represents bearing friction, and other parasitic torques on the rotating shaft in a gas turbine. The work associated with this needs to be provided by the turbine, but does not go into driving the compressor. The ramjet has no shaft, and hence does not encounter this. 2A.10 How is thrust created in a ramjet? You can look at thrust in several ways. One is through the integral form of the momentum equation, which relates thrust to the difference between exit and inlet velocities, multiplied by the mass flow. Another way, however, is to look at the forces on the ramjet structure, basically the summation of pressure forces on all the surfaces. I attempted to do this in class using the turbojet with an afterburner. For the ramjet, from the same considerations, we would have an exit nozzle that was larger in diameter than the inlet so that the structural area on which there is a force in the retarding direction is smaller than the area on which there is a force in the thrust direction. 2A.11 Why don't we like the numbers 1 and 2 for the stations? Why do we go 0-3? A common convention in the industry is that station 0 is far upstream, station 1 is after the shock in the inlet (if there is one), station 2 is at inlet to the compressor (after the inlet/diffuser) and station 3 is after the compressor. In class, when we examined the ramjet we considered no changes in stagnation pressure between 0 and 2, so I have used 0 as the initial state for the compression process. It would be more precise to differentiate between stations 0 and 2, and I will do this where appropriate. 2A.12 For the Brayton cycle efficiency, why does T3=Tt0? The ramjet is operating as a Brayton cycle where ηb= 1 – Tinlet / T compressor exit. For the ramjet discussed in class the inlet temperature is T0 and since there is no compressor (no moving parts) the only compression we get is from diffusion. We assumed isentropic diffusion in the diffuser and found for very low Mach numbers that the diffuser exit or combustor inlet temperature T3 is Tt3. From first law we know that for a steady, adiabatic flow where no work is done the stagnation enthalpy stays constant. Assuming perfect gas we thus get Tt0 = Tt3 = T3. So we can write for the ramjet thermal efficiency ηb= 1 – T0 / Tt0 . 2A.13 How can we idealize fuel addition as heat addition? The validity of an approximation rests on what the answer is going to be used for. We are seeking basically only one item concerning combustor exit conditions, namely the exit temperature or the exit enthalpy. The final state is independent of how we add the heat, and depends only on whether we add the heat. If it is done from an electrical heater or from combustion, and if we neglect the change in the constitution of the gas due to the combustion products (most of the gas is nitrogen) the enthalpy rise is the same no matter how the temperature rise is achieved. 2A.14a What is the relation between h t4 − h t3 = f∆h f and the existence of the maximum value of Tt4? The two are very different physical statements. The first is the SFEE (steady flow energy equation) plus the approximation that inlet and exit mass flows to the control volume are the same. The heat received within the volume is represented by the quantity f∆h f , where ∆h f is the heat liberated per kilogram of fuel. The second statement is a representation of the fact that the degree of completion of the reaction in the combustor depends on temperature, so that even though the inlet temperature increases strongly as the Mach number increases, the combustor exit temperature does not change greatly. This is an attempt to represent a complex physical process (or set of processes) in an approximate manner, not a law of nature. 2A.14b Why didn’t we have a 2s point for the Brayton cycle with non-ideal components? If we didn’t, we should have, or I should at least have marked the point at which the compressor exit would be if the compression process was isentropic. 2A.15 What is the variable fstoich? fstoich is the fuel-to-air ratio for stoichiometric combustion, or in other words the fuel-toair ratio for a chemically correct combustion process during which all fuel is burnt. 2A.16 How can T4 be the maximum turbine inlet temperature? T0 I agree that the T4/T0 is a temperature ratio. If we assume constant ambient temperature then this ratio reflects the maximum cycle temperature. The main point was to emphasize that the higher your turbine inlet temperature the higher your power and efficiency levels. 2A.17 When there are losses in the turbine that shift the expansion in T-s diagram to the right, does this mean there is more work than ideal since the area is greater? We have to be careful when looking at the area enclosed by a cycle or underneath a path in the T-s diagram. Only for a reversible cycle, the area enclosed is the work done by the cycle (see notes page 1C-5). Looking at the Brayton cycle with losses in compressor and turbine the net work is the difference between the heat absorbed and the heat rejected (from 1st law). The heat absorbed can be found by integrating TdS = dQ along the heat addition process. The heat rejected during the cycle with losses in compressor and turbine is larger than in the ideal cycle (look at the area underneath the path where heat is rejected, this area is larger than when there are no losses dsirrev = 0 – see also muddy point 1C.1). So we get less net work if irreversibilities are present. It is sometimes easier to look at work and heat (especially shaft work for turbines and compressors) in the h-s diagram because the enthalpy difference between two states directly reflects the shaft work (remember, enthalpy includes the flow work!) and / or heat transfer. 2A.18a For an afterburning engine, why must the nozzle throat area increase if the temperature of the fluid is increased? The Mach number of the flow is unity at the throat with and without the afterburner lit. The ratio of static pressure to stagnation pressure at the throat is thus the same with and without the afterburner lit. The ratio of static temperature to stagnation temperature at the throat is thus the same with and without the afterburner lit. Tt P γ −1 2 γ − 1 2 γ /( = 1+ M ; t = 1 + M 2 2 T P The flow through the throat is P γRT Athroat . ṁ = ρcAthroat = ρaAthroat = RT The flow through the throat thus scales as P Athroat ˙ mA / B T A/ B = P m˙ noA / B A T throat noA / B γ −1 ) From what we have said, however, the pressure at the throat is the same in both cases. Also, we wish to have the mass flow the same in both cases in order to have the engine operate at near design conditions. Putting these all together, plus use of the idea that the ratio of stagnation to static temperature at the throat is the same for both cases gives the relation Athroat A / B Tt A / B = AthroatnoA / B TtnoA / B The necessary area to pass the flow is proportional to the square root of the stagnation temperature. If too much fuel is put into the afterburner, the increase in area cannot be met and the flow will decrease. This can stall the engine, a serious consequence for a single engine fighter. 2A.18b Why doesn’t the pressure in the afterburner go up if heat is added? From discussions after lecture, the main point here seems to be that the process of heat addition in the afterburner, or the combustor, is not the same as heat addition to a gas in a box. In that case the density (mass/volume) would be constant and, from P = ρRT , increasing the temperature would increase the pressure. In a combustor, the geometry is such that the pressure is approximately constant; this happens because the fluid has the freedom to expand so the density decreases. From the equation P = ρRT if the temperature goes up, the density must go down. 2A.18c Why is the flow in the nozzle choked? As seen in Unified, choking occurs when the stagnation to static pressure ratio ( Pt / P) gets to a certain value, 1.89 for gas with γ of 1.4. Almost all jet aircraft operate at flight conditions such that this is achieved. If you are not comfortable with the way in which the concepts of choking are laid out in the Unified notes, please see me and I can give some references. 2A.18d What’s the point of having a throat if it creates a retarding force? As shown in Unified, to accelerate the flow from subsonic to supersonic, i.e., to create the high velocities associated with high thrust, one must have a convergingdiverging nozzle, and hence a throat. 2A.18e Why isn’t the stagnation temperature conserved in this steady flow? Heat is added in the afterburner, so the stagnation temperature increases. 2.B Power Cycles with Two-Phase Media (Vapor Power Cycles) [SB&VW – Chapter 3, Chapter 11, Sections 11.1 to 11.7] In this section, we examine cycles that use two-phase media as the working fluid. These can be combined with gas turbine cycles to provide combined cycles which have higher efficiency than either alone. They can also be used by themselves to provide power sources for both terrestrial and space applications. The topics to be covered are: i) Behavior of two-phase systems: equilibrium, pressure temperature relations ii) Carnot cycles with two-phase media iii) Rankine cycles iv) Combined cycles 2.B.1 Behavior of Two-Phase Systems The definition of a phase, as given by SB&VW, is “a quantity of matter that is homogeneous throughout”. Common examples of systems that contain more than one phase are a liquid and its vapor and a glass of ice water. A system which has three phases is a container with ice, water, and water vapor. We wish to find the relations between phases and the relations that describe the change of phase (from solid to liquid, or from liquid to vapor) of a pure substance, including the work done and the heat transfer. To start we consider a system consisting of a liquid and its vapor in equilibrium, which are enclosed in a container under a moveable piston, as shown in Figure 2B-1. The system is maintained at constant temperature through contact with a heat reservoir at temperature T, so there can be heat transfer to or from the system. Water vapor Water vapor Liquid water Liquid water (a) (b) (c) Figure 2B-1: Two-phase system in contact with constant temperature heat reservoir For a pure substance, as shown at the right, there is a one-to-one correspondence between the temperature at which vaporization occurs and the pressure. These values are called the saturation pressure and saturation temperature (see Ch. 3 in SB&VW). P-T relation for liquid-vapor system 2B-1 This means there is an additional constraint for a liquid-vapor mixture, in addition to the equation of state. The consequence is that we only need to specify one variable to determine the state of the system. For example, if we specify T then P is set. In summary, for two phases in equilibrium, P = P(T ) . If both phases are present, any quasi-static process at constant T is also at constant P. Let us examine the pressure-volume behavior of a liquid-vapor system at constant temperature. For a single-phase perfect gas we know that the curve would be Pv = constant. For the two-phase system the curve looks quite different, as indicated in Figure 2B-2. D Critical point Liquid phase Pressure, P Ga sp has e Criti cal i Mixture of liquid and vapor Vapo r soth erm phas e C Liquid saturation curve B A Vapor saturation curve Volume, V Figure 2B-2 – P-v diagram for two-phase system showing isotherms Several features of the figure should be noted. First, there is a region in which liquid and vapor can coexist. This is roughly dome-shaped and is thus often referred to as the “vapor dome”. Outside of this regime, the equilibrium state will be a single phase. The regions of the diagram in which the system will be in the liquid and vapor phases respectively are indicated. Second is the steepness of the isotherms in the liquid phase, due to the small compressibility of most liquids. Third, the behavior of isotherms at temperatures below the “critical point” (see below) in the region to the right of the vapor dome approach those of an ideal gas as the pressure decreases and the ideal gas relation is a good approximation in this region. The behavior shown is found for all the isotherms that go through the vapor dome. At a high enough temperature, specifically at a temperature corresponding to the pressure at the peak of the vapor dome, there is no transition from liquid to vapor and the fluid goes continuously from a liquid-like behavior to a gas-type behavior. This behavior is unfamiliar, mainly because the temperatures and pressures are not ones that we typically experience; for water the critical temperature is 374oC and the associated critical pressure is 220 atmospheres. 2B-2 There is a distinct nomenclature used for systems with more than one phase. In this, the terms “vapor” and “gas” seem to be used interchangeably. In the zone where both liquid and vapor exist, there are two bounding situations. When the last trace of vapor condenses, the state becomes saturated liquid. When the last trace of liquid evaporates the state becomes saturated vapor (or dry vapor). If we put heat into a saturated vapor it is referred to as superheated vapor. Nitrogen at room temperature and pressure (at one atmosphere the vaporization temperature of nitrogen is 77 K) is a superheated vapor. Figure 2B-3 available from: Sonntag, Borgnakke and Van Wylen, Fundamentals of Thermodynamics, 5th Ed., John Wiley & Sons. Figure 3.3, p35. Figure 2B-3: Constant pressure curves in T-v coordinates showing vapor dome Figure 2B-3 shows lines of constant pressure in temperature-volume coordinates. Inside the vapor dome the constant pressure lines are also lines of constant temperature. It is useful to describe the situations encountered as we decrease the pressure or equivalently increase the specific volume, starting from a high pressure-low specific volume state (the upper left-hand side of the isotherm in Figure 2B-2). The behavior in this region is liquid-like with very little compressibility. As the pressure is decreased, the volume changes little until the boundary of the vapor dome is reached. Once this occurs, however, the pressure is fixed because the temperature is constant. As the piston is withdrawn, the specific volume increases through more liquid evaporating and more vapor being produced. During this process, since the expansion is isothermal (we specified that it was), heat is transferred to the system. The specific volume will increase at constant pressure until the right hand boundary of the vapor dome is reached. At this point, all the liquid will have been transformed into vapor and the system again behaves as a single-phase fluid. For water at temperatures near room temperature, the behavior would be essentially that of a perfect gas in this region. To the right of the vapor dome, as mentioned above, the behavior is qualitatively like that of a perfect gas. Referring to Figure 2B-4, we define notation to be used in what follows. The states a and c denote the conditions at which all the fluid is in the liquid state and the gaseous state respectively. 2B-3 Standard-liquid line Critical point Saturated-vapor line Mixture "state" Liquid c a Gas (vapor) b State of vapor in mixture T (constant T line) State of liquid in mixture Liquid + gas vf vg v v Figure 2B-4: Specific volumes at constant temperature and states within the vapor dome in a liquid-vapor system The specific volumes corresponding to these states are: v f - specific volume of liquid phase vg - specific volume of gas phase For conditions corresponding to specific volumes between these two values, i.e., for state b, the system would exist with part of the mass in a liquid state and part of the mass in a gaseous (vapor) state. The average specific volume for this condition is v - average specific volume of two-phase system. We can relate the average specific volume to the specific volumes for liquid and vapor and the mass that exists in the two phases as follows. The total mass of the system is given by total mass = m = liquid mass + vapor mass = m f + mg . The volume of the system is Volume of liquid = V f = m f v f Volume of vapor = Vg = mg vg Total volume = V = m f v f + mg vg . The average specific volume, v , is the ratio of the total volume to the total mass of the system v= m f v f + mg vg m f + mg = average specific volume. 2B-4 The fraction of the total mass in the vapor phase is called quality, and denoted by X. X= mg m f + mg = quality of a liquid-vapor system. In terms of the quality and specific volumes, the average specific volume can be expressed as: v = X ⋅ vg + (1 − X ) ⋅ v f In reference to Figure 2B-5, ab = v − v f , ac = vg − v f . ab v − v f = = X = quality . ac vg − v f p T b a c T L L-V V T vf (a) v vg v (b) Figure 2B-5: Liquid vapor equilibrium in a two-phase medium 2.B.2 - Work and Heat Transfer with Two-Phase Media We examine the work and heat transfer in quasi-static processes with two-phase systems. For definiteness, consider the system to be a liquid-vapor mixture in a container whose volume can be varied through movement of a piston, as shown in Figure 2B-5. The system is kept at constant temperature through contact with a heat reservoir at temperature T. The pressure is thus also constant, but the volume, V, can change. For a fixed mass, the volume is proportional to the specific volume v so that point b in Figure 2B-5 must move to the left or the right as V changes. This implies that the amount of mass in each of the two phases, and hence the quality, also changes because mass is transferred from one phase to the other. We wish to find the heat and work transfer associated with the change in mass in each phase. The change in volume can be related to the changes in mass in the two phases as, dV = vg dmg + v f dm f . 2B-5 The system mass is constant ( m = m f + mg = constant ) so that for any changes dm = 0 = dm f + dmg . We can define the quantity dm fg dm fg = dmg = - dm f = mass transferred from liquid to vapor. In terms of dm fg the volume change of the system is ( ) dV = vg − v f dm fg . The work done is given by dW = PdV = P vg − v f dm fg . ( ) The change in internal energy, ∆U , can be found as follows. The internal energy of the system can be expressed in terms of the mass in each phase and the specific internal energy (internal energy per unit mass, u) of the phase as, U = u f m f + ug mg ( ) dU = u f dm f + ug dmg = ug − u f dm fg . Note that the specific internal energy can be expressed in a similar way as the specific volume in terms of the quality and the specific enthalpy of each phase: u = X ⋅ ug + (1 − X ) ⋅ u f Writing the first law for this process: dQ = dU + dW = ug − u f dm fg + P vg − v f dm fg . ( ) ( ) = [(u + Pv ) − (u + Pv )]dm = (h − h )dm . g g g f f f fg fg The heat needed for the transfer of mass is proportional to the difference in specific enthalpy between vapor and liquid. The pressure and temperature are constant, so that the specific internal energy and the specific enthalpy for the liquid phase and the gas phase are also constant. For a finite change in mass from liquid to vapor, m fg , therefore, the quantity of heat needed is ( ) Q = hg − h f m fg = ∆H (enthalpy change) . 2B-6 The heat needed per unit mass, q, for transformation between the two phases is q= ( ) Q = hg − h f = h fg . m fg The notation h fg refers to the specific enthalpy change between the liquid state and the vapor state. The expression for the amount of heat needed, q, is a particular case of the general result that in any reversible process at constant pressure, the heat flowing into, or out of, the system is equal to the enthalpy change. Heat is absorbed if the change is from solid to liquid (heat of fusion), liquid to vapor (heat of vaporization), or solid to vapor (heat of sublimation). A numerical example is furnished by the vaporization of water at 100oC: i) How much heat is needed per unit mass of fluid vaporized? ii) How much work is done per unit mass of fluid vaporized? iii) What is the change in internal energy per unit mass of fluid vaporized?. In addressing these questions, we make use of the fact that problems involving heat and work exchanges in two-phase media are important enough that the values of the specific thermodynamic properties that characterize these transformations have been computed for many different working fluids. The values are given in SB&VW in Tables B.1.1 and B.1.2 for water at saturated conditions and in Tables B.1.3, B.1.4, and B.1.5 for other conditions, as well as for other working fluids. From these: - At 100oC, the vapor pressure is 0.1013 MPa, - The specific enthalpy of the vapor, hg , is 2676 kJ/kg and the specific enthalpy of the liquid, h f , is 419 kJ/kg - The difference in enthalpy between liquid and vapor, h fg , occurs often enough so that it is tabulated also. This is 2257 kJ/kg, - The specific volume of the vapor is 1.6729 m3/kg and the specific volume of the liquid is 0.001044. The heat input to the system is the change in enthalpy between liquid and vapor, h fg , and is equal to 2.257 x 106 J/kg. ( ) The work done is P vg − v f which has a value of ( ) P vg − v f =0.1013 x 106 x [1.629 – 0.001044] =0.169 x 106 J/kg. The change in internal energy per unit mass ( u fg ) can be found from ∆u = q − w or from the tabulated values as 2.088 x 106 J/kg. This is much larger than the work done. Most of the heat input is used to change the internal energy rather than appearing as work. Muddy points 2B-7 For the vapor dome, is there vapor and liquid inside the dome and outside is it just liquid or just gas? Is it interchangeable? Is it true for the plasma phase? (MP 2B.1) What is hfg ? How do we find it? (MP 2B.2) Reasoning behind the slopes for T=cst lines in the P-V diagram. (MP 2B.3) For a constant pressure heat addition, why is q=∆h? (MP 2B.4) What is latent heat? (MP 2B.5) Why is U a function of x? (MP 2B.6) 2.B.3 The Carnot Cycle as a Two-Phase Power Cycle A Carnot cycle that uses a two-phase fluid as the working medium is shown below in Figure 2B-6. Figure 2B-6a gives the cycle in P-v coordinates, 2B-6b in T-s coordinates, and 2B-6c in h-s coordinates. The boundary of the region in which there is liquid and vapor both present (the vapor dome) is also indicated. Note that the form of the cycle is different in the T-s and h-s representation; it is only for a perfect gas with constant specific heats that cycles in the two coordinate representations have the same shapes. T1 T2 p a p2 b T2 p1 c d T1 v (a) p-v diagram T b h a T2 b a T1 g c d f s1 h e s2 c d s (b) T-s diagram s (c) h-s diagram Figure 2B-6: Carnot cycle with two-phase medium. (a) cycle in P-v coordinates, (b) cycle in T-s coordinates, (c) cycle in h-s coordinates The processes in the cycle are as follows: i) Start at state a with saturated liquid (all of mass in liquid condition). Carry out a reversible isothermal expansion to b (a b) until all the liquid is vaporized. During this process a quantity of heat q H per unit mass is received from the heat source at temperature T2 . ii) Reversible adiabatic (i.e., isentropic) expansion (b c) lowers the temperature to T1. Generally state c will be in the region where there is both liquid and vapor. 2B-8 iii) iv) Isothermal compression (c d) at T1 to state d. During this compression, heat q L per unit mass is rejected to the source at T1. Reversible adiabatic (i.e., isentropic) compression (d a) in which the vapor condenses to liquid and the state returns to a. In the T-s diagram the heat received, q H , is abef and the heat rejected, q L , is dcef. The net work is represented by abcd. The thermal efficiency is given by η= wnet Area abcd T = = 1− 1 . Area abef qH T2 In the h-s diagram, the isentropic processes are vertical lines as in the T-s diagram. The isotherms in the former, however, are not horizontal as they are in the latter. To see their shape we note that for these two-phase processes the isotherms are also lines of constant pressure (isobars), since P = P(T). The combined first and second law is dp Tds = dh − . ρ For a constant pressure reversible process, dqrev = Tds = dh . The slope of a constant pressure line in h-s coordinates is thus, ∂h = T = constant ; slope of constant pressure line for two-phase medium. ∂s P The heat received and rejected per unit mass is given in terms of the enthalpy at the different states as, q H = hb − ha q L = hd − hc . (In accord with our convention this is less than zero.) The thermal efficiency is w q + qL (hb − ha ) + (hd − hc ) η = net = H = , qH qH (hb − ha ) or, in terms of the work done during the isentropic compression and expansion processes, which correspond to the shaft work done on the fluid and received by the fluid, η= (hb − hc ) − (ha − hd ) . (hb − ha ) Example: Carnot steam cycle: Heat source temperature = 300oC Heat sink temperature = 20oC What is the (i) thermal efficiency and (ii) ratio of turbine work to compression (pump) work if a) all processes are reversible? b) the turbine and the pump have adiabatic efficiencies of 0.8? 2B-9 Neglect the changes in kinetic energy at inlet and outlet of the turbine and pump. a) For the reversible cycle, ηthermal = ηCarnot = 1 − = 1− T1 T2 293 = 0.489 573 To find the work in the pump (compression process) or in the turbine, we need to find the enthalpy changes between states b and c, ∆hbc , and the change between a and d, ∆had . To obtain these the approach is to use the fact that s = constant during the expansion to find the quality at state c and then, knowing the quality, calculate the enthalpy as h = Xhg + (1 − X )h f . We know the conditions at state b, where the fluid is all vapor, i.e., we know Tb , hb , sb : ( ) ( ) (300 C) = s (300 C) = 5.7045 kJ/kg - K hb = hvapor 300 o C = hg 300 o C = 2749 kJ/kg sb = svapor o o g sb = sc in the isentropic expansion process. We now need to find the quality at state c, Xc . Using the definition of quality given in Section 2.B.1, and noting that sc = Xc sg + (1 − Xc )s f , we obtain, Xc = sc − s f (Tc ) sg (Tc ) − s f (Tc ) = sc − s f (Tc ) s fg (Tc ) . The quantity sc is the mass-weighted entropy at state c, which is at temperature Tc. The quantity s f (Tc ) is the entropy of the liquid at temperature Tc . The quantity sg (Tc ) is the entropy of the gas (vapor) at temperature Tc . The quantity ∆s fg (Tc ) = ∆sliquid →gas at Tc . We know: sc = sb = 5.7045 kJ/kg - K s fg = 8.3706 kJ/kg - K s f = 0.2966 kJ/kg - K . The quality at state c is thus, Xc = 5.7045 − 0.2966 = 0.646 . 8.3706 The enthalpy at state c is, 2B-10 hc = Xc hg + (1 − Xc )h f at Tc . Substituting the values, hc = 0.646 × 2538.1 + 0.354 × 83.96 kJ/kg =1669.4 kJ/kg. The turbine work/unit mass is the difference between the enthalpy at state b and state c, hb − hc = wturbine = 2749 − 1669.4 = 1079.6 kJ/kg . We can apply a similar process to find the conditions at state d: Xd = sd − s f (Td ) sg (Td ) − s f (Td ) = sc − s f (Td ) s fg (Td ) . We have given that Tc = Td . Also sd = sa = s f at 300 o C . The quality at state d is Xd = 3.253 − 0.2966 = 0.353 < Xc 8.3706 The enthalpy at state d is hd = X d hg + (1 − X d )h f = 0.353 x 2538.1 + 0.647 x 83.96 = 950.8 kJ/kg. The work of compression (pump work) is ∆had = ha − hd . Substituting the numerical values, ∆had = 1344- 950.8 = 393.3 kJ/kg. wturbine = 2.75 wcompression We can check the efficiency by computing the ratio of net work ( wnet = wturbine − wcompression ) to the heat input ( Ta s fg ). Doing this gives, not surprisingly, the same value as the Carnot equation. The ratio of turbine work to compression work (pump work) is b) Efficiency and work ratio for a cycle with adiabatic efficiencies of pump and turbine both equal to 0.8 (non-ideal components). We can find the turbine work using the definition of turbine and compressor adiabatic efficiencies. The relation between the enthalpy changes is wturbine = hb − hc′ = ηturbine (hb − hc ) = actual turbine work received. Substituting the numerical values, the turbine work per unit mass is 863.7 kJ/kg. 2B-11 For the compression process, we use the definition of compressor (or pump) adiabatic efficiency: 1 wcompression = ha′ − hd = (ha − hd ) ηcompression = actual work to achieve given pressure difference ( ) = 491.6 kJ/kg. The value of the enthalpy at state a ′ is 1442.4 kJ/kg. The thermal efficiency is given by w − wcompression wnet = turbine ηthermal = heat input heat input (h − hc ′ ) − (ha ′ − hd ) . = b (hb − ha ′ ) Substituting the numerical values, we obtain for the thermal efficiency with non-ideal components, ηthermal = 0.285 . A question arises as to whether the Carnot cycle can be practically applied for power generation. The heat absorbed and the heat rejected both take place at constant temperature and pressure within the two-phase region. These can be closely approximated by a boiler for the heat addition process and a condenser for the heat rejection. Further, an efficient turbine can produce a reasonable approach to reversible adiabatic expansion, because the steam is expanded with only small losses. The difficulty occurs in the compression part of the cycle. If compression is carried out slowly, there is equilibrium between the liquid and the vapor, but the rate of power generation may be lower than desired and there can be appreciable heat transfer to the surroundings. Rapid compression will result in the two phases coming to very different temperatures (the liquid temperature rises very little during the compression whereas the vapor phase temperature changes considerably). Equilibrium between the two phases cannot be maintained and the approximation of reversibility is not reasonable. Another circumstance is that in a Carnot cycle all the heat is added at the same temperature. For high efficiency we need to do this at a higher temperature than the critical point, so that the heat addition no longer takes place in the two-phase region. Isothermal heat addition under this circumstance is difficult to accomplish. Also, if the heat source and the cycle are considered together, the products of combustion which provide the heat can be cooled only to the highest temperature of the cycle. The source will thus be at varying temperature while the system requires constant temperature heat addition, so there will be irreversible heat transfer. In summary, the practical application of the Carnot cycle is limited because of the inefficient compression process, the low work per cycle, the upper limit on temperature for operation in the two-phase flow regime, and the irreversibility in the heat transfer from the heat source. In the next section, we examine the Rankine cycle, which is much more compatible with the characteristics of two-phase media and available machinery for carrying out the processes. 2B-12 Muddy points What is the reason for studying two-phase cycles? (MP 2B.7) How did you get thermal efficiency? How does a boiler work? (MP 2B.8) 2.B.4 Rankine Power Cycles A schematic of the components of a Rankine cycle is shown in Figure 2B-7. The cycle is shown on P-v, T-s, and h-s coordinates in Figure 2B-8. The processes in the Rankine cycle are as follows: i) d e: Cold liquid at initial temperature T1 is pressurized reversibly to a high pressure by a pump. In this, the volume changes slightly. ii) e a: Reversible constant pressure heating in a boiler to temperature T2 iii) a b: Heat added at constant temperature T2 (constant pressure), with transition of liquid to vapor iv) b c: Isentropic expansion through a turbine. The quality decreases from unity at point b to Xc < 1 v) c d: Liquid-vapor mixture condensed at temperature T1 by extracting heat. Figure 2B-7 Available from: Moran and Shapiro, Fundamentals of Engineering Thermodynamics, 4th Ed. John Wiley & Sons. Figure E8.1 p.328 Figure 2B-7: Rankine power cycle with two-phase working fluid [Moran and Shapiro, Fundamentals of Engineering Thermodynamics] 2B-13 p T1 T2 e p2 a b T2 p1 c c′ d T1 v (a) p-v coordinates T a T2 T1 b h b e e c c′ d a c c′ d s s (b) T-s coordinates (c) h-s coordinates Figure 2B-8: Rankine cycle diagram. (a) P-v coordinates, (b) T-s coordinates, (c) h-s coordinates. Stations correspond to those in Figure 2B-7 In the Rankine cycle, the mean temperature at which heat is supplied is less than the maximum temperature, T2 , so that the efficiency is less than that of a Carnot cycle working between the same maximum and minimum temperatures. The heat absorption takes place at constant pressure over eab, but only the part ab is isothermal. The heat rejected occurs over cd; this is at both constant temperature and pressure. To examine the efficiency of the Rankine cycle, we define a mean effective temperature, Tm in terms of the heat exchanged and the entropy differences: q H = Tm2 ∆s2 q L = Tm1 ∆s1. The thermal efficiency of the cycle is ηthermal = Tm2 (sb − se ) − Tm1 (sc − sd ) Tm2 (sb − se ) . The compression and expansion processes are isentropic, so the entropy differences are related by 2B-14 s b − se = sc − s d . The thermal efficiency can be written in terms of the mean effective temperatures as, ηthermal = 1 − Tm1 Tm2 . For the Rankine cycle, Tm1 ≈ T1 , Tm2 < T2 . From Equation (B.4.1), we see not only the reason that the cycle efficiency is less than that of a Carnot cycle, but the direction to move in terms of cycle design (increased Tm2 ) if we wish to increase the efficiency. There are several features that should be noted about Figure 2B-8 and the Rankine cycle in general: i) The T-s and the h-s diagrams are not similar in shape, as they were with the perfect gas. The slope of a constant pressure reversible heat addition line is, as derived in Section 1.C.4, ∂h = T . In the two-phase region, constant pressure means also constant temperature, ∂s p so the slope of the constant pressure heat addition line is constant and the line is straight. ii) The effect of irreversibilities is represented by the dashed line from b to c ′ . Irreversible behavior during the expansion results in a value of entropy. sc′ .at the end state of the expansion that is higher than sc . The enthalpy at the end of the expansion (the turbine exit) is thus higher for the irreversible process than for the reversible process, and, as seen for the Brayton cycle, the turbine work thus lower in the irreversible case. iii) The Rankine cycle is less efficient than the Carnot cycle for given maximum and minimum temperatures, but, as said earlier, it is more effective as a practical power production device. Muddy points Where does degrees Rankine come from? Related to Rankine cycles? (MP 2B 9) 2.B.5: Enhancements of, and Effect of Design Parameters on, Rankine Cycles The basic Rankine cycle can be enhanced through processes such as superheating and reheat. Diagrams for a Rankine cycle with superheating are given in Figure 2B-9. The heat addition is continued past the point of vapor saturation, in other words the vapor is heated so that its temperature is higher than the saturation temperature associated with Pa ( = Pb = Pc = Pd ) . This 2B-15 p T1 T2 a p2 p1 T3 d c b f e′ T3 T2 e T1 v (a) T T3 d h d c b T2 c a b a T1 f e′ e f e′ e s (b) s (c) Figure 2B-9: Rankine cycle with superheating does several things. First, it increases the mean temperature at which heat is added, Tm2 , thus increasing the efficiency of the cycle (see Equation B.4.1). Second is that the quality of the twophase mixture during the expansion is higher with superheating, so that there is less moisture content in the mixture as it flows through the turbine. (The moisture content at e is less than that at e ′ .) This is an advantage in terms of decreasing the mechanical deterioration of the blading. The heat exchanges in the superheated cycle are; Along abcd, which is a constant pressure (isobaric) process: q2 = hd − ha . Along ef: q1 = h f − he , (< 0). The thermal efficiency of the ideal Rankine cycle with superheating is ηthermal = ( hd − ha − he − h f hd − ha ) This can be expressed explicitly in terms of turbine work and compression (pump) work as: 2B-16 ηthermal = ( hd − he − ha − h f hd − ha ). Compared to the basic cycle, superheating has increased the turbine work, increased the mean temperature at which heat is received, Tm2 , and increased the cycle efficiency. A comparison of the Carnot cycle and the Rankine cycle with superheat is given in Figure 2B-10. The maximum and minimum temperatures are the same, but the average temperature at which heat is absorbed is lower for the Rankine cycle. T Isothermal Isentropic g T1 d g d = Carnot f c b a T2 e d b e f a f =Rankine c e s Figure 2B-10: Comparison of Rankine cycle with superheat and Carnot cycle To alleviate the problem of having moisture in the turbine, one can heat again after an initial expansion in a turbine, as shown in Figure 2B-11, which gives a schematic of a Rankine cycle for space power application. This process is known as reheat. The main practical advantage of reheat (and of superheating) is the decrease in moisture content in the turbine because most of the heat addition in the cycle occurs in the vaporization part of the heat addition process. We can also examine the effect of variations in design parameters on the Rankine cycle. Consider first the changes in cycle output due to a decrease in exit pressure. In terms of the cycle shown in Figure 2B-12, the exit pressure would be decreased from Pexit to ( Pexit − dPexit ) . The original cycle is abcdea, and the modified cycle is abfgha. The consequences are that the cycle work, which is 2B-17 Turbine 1st stage Electric power 2nd stage Generator 1′ 2 Nuclear heat source 5 3 Reheat Q′ 2′ Condensing Radiator 4′ Temperature (T) 1′ 2′ 5 2 4′ 3 4 Qreject Win Entropy (s) 4 Rankine cycle with reheat Pump Figure 2B-11: Rankine cycle with superheating and reheat for space power application T p4 3 2′ 2 p4 ′ 4 1 1′ 4′ a′ a b s Figure 2B-12: Effect of exit pressure on Rankine cycle efficiency the integral of Tds around the cycle, is increased. In addition, as drawn, although the levels of the mean temperature at which the heat is absorbed and rejected both decrease, the largest change in the mean temperature of the heat rejection, so that the thermal efficiency increases. Another design parameter is the maximum cycle pressure. Figure 2B-13 shows comparison of two cycles with different maximum pressure but the same maximum temperature, which is set by materials properties. The average temperature at which the heat is supplied for the cycle with a higher maximum pressure is increased over the original cycle, so that the efficiency increases. 2B-18 T 3′ 2′ 2 1 4′ a 3 4 b′ b s Figure 2B-13: Effect of maximum boiler pressure on Rankine cycle efficiency Muddy points Why do we look at the ratio of pump (compression) work to turbine work? We did not do that for the Brayton cycle. (MP 2B.10) Shouldn't the efficiency of the super/re-heated Rankine cycle be larger because its area is greater? (MP 2B.11) Why can't we harness the energy in the warm water after condensing the steam in a power plant? (MP 2B.12) 2.B.6 Combined Cycles in Stationary Gas Turbine for Power Production The turbine entry temperature in a gas turbine (Brayton) cycle is considerably higher than the peak steam temperature. Depending on the compression ratio of the gas turbine, the turbine exhaust temperature may be high enough to permit efficient generation of steam using the “waste heat” from the gas turbine. A configuration such as this is known as a gas turbine-steam combined cycle power plant. The cycle is illustrated in Figure 2B-14. Figure 2B-14 available from: Kerrebrock, Aircraft Engines and Gas Turbines, 2nd Ed. MIT Press. Figure 1.10, p14. Figure 2B-14: Gas turbine-steam combined cycle [Kerrebrock, Aircraft Engines and Gas Turbines] The heat input to the combined cycle is the same as that for the gas turbine, but the work output is larger (by the work of the Rankine cycle steam turbine). A schematic of the overall heat engine, 2B-19 which can be thought of as composed of an upper and a lower heat engine in series, is given in Figure 2B-15. The upper engine is the gas turbine (Brayton cycle) which expels heat to the lower Figure 2B-15 drawn from: Lee Langston. Global Gas Turbine News, ASME International Gas Turbine Institute. Figure 2B-15: Schematic of combine cycle using gas turbine (Brayton cycle) and steam turbine (Rankine cycle) [Langston] engine, the steam turbine (Rankine cycle). The overall efficiency of the combined cycle can be derived as follows. We denote the heat received by the gas turbine as Qin and the heat rejected to the atmosphere as Qout . The heat out of the gas turbine is denoted as Q1 . The hot exhaust gases from the gas turbine pass through a heat exchanger where they are used as the heat source for the two-phase Rankine cycle, so that Q1 is also the heat input to the steam cycle. The overall combined cycle efficiency is ηCC = W WB + WR = , Qin Qin where the subscripts refer to combined cycle (CC), Brayton cycle (B) and Rankine cycle (R) respectively. From the first law, the overall efficiency can be expressed in terms of the heat inputs and heat rejections of the two cycles as (using the quantity Q1 to denote the magnitude of the heat transferred): Qin − Q1 + ( Q 1 − Qout ) Q1 Qout Q1 . ηCC = = 1 − + 1 − Qin Q1 Qin Qin 2B-20 The first square bracket term on the right hand side is the Brayton cycle efficiency, η B , the second is the Rankine cycle efficiency, η R , and the term in parentheses is (1- η B ). The combined cycle efficiency can thus be written as, ηCC = η B + η R − η Bη R ; Combined cycle efficiency. (B.6.1) Equation (B.6.1) gives insight into why combined cycles are so successful. Suppose that the gas turbine cycle has an efficiency of 40%, which is a representative value for current Brayton cycle gas turbines, and the Rankine cycle has an efficiency of 30%. The combined cycle efficiency would be 58%, which is a very large increase over either of the two simple cycles. Some representative efficiencies and power outputs for different cycles are shown in Figure 2B-16. Figure 2B-16 drawn from: Dominic Bartol, Keynote talk, 1997 International Gas Turbine Institute (IGTI) Turbo Expo. Figure 2B-16: Comparison of efficiency and power output of various power products [Bartol (1997)] 2.B.7 Some Overall Comments on Thermodynamic Cycles i) There are many different power and propulsion cycles, and we have only looked at a few of these. Many other cycles have been devised in the search for ways to increase efficiency and power in practical devices. ii) We can view a given cycle in term of elementary Carnot cycles, as sketched in the figure on the right. This shows that the efficiency of any other cycle operating between two given temperatures will be less than that of a Carnot cycle. Tmax T Tmin s 2B-21 iii) If we view the thermal efficiency as Theat rejected Average , ηthermal = 1 − (Theat absorbed ) Average ( ) (derived in Section 2.B.4), this means that we should accept heat at a high temperature and reject it at a low temperature for high efficiency. This objective must be tempered by considerations of practical application. iv) The cycle diagrams in T-s and h-s coordinates will only be similar if the working medium is an ideal gas. For other media (for example, a two-phase mixture) they will look different. v) Combined cycles make use of the rejected heat from a “topping” cycle as heat source for a “bottoming” cycle. The overall efficiency is higher than the efficiency of either cycle. 2B-22 Muddiest Points on Part 2B 2B.1 For the vapor dome, is there vapor and liquid inside the dome and outside is it just liquid or just gas? Is it interchangeable? Is it true for the plasma phase? The vapor dome separates the two-phase region from the single-phase region. Inside, we have a mixture of liquid and vapor. The peak of the vapor dome is called the critical point. The left-hand side leg of the vapor dome (from the critical point) is called the saturated liquid line along which the quality x is zero (purely liquid). The right-hand side leg is denoted the saturated vapor line and the quality x is one (purely vapor). For further details see the notes. Heating of a solid or liquid substance leads to phase transition to a liquid or gaseous state, respectively. This takes place at a constant temperature for a given pressure, and requires an amount of energy known as latent heat. On the other hand, the transition from a gas to an ionized gas, i.e., plasma, is not a phase transition, since it occurs gradually with increasing temperature. During the process, a molecular gas dissociates first into an atomic gas which, with increasing temperature, is ionized. Resulting plasma consists of a mixture of neutral particles, positive ions (atoms or molecules that have lost one or more electrons), and negative electrons. 2B.2 What is hfg ? How do we find it? The quantity h fg represents the specific enthalpy change between the liquid and vapor phases of a substance at constant temperature, and thus constant pressure, and thus constant temperature. It is therefore the heat input, per unit mass, to vaporize a kilogram of liquid. See the notes, Section 2.B.2. 2B.3 Reasoning behind the slopes for T=cst lines in the P-V diagram. The slope of an isotherm in the gaseous phase (to the right of the vapor dome) is similar to the slope we found for the isotherm of an ideal gas (PV=const). Inside the vapor dome pressure and temperature are directly related to one another (P = P(T), vapor pressure curve) such that an isotherm is a horizontal line (isobar inside the vapor dome). In the liquid phase the isotherms are very steep lines, because for liquids the volume is about constant (very low compressibility). 2B.4 For a constant pressure heat addition, why is q=∆h? The combined first and second law is dh = Tds + vdP . For a reversible constant pressure process, dP = 0, and the heat input, dq, is TdS. Thus for a reversible constant pressure process, the answer is yes. For an irreversible process we can say from the steady flow energy equation: dq = dht = dh + cdc . For a steady flow, the one-dimensional momentum equation is dP Fviscous + , ρ ρ where Fviscous represents the viscous forces in an irreversible flow. Combining these two expressions, and using dP = 0 (the condition of constant pressure) gives F dq = dh + cdc = dh + viscous . ρ Without going into any detail concerning the form of the viscous forces, this equation shows that the equality between heat input and enthalpy change does not hold for general irreversible flow processes at constant pressure. cdc = − 2B.5 What is latent heat? Latent heat is a term for the enthalpy change needed for vaporization. 2B.6 Why is U a function of x? Inside the vapor dome we have a mixture of liquid and vapor. The internal energy U of the system (liquid and vapor) can be expressed in terms of the mass in each phase and the specific internal energy u of each phase as, U = uf mf + ug mg. Introducing the quality x as the fraction of the total mass in the vapor phase x = mg/(mg+mf) we can write U = (1 – x) uf + x ug. Since the specific energy of the saturated liquid and the saturated vapor are functions of temperature the internal energy U of the two-phase system is a function of x and T (see also notes on page 2B-6). 2B.7 What is the reason for studying two-phase cycles? Their immense practical utility in a number of industrial devices and their intrinsic interest as applications of the basic principles. 2B.8 How did you get thermal efficiency? How does a boiler work? The thermal efficiency is, as previously, the net work done divided by the heat input. Using the first law for a control volume we can write both of these quantities in terms of the enthalpy at different states of the cycle. For the steam cycles discussed in class, a boiler is a large (as in the viewgraph of the Mitsubishi power plant) structure with a lot of tubes running through it. The water (or whatever medium is used in the cycle) runs through the tubes. Hot gases wash over the outside of the tubes. The hot gases could be from a combustor or from the exit of a gas turbine. 2B.9 Where does degrees Rankine come from? Related to Rankine cycles? I think the answer is yes, although I do not know for sure. If so, this is the same Rankine who has his name on the Rankine-Hugoniot conditions across a shock wave. 2B.10 Why do we look at the ratio of pump (compression) work to turbine work? We did not do that for the Brayton cycle. If the ratio of compression work to turbine work were close to unity for an ideal cycle, small changes in component efficiencies would have large effects on cycle efficiency and work. For the Rankine cycle this is not true. (The effect of pump efficiency on Rankine cycle efficiency is clearly small in the class example.) For the Brayton cycle, where the net work is the difference of two numbers which are of (relatively) similar sizes, the effect of compressor and turbine efficiency on cycle efficiency can be much larger. I used the word “sensitive” and the meaning was that the cycle performance responded strongly to changes in the compressor and turbine behavior. 2B.11 Shouldn't the efficiency of the super/re-heated Rankine cycle be larger because its area is greater? The area enclosed by an ideal cycle in a T-s diagram is the net work done, but it does not tell you about efficiency. We saw that for example when we looked at the Brayton cycle for the condition of maximum work, rather than maximum efficiency (Section 1.A.4 in the notes). 2B.12 Why can't we harness the energy in the warm water after condensing the steam in a power plant? Let's assume the temperature of the warm water after condensing the steam is at a temperature of about 30 to 40 degrees C. If we consider running a heat engine between this heat reservoir (say 35 degrees C) and the surroundings at 20 degrees C, we would get an ideal thermal efficiency of about 5%. In other words, the available useful work is relatively small if we considered the lower heat reservoir to be the surroundings. In general, there is a property that only depends on state variables called availability. The change in availability gives the maximum work between two states, where one state is referred to the surroundings (dead state). Part 2.C: Introduction to Thermochemistry [SB&VW-14.1-14.6] Until now, we have specified the heat given to the devices analyzed, and not concerned ourselves with how this heat might be produced. In this section, we examine the issue of how we obtain the heat needed for work production. For the most part, this is from converting chemical energy into heat, so the discussion will be on reacting mixtures of gas which are involved in chemical combustion processes. The topic addressed is “thermochemistry”, which is the combining of thermodynamics with chemistry to predict such items as how much heat is released from a chemical reaction. This is the “Q” or “q” that we have used in the cycle analysis. The principal components of the approach are use of a chemical balance plus the steady flow energy equation (SFEE) which equates the sum of shaft work (from) and heat transfer (to) a control volume to the difference in control volume inlet and exit enthalpy fluxes. 2.C.1 Fuels There are a wide variety of fuels used for aerospace power and propulsion. A primary one is jet fuel (octane, essentially kerosene) which has the chemical formula C 8 H 18 . Other fuels we consider are hydrogen (H 2 ) and methane ( CH 4 ). The chemical process in which a fuel, for example methane, is burned consists of (on a very basic level—there are many intermediate reactions that need to be accounted for when computations of the combustion process are carried out): CH 4 + 2O 2 → CO 2 + 2H 2 O . (Reactants) (Products) The reactions we describe are carried out in air, which can be approximated as 21% O 2 and 79% N 2 . This composition is referred to as “theoretical air”. There are other components of air (for example Argon, which is roughly 1%), but the results given using the theoretical air approximation are more than adequate for our purposes. With this definition, for each mole of O 2 , 3.76 (79/21) moles of N 2 are involved: CH 4 + 2O 2 + 2(3.76)N 2 → CO 2 + 2H 2 O + 7.52N 2 Even if the nitrogen is not part of the combustion process, it leaves the combustion chamber at the same temperature as the other products, and this change in state (change in enthalpy) needs to be accounted for in the steady flow energy equation. At the high temperatures achieved in internal combustion engines (aircraft and automobile) reaction does occur between the nitrogen and oxygen, which gives rise to oxides of nitrogen, although we will not consider these reactions. The condition at which the mixture of fuel and air is such that both completely participate in the reaction is called stoichiometric. In gas turbines, excess air is often used so that the 2C-1 temperatures of the gas exiting the combustor is kept to within desired limits (see Figures A-8, A9, A-11 in Part 1 for data on these limits.) Muddy points Why is there 3.76 N2? (MP 2C.1) What is the most effective way to solve for the number of moles in the reactions? (MP 2C.2) 2.C.2 Fuel-Air Ratio The reaction for aeroengine fuel at stoichiometric conditions is: C 8 H18 +12.5 O 2 +12.5 (3.76) N 2 → 8 CO 2 + 9 H 2 O + 47.0 N 2 On a molar basis, the ratio of fuel to air is [1/(12.5+47.0)] = 1/59.5 = 0.0167. To find the ratio on a mass flow basis, which is the way in which the aeroengine industry discusses it, we need to “weight” the molar proportions by the molecular weight of the components. The fuel molecular weight is 114, the oxygen molecular weight is 32 and the nitrogen molecular weight is (approximately) 28. The fuel/air ratio on a mass flow basis is thus 1 ×114 Fuel-air ratio = = 0.0664 12.5 × 32 +12.5 × 3.76 × 28 If we used the actual constituents of air we would get 0.0667, a value about 0.5% different. Muddy points Do we always assume 100% complete combustion? How good an approximation is this? (MP 2C.3) 2.C.3 Enthalpy of formation The systems we have worked with until now have been of fixed chemical composition. Because of this, we could use thermodynamic properties relative to an arbitrary base, since all comparisons could be made with respect to the chosen base. For example, the specific energy u f ( 0.01o C) = 0.0 for steam. If there are no changes in composition, and only changes in properties of given substances, this is adequate. If there are changes in composition, however, we need to have a reference state so there is consistency for different substances. The convention used is that the reference state is a temperature of 25oC (298 K) and a pressure of 0.1 MPa. (These are roughly room conditions.) At these reference conditions, the enthalpy of the elements (oxygen, hydrogen, nitrogen, carbon, etc.) is taken as zero. The results of a combustion process can be diagrammed as below. The reactants enter at standard conditions; the combustion (reaction) takes place in the volume indicated. Downstream of the reaction zone there is an appropriate amount of heat transfer with the surroundings so that the products leave at the standard conditions. For the reaction of carbon and oxygen to produce CO2, the heat that has to be extracted is QCV = −393, 522 kJ/kmole ; this is heat that comes out of the control volume. 2C-2 C + 02 C02 1 kmole C 1 kmole C02 25o C, 0.1 MPa 25o C, 0.1 MPa Volume 1 kmole C02 Qcv= -393,522 KJ, heat is out of control volume Figure C-1: Constant pressure combustion There is no shaft work done in the control volume and the first law for the control volume (SFEE) reduces to: mass flow of enthalpy in + rate of heat addition = mass flow of enthalpy out. We can write this statement in the form ∑ m˙ i hi + Q̇CV = ∑ m˙ e he R (C.3.1) P In Eq. (C.3.1) the subscripts “R” and “P” on the summations refer to the reactants (R) and products (P) respectively. The subscripts on the mass flow rates and enthalpies refer to all of the components at inlet and at exit. The relation in terms of mass flows can be written in molar form, which is often more convenient for reacting flow problems, by using the molecular weight, Mi , to define the molar mass flow rate, ṅi , and molar enthalpy, hi , for any individual ith (or eth) component as ṅi = ṁi / Mi ; mass flow rate in terms of kmoles/sec hi = Mi hi ; enthalpy per kmole The SFEE is, in these terms, ∑ n˙i hi + Q̇CV = ∑ n˙ e he . R (C.3.2) P The statements that have been made do not necessarily need to be viewed in the context of flow processes. Suppose we have one unit of C and one unit of O2 at the initial conditions and we carry out a constant pressure reaction at ambient pressure, Pamb . If so, 2C-3 U final − Uinitial = Q − W ( ) = QCV − Pamb V final − Vinitial , since Pi = Pf = Pamb . Combining terms, U final + Pfinal V final − (Uinitial + Pinitial Vinitial ) = QCV , or, H final − Hinitial = QCV . In terms of the numbers of moles and the specific enthalpy this is ∑ ni hi + QCV = ∑ ne he R (C.3.3) P The enthalpy of CO2, at 25oC and 0.1 MPa, with reference to a base where the enthalpy of the elements is zero, is called the enthalpy of formation and denoted by h fo . Values of the heat of formation for a number of substances are given in Table A.9 in SB&VW. The enthalpies of the reactants and products for the formation of CO2 are: hO2 = hC = 0 ( ) For one kmole: QCV = ∑ ne he = H P = h of P CO2 = −393, 522 kJ/kmole. The enthalpy of CO2 in any other state (T,P)is given by ( ) hT ,P = h 0f 298K,0.1 MPa + ( ∆h ) 298K,0.1 Mpa→T , P . These descriptions can be applied to any compound. For elements or compounds that exist in more than one state at the reference conditions (for example, carbon exists as diamond and as graphite), we also need to specify the state. Note that there is a minus sign for the heat of formation. The heat transfer is out of the control volume and is thus negative by our convention. This means that helements > hCO2 = −393, 522 kJ . Muddy points Is the enthalpy of formation equal to the heat transfer out of the combustion during the formation reaction? (MP 2C.4) Are the enthalpies of H2 and H (monoatomic hydrogen) both zero at 298K? (MP 2C.5) 2C-4 2.C.4 First Law Analysis of Reacting Systems The form of the first law for the control volume is (there is no shaft work): ∑ n˙i hi + Q̇CV = ∑ n˙ e he . R P This is given in terms of the moles of the different constituents, and it reduces to the more familiar form for a single fluid (say air) with no reactions occurring. We need to specify one parameter as the basis of the solution; 1 kmole of fuel, 1 kmole of air, 1 kmole total, etc. We use 1 kmole of fuel as the basic unit and examine the burning of hydrogen. H2 + 20 2 → 2H2 O The reactants and the products are both taken to be at 0.1MPa and 25oC, so the inlet and exit P and T are specified. The control volume is the combustion chamber. There is no shaft work done and the SFEE is in the form of Equation (C.1.2). The enthalpy of the entering gas is zero for both the hydrogen and the oxygen (elements have enthalpies defined as zero at the reference state). If the exit products are in the gaseous state, the exit enthalpy is therefore related to the enthalpy of formation of the product by: ṅe h H2O e H2O = ṅe H2O (h ) o f H O( g ) 2 = 2 x (- 241,827)kJ = - 483,654 kJ; gaseous state at exit. If the water is in a liquid state at the exit of the process: ṅe he H2O = ṅe h fo H2O H2O ( ) H2O ( l ) = 2 x (- 285,783) kJ = - 571, 676. There is more heat given up if the products emerge as liquid. The difference between the two values is the enthalpy needed to turn the liquid into gas at 25oC: h fg = 2442 kJ/kmole. A more complex example is provided by the burning of methane (natural gas) in oxygen, producing . CH 4 + 2O 2 → CO 2 + 2H 2 O(l ) The components in this reaction equation are three ideal gases (methane, oxygen, and CO2) and liquid water. We again specify that the inlet and exit states are at the reference conditions so that: ( ) = (h ) o ∑ ni hi = h f R ∑ ne he P QCV CH4 o f CO 2 = −74, 873 kJ ( ) + 2 h fo H2O ( l ) =-393,522 + 2(-285,838) = -965,198 kJ = −965,198 kJ – (-74,873) = -890,325 kJ. 2C-5 Suppose the substances which comprise the reactants and the products are not at 25oC and 0.1MPa. If so, the expression that connects the reactants and products is; o (C.4.1) . QCV + ∑ ni h fo + ∆ h = n h + ∆ h ∑ e f { { P R Between Between Te , Pe and reference Ti , Pi and reference i e conditions conditions Equation (C.4.1) shows that we must compute the enthalpy difference ∆h between the reference conditions and the given state if the inlet or exit conditions are not the reference pressure and temperature. There are different levels of approximation for the computation: (a) assume the specific heat is constant over the range at some average value, (b) use the polynomial expressions (Table A.6) in the integral, and (c) use tabulated values. The first is the simplest and the crudest. Combustion processes often involve changes of a thousand degrees or more and, as Figure C-2 shows, the specific heat for some gases can change by a factor of two or more over this range, although the changes for air are more modest. This means that, depending on the accuracy desired, one may need to consider the temperature dependence of the specific heat in computing ∆h . Figure C-2 available from: Sonntag, Bognakke and Van Wylen, Fundamentals of Thermodynamics, 6th Ed., John Wiley & Sons. Figure 5.11 p.137. Figure C-2: Specific heat as a function of temperature [from SB&VW] Muddy points When doing cycle analysis, do we have to consider combustion products and their effect on specific heat ratio (γ is not 1.4)? (MP 2C.6) 2C-6 2.C.5 Adiabatic Flame Temperature For a combustion process that takes place adiabatically with no shaft work, the temperature of the products is referred to as the adiabatic flame temperature. This is the maximum temperature that can be achieved for given reactants. Heat transfer, incomplete combustion, and dissociation, all result in lower temperature. The maximum adiabatic flame temperature for a given fuel and oxidizer combination occurs with a stoichiometric mixture (correct proportions such that all fuel and all oxidizer are consumed). The amount of excess air can be tailored as part of the design to control the adiabatic flame temperature. The considerable distance between present temperatures in a gas turbine engine and the maximum adiabatic flame temperature at stoichiometric conditions is shown in Figure A-11 of Part 1, based on a compressor exit temperature of 1200oF (922 K). An initial view of the concept of adiabatic flame temperature is provided by examining two reacting gases, at a given pressure, and asking what the end temperature is. The f = Final state process is shown schematically at the right, ∆ha where temperature is plotted versus the Actual path percentage completion of the reaction. ∆h2 T The initial state is i and the final state is f, 2 with the final state at a higher Constant P 1 ∆h1 temperature than the initial state. The State i 2 solid line in the figure shows a Constant P representation of the “actual” process. 0 100% Percentage To see how we would arrive at the final completion state the dashed lines break the state of reaction change into two parts. Process (1) is reaction at constant T and P. To carry out such a process, we would need to extract heat. Suppose the total amount of heat extracted Schematic of adiabatic flame temperature per unit mass is q1 . The relation between the enthalpy changes in Process (1) is ( ) h2 − hi = − q1 = h of unit mass where q1 is the “heat of reaction”. For Process (2), we put this amount back into the products to raise their temperature to the final level. For this process, h f − h2 = q1, or, if we can approximate the specific heat as constant (using ( ) some appropriate average value) c pav. T f − T2 = q1 . For the overall process there is no work done and no heat exchanged so that the difference in enthalpy between initial and final states is zero: ∆h1 + ∆h2 = ∆hadiabatic = 0 . The temperature change during this second process is therefore given by (approximately) 2C-7 (T f ) − T2 = q1 c pav. = (h ) o f unit mass . c pav. (C.5.1) The value of the adiabatic flame temperature given in Equation (C.5.1) is for 100% completion of the reaction. In reality, as the temperature increases, the tendency is for the degree of reaction to be less than 100%. For example, for the combustion of hydrogen and oxygen, at high temperatures the combustion product (water) dissociates back into the simpler elemental reactants. The degree of reaction is thus itself a function of temperature that needs to be computed. We used this idea in discussing the stoichiometric ramjet, when we said that the maximum temperature was independent of flight Mach number and hence of inlet stagnation temperature. It is also to be emphasized that the idea of a constant (average) specific heat, c p av. , is for illustration and not inherently part of the definition of adiabatic flame temperature. An example computation of adiabatic flame temperature is furnished by the combustion of liquid octane at 25oC with 400% theoretical air. The reaction is [ ] C 8 H18 (l ) +12.5O 2 +12.5(3.76N 2 ) + 3 12.5O 2 +12.5(3.76N 2 ) → 8CO 2 + 9H 2 O(g) + 37.5O 2 +188N 2 . For an adiabatic process ( o ∑ ni h f + ∆h R ) i ( ) = ∑ ne h fo + ∆h . P e (C.5.2) At adiabatic flame temperature We can again think of the general process in steps: a) Bring reactants to 25oC the term ( ∆h ) from the initial temperature, using whatever [ i ] heat transfer, qa , is needed. In this example we do not need step (i) because we are already at the reference temperature. . There will be some heat transfer in this step, b) Reaction at 25oC - the term h of reactants→ products qb , out of the combustor. c) Put back heat qa + qb into the products of combustion. The resulting temperature is the adiabatic flame temperature. ( ) In the present case Equation (C.1.6) is, explicitly: h fo C 8H18 (l ) = 8h fo CO + 9h fo H O + {∆hCO2 + 9∆hH2O + 37.5∆hO2 +188∆hN2 } 2 2 We can examine the terms in the SFEE separately, starting with the heat of formation terms, h fo : 8h fo CO + 9h fo H 2 2O − h fo C 8H18 (l ) = 8 (-393,522) + 9 (-241,827) – (-249,952) 2C-8 = -5.075 X 106 kJ/kmole. The exit state at the adiabatic flame temperature is specified by: 6 ∑ ne ∆he = 5.075 X 10 kJ/kmole P We find the adiabatic flame temperature in three ways, approximate solution using an average value of c p , a more accurate one using the tabulated evolution of cp with temperature and a more precise solution using the tabulated values for gas enthalpy in Table A.8 of SB&VW. a) Approximate solution using “average” values of specific heat: From Figure C-2 we can use the values at 500K as representative. These are: Gas c p (kJ/kmole) CO2 H2 O O2 N2 45 35 30 30. Using ∆h = c p"ave" ∆T , {( ∑ ne ∆he = ∆T 8 c p P ) CO2 ( ) + 9 cp H 2O ( ) + 37.5 c p O2 ( ) +188 c p N2 } where ∆T = T final − 25 o C = T final − 298 K ∑ ne ∆he = 7440 ∆T kJ P ∆T = 682 ⇒ T final = 980 K . b) Solution for adiabatic flame temperature using evolutions of specific heats with temperature Tables give the following evolutions of specific heats with temperature: Gas Evolution of cp/ R with T (kJ/kmol) 2.401+8.735.10-3xT-6.607.10-6xT2+2.002.10-9xT3 CO2 H2 O 4.070-1.108.10-3xT+4.152.10-6xT2-2.964.10-9xT3+0.807.10-12xT4 O2 3.626-1.878.10-3xT+7.055.10-6xT2-6.764.10-9xT3+2.156.10-12xT4 N2 3.675-1.208.10-3xT+2.324.10-6xT2-0.632.10-9xT3-0.226.10-12xT4 Using ∆h = Tf in K (T).dT and the same equation as above, we obtain: ∫ c p 298 K Tf= 899 K c) Solution for adiabatic flame temperature using tabulated values for gas enthalpy: T= 900 K T=1000K ∆hCO2 28,041 33,405 ∆hH2O 21,924 25,978 ∆hO2 19,246 22,707 2C-9 ∆hN2 18,221 kJ/kmole 21,460 kJ/kmole Plugging in the numbers shows the answer is between these two conditions. Linearly interpolating gives a value of T final = 962 K . Muddy points Does "adiabatic flame temperature" assume 100% combustion? (MP 2C.7) What part of the computation for adiabatic flame temperature involves iteration? (MP 2C.8) 2C-10 Muddiest points on part 2C 2C.1 Why is there 3.76 N2? This is to represent the components other than oxygen that are in air. From SB&VW, page 525, “The assumption that air is 21% oxygen and 79% nitrogen by volume leads to the conclusion that for each mole of oxygen, 79/21 =3.76 moles of nitrogen are involved.” 2C.2 What is the most effective way to solve for the number of moles in the reactions? What we are doing is basically counting atoms on both sides of the reactants�products statement. See Section 14.2 of SB&VW for a description of the combustion process and for going from ratios in terms of moles to ratios in terms of mass. 2C.3 Do we always assume 100% complete combustion? How good an approximation is this? In the problems we do, we will only consider 100% combustion. approximation for the range of problems that we address. It is a good 2C.4 Is the enthalpy of formation equal to the heat transfer out of the combustion during the formation reaction? As defined, the enthalpy of formation relates to a process in which the initial and final states are at the same temperature. If there is combustion in between, heat will have to be removed for this condition to occur. The enthalpy of formation is equal to the negative of the magnitude of the heat outflow. If we consider the combustion as occurring in a control volume, then per kmole h fo = −Qcv , where Qcv is the heat transfer out of the control volume per kmole. This is not in accord with our convention, and if you please feel free to transform it back into the notation we have used before. (I find that if I do this there are too many minus signs to keep track of easily.) 2C.5 Are the enthalpies of H2 and H (monoatomic hydrogen) both zero at 298K? The enthalpies of the elements are taken as zero at 298 and 0.1 MPa. In some cases there are more than one form of the element. In that case the form chosen to have the value of zero is that which is chemically stable at the reference state. The other forms then have an enthalpy which is consistent with the reaction that produces this form of the element. For hydrogen H2 has zero enthalpy at the reference conditions and H has an enthalpy of 217,999 kJ/kmole (see Table A.8 in SB&VW), consistent with the idea that energy has to be supplied to break the molecule apart. 2C.6 When doing a cycle analysis, do we have to consider combustion products and their effect on specific heat ratio (γ is not 1.4)? The specific heat ratio does depend on combustion products but the effect is not large because the fuel air ratio is small. For example, for conditions of fuel air ratio ).034, the specific heat ratio at room temperature is about 1.38. A larger variation encountered in practice is with temperature; for a temperature of 1750K the specific heat ratio of pure air is 1.3. 2C.7 Does "adiabatic flame temperature" assume 100% combustion? Yes. This is the maximum temperature that could be produced. Incomplete combustion will lower the temperature, as will heat transfer out of the combustion region. 2C.8 What part of the computation for adiabatic flame temperature involves iteration? If the specific heat is not a simple analytic function of temperature (i.e., suppose it is known only in tabular form), we cannot get a closed form solution for the adiabatic flame temperature. We can, however, readily solve the enthalpy balance (SFEE) numerically (this is where the iteration comes in) to find at what temperature the products have to come out to have the same enthalpy (including the enthalpy of formation) as the reactants. We did not do this calculation yet, but we will do it to show what the iteration is all about. Remember that the assumption of constant specific heat is just that, an assumption. While this is an excellent assumption for many practical problems, if the precision of the answer needed is very high, or if the range of temperatures is large (see Figure 5.11 in SB&VW), then we cannot assume constant specific heat. PART 3 INTRODUCTION TO ENGINEERING HEAT TRANSFER Introduction to Engineering Heat Transfer These notes provide an introduction to engineering heat transfer. Heat transfer processes set limits to the performance of aerospace components and systems and the subject is one of an enormous range of application. The notes are intended to describe the three types of heat transfer and provide basic tools to enable the readers to estimate the magnitude of heat transfer rates in realistic aerospace applications. There are also a number of excellent texts on the subject; some accessible references which expand the discussion in the notes are listen in the bibliography. HT-1 Table of Tables Table 2.1: Thermal conductivity at room temperature for some metals and non-metals ............. HT-7 Table 2.2: Utility of plane slab approximation..........................................................................HT-17 Table 9.1: Total emittances for different surfaces [from: A Heat Transfer Textbook, J. Lienhard ]HT-63 HT-2 Table of Figures Figure 1.1: Conduction heat transfer ......................................................................................... HT-5 Figure 2.1: Heat transfer along a bar ......................................................................................... HT-6 Figure 2.2: One-dimensional heat conduction ........................................................................... HT-8 Figure 2.3: Temperature boundary conditions for a slab............................................................ HT-9 Figure 2.4: Temperature distribution through a slab .................................................................HT-10 Figure 2.5: Heat transfer across a composite slab (series thermal resistance) ............................HT-11 Figure 2.6: Heat transfer for a wall with dissimilar materials (Parallel thermal resistance)........HT-12 Figure 2.7: Heat transfer through an insulated wall ..................................................................HT-11 Figure 2.8: Temperature distribution through an insulated wall ................................................HT-13 Figure 2.9: Cylindrical shell geometry notation........................................................................HT-14 Figure 2.10: Spherical shell......................................................................................................HT-17 Figure 3.1: Turbine blade heat transfer configuration ...............................................................HT-18 Figure 3.2: Temperature and velocity distributions near a surface. ...........................................HT-19 Figure 3.3: Velocity profile near a surface................................................................................HT-20 Figure 3.4: Momentum and energy exchange in turbulent flow. ...............................................HT-20 Figure 3.5: Heat exchanger configurations ...............................................................................HT-23 Figure 3.6: Wall with convective heat transfer .........................................................................HT-25 Figure 3.7: Cylinder in a flowing fluid .....................................................................................HT-26 Figure 3.8: Critical radius of insulation ....................................................................................HT-29 Figure 3.9: Effect of the Biot Number [hL / kbody] on the temperature distributions in the solid and in the fluid for convective cooling of a body. Note that kbody is the thermal conductivity of the body, not of the fluid.........................................................................................................HT-31 Figure 3.10: Temperature distribution in a convectively cooled cylinder for different values of Biot number, Bi; r2 / r1 = 2 [from: A Heat Transfer Textbook, John H. Lienhard] .....................HT-32 Figure 4.1: Slab with heat sources (a) overall configuration, (b) elementary slice.....................HT-32 Figure 4.2: Temperature distribution for slab with distributed heat sources ..............................HT-34 Figure 5.1: Geometry of heat transfer fin .................................................................................HT-35 Figure 5.2: Element of fin showing heat transfer ......................................................................HT-36 Figure 5.3: The temperature distribution, tip temperature, and heat flux in a straight onedimensional fin with the tip insulated. [From: Lienhard, A Heat Transfer Textbook, PrenticeHall publishers].................................................................................................................HT-40 Figure 6.1: Temperature variation in an object cooled by a flowing fluid .................................HT-41 Figure 6.2: Voltage change in an R-C circuit............................................................................HT-42 Figure 8.1: Concentric tube heat exchangers. (a) Parallel flow. (b) Counterflow.......................HT-44 Figure 8.2: Cross-flow heat exchangers. (a) Finned with both fluids unmixed. (b) Unfinned with one fluid mixed and the other unmixed ....................................................................................HT-45 Figure 8.3: Geometry for heat transfer between two fluids .......................................................HT-45 Figure 8.4: Counterflow heat exchanger...................................................................................HT-46 Figure 8.5: Fluid temperature distribution along the tube with uniform wall temperature .........HT-46 Figure 9.1: Radiation Surface Properties ..................................................................................HT-52 Figure 9.2: Emissive power of a black body at several temperatures - predicted and observed..HT-53 Figure 9.3: A cavity with a small hole (approximates a black body) .........................................HT-54 Figure 9.4: A small black body inside a cavity .........................................................................HT-54 Figure 9.5: Path of a photon between two gray surfaces ...........................................................HT-55 HT-3 Figure 9.6: Thermocouple used to measure temperature...........................................................HT-59 Figure 9.7: Effect of radiation heat transfer on measured temperature. .....................................HT-59 Figure 9.8: Shielding a thermocouple to reduce radiation heat transfer error ............................HT-60 Figure 9.9: Radiation between two bodies................................................................................HT-60 Figure 9.10: Radiation between two arbitrary surfaces .............................................................HT-61 Figure 9.11: Radiation heat transfer for concentric cylinders or spheres ...................................HT-62 Figure 9.12: View Factors for Three - Dimensional Geometries [from: Fundamentals of Heat Transfer, F.P. Incropera and D.P. DeWitt, John Wiley and Sons] ......................................HT-64 Figure 9.13: Fig. 13.4--View factor for aligned parallel rectangles [from: Fundamentals of Heat Transfer, F.P. Incropera and D.P. DeWitt, John Wiley and Sons] ......................................HT-65 Figure 9.14: Fig 13.5--View factor for coaxial parallel disk [from: Fundamentals of Heat Transfer, F.P. Incropera and D.P. DeWitt, John Wiley and Sons] .....................................................HT-65 Figure 9.15: Fig 13.6--View factor for perpendicular rectangles with a common edge .............HT-66 HT-4 1.0 Heat Transfer Modes Heat transfer processes are classified into three types. The first is conduction, which is defined as transfer of heat occurring through intervening matter without bulk motion of the matter. Figure 1.1 shows the process pictorially. A solid (a block of metal, say) has one surface at a high temperature and one at a lower temperature. This type of heat conduction can occur, for example, through a turbine blade in a jet engine. The outside surface, which is exposed to gases from the combustor, is at a higher temperature than the inside surface, which has cooling air next to it. The level of the wall temperature is critical for a turbine blade. Thigh Tlow Heat “flows” to right ( q& ) Solid Figure 1.1: Conduction heat transfer The second heat transfer process is convection, or heat transfer due to a flowing fluid. The fluid can be a gas or a liquid; both have applications in aerospace technology. In convection heat transfer, the heat is moved through bulk transfer of a non-uniform temperature fluid. The third process is radiation or transmission of energy through space without the necessary presence of matter. Radiation is the only method for heat transfer in space. Radiation can be important even in situations in which there is an intervening medium; a familiar example is the heat transfer from a glowing piece of metal or from a fire. Muddy points How do we quantify the contribution of each mode of heat transfer in a given situation? (MP HT.1) 2.0 Conduction Heat Transfer We will start by examining conduction heat transfer. We must first determine how to relate the heat transfer to other properties (either mechanical, thermal, or geometrical). The answer to this is rooted in experiment, but it can be motivated by considering heat flow along a "bar" between two heat reservoirs at TA, TB as shown in Figure 2.1. It is plausible that the heat transfer rate Q& , is a HT-5 function of the temperature of the two reservoirs, the bar geometry and the bar properties. (Are there other factors that should be considered? If so, what?). This can be expressed as Q& = f1 (TA , TB , bar geometry, bar properties) (2.1) It also seems reasonable to postulate that Q& should depend on the temperature difference TA - TB. If TA – TB is zero, then the heat transfer should also be zero. The temperature dependence can therefore be expressed as Q& = f2 [ (TA - TB), TA, bar geometry, bar properties] TA (2.2) TB Q& L Figure 2.1: Heat transfer along a bar An argument for the general form of f2 can be made from physical considerations. One requirement, as said, is f2 = 0 if TA = TB. Using a MacLaurin series expansion, as follows: ∂f ∆T + L ∂( ∆T) 0 f( ∆T) = f(0) + (2.3) If we define ∆T = TA – TB and f = f2, we find that (for small TA – TB), ⋅ f 2 (TA − TB ) = Q = f 2 (0) + ∂f 2 ∂(TA − TB ) T A −T B =0 (TA − TB ) + L. (2.4) We know that f2(0) = 0 . The derivative evaluated at TA = TB (thermal equilibrium) is a measurable ⋅ ∂f 2 property of the bar. In addition, we know that Q > 0 if TA > TB or > 0 . It also seems ∂ TA − TB reasonable that if we had two bars of the same area, we would have twice the heat transfer, so that we can postulate that Q& is proportional to the area. Finally, although the argument is by no means rigorous, experience leads us to believe that as L increases Q& should get smaller. All of these lead to the generalization (made by Fourier in 1807) that, for the bar, the derivative in equation (2.4) has the form ( HT-6 ) ∂f 2 ∂ TA − TB ( ) T A −T B =0 = kA . L (2.5) In equation (2.5), k is a proportionality factor that is a function of the material and the temperature, A is the cross-sectional area and L is the length of the bar. In the limit for any temperature difference ∆T across a length ∆x as both L, TA - TB → 0, we can say (T − TB ) (T − TA ) dT . = − kA B = − kA Q& = kA A dx L L (2.6) A more useful quantity to work with is the heat transfer per unit area, defined as Q& = q& . A (2.7) The quantity q& is called the heat flux and its units are Watts/m2. The expression in (2.6) can be written in terms of heat flux as q& = − k dT . dx (2.8) Equation 2.8 is the one-dimensional form of Fourier's law of heat conduction. The proportionality constant k is called the thermal conductivity. Its units are W / m-K. Thermal conductivity is a well-tabulated property for a large number of materials. Some values for familiar materials are given in Table 1; others can be found in the references. The thermal conductivity is a function of temperature and the values shown in Table 1 are for room temperature. Table 2.1: Thermal conductivity at room temperature for some metals and non-metals Metals k [W/m-K] Non-metals k [W/m-K] H20 0.6 Ag 420 Air 0.026 Cu 390 Engine oil 0.15 HT-7 Al 200 H2 0.18 Fe 70 Brick 0.4 -0 .5 Steel 50 Wood Cork 0.2 0.04 2.1 Steady-State One-Dimensional Conduction Insulated (no heat transfer) Q& (x ) Q& (x + dx ) dx x Figure 2.2: One-dimensional heat conduction For one-dimensional heat conduction (temperature depending on one variable only), we can devise a basic description of the process. The first law in control volume form (steady flow energy equation) with no shaft work and no mass flow reduces to the statement that ΣQ& for all surfaces = 0 (no heat transfer on top or bottom of figure 2.2). From equation (2.8), the heat transfer rate in at the left (at x) is dT Q̇( x) = −k⎛ A ⎞ . ⎝ dx ⎠ x (2.9) The heat transfer rate on the right is ˙ ˙ ( x + dx) = Q ˙ ( x) + dQ dx + L. Q dx x Using the conditions on the overall heat flow and the expressions in (2.9) and (2.10) ˙ ˙ ( x) − ⎛⎜Q ˙ ( x) + dQ ( x)dx + L⎞⎟ = 0 . Q ⎝ ⎠ dx (2.10) (2.11) Taking the limit as dx approaches zero we obtain ˙ ( x) dQ = 0, dx (2.12a) or HT-8 d ⎛ dT ⎞ ⎜ kA ⎟ = 0 . dx ⎝ dx ⎠ (2.12b) If k is constant (i.e. if the properties of the bar are independent of temperature), this reduces to d ⎛ dT ⎞ ⎜A ⎟ =0 dx ⎝ dx ⎠ (2.13a) or (using the chain rule) 2 ⎛ 1 dA ⎞ dT +⎜ = 0. ⎟ ⎝ A dx ⎠ dx dx d T (2.13b) 2 Equations (2.13a) or (2.13b) describe the temperature field for quasi-one-dimensional steady state (no time dependence) heat transfer. We now apply this to some examples. Example 2.1: Heat transfer through a plane slab T = T1 T = T2 Slab x=0 x=L x Figure 2.3: Temperature boundary conditions for a slab For this configuration, the area is not a function of x, i.e. A = constant. Equation (2.13) thus became d 2T =0. dx 2 (2.14) Equation (2.14) can be integrated immediately to yield dT =a dx (2.15) HT-9 T = ax + b . and (2.16) Equation (2.16) is an expression for the temperature field where a and b are constants of integration. For a second order equation, such as (2.14), we need two boundary conditions to determine a and b. One such set of boundary conditions can be the specification of the temperatures at both sides of the slab as shown in Figure 2.3, say T (0) = T1; T (L) = T2. The condition T (0) = T1 implies that b = T1. The condition T2 = T (L) implies that T2 = aL + T1, or T −T a= 2 1 . L With these expressions for a and b the temperature distribution can be written as ⎛T −T ⎞ T x = T1 + ⎜ 2 1 ⎟ x . ⎝ L ⎠ () (2.17) This linear variation in temperature is shown in Figure 2.4 for a situation in which T1 > T2. T T1 T2 x Figure 2.4: Temperature distribution through a slab The heat flux q& is also of interest. This is given by q& = − k (T − T ) dT = − k 2 1 = constant . dx L (2.18) Muddy points How specific do we need to be about when the one-dimensional assumption is valid? Is it enough to say that dA/dx is small? (MP HT.2) Why is the thermal conductivity of light gases such as helium (monoatomic) or hydrogen (diatomic) much higher than heavier gases such as argon (monoatomic) or nitrogen (diatomic)? (MP HT.3) HT-10 2.2 Thermal Resistance Circuits There is an electrical analogy with conduction heat transfer that can be exploited in problem solving. The analog of Q& is current, and the analog of the temperature difference, T1 - T2, is voltage difference. From this perspective the slab is a pure resistance to heat transfer and we can define T − T2 Q& = 1 R (2.19) where R = L/kA, the thermal resistance. The thermal resistance R increases as L increases, as A decreases, and as k decreases. The concept of a thermal resistance circuit allows ready analysis of problems such as a composite slab (composite planar heat transfer surface). In the composite slab shown in Figure 2.5, the heat flux is constant with x. The resistances are in series and sum to R = R1 + R2. If TL is the temperature at the left, and TR is the temperature at the right, the heat transfer rate is given by T − TR TL − TR Q& = L = . R R1 + R2 (2.20) x TL TR 1 2 Q& R1 R2 Figure 2.5: Heat transfer across a composite slab (series thermal resistance) Another example is a wall with a dissimilar material such as a bolt in an insulating layer. In this case, the heat transfer resistances are in parallel. Figure 2.6 shows the physical configuration, the heat transfer paths and the thermal resistance circuit. HT-11 k1 R1 model Q& k2 R2 k1 Figure 2.6: Heat transfer for a wall with dissimilar materials (Parallel thermal resistance) For this situation, the total heat flux Q& is made up of the heat flux in the two parallel paths: Q& = Q& + Q& with the total resistance given by: 1 2 1 1 1 = + . R R1 R2 (2.21) More complex configurations can also be examined; for example, a brick wall with insulation on both sides. Brick 0.1 m R1 R2 R3 T4 = 10 °C T1 = 150 °C T2 T1 T3 T2 T3 T4 Insulation 0.03 m Figure 2.7: Heat transfer through an insulated wall The overall thermal resistance is given by R = R1 + R2 + R3 = L1 L L + 2 + 3 k1 A1 k 2 A2 k 3 A3 . Some representative values for the brick and insulation thermal conductivity are: HT-12 (2.22) kbrick = k2 = 0.7 W/m-K kinsulation = k1 = k3 = 0.07 W/m-K Using these values, and noting that A1 = A2 = A3 = A, we obtain: AR1 = AR3 = AR2 = L1 0.03 m = = 0.42 m 2 K/W k1 0.07 W/m K L2 0.1 m = = 0.14 m 2 K/W . k 2 0.7 W/m K This is a series circuit so q& = T − T4 Q& 140 K = constant throughout = 1 = = 142 W/m 2 2 A RA 0.98 m K/W 1.0 1 2 3 4 T − T4 T1 − T4 x 0 Figure 2.8: Temperature distribution through an insulated wall The temperature is continuous in the wall and the intermediate temperatures can be found from applying the resistance equation across each slab, since Q& is constant across the slab. For example, to find T2: q& = T1 − T2 = 142 W/m 2 R1 A This yields T1 – T2 = 60 K or T2 = 90 °C. The same procedure gives T3 = 70 °C. As sketched in Figure 2.8, the larger drop is across the insulating layer even though the brick layer is much thicker. Muddy points What do you mean by continuous? (MP HT.4) Why is temperature continuous in the composite wall problem? Why is it continuous at the interface between two materials? (MP HT.5) HT-13 Why is the temperature gradient dT/dx not continuous? (MP HT.6) Why is ∆T the same for the two elements in a parallel thermal circuit? Doesn't the relative area of the bolt to the wood matter? (MP HT.7) 2.3 Steady Quasi-One-Dimensional Heat Flow in Non-Planar Geometry The quasi one-dimensional equation that has been developed can also be applied to non-planar geometries. An important case is a cylindrical shell, a geometry often encountered in situations where fluids are pumped and heat is transferred. The configuration is shown in Figure 2.9. control volume r1 r1 r2 r2 Figure 2.9: Cylindrical shell geometry notation For a steady axisymmetric configuration, the temperature depends only on a single coordinate (r) and Equation (2.12b) can be written as k d ⎛ dT ⎞ ⎜A r ⎟ =0 dr ⎝ dr ⎠ () (2.23) or, since A = 2π r, d ⎛ dT ⎞ ⎜r ⎟ = 0. dr ⎝ dr ⎠ (2.24) The steady-flow energy equation (no flow, no work) tells us that Q& in = Q& out or dQ& =0 dr (2.25) The heat transfer rate per unit length is given by ⋅ Q = − k ⋅ 2π r dT . dr HT-14 Equation (2.24) is a second order differential equation for T. Integrating this equation once gives r dT =a. dr (2.26) where a is a constant of integration. Equation (2.26) can be written as dT = a dr r (2.27) where both sides of equation (2.27) are exact differentials. It is useful to cast this equation in terms of a dimensionless normalized spatial variable so we can deal with quantities of order unity. To do this, divide through by the inner radius, r1 dT = a d (r / r1 ) (r / r1 ) (2.28) Integrating (2.28) yields ⎛r⎞ T = a ln ⎜ ⎟ + b . ⎝ r1 ⎠ (2.29) To find the constants of integration a and b, boundary conditions are needed. These will be taken to be known temperatures T1 and T2 at r1 and r2 respectively. Applying T = T1 at r = r1 gives T1 = b. Applying T = T2 at r = r2 yields r T2 = a ln 2 + T1 , r1 or a= T2 − T1 . ln(r2 / r1 ) The temperature distribution is thus T = (T2 − T1 ) ln(r / r1 ) + T1 . ln(r2 / r1 ) (2.30) As said, it is generally useful to put expressions such as (2.30) into non-dimensional and normalized form so that we can deal with numbers of order unity (this also helps in checking whether results are consistent). If convenient, having an answer that goes to zero at one limit is also useful from the perspective of ensuring the answer makes sense. Equation (2.30) can be put in nondimensional form as HT-15 T − T1 ln(r / r1 ) = . T2 − T1 ln(r2 / r1 ) (2.31) The heat transfer rate, Q& , is given by (T − T1 ) 1 = 2π k (T1 − T2 ) dT Q& = − kA = − 2π r1k 2 dr ln(r2 / r1 ) ln(r2 / r1 ) r1 per unit length. The thermal resistance R is given by R= ln(r2 / r1 ) 2πk (2.32) T − T2 . Q& = 1 R The cylindrical geometry can be viewed as a limiting case of the planar slab problem. To r −r make the connection, consider the case when 2 1 << 1 . From the series expansion for ln (1 + x) r1 we recall that ( 2 ) 3 x x ln 1+ x ≈ x + +K 2 3 (2.33) (Look it up, try it numerically, or use the binomial theorem on the series below and integrate term by 1 term. = 1− x + x2 +K) 1+ x The logarithms in Equation (2.31) can thus be written as ⎛ r − r1 ⎞ r − r1 ln ⎜ 1 + and ⎟ ≅ r1 ⎠ r1 ⎝ ln r2 r2 − r1 ≅ r1 r1 (2.34) in the limit of (r2 – r1) << r1. Using these expressions in equation (2.30) gives T = (T2 − T1 ) (r − r1 ) + T (r2 − r1 ) 1 . (2.35) With the substitution of r – r1 = x, and r2 – r1 = L we obtain T = T1 + (T2 − T1 ) x L (2.36) HT-16 which is the same as equation (2.17). The plane slab is thus the limiting case of the cylinder if (r r1) / r << 1, where the heat transfer can be regarded as taking place in (approximately) a planar slab. ln (1 + x ) To see when this is appropriate, consider the expansion , which is the ratio of heat flux for a x cylinder and a plane slab. Table 2.2: Utility of plane slab approximation x .1 .2 .3 .4 .5 ln (1 + x ) x .95 .91 .87 .84 .81 For < 10% error, the ratio of thickness to inner radius should be less than 0.2, and for 20% error, the thickness to inner radius should be less than 0.5. A second example is the spherical shell with specified temperatures T (r1) = T1 and T (r2) = T2, as sketched in Figure 2.10. T2 T1 r1 r2 Figure 2.10: Spherical shell The area is now A(r ) = 4πr 2 , so the equation for the temperature field is d ⎛ 2 dT ⎞ ⎜r ⎟ = 0. dr ⎝ dr ⎠ (2.37) Integrating equation (2.37) once yields dT = a/ r2. dr (2.38) Integrating again gives HT-17 T =− a +b r or, normalizing the spatial variable T= a′ +b (r / r1 ) (2.39) where a′ and b are constants of integration. As before, we specify the temperatures at r = r1 and r = r2. Use of the first boundary condition gives T (r1 ) = T1 = a′ + b . Applying the second boundary condition gives T (r2 ) = T2 = a′ +b (r2 / r1 ) Solving for a′ and b, a′ = T1 − T2 1 − r1 / r2 (2.40) T − T2 b = T1 − 1 . 1 − r1 / r2 In non-dimensional form the temperature distribution is thus: T1 − T 1 − (r1 / r ) = T1 − T2 1 − (r1 / r2 ) (2.41) HT-18 3.0 Convective Heat Transfer The second type of heat transfer to be examined is convection, where a key problem is determining the boundary conditions at a surface exposed to a flowing fluid. An example is the wall temperature in a turbine blade because turbine temperatures are critical as far as creep (and thus blade) life. A view of the problem is given in Figure 3.1, which shows a cross-sectional view of a turbine blade. There are three different types of cooling indicated, all meant to ensure that the metal is kept at a temperature much lower than that of the combustor exit flow in which the turbine blade operates. In this case, the turbine wall temperature is not known and must be found as part of the solution to the problem. Figure 3.1: Turbine blade heat transfer configuration To find the turbine wall temperature, we need to analyze convective heat transfer, which means we need to examine some features of the fluid motion near a surface. The conditions near a surface are illustrated schematically in Figure 3.2. y y c∞ Ve locity distribution; c = 0 at surface δ′ c (velocity) T T∞ Tw Figure 3.2: Temperature and velocity distributions near a surface. HT-19 In a region of thickness δ′, there is a thin "film" of slowly moving fluid through which most of the temperature difference occurs. Outside this layer, T is roughly uniform (this defines δ′). The heat flux can thus be expressed as ⋅ ( Q k Tw − T∞ q= = A d′ ⋅ ) (3.1) It cannot be emphasized enough that this is a very crude picture. The general concept, however, is correct, in that close to the wall, there is a thin layer in which heat is transferred basically by conduction. Outside of this region is high mixing. The difficulty is that the thickness of the layer is not a fluid property. It depends on velocity (Reynolds number), structure of the wall surface, pressure gradient and Mach number. Generally δ′ is not known and needs to be found and it is customary to calculate the heat transfer using [kfluid / δ′]. This quantity has the symbol h and is known as the convective heat transfer coefficient. The units of h are W/m2K. The convective heat transfer coefficient is defined by ⋅ ⋅ ( Q q = = h Tw − T∞ A ) (3.2) Equation 3.2 is often called Newton’s Law of Cooling. For many situations of practical interest, the quantity h is still known mainly through experiments. Muddy points How do we know that δ' is not a fluid property? (MP HT.8) 3.1 The Reynolds Analogy We describe the physical mechanism for the heat transfer coefficient in a turbulent boundary layer because most aerospace vehicle applications have turbulent boundary layers. The treatment closely follows that in Eckert and Drake (1959). Very near the wall, the fluid motion is smooth and laminar, and molecular conduction and shear are important. The shear stress, τ, at a plane is given by dc dT . The latter is the same µ = τ (where µ is the dynamic viscosity), and the heat flux by q& = − k dy dy expression that was used for a solid. The boundary layer is a region in which the velocity is lower than the free stream as shown in Figures 3.2 and 3.3. In a turbulent boundary layer, the dominant mechanisms of shear stress and heat transfer change in nature as one moves away from the wall. HT-20 plane Figure 3.3: Velocity profile near a surface As one moves away from the wall (but still in the boundary layer), the flow is turbulent. The fluid particles move in random directions and the transfer of momentum and energy is mainly through interchange of fluid particles, shown schematically in Figure 3.4. m′ cp T′ 2 2 a a 1 1 m′ cp T Figure 3.4: Momentum and energy exchanges in turbulent flow. With reference to Figure 3.4, because of the turbulent velocity field, a fluid mass m′ penetrates the plane aa per unit time and unit area. In steady flow, the same amount crosses aa from the other side. Fluid moving up transports heat m′ cp T. Fluid moving down transports m′ cp T′ downwards. If T > T′, there is a turbulent downwards heat flow, q̇t , given by q̇t = m′c p (T ′ − T ) that results. Fluid moving up also has momentum m′ c and fluid moving down has momentum m′ c′. The net flux of momentum down per unit area and time is therefore m′ (c′ - c). This net flux of momentum per unit area and time is a force per unit area or stress, given by t t = m' (c′ − c) (3.3) Based on these considerations, the relation between heat flux and shear stress at plane aa is HT-21 T′− T⎞ q̇t = τ t c p⎛ ⎝ c′ − c ⎠ (3.4) or (again approximately) q̇t = τ t c p dT dc (3.5) since the locations of planes 1-1 and 2-2 are arbitrary. For the laminar region, the heat flux towards the wall is q̇ = τ k dT µ dc The same relationship is applicable in laminar or turbulent flow if k = c p or, expressed slightly µ differently, cp k = µ/ρ υ = =1 k / ρc p α where υ is the kinematic viscosity, and α is the thermal diffusivity. The quantity µcp/k is known as the Prandtl number (Pr), after the man who first presented the idea of the boundary layer and was one of the pioneers of modern fluid mechanics. For gases, Prandtl numbers are in fact close to unity and for air Pr = 0.71 at room temperature. The Prandtl number varies little over a wide range of temperatures; approximately 3% from 300-2000 K. We want a relation between the values at the wall (at which T = Tw and c = 0) and those in the free stream. To get this, we integrate the expression for dT from the wall to the free stream dT = 1 q̇ dc cp τ (3.6) where the relation between heat transfer and shear stress has been taken as the same for both the laminar and the turbulent portions of the boundary layer. The assumption being made is that the mechanisms of heat and momentum transfer are similar. Equation (3.6) can be integrated from the wall to the freestream (conditions "at ∞"): ∞ ∫ w where 1 dT = cp ∞ ∫ ⎛⎝τq̇⎞⎠dc (3.7) w q̇ and cp are assumed constant. τ HT-22 Carrying out the integration yields q̇ c T∞ − Tw = w ∞ τw c p (3.8) where c∞ is the velocity and cp is the specific heat. In equation (3.8), q& w is the heat flux to the wall and τw is the shear stress at the wall. The relation between skin friction (shear stress) at the wall and heat transfer is thus q̇w τ = w 2. ρ∞c p (T∞ − Tw )c∞ ρ∞c∞ The quantity (3.9) τw is known as the skin friction coefficient and is denoted by Cf . The skin 2 1/2ρ∞c∞ friction coefficient has been tabulated (or computed) for a large number of situations. If we define a non-dimensional quantity q̇w h(T∞ − Tw ) h = St, = = ρ∞c p (T∞ − Tw )c∞ ρ∞c p (T∞ − Tw )c∞ ρ∞c pc∞ known as the Stanton Number, we can write an expression for the heat transfer coefficient, h as h ≈ ρ∞c pc∞ Cf . 2 (3.10) Equation (3.10) provides a useful estimate of h, or q& w , based on knowing the skin friction, or drag. The direct relationship between the Stanton Number and the skin friction coefficient is St = Cf 2 The relation between the heat transfer and the skin friction coefficient q̇w ≈ τ wc p (Tw − T∞ ) c∞ is known as the Reynolds analogy between shear stress and heat transfer. The Reynolds analogy is extremely useful in obtaining a first approximation for heat transfer in situations in which the shear stress is "known". HT-23 An example of the use of the Reynolds analogy is in analysis of a heat exchanger. One type of heat exchanger has an array of tubes with one fluid flowing inside and another fluid flowing outside, with the objective of transferring heat between them. To begin, we need to examine the flow resistance of a tube. For fully developed flow in a tube, it is more appropriate to use an average velocity c and a bulk temperature TB . Thus, an approximate relation for the heat transfer is T −T q̇w ≈ τ wc p B w . c (3.11) The fluid resistance (drag) is all due to shear forces and is given by τw Aw = D, where Aw is the tube “wetted” area (perimeter x length). The total heat transfer, Q& , is q& w A w, so that T − TW Q& = Dc p B c (3.12) The power, P, to drive the flow through a resistance is given by the product of the drag and the velocity, Dc , so that Q̇ c p (TB − Tw ) = P c 2 (3.13) The mass flow rate is given by m& = ρc A where A is the cross sectional area. For given mass flow rate and overall heat transfer rate, the power scales as c2 or as 1/A2, i.e. ˙˙2 1 Qm P∝ 2 ρ c p (TB − Tw ) A2 (3.14) Equations (3.13) and (3.14) show that to decrease the power dissipated, we need to decrease c , which can be accomplished by increasing the cross-sectional area. Two possible heat exchanger configurations are sketched in Figure 3.5; the one on the right will have a lower loss. heat exchanger heat exchanger m⋅ diffuser high loss lower loss Figure 3.5: Heat exchanger configurations HT-24 To recap, there is an approximate relation between skin friction (momentum flux to the wall) and heat transfer called the Reynolds analogy that provides a useful way to estimate heat transfer rates in situations in which the skin friction is known. The relation is expressed by St = Cf 2 (3.15a) or or heat flux to wall momentum flux to wall = convected heat flux convected momentum flux (3.15b) τ q̇w ≈ w2 ρ∞c∞c p (T∞ − Tw ) ρ∞c∞ (3.15c) The Reynolds analogy can be used to give information about scaling of various effects as well as initial estimates for heat transfer. It is emphasized that it is a useful tool based on a hypothesis about the mechanism of heat transfer and shear stress and not a physical law. Muddy points What is the "analogy" that we are discussing? Is it that the equations are similar? (MP HT.9) In what situations does the Reynolds analogy "not work"? (MP HT.10) 3.2 Combined Conduction and Convection We can now analyze problems in which both conduction and convection occur, starting with a wall cooled by flowing fluid on each side. As discussed, a description of the convective heat transfer can be given explicitly as ˙ Q = q˙ = h(Tw − T∞ ) A (3.16) This could represent a model of a turbine blade with internal cooling. Figure 3.6 shows the configuration. HT-25 δ 1′ δ 2′ T2 Tw2 T Tw1 T1 L Figure 3.6: Wall with convective heat transfer The heat transfer in fluid 1 is given by Q̇ = h1(Tw1 − T1) , A which is the heat transfer per unit area to the fluid. The heat transfer in fluid (2) is similarly given by Q̇ = h2 (T2 − Tw2 ) . A Across the wall, we have Q̇ k = (Tw2 − Tw1) . A L The quantity Q& /A is the same in all of these expressions. Putting them all together to write the known overall temperature drop yields a relation between heat transfer and overall temperature drop, T2 – T1 : T2 − T1 = (T2 − Tw2 ) + (Tw2 − Tw1) + (Tw1 − T1) = We can define a thermal resistance, R, as before, such that Q̇ = (T2 − T1) R , HT-26 Q̇ ⎡ 1 L 1 ⎤ + + . A ⎣⎢ h1 k h2 ⎦⎥ (3.17) where R is given by R= 1 L 1 + + . h1 A Ak h2 A (3.18) Equation (3.18) is the thermal resistance for a solid wall with convection heat transfer on each side. For a turbine blade in a gas turbine engine, cooling is a critical consideration. In terms of Figure 3.6, T2 is the combustor exit (turbine inlet) temperature and T1 is the temperature at the compressor exit. We wish to find T w 2 because this is the highest metal temperature. From (3.17), the wall temperature can be written as Tw2 = T2 - Q̇ T −T 1 = T2 − 2 1 Ah2 R Ah2 (3.19) Using the expression for the thermal resistance, the wall temperatures can be expressed in terms of heat transfer coefficients and wall properties as Tw2 = T2 − T2 − T1 h2 Lh2 + +1 h1 k (3.20) Equation (3.20) provides some basic design guidelines. The goal is to have a low value of T w 2 . This means h1/h2 should be large, k should be large (but we may not have much flexibility in choice of material) and L should be small. One way to achieve the first of these is to have h2 low (for example, to flow cooling air out as in Figure 3.1 to shield the surface). A second example of combined conduction and convection is given by a cylinder exposed to a flowing fluid. The geometry is shown in Figure 3.7. T∞ rr11 c∞ r2 Figure 3.7: Cylinder in a flowing fluid HT-27 For the cylinder the heat flux at the outer surface is given by q˙ = ˙ Q = h(Tw − T∞ ) at r = r2 A The boundary condition at the inner surface could either be a heat flux condition or a temperature specification; we use the latter to simplify the algebra. Thus, T = T1 at r = r1 . This is a model for the heat transfer in a pipe of radius r1 surrounded by insulation of thickness r2 - r1. The solution for a cylindrical region was given in Section 2.3 as T ( r ) = a ln r +b r1 Use of the boundary condition T (r1) = T1 yields b = T1. At the interface between the cylinder and the fluid, r = r2, the temperature and the heat flow are continuous. (Question: Why is this? How would you argue the point?) q̇ = −k dT a r ⎡⎛ ⎤ ⎞ = −k = h ⎜ aln 2 + T1 ⎟ − T∞ ⎢⎣⎝ ⎥⎦ ⎠ dr r2 r1 heat flux just inside cylinder (3.21) surface heat flux to fluid Plugging the form of the temperature distribution in the cylinder into Equation (3.21) yields r⎞ ⎛k −a⎜ + hln 2 ⎟ = h(T1 − T∞ ) . ⎝ r2 r1 ⎠ The constant of integration, a, is a= ( − h T1 − T∞ ) = − (T1 − T∞ ) k r + h ln 2 r2 r1 k r + ln 2 hr2 r1 , and the expression for the temperature is, in normalized non-dimensional form ( ) ( ) ln r/r1 T1 − T = . T1 − T∞ k + ln r2 /r1 hr2 (3.22) HT-28 The heat flow per unit length, Q̇ , is given by Q̇ = 2π (T1 − T∞ )k k + ln(r2 /r1) hr2 (3.23) The units in Equation (3.23) are W / m-s. A problem of interest is choosing the thickness of insulation to minimize the heat loss for a fixed temperature difference T1 - T ∞ between the inside of the pipe and the flowing fluid far away from the pipe. (T1 - T∞ is the driving temperature distribution for the pipe). To understand the behavior of the heat transfer we examine the denominator in Equation (3.23) as r2 varies. The thickness of insulation that gives maximum heat transfer is given by d ⎛ k r⎞ + ln 2 ⎟ = 0 ⎜ dr2 ⎝ hr2 r1 ⎠ (3.24) (Question: How do we know this is a maximum?) From Equation (3.24), the value of r2 for maximum Q& is thus (r2)maximum heat transfer = k/h. (3.25) If r2 is less than this, we can add insulation and increase heat loss. To understand why this occurs, consider Figure 3.8, which shows a schematic of the thermal resistance and the heat transfer. As r2 increases from a value less than r2 = k/h, two effects take place. First, the thickness of the insulation increases, tending to drop the heat transfer because the temperature gradient decreases. Secondly, the area of the outside surface of the insulation increases, tending to increase the heat transfer. The second of these is (loosely) associated with the k/hr2 term, the first with the ln(r2/r1) term. There are thus two competing effects which combine to give a maximum Q̇ at r2 = k/h. HT-29 r ln r2 1 R k hr2 ⋅ Q r2 k h Figure 3.8: Critical radius of insulation Muddy points 1 , what is A? (MP HT.11) h.A It seems that we have simplified convection a lot. Is finding the heat transfer coefficient, h, really difficult? (MP HT.12) What does the "K" in the contact resistance formula stand for? (MP HT.13) In the equation for the temperature in a cylinder (3.22), what is "r"? (MP HT.14) In the expression 3.3 Dimensionless Numbers and Analysis of Results Phenomena in fluid flow and heat transfer depend on dimensionless parameters. The Mach number and the Reynolds number are two you have already seen. These parameters give information as to the relevant flow regimes of a given solution. Casting equations in dimensionless form helps show the generality of application to a broad class of situations (rather than just one set of dimensional parameters). It is generally good practice to use non-dimensional numbers, forms of equations, and results presentation whenever possible. The results for heat transfer from the cylinder are already in dimensionless form but we can carry the idea even further. For the cylinder: ( ) ( ) ln r/r1 T - T1 = T' < T1 k/hr2 + ln r2 /r1 The parameter hr2 or (3.26) hL , where L is a relevant length for the particular problem of interest, is k k called the Biot number denoted by Bi. In terms of this parameter, HT-30 ( ) ( ) ln r/r1 T - T1 = T∞ − T1 1 + ln r2 /r1 Bi (3.27) The size of the Biot number gives a key to the regimes in which different features are dominant. For Bi >> 1 the convection heat transfer process offers little resistance to heat transfer. There is thus only a small ∆T outside (i.e. T (r2) close to T∞) compared to the ∆T through the solid with a limiting behavior of T − T1 lnr/r1 = T∞ − T lnr2 /r1 as Bi goes to infinity. This is much like the situation with an external temperature specified. For Bi << 1 the conduction heat transfer process offers little resistance to heat transfer. The temperature difference in the body (i.e. from r1 to r2) is small compared to the external temperature difference, T1 - T∞ . In this situation, the limiting case is T − T1 = Bi ln r/r1 T∞ − T1 ( ) In this regime there is approximately uniform temperature in the cylinder. The size of the Biot number thus indicates the regimes where the different effects become important. Figure 3.9 shows the general effect of Biot number on temperature distribution. Figure 3.10 is a plot of the temperature distribution in the cylinder for values of Bi = 0.1, 1.0 and 10.0. HT-31 Figure 3.9: Effect of the Biot Number [hL / kbody] on the temperature distributions in the solid and in the fluid for convective cooling of a body. Note that kbody is the thermal conductivity of the body, not of the fluid. [adapted from: A Heat Transfer Textbook, John H. Lienhard, Prentice-Hall Publishers, 1980] HT-32 Figure 3.10: Temperature distribution in a convectively cooled cylinder for different values of Biot number, Bi; r2 / r1 = 2 [from: A Heat Transfer Textbook, John H. Lienhard] 4.0 Temperature Distributions in the Presence of Heat Sources There are a number of situations in which there are sources of heat in the domain of interest. Examples are: 1) Electrical heaters where electrical energy is converted resistively into heat 2) Nuclear power supplies 3) Propellants where chemical energy is the source These situations can be analyzed by looking at a model problem of a slab with heat sources α (W/m3) distributed throughout. We take the outside walls to be at temperature Tw. and we will determine the maximum internal temperature. Slice at x, x+dx Tw Tw heat sources W α 3 m q& + q& x x dq& dx dx x + dx (b) (a) Figure 4.1: Slab with heat sources (a) overall configuration, (b) elementary slice With reference to Figure 4.1(b), a steady-state energy balance yields an equation for the heat flux, q& . HT-33 ⋅ ⎞ ⎛⋅ d q ⎜ q + adx − q + dx ⎟ = 0 dx ⎟ ⎜ ⎠ ⎝ ⋅ (4.1) ⋅ or dq = α. dx (4.2) There is a change in heat flux with x due to the presence of the heat sources. The equation for the temperature is d 2T +α / k = 0 dx 2 (4.3) Equation (4.3) can be integrated once, α dT =− x+a dx k (4.4) and again to give T =− α 2 x + ax + b 2k (4.5) where a and b are constants of integration. The boundary conditions imposed are T (0 ) = T (L ) = Tw . αL Substituting these into Equation (4.5) gives b = Tw and a = . The temperature distribution is thus 2k T =− αx 2 α + Lx + TW . 2k 2k (4.6) Writing (4.6) in a normalized, non-dimensional fashion gives a form that exhibits in a more useful manner the way in which the different parameters enter the problem: 2 1⎛ x x ⎞ = ⎜ − 2⎟ 2 2⎝L L ⎠ αL /k T − TW (4.7) This distribution is sketched in Figure 4.2. It is symmetric about the mid-plane at x = the energy due to the sources exiting the slab on each side. HT-34 L , with half 2 ⎛ T − TW ⎞ ⎜ 2 ⎟ ⎝ aL /k ⎠ 0.125 X 0 0.5 1.0 L Figure 4.2:Temperature distribution for slab with distributed heat sources The heat flux at the side of the slab, x = 0, can be found by differentiating the temperature 2 dT _L £ 1 ¥ distribution and evaluating at x = 0 : < k = <k ² ´ = <_L/2 . dx x = 0 2K ¤ L ¦ This is half of the total heat generated within the slab. The magnitude of the heat flux is the same at x = L, although the direction is opposite. A related problem would be one in which there were heat flux (rather than temperature) boundary conditions at x = 0 and x = L, so that Tw is not known. We again determine the maximum temperature. At x = 0 and L, the heat flux and temperature are continuous so <k ( ) dT = h T < T' at x = 0,L. dx (4.8) Referring to the temperature distribution of Equation (4.6) gives for the two terms in Equation (4.8), _x dT = k £ - + a¥ = ( <_x + ka) ¤ k ¦ dx £ _x 2 ¥ h(T < T' ) = h ² < + _x + b < T' ´ ¤ k ¦ k (4.9) (4.10) Evaluating (4.10) at x = 0 and L allows determination of the two constants a and b. This is left as an exercise for the reader. Muddy points For an electric heated strip embedded between two layers, what would the temperature distribution be if the two side temperatures were not equal? (MP HT.15) HT-35 5.0 Heat Transfer From a Fin Fins are used in a large number of applications to increase the heat transfer from surfaces. Typically, the fin material has a high thermal conductivity. The fin is exposed to a flowing fluid, which cools or heats it, with the high thermal conductivity allowing increased heat being conducted from the wall through the fin. The design of cooling fins is encountered in many situations and we thus examine heat transfer in a fin as a way of defining some criteria for design. A model configuration is shown in Figure 5.1. The fin is of length L. The other parameters of the problem are indicated. The fluid has velocity c∞ and temperature T∞. Fin T0 (wall) y x x=0 (wall) x=L T∞ , c ∞ Figure 5.1: Geometry of heat transfer fin We assume (using the Reynolds analogy or other approach) that the heat transfer coefficient for the fin is known and has the value h. The end of the fin can have a different heat transfer coefficient, which we can call hL. The approach taken will be quasi-one-dimensional, in that the temperature in the fin will be assumed to be a function of x only. This may seem a drastic simplification, and it needs some explanation. With a fin cross-section equal to A and a perimeter P, the characteristic dimension in the transverse direction is A / P (For a circular fin, for example, A / P = r / 2). The regime of interest will be taken h(A / P ) to be that for which the Biot number is much less than unity, Bi = << 1 , which is a realistic k approximation in practice. The physical content of this approximation can be seen from the following. Heat transfer per unit area out of the fin to the fluid is roughly of magnitude ~ h(Tw - T∞ ) per unit area . The heat transfer per unit area within the fin in the transverse direction is (again in the same approximate terms) HT-36 (T1 − Tw ) , where T1 is an internal temperature. These two quantities must be of the same A/ P T −T A/ P magnitude. If h << 1 , then 1 w << 1 . In other words, if Bi << 1, there is a much larger Tw − T∞ k capability for heat transfer per unit area across the fin than there is between the fin and the fluid, and thus little variation in temperature inside the fin in the transverse direction. To emphasize the point, consider the limiting case of zero heat transfer to the fluid i.e., an insulated fin. Under these conditions, the temperature within the fin would be uniform and equal to the wall temperature. k If there is little variation in temperature across the fin, an appropriate model is to say that the temperature within the fin is a function of x only, T = T(x), and use a quasi-one-dimensional approach. To do this, consider an element, dx, of the fin as shown in Figure 5.2. There is heat flow dQ& of magnitude Q& in at the left-hand side and heat flow out of magnitude Q& out = Q& in + dx at the dx right hand side. There is also heat transfer around the perimeter on the top, bottom, and sides of the fin. From a quasi-one-dimensional point of view, this is a situation similar to that with internal heat sources, but here, for a cooling fin, in each elemental slice of thickness dx there is essentially a heat sink of magnitude Pdxh T − T , where Pdx is the area for heat transfer to the fluid. ∞ ( ) Q& in Q& out x x+dx Figure 5.2:Element of fin showing heat transfer The heat balance for the element in Figure 5.2 can be written in terms of the heat flux using Q& = q&A , for a fin of constant area: dq˙ q˙ A = Ph(T − T∞ )dx + ⎛q˙ A + dxA⎞ . ⎝ ⎠ dx (5.1) From Equation (5.1) we obtain A dq̇ + Ph(T − T∞ ) = 0 . dx (5.2) In terms of the temperature distribution, T(x): 2 d T dx 2 − ( ) Ph T − T∞ = 0 . Ak (5.3) HT-37 The quantity of interest is the temperature difference (T - T∞ ), and we can change variables to put Equation (5.3) in terms of this quantity using the substitution d dT . (T − T∞ ) = dx dx (5.4) Equation (5.3) can therefore be written as d 2 dx 2 (T − T∞ ) − hP (T − T∞ ) = 0 . Ak (5.5) Equation (5.5) describes the temperature variation along the fin. It is a second order equation and needs two boundary conditions. The first of these is that the temperature at the end of the fin that joins the wall is equal to the wall temperature. (Does this sound plausible? Why or why not?) (T − T∞ )x = 0 = T0 − T∞ . (5.6) The second boundary condition is at the other end of the fin. We will assume that the heat transfer from this end is negligible1. The boundary condition at x = L is ( d T − T∞ dx ) x=L = 0. (5.7) The last step is to work in terms of non-dimensional variables to obtain a more compact description. ~ ~ T - T∞ as ∆ T , where the values of ∆ T range from zero to one and ξ = x / L , In this we define T0 − T∞ where ξ also ranges over zero to one. The relation between derivatives that is needed to cast the equation in terms of ξ is d dξ d 1 d . = = dx dx dξ L dξ 1 Equation (5.5) can be written in this dimensionless form as ~ d 2 ∆ T ⎛ hP 2 ⎞ ~ L ∆ T = 0. − ⎝ kA ⎠ dξ 2 (5.8) There is one non-dimensional parameter in Equation (5.8), which we will call m and define by 1 Note: We don’t need to make this assumption, and if we were looking at the problem in detail we would solve it numerically and not worry about whether an analytic solution existed. In the present case, developing the analytic solution is useful in presenting the structure of the solution as well as the numbers, so we resort to the mild fiction of no heat transfer at the fin end. We need to assess, after all is said and done, whether this is appropriate or not. HT-38 m2 L 2= hP L 2 . kA (5.9) The equation for the temperature distribution we have obtained is ~ 2 d ∆T dξ 2 ~ 2 2 − m L ∆T = 0. (5.10) This second order equation has the solution ~ mLξ − mLξ ∆ T = ae + be . (5.11) (Try it and see). The boundary condition at ξ = 0 is ~ ∆ T(0) = a + b = 1 . (5.12a) The boundary condition at ξ = 1 is that the temperature gradient is zero or ~ d∆ T −m m (L) = mLae − mLbe = 0 . dξ (5.12b) ~ Solving the two equations given by the boundary conditions for a and b gives an expression for ∆ T x −x ⎛ e +e ⎞ in terms of the hyperbolic cosine or cosh: ⎜ cosh x = 2 ⎟⎠ ⎝ ~ ∆T = ( cosh mL 1 − ξ ) (5.13) cosh mL This is the solution to Equation (5.8) for a fin with no heat transfer at the tip. In terms of the actual heat transfer parameters it is written as ⎛ cosh ⎜ ⎛ 1 − ⎝⎝ T − T∞ = T0 − T∞ ⎛ cosh ⎜ ⎝ x ⎞ hP ⎞ L L ⎠ kA ⎟⎠ . hP ⎞ L kA ⎟⎠ HT-39 (5.14) The amount of heat removed from wall due to the fin, which is the quantity of interest, can be found by differentiating the temperature and evaluating the derivative at the wall, x = 0: ⋅ Q = − kA ( d T − T∞ dx ) x =0 (5.15) or ⋅ ~ QL dD T =− dξ kA T0 − T∞ ( ) ξ =0 = mLsinh mL = mL tanh mL cosh mL (5.16) ⋅ Q ( kAhP T0 − T∞ ) = tanh mL (5.16a) The solution is plotted in Figure 5.3, which is taken from the book by Lienhard. Several features of the solution should be noted. First, one does not need fins which have a length such that m is much greater than 3. Second, the assumption about no heat transfer at the end begins to be inappropriate as m gets smaller than 3, so for very short fins the simple expression above would not be a good estimate. We will see below how large the error is. HT-40 Heat flow cannot be noticeably improved by lengthening the fin beyond L = 3 / m Dimensionless heat flow into the fin, Q 1 cosh mL hPkA ( T0 − T∞ ) tanh mL The temperature excess at the tip is less than 1.4%, beyond L = 5 / m Dimensionless temperature at tip (T - T∞ ) / (T0 - T∞ ) mL - 1 (a very stumpy fin) Dimensionless temperature (T - T∞ ) / (T0 - T∞ ) mL = 2 mL = 5 (a long "over designed" fin) mL = 3 Dimensionless axial position ξ = x / L Figure 5.3:The temperature distribution, tip temperature, and heat flux in a straight onedimensional fin with the tip insulated. [Adapted from: Lienhard, A Heat Transfer Textbook, PrenticeHall publishers] Muddy points Why did you change the variable and write the derivative d 2 (T - T∞ ) d2T as in the equation dx 2 dx 2 for heat transfer in the fin? (MP HT.16) What types of devices use heat transfer fins? (MP HT.17) Why did the Stegosaurus have cooling fins? Could the stegosaurus have "heating fins"? (MP HT.18) HT-41 6.0 Transient Heat Transfer (Convective Cooling or Heating) All the heat transfer problems we have examined have been steady state, but there are often circumstances in which the transient response to heat transfer is critical. An example is the heating up of gas turbine compressors as they are brought up to speed during take-off. The disks that hold the blades are large and take a long time to come to temperature, while the casing is thin and in the path of high velocity compressor flow and thus comes to temperature rapidly. The result is that the case expands away from the blade tips, sometimes enough to cause serious difficulties with aerodynamic performance. To introduce the topic as well as to increase familiarity with modeling of heat transfer problems, we examine a lumped parameter analysis of an object cooled by a stream. This will allow us to see what the relevant non-dimensional parameters are and, at least in a quantitative fashion, how more complex heat transfer objects will behave. We want to view the object as a "lump" described by a single parameter. We need to determine when this type of analysis would be appropriate. To address this, consider the temperature difference T1 - Tw between two locations in the object, as shown in Figure 6.1. Tw c∞ T1 T∞ object Figure 6.1:Temperature variation in an object cooled by a flowing fluid If the heat transfer within the body and from the body to the fluid are of the same magnitude, ( ) h Tw − T∞ ≈ ( k T −T L 1 w ) (6.1) where L is a relevant length scale, say half the thickness of the object. The ratio of the temperature difference is T1 − Tw hL ≈ Tw − T∞ k (6.2) If the Biot number is small the ratio of temperature differences described in Equation (6.2) is also T1 − Tw << 1. We can thus say T1 − Tw << Tw − T∞ and neglect the temperature non-uniformity Tw − T∞ within the object. ( ( ) ) ( ) HT-42 The approximation made is to view the object as having a spatially uniform temperature that is a function of time only. Explicitly, T = T(t). The first law applied to the object is (using the fact that for solids cp = cv = c) ⋅ Q in = ρVc dT dt (6.3) where ρ is the density of the object and V is its volume. In terms of heat transferred to the fluid, dT . The rate of heat transfer to the fluid is Ah (T - T∞), so the expression for the time Q& out = − ρVc dt evolution of the temperature is ( ) Ah T − T∞ = −ρVc dT . dt (6.4) The initial temperature, T(0), is equal to some known value, which we can call Ti . Using this, ⎛ T − T∞ ⎞ Equation (6.4) can be written in terms of a non-dimensional temperature difference ⎜ ⎟, ⎝ Ti − T∞ ⎠ d ⎛ T − T∞ ⎞ hA ⎛ T − T∞ ⎞ ⎜ ⎟+ ⎜ ⎟ =0 dt ⎝ Ti − T∞ ⎠ ρVc ⎝ Ti − T∞ ⎠ (6.5) At time t = 0, this non-dimensional quantity is equal to one. Equation (6.5) is an equation you have ⎛ dx x ⎞ seen before, ⎜ + = 0⎟ which has the solution x = ae − t / τ . For the present problem the form is ⎝ dt τ ⎠ T − T∞ − hAt/pVc = ae . Ti − T∞ (6.6) The constant a can be seen to be equal to unity to satisfy the initial condition. This form of equation ρVc . implies that the solution has a heat transfer "time constant" given by τ = hA The time constant,τ, is in accord with our "intuition"; high density, large volume, or high specific heat all tend to increase the time constant, while high heat transfer coefficient and large area will tend to decrease the time constant. This is the same form of equation and the same behavior you have seen for the R-C circuit, as shown schematically below. The time dependence of the voltage in dE E the R-C circuit when the switch is opened suddenly is given by the equation + = 0 . There dt RC are, in fact, a number of physical processes which have (or can be modeled as having) this type of exponentially decaying behavior. HT-43 C E0 R Figure 6.2: Voltage change in an R-C circuit Muddy points • dT (6.3), what is c? (MP HT.19) dt In the lumped parameter transient heat transfer problem, does a high density "slow down" heat transfer? (MP HT.20) In equation Q in = ρ.V.c. 7.0 Some Considerations in Modeling Complex Physical Processes In Sections 5 and 6, a number of assumptions were made about the processes that we were attempting to describe. These are all part of the general approach to modeling of physical systems. The main idea is that for engineering systems, one almost always cannot compute the process exactly, especially for fluid flow problems. At some level of detail, one generally needs to model, i.e. to define some plausible behavior for attributes of the system that will not be computed. Modeling can span an enormous range from the level of our assumption of uniform temperature within the solid object to a complex model for the small scale turbulent eddies in the flow past a compressor blade. In carrying out such modeling, it is critical to have a clear idea of just what the assumptions really mean, as well as the fidelity that we ascribe to the descriptions of actual physical phenomena, and we thus look at the statements we have made in this context. One assertion made was that because ( ) ( hL k T − T ≈ h TW − T∞ L c w k << 1 and on the basis of a heat balance, ) we could assume (Tbody interior − Tw ) << 1 (Tw − T∞ ) HT-44 Based on this, we said that Tbody is approximately uniform and Tw ≈ Tbody interior . Another aspect is hA that setting << 1 erases any geometrical detail of the fin cross section. The only place where P Pk and A enter the problem is in a non-dimensional combination. A third assumption, made in the fin problem, was that the heat transfer at the far end can be neglected. The solution including this hL L hL L is an axial Biot number is given as Equation (7.1). If the quantity is small, effect, where k k you can see that Equation (7.1) reduces to the previous result (5.13). ( ) ( ( ) ) ( T − T∞ cosh mL 1 − ξ + Biaxial / m sinh mL 1 − ξ = T0 − T∞ cosh mL + Biaxal / mL sinh mL ) (7.1) and ⋅ Q ( kAhP T0 − T∞ 8.0 ) = (Biaxial /mL) + tanh mL Bi 1 + axial tanh mL mL (7.2) Heat Exchangers The general function of a heat exchanger is to transfer heat from one fluid to another. The basic component of a heat exchanger can be viewed as a tube with one fluid running through it and another fluid flowing by on the outside. There are thus three heat transfer operations that need to be described: 1) Convective heat transfer from fluid to the inner wall of the tube 2) Conductive heat transfer through the tube wall 3) Convective heat transfer from the outer tube wall to the outside fluid Heat exchangers are typically classified according to flow arrangement and type of construction. The simplest heat exchanger is one for which the hot and cold fluids move in the same or opposite directions in a concentric tube (or double-pipe) construction. In the parallel-flow arrangement of Figure 8.1a, the hot and cold fluids enter at the same end, flow in the same direction, and leave at the same end. In the counterflow arrangement of Figure 8.1b, the fluids enter at opposite ends, flow in opposite directions, and leave at opposite ends. Alternatively, the fluids may be in cross flow (perpendicular to each other), as shown by the finned and unfinned tubular heat exchangers of Figure 8.2. The two configurations differ according to whether the fluid moving over the tubes is unmixed or mixed. In Figure 8.2a, the fluid is said to be unmixed because the fins prevent motion in a direction (y) that is transverse to the main-flow direction (x). In this case the fluid temperature varies with x and y. In contrast, for the unfinned tube bundle of Figure 8.2b, fluid motion, hence mixing, in the transverse direction is possible, and temperature variations are primarily in the main-flow direction. Since the tube flow is unmixed, HT-45 both fluids are unmixed in the finned exchanger, while one fluid is mixed and the other unmixed in the unfinned exchanger. (a) (b) Figure 8.1: Concentric tubes heat exchangers. (a) Parallel flow. (b) Counterflow x y Cross flow T = f(x,y) Cross flow T = f(x) Tube flow (a) Tube flow (b) Figure 8.2: Cross-flow heat exchangers. (a) Finned with both fluids unmixed. (b) Unfinned with one fluid mixed and the other unmixed To develop the methodology for heat exchanger analysis and design, we look at the problem of heat transfer from a fluid inside a tube to another fluid outside. HT-46 r1 r2 TA TB T1 T2 Figure 8.3: Geometry for heat transfer between two fluids We examined this problem before in Section 3.2 and found that the heat transfer rate per unit length is given by Q& = 2πk (T A − TB ) r k k + ln 2 + r1 h1 r1 r2 h2 (8.1) It is useful to define an overall heat transfer coefficient h0 per unit length as Q& = 2πr2 h0 (TA − TB ). (8.2) From (8.1) and (8.2) the overall heat transfer coefficient, h0 , is r r r 1 1 = 2 + 2 ln 2 + . h0 r1h1 k r1 h2 (8.3) We will make use of this in what follows. A schematic of a counterflow heat exchanger is shown in Figure 8.4. We wish to know the temperature distribution along the tube and the amount of heat transferred. HT-47 Ta2 Tb1 Tb2 Ta1 Figure 8.4: Counterflow heat exchanger To address this we start by considering the general case of axial variation of temperature in a tube with wall at uniform temperature T0 and a fluid flowing inside the tube. T1 T2 x=0 x=L dx ∆ T2 T1 ∆ T1 T0 cooling T2 heating Figure 8.5: Fluid temperature distribution along the tube with uniform wall temperature The objective is to find the mean temperature of the fluid at x, T(x), in the case where fluid comes in at x = 0 with temperature T1 and leaves at x = L with temperature T2 . The expected distribution for heating and cooling are sketched in Figure 8.5. For heating (T0 > T), the heat flow from the pipe wall in a length dx is ⋅ q πDdx = hπD(T0 − T)dx where D is the pipe diameter. The heat given to the fluid (the change in enthalpy) is given by 2 ρumc p ⋅ πD dT = m c p dT 4 HT-48 where ρ is the density of the fluid, um is the mean velocity of the fluid, cp is the specific heat of the ⋅ fluid and m is the mass flow rate of the fluid. Setting the last two expressions equal and integrating from the start of the pipe, we find T ∫ T1 x dT 4h = dx . ρumc p D T0 − T ∫ (8.4) 0 Carrying out the integration, 4hx = ρumc p D T ∫ T1 ( ) T d T0 − T dT =− = − ln T0 − T T0 − T T0 − T ∫ T1 ( ) T1 , T (8.5) i.e. ⎛T −T ⎞ 4hx . ln ⎜ 0 ⎟=− ρumc p D ⎝ T0 − T1 ⎠ (8.6) Equation (8.6) can be written as T0 − T = e −β x T0 − T1 (8.7) where β= 4h πhD = ⋅ ρumc p D m cp . (8.8) This is temperature distribution along the pipe. The exit temperature at x = L is πhDL − T0 − T2 m& c =e p T0 − T1 (8.9) The total heat transfer to the wall all along the pipe is ⋅ ⋅ ( ) Q = m c p T1 − T2 . From Equation (8.9), HT-49 ⋅ m cp = hπDL . ⎛ T0 − T1 ⎞ ln ⎜ ⎟ ⎝ T0 − T2 ⎠ The total rate of heat transfer is therefore ⋅ Q= ( hπDL T1 − T2 T1 − T0 T2 − T0 ln ), (8.10) ⋅ or Q = hπDL∆TLM where ∆TLM is the logarithmic mean temperature difference, defined as ∆TLM = T2 − T1 ∆T − ∆T2 = 1 . T0 − T2 ⎛ ⎞ T ∆ 1 ln ln⎜ ⎟ T0 − T1 ⎝ ∆T2 ⎠ (8.11) The concept of a logarithmic mean temperature difference is useful in the analysis of heat exchangers. We can define a logarithmic mean temperature difference for a counterflow heat exchanger as follows. With reference to Figure 8.4, an overall heat balance between the two counterflowing streams is ⋅ ⋅ ( ) ⋅ ( ) Q = m a c pa Ta1 − Ta2 = m b c pb Tb2 − Tb1 . (8.12) ⋅ From a local heat balance, the heat given up by stream a in length dx is − m a c pa dTa . (There is a ⋅ negative sign since Ta decreases). The heat taken up by stream b is − m b c pb dTb . (There is a negative sign because Tb decreases as x increases). The local heat balance is: ⋅ ⋅ ⋅ ⋅ − m a c pa dTa = − mb c pb dTb = q dA = q πDdx (8.13) Solving (8.13) for dTa and dTb, we find: ⋅ dTa = − q dA ⋅ m a c pa ⋅ ; dTb = − q dA ⋅ m b c pb HT-50 (8.14) ⎛ ⎞ ⎛ 1 1 1 ⎟ ⋅ 1⎞ ⋅ ⎜ d Ta − Tb = d∆T = − ⋅ − ⋅ q dA = −⎜ − ⎟ q πDdx ⎜ ⎟ ⎝ Wa Wb ⎠ ⎝ m a c pa m b c pb ⎠ ( ) (8.15) where W = m& c p . Also, q& = h0 ∆T where h0 is the overall heat transfer coefficient. We can then say: ⎛ 1 d∆T 1⎞ = − h0 πD⎜ − ⎟ dx . ∆T ⎝ Wa Wb ⎠ Integrating from x = 0 to x = L gives ⎛ T − Tb1 ⎞ ⎛ 1 1⎞ ln ⎜ a2 − ⎟. ⎟ = − h0 πDL⎜ ⎝ Ta1 − Tb2 ⎠ ⎝ Wa Wb ⎠ (8.16) Equation (8.16) can also be written as: Ta 2 − Tb1 = e −α Ta1 − Tb 2 (8.17) where ⎛ 1 1⎞ α = h0 πDL⎜ − ⎟ ⎝ Wa Wb ⎠ We know that ⋅ Ta1 - Ta2 = Q / Wa ⋅ Tb2 - Tb1 = Q / Wb . (8.18) Thus (Ta1 − Tb2 ) − (Ta2 − Tb1 ) = Q⎛⎜⎝ W1 ⋅ a Solving for the total heat transfer: ⋅ Ta1 − Tb2 − Ta2 − Tb1 Q= ⎛ 1 1⎞ − ⎟ ⎜ ⎝ Wa Wb ⎠ ( ) ( − 1⎞ ⎟. Wb ⎠ ) (8.19) (8.20) ⎛ 1 1⎞ − ⎟ in terms of other parameters as Rearranging (8.16) allows us to express ⎜ ⎝ Wa Wb ⎠ HT-51 ⎛ T − Tb2 ⎞ ln ⎜ a2 ⎟ ⎛ 1 ⎝ Ta1 − Tb2 ⎠ 1⎞ − ⎟=− . ⎜ h0 πDL ⎝ Wa Wb ⎠ (8.21) Substituting (8.21) into (8.20) we obtain a final expression for the total heat transfer for a counterflow heat exchanger: ( ) ( ⋅ Ta1 − Tb2 − Ta2 − Tb1 Q = h0 πDL ⎛T −T ⎞ ln ⎜ a1 b2 ⎟ ⎝ Ta2 − Tb1 ⎠ ) (8.22) or ⋅ Q = h0 πDL∆TLM (8.23) 8.1 Efficiency of a Counterflow Heat Exchanger Suppose we know the two inlet temperatures Ta1, Tb1, and we need to find the outlet temperatures. From (8.17) Ta2 - Tb1 = (Ta1 - Tb2) e-α Ta2 - Ta1 = Tb1 - Ta1 + (Ta1 - Tb2 ) e-α Rearranging (8.18), Tb 2 = Tb1 + Wa (Ta1 − Ta 2 ) Wb Thus Ta2 - Ta1 = Tb1 - Ta1 + (Ta1 - Tb1 ) e-α - ( ) Wa −α Ta1 − Ta2 e Wb −α a −α ⎞ e ⎟ = (Ta1 − Tb1 )(1 − e ) (Ta1 − Ta2 )⎛⎜⎝ 1 − W W ⎠ b or (Ta1 − Ta 2 )= η (Ta1 − Tb1 ) (8.24) where η is the efficiency of a counterflow heat exchanger: HT-52 η= −α −α 1− e 1− e = ⋅ W −α 1− a e m a c pa − α 1− ⋅ e Wb c pb mb (8.25) From (8.24) and (8.25) we can find outlet temperatures Ta2 and Tb2: ⋅ Tb2 − Tb1 = ⋅ m a c pa ⋅ m b c pb m a c pa (Ta1 − Ta2 ) = ⋅ m b c pb ( η Ta1 − Tb2 We examine three examples. i) ⋅ ⋅ m b c pb > m a c pa ∆T can approach zero at cold end ⎛ ⎞ 1 1 ⎟ ⎜ η → 1 as h0 , surface area , πDLh0 ⋅ − ⋅ ⎜ ⎟ ⎝ m a c pa m b c pb ⎠ T −T Maximum value of ratio a1 a 2 = 1 Ta1 − Tb1 ⋅ T − Tb1 m a c pa = Maximum value of ratio b2 Ta1 − Tb1 m⋅ b c pb ii) ⋅ ⋅ m b c pb < m a c pa ⋅ α is negative , η → m b c pb ⋅ m a c pa as [ ] → ∞ (Wb < Wa ) ⋅ T − Ta2 m b c pb = Maximum value of a1 Ta1 − Tb1 m⋅ a c pa Tb 2 − Tb1 Maximum value of =1 Ta1 − Tb 2 iii) ⋅ ⋅ m a c pa = m b c pb d (Ta − Tb ) = 0 temperature difference remains uniform, η = 1 HT-53 ) 9.0 Radiation Heat Transfer (Heat transfer by thermal radiation) All bodies radiate energy in the form of photons moving in a random direction, with random phase and frequency. When radiated photons reach another surface, they may either be absorbed, reflected or transmitted. The behavior of a surface with radiation incident upon it can be described by the following quantities: α = absorptance - fraction of incident radiation absorbed ρ = reflectance - fraction of incident radiation reflected τ = transmittance – fraction of incident radiation transmitted Figure 9.1 shows these processes graphically. Radiation absorbed, α Incident radiation Radiation transmitted, τ Radiation reflected, ρ Figure 9.1: Radiation Surface Properties From energy considerations the three coefficients must sum to unity α+ρ+τ=1 (9.1) Reflective energy may be either diffuse or specular (mirror-like). Diffuse reflections are independent of the incident radiation angle. For specular reflections, the reflection angle equals the angle of incidence. 9.1 Ideal Radiators An ideal thermal radiator is called a "black body". It has several properties: 1) It has α = 1, and absorbs all radiation incident on it. 2) The energy radiated per unit area is Eb = σT4 where σ is the Stefan-Boltzmann constant, σ = 5.67 x 10-8 W 2 m K (9.2) 4 HT-54 The units of Eb are therefore W m 2 . Monochromatic emissive power, eλ kW/m -micron The energy of a black body, E b , is distributed over a range of wavelengths of radiation. We can dE ∆Eb define eλ = b ≈ , the energy radiated per unit area for a range of wavelengths of width ∆λ . dλ ∆λ The behavior of eλ is given in Figure 9.2. Figure 9.2: Emissive power of a black body at several temperatures - predicted and observed (λT )eλ max = 0.2898 cm K [adapted from: A Heat Transfer Textbook by Lienhard, J.] The distribution of eλ varies with temperature. The quantity λT at the condition where eλ is a maximum is given by (λT )eλmax = 0.2898 cm K. As T increases, the wavelength for maximum energy emission shifts to shorter values. The frequency of the radiation, f, is given by f = c/λ so high energy means short wavelengths and high frequency. HT-55 Figure 9.3: A cavity with a small hole (approximates a black body) A physical realization of a black body is a cavity with a small hole. There are many reflections and absorptions. Very few entering photons (light rays) will get out. The inside of the cavity has radiation which is homogeneous and isotropic (the same in any direction, uniform everywhere). Suppose we put a small black body inside the cavity as seen in Figure 9.4. The cavity and the black body are both at the same temperature. Temperature T Cavity Black body Figure 9.4: A small black body inside a cavity The radiant energy absorbed by the black body per second and per m2 is αΒH, where H is the irradiance, the radiant energy falling on any surface inside the cavity. The radiant energy emitted by the black body is EB. Since αB = 1 for a black body, H = EB. The irradiance within a cavity whose walls are at temperature T is therefore equal to the radiant emittance of a black body at the same temperature and irradiance is a function of temperature only. 9.2 Kirchhoff's Law and "Real Bodies" Real bodies radiate less effectively than black bodies. The measurement of this is the emittance, ε , defined by HT-56 Emittance: ← radiation from real body at T ε= E E b ← radiation from black body at T Values of emittance vary greatly for different materials. They are near unity for rough surfaces such as ceramics or oxidized metals, and roughly 0.02 for polished metals or silvered reflectors. A table of emittances for different substances is given at the end of this section as Table 9.1, taken from the book by Lienhard. The level of the emittance can be related to the absorptance using the following arguments. Suppose we have a small non-black body in the cavity. The power absorbed per unit area is equal to αH. The power emitted is equal to E. An energy balance gives E = Ebε = α H =α Eb.. Thus E =α =ε Eb (9.3) Equation (9.3), the relation α = ε , is known as Kirchhoff's Law. It implies that good radiators are good absorbers. It was derived for the case when Tbody = Tsurroundings (cavity) and is not strictly true for all circumstances when the temperature of the body and the cavity are different, but it is true if αλ = α ,ελ = ε, so the absorptance and emittance are not functions of λ . This situation describes a "gray body". Also, since αλ , ελ are properties of the surface, αλ = ελ . 9.3 Radiation Heat Transfer Between Planar Surfaces ρ1ρ2α 2 E1 α 2 E1 Surface 2 ρ1ρ 2 2 E1 ρ2 E1 2 E1 ρ1ρ2 E1 α1ρ2 E1 2 ρ1 ρ2 E1 2 ρ1ρ2 α 2 E1 Surface 1 Figure 9.5: Path of a photon between two gray surfaces Consider the two infinite gray surfaces shown in Figure 9.5. We suppose that the surfaces are thick enough so that, α + ρ = 1 (no radiation transmitted so transmittance = 0). Consider a photon emitted from Surface 1 (remembering that the reflectance ρ = 1 - α): Surface 1 emits Surface 2 absorbs E1 E1 α 2 HT-57 Surface 2 reflects Surface 1 absorbs Surface 1 reflects Surface 2 absorbs Surface 2 reflects Surface 1 absorbs E1 (1 − α 2 ) E1 (1 − α 2 )α1 E1 (1 − α 2 )(1 − α1 ) E1 (1 − α 2 )(1 − α1 )α 2 E1 (1 − α 2 )(1 − α1 )(1 − α 2 ) E1 (1 − α 2 )(1 − α1 )(1 − α 2 )α1 The same can be said for a photon emitted from Surface 2: Surface 2 emits Surface 1 absorbs Surface 1 reflects Surface 2 absorbs Surface 2 reflects etc…… E2 E 2α1 E 2 (1 − α1 ) E 2 (1 − α1 )α 2 E2 (1 − α 1 )(1 − α 2 ) We can add up all the energy E 1 absorbed in 1 and all the energy E 2 absorbed in 2. In doing the bookkeeping, it is helpful to define β = (1 − α1 )(1 − α 2 ) . The energy E 1 absorbed in 1 is E1 (1 − α 2 )α1 + E1 (1 − α 2 )α1 (1 − α 2 )(1 − α1 )+ K This is equal to ( ) E1 (1 − α 2 )α 1 1 + β + β 2 + ... . However 1 −1 = (1 − β ) = 1 + β + β 2 + .... 1− β We thus observe that the radiation absorbed by surface 1 can be written as E1 (1 − α 2 )α 1 . 1− β Likewise E2 (1 − α1 )α 2 is the radiation generated at 2 and absorbed there as well. 1− β Putting this all together we find that ( ) ⎛ E2 1 − α 1 α 2 ⎞ E α 2 1 E2 − ⎜ ⎟ = 1−β β 1 − ⎠ ⎝ is absorbed by 1. The net heat flux from 1 to 2 is HT-58 ⋅ q net 1 to 2 ( ) ( ) − E2α1 E1 − E1 1 − α1 − α 2 + α1α2 − E1α1 + E1α1α2 − E2α1 = 1− β 1 − 1 − α1 − α 2 + α1α2 E1α 2 − E2α1 . α1 + α 2 − α1α2 (9.4) = E1 − E1 1 − α 2 α1 1− β ( ) or ⋅ q net 1 to 2 = If T1 = T2 , we would have q& = 0, so from Equation (9.4) E1 E 2 = = f (T) . α1 α 2 If body 2 is black, α2 = 1, and E2 = σT4. E1 = σT 4 α1 ε 1σT 4 = σT 4 α1 Therefore, again, ε 1 = α1 for any gray surface (Kirchhoff's Law). Using Kirchhoff's Law we find, 4 ⋅ q net 1 to 2 4 ε σT ε − ε2σT2 ε1 = 1 1 2 ε1 + ε2 − ε1ε2 or, as the final expression for heat transfer between gray, planar, surfaces: ⋅ q net 1 to 2 4 4 σ⎛⎝ T1 − T2 ⎞⎠ . = 1 1 + −1 ε1 ε2 (9.5) HT-59 Example 1: Use of a thermos bottle to reduce heat transfer Silvered Walls Inside of thermos (hot fluid) Outside of thermos (Cold) 1 2 ε1 = ε2 = 0.02 for silvered walls T1 = 100 °C = 373 K T2 = 20 °C = 293 K ⋅ q1 to 2 4 4 σ⎛⎝ T1 − T2 ⎞⎠ 1100 − 420 W = = 6.9 2 = 1 1 99 m + −1 0.02 0.02 For the same ∆T, if we had cork insulation with k = 0.04, what thickness would be needed? ⋅ ( )( ) 0.04 80 k∆T k∆T so a thickness L = ⋅ = = 0.47 m would be needed. The thermos is indeed a L 6.9 q good insulator. q= Example 2: Temperature measurement error due to radiation heat transfer Thermocouples (see Figure 9.6) are commonly used to measure temperature. There can be errors due to heat transfer by radiation. Consider a black thermocouple in a chamber with black walls. HT-60 T1 metal 1 metal 2 Measured Voltage T2 Note: The measured voltage is related to the difference between T1 and T2 (the latter is a known temperature). Figure 9.6: Thermocouple used to measure temperature Suppose the air is at 20 °C, the walls are at 100 °C, and the convective heat transfer coefficient is W h = 15 2 . m K What temperature does the thermocouple read? We use a heat (energy) balance on the control surface shown in Figure 9.7. The heat balance states that heat convected away is equal to heat radiated into the thermocouple in steady state. (Conduction heat transfer along the thermocouple wires is neglected here, although it would be included for accurate measurements.) Heat in (radiation) Ttc Heat out (convection) Control volume Tair Twall Figure 9.7: Effect of radiation heat transfer on measured temperature. HT-61 The heat balance is ( ) 4 4 hA Ttc − Tair = σA⎛⎝ Twall − Ttc ⎞⎠ (9.6) where A is the area of the thermocouple. Substituting the numerical values gives ( ) 4 -8 4 15 Ttc − 293 = 5.67 x 10 ⎛⎝ 373 − Ttc ⎞⎠ from which we find Ttc = 51 °C = 324 K. The thermocouple thus sees a higher temperature than the air. We could reduce this error by shielding the thermocouple as shown in Figure 9.8. Radiation shield < 100 ϒC 100 ϒC Figure 9.8: Shielding a thermocouple to reduce radiation heat transfer error Muddy points Which bodies does the radiation heat transfer occur between in the thermocouple?(MP HT.21) Still muddy about thermocouples. (MP HT.22) Why does increasing the local flow velocity decrease the temperature error for the thermocouple? (MP HT.23) 9.4 Radiation Heat Transfer Between Black Surfaces of Arbitrary Geometry In general, for any two objects in space, a given object 1 radiates to object 2, and to other places as well, as shown in Figure 9.9. HT-62 elsewhere Qnet T2 Figure 9.9: Radiation between two bodies We want a general expression for energy interchange between two surfaces at different temperatures. This is given by the radiation shape factor or view factor, Fi - j. For the situation in Figure 9.10 F1-2 = fraction of energy leaving 1 which reaches 2 F2 -1 = fraction of energy leaving 2 which reaches 1 F1-2 , F2 -1 are functions of geometry only A 2 , T2 radiation Body 2 A 1 , T1 Body 1 Figure 9.10: Radiation between two arbitrary surfaces For body 1, we know that Eb is the emissive power of a black body, so the energy leaving body 1 is Eb1 A1. The energy leaving body 1 and arriving (and being absorbed) at body 2 is Eb1 A1 F1-2. The energy leaving body 2 and being absorbed at body 1 is Eb2 A2 F2-1. The net energy interchange from body 1 to body 2 is ⋅ Eb1 A1 F1-2 - Eb2 A2 F2-1 = Q1− 2 . (9.7) Suppose both surfaces are at the same temperature so there is no net heat exchange. If so, Eb1 A1 F1-2 - Eb2 A2 F2-1 = 0, but also Eb1 = Eb2. Thus A1 F1-2 = A2 F2-1. (9.8) HT-63 Equation (9.7) is the shape factor reciprocity relation. The net heat exchange between the two surfaces is ⋅ ( Q1− 2 = A1F1− 2 Eb1 − Eb2 ) [or A2F2−1(Eb1 − Eb2 )] Example: Concentric cylinders or concentric spheres T2 T1 Figure 9.11: Radiation heat transfer for concentric cylinders or spheres The net heat transfer from surface 1 to surface 2 is ⋅ ( ) Q1− 2 = A1F1− 2 Eb1 − Eb2 . We know that F1-2 = 1, i.e. that all of the energy emitted by 1 gets to 2. Thus ⋅ ( Q1− 2 = A1 Eb1 − Eb2 ) This can be used to find the net heat transfer from 2 to 1. ⋅ ( ) ( ) ( Q 2 −1 = A2 F2 −1 Eb2 − Eb1 = A1F1− 2 Eb2 − Eb1 = A Eb2 − Eb1 1 ) View factors for other configurations can be found analytically or numerically. Shape factors are given in textbooks and reports (they are tabulated somewhat like Laplace transforms), and examples of the analytical forms and numerical values of shape factors for some basic engineering configurations are given in Figures 9.12 through 9.15, taken from the book by Incropera and DeWitt. HT-64 Surface Metals Temperature ( o C) H Aluminum Polished, 98% pure Surface Asbestos 200 - 600 Commercial sheet 90 Heavily oxidized 90 - 540 0.04 - 0.06 260 H 40 0.93 - 0.97 40 0.93 Brick 0.09 Red, Rough 0.2 - 0.33 Silica Brass Highly polished Nonmetals Temperature ( o C) 980 0.8 - 0.85 Fireclay 9980 0.75 0.59 0.03 Ordinary refractory 1090 Dull plate 40 - 260 0.22 Magnesite refractory 980 0.38 Oxidized 40 - 260 0.46 - 0.56 White refractory 1090 0.29 Highly polished electrolytic 90 0.02 Filament Slightly polished, to dull 40 0.12 - 0.15 Lampsoot 40 0.95 40 0.94 Copper Black oxidized Carbon 1040 - 1430 0.53 40 0.76 Concrete, rough 90 - 600 0.02 - 0.035 Glass 150 - 480 0.14 - 0.32 Quartz glass (2mm) 260 - 540 0.96 - 0.66 Steel, polished 40 - 260 0.07 - 0.1 Pyrex 260 - 540 0.94 - 0.74 Sheet steel, rolled 40 0.66 Sheet steel, strong rough oxide 40 0.8 Gold: pure, polished Iron and Steel Mild steel, polished Smooth Gypsum Ice 40 0.94 40 0.8 - 0.9 0 0.97 - 0.98 Cast iron, oxidized 40 - 260 0.57 - 0.66 Limestone 40 - 260 0.95 - 0.83 Iron, rusted 40 0.61 - 0.85 Marble 40 0.93 - 0.95 Wrought iron, smooth 40 0.35 Mica 40 0.75 Wrought iron, dull oxidized 20 - 360 0.94 Paints 40 Stainless, polished 0.07 - 0.17 Black gloss 40 230 - 900 0.5 - 0.7 White paint 40 Lacquer 40 0.8 - 0.95 Polished 40 - 260 0.05 - 0.08 Various oil paints 40 0.92 - 0.96 Oxidized 40 - 200 0.63 Red lead 90 0.93 Stainless, after repeated heating Lead Mercury: pure, clean 40 - 90 0.1 - 0.12 Platinum Pure, polished plate Oxidized at 590oC Drawn wire and strips Silver Tin 200 - 590 260 - 590 40 - 1370 200 40 - 90 0.05 - 0.1 0.07 - 0.11 0.04 - 0.19 Filament 540 - 1090 2760 Paper White 40 0.95 - 0.98 Other colors 40 0.92 - 0.94 Roofing 40 0.91 40 - 260 0.92 Plaster, rough lime 0.01 - 0.04 Quartz 0.05 Rubber 40 0.86 - 0.94 Tungsten Filament 0.9 0.89 - 0.97 100 - 1000 0.89 - 0.58 Snow 10 - 20 0.82 0.11 - 0.16 Water, thickness 0.1mm 40 0.96 0.39 Wood 40 0.8 - 0.9 20 0.9 Oak, planed Table 9.1:Total emittances for different surfaces [Adapted from: A Heat Transfer Textbook, J. Lienhard] Figure 9.12: View Factors for Three - Dimensional Geometries Figure 9.13: Fig. 13.4--View factor for aligned parallel rectangles Figure 9.14: Fig 13.5--View factor for coaxial parallel disk Figure 9.15: Fig 13.6--View factor for perpendicular rectangles with a common edge [from: Fundamentals of Heat Transfer, F.P. Incropera and D.P. DeWitt, John Wiley and Sons] HT-66 Heat Transfer References Eckert, E. R. G. and Drake, R. M., Heat and Mass Transfer, McGraw Hill Book Company, 1959. Eckert, E. R. G. and Drake, R. M., Analysis of Heat and Mass Transfer, Hemisphere Pub. Corp., 1987. Incropera, F. P. and Dewitt, D. P., Fundamentals of Heat and Mass Transfer, Wiley, 1990. Lienhard, J. H., A Heat Transfer Textbook, Prentice-Hall, 1987. Mills, A. F., Heat Transfer, Irwin, 1992. HT-67 Muddiest points on heat transfer HT.1 How do we quantify the contribution of each mode of heat transfer in a given situation? Developing the insight necessary to address the important parts of a complex situation (such as turbine heat transfer) and downplay (neglect or treat approximately) the other aspects is part of having a solid working knowledge of the fundamentals. This is an important issue, because otherwise every problem will seem very complex. One way to sort out what is important is to make order of magnitude estimates (similar to what we did to answer when the one-dimensional heat transfer approximation was appropriate) to see whether all three modes have to be considered. Sometimes one can rule out one or two modes on the basis of the problem statement. For example if there were a vacuum between the two surfaces in the thermos bottle, we would not have to consider convection, but often the situation is more subtle. HT.2 How specific do we need to be about when the one-dimensional assumption is valid? Is it enough to say that dA/dx is small? The answer really is “be specific enough to enable one to have faith in the analysis or at least some idea of how good the answer is”. This is a challenge that comes up a great deal. For now, if we say that A is an area defined per unit depth normal to the blackboard then saying dA/dx is small, which is a statement involving a non-dimensional parameter, is a reasonable criterion. HT.3 Why is the thermal conductivity of light gases such as helium (monoatomic) or hydrogen (diatomic) much higher than heavier gases such as argon (monoatomic) or nitrogen (diatomic)? To answer this, we need some basics of the kinetic theory of gases. A reference for this is Castellan, Physical Chemistry, Benjamin/Cummings Publishers, 1986. Two factor contribute, the collision cross section and the average molecular velocity. For the gases mentioned above the dominant factor appears to be the velocity. The kinetic energy per molecule at a given temperature is the same and so the lower the molecular weight the higher the average molecular velocity. HT.4 What do you mean by continuous? The meaning is similar to the definition you have seen in the math subjects. A way to state it is that the function at a given location has the same value as we approach the location independent of the direction we approach from. To say this in a more physical manner, the temperature as a function of x has the same value at x = a, say, whether we approach location a from the left or from the right. In terms of what the function looks like, it will have no “jumps” or discontinuous (step) changes in value. HT.5 Why is temperature continuous in the composite wall problem? Why is it continuous at the interface between two materials? We can argue this point by supposing T were not continuous, i.e., there was a different temperature on the two sides of an interface between two materials, with the interface truly a sharp plane. If so, there would be a finite temperature difference across an infinitesimal distance, leading to a very large (infinite in the limit) heat flux which would immediately erase the temperature difference. This argument could also be applied to any location inside a solid of uniform composition, guaranteeing that the temperature is continuous in each material. HT.6 Why is the temperature gradient dT/dx not continuous? As derived in class, across an interface the heat flux is continuous. From the first law, for a thin control volume that encloses the interface the net heat flow into the control volume is zero. As sketched below, the contribution from the heat flux at the upper and lower ends of the control volume is negligible so the heat flux in one side must be the same as the flux out of the other. The heat flux, however, is related to the temperature gradient by dT r q = − k∇T , or, for one-dimension, q x = − k ⎛ ⎞ . If the heat flux is continuous, and if ⎝ dx ⎠ the thermal conductivity, k, is not the same in the two materials, then dT/dx is not continuous. Material A Material B Control volume qin qout Interface HT.7 Why is ∆T the same for the two elements in a parallel thermal circuit? Doesn't the relative area of the bolt to the wood matter? In terms of the bolt through the wood wall, the approximation made is that the bolt and the wood are both exposed to the same conditions at the two sides of the wall. The relative areas of the bolt and the wood indeed do matter. Suppose we consider a square meter area of wood without bolts. It has a certain heat resistance. If we now add bolts to Lbolt . If there are n bolts, they will be kbolt Abolt in parallel, and the effective area will be n times the area for one bolt. The thermal ⎛ ⎞ resistance of the wood will be increased by a factor of ⎜ Awithout / Awith ⎟ , which is larger ⎝ bolts n bolts ⎠ than unity. the wall, the resistance of each bolt is Rbolt = In summary, the amount by which the heat transfer is increased depends on the fractional area with low thermal resistance compared to the fractional area with high thermal resistance. HT.8 How do we know that δ' is not a fluid property? ⎛ δ ′⎞ The term ⎜ ⎟ represents the resistance to heat transfer for a unit area. The resistance to ⎝k⎠ heat transfer per unit area (1/heat transfer coefficient) can be computed for cases of laminar flow, or measured experimentally where we cannot compute it, and it is found that for the same state variables, it can have a range of values of several orders of magnitude depending on the parameters I described (Reynolds number, surface condition, surface shape…). Put another way, if the value of the resistance is affected by the surface condition (smooth, bumpy, corrugated, etc.) how can the resistance be just a property of the fluid? HT.9 What is the "analogy" that we are discussing? Is it that the equations are similar? While the equations are similar, the concepts are deeper than that. The analogy is drawn between the heat transfer process (transfer of heat represented by heat flux) and the momentum transfer process (transfer of momentum represented by shear stress) HT.10 In what situations does the Reynolds analogy "not work"? The Reynolds Analogy is just that. It is not a law of nature, but rather a plausible hypothesis that allows useful estimation of the heat transfer coefficients in many situations in which little or no explicit heat transfer information exists. In the form we have derived it, the Reynolds Analogy is appropriate for use in air, but it is not strictly µc p applicable if there are pressure gradients, or if the Prandtl number ( Pr = ) is not k unity. However, the conceptual framework provided by the analogy has been found useful enough that the analogy has been extended (in a more complex form, as briefly discussed in class) for application to these situations. HT.11 In the expression 1 , what is A? h.A A is the area normal to the heat flow. For a turbine blade, for example, it would be the outer surface area of the blade. HT.12 It seems that we have simplified convection a lot. Is finding the heat transfer coefficient, h, really difficult? We have indeed “simplified convection a lot”. We will look at heat transfer by convection in more depth in a few lectures, but to answer the question in a few sentences, finding the heat transfer coefficient is a difficult problem, because it necessitates determining the fluid dynamics; the latter is key to predictions of heat transfer. Even with present computational power, calculating the flow around aerospace devices with the accuracy needed to be confident about heat transfer coefficients is not by any means a “standard” calculation. For some circumstances, it is still beyond the state of the art. We will concentrate on describing (i) the basic mechanisms of convective heat transfer and (ii) ways of estimating the heat transfer coefficient from known fluid dynamic information. HT.13 What does the "K" in the contact resistance formula stand for? ∆Tdriving The definition of the resistance comes out of Q& = . The units of Q& are Rthermal Watts/meter2, so the units of Rthermal are [degrees x meters2/Watts]. The K is thus the symbol for Kelvin, or degree centigrade. HT.14 In the equation for the temperature in a cylinder (3.22), what is "r"? The variable r denotes the radial coordinate, in other words the location of the point at which we want to know the temperature. HT.15 For an electric heated strip embedded between two layers, what would the temperature distribution be if the two side temperatures were not equal? If the two temperatures on the outer surfaces of the composite layer were T1 and T2 , the heat flux from the two sides of the electrical heater would be different, but the sum of the two heat fluxes would still be equal to the heat generated per unit area and unit time. In fact heat could be coming into one side of the heater if one of the temperatures were high enough. (Sketch this out and prove it to yourself.) The temperature distribution would be linear from the heater temperature to the two surface temperatures. If this example is not clear, please come and see me. d 2 (T - T∞ ) d2T HT.16 Why did you change the variable and write the derivative as in dx 2 dx 2 the equation for heat transfer in the fin? The heat balance and derivation of the equation for temperature (5.3) is given in Section 5.0 of the notes. This is d 2T Ph (5.3) − (T − T∞ ) = 0 2 Ak dx It is not necessary to change variables, and one could solve (5.3) as is d 2T Ph (or in the form T = T∞ ). However, since: (1) the reference temperature is T∞ , − 2 Ak dx and what is of interest is the difference T − T∞ , (2) making the substitution results in a simpler form of the equation to be solved, and (3) the derivative of T − T∞ is the same as the derivative of T , I found it convenient to put it the equation in the form of Eq. (5.5), d 2 (T − T∞ ) Ph (5.5) − (T − T∞ ) = 0 . 2 Ak dx HT.17 What types of devices use heat transfer fins? A number of types of heat exchangers use fins. Examples of the use of fins you may have seen are cooling fins on motorcycle engine heads, cooling fins on electric power transformers, or cooling fins on air conditioners. HT.18 Why did the Stegosaurus have cooling fins? Could the stegosaurus have "heating fins"? My knowledge of this issue extends only to reading about it in the text by Lienhard (see reference in notes). The journal article referenced in that text is: Farlow, J. O., Thompson, C. V., and Rosner, D.E., “Plates of the Dinosaur Stegosaurus: Forced Convection Heat Loss Fins?, Science, vol. 192, no. 4244, 1976, pp. 1123-1125 and cover. • HT.19 In equation Q in = ρ.V.c. dT (6.3), what is c? dt c is the specific heat for a unit mass. We don’t have to differentiate between the two specific heats for a solid because the volume changes are very small, unlike a gas. HT.20 In the lumped parameter transient heat transfer problem, does a high density "slow down" heat transfer? It doesn’t. The high density slows down the rate at which the object changes temperature; high density means more “heat capacity”. HT.21 Which bodies does the radiation heat transfer occur between in the thermocouple? The radiation heat transfer was between the walls (or more generally the boundaries of the duct) and the thermocouple. The boundaries may not be at the same temperature of the flowing fluid, for example in a turbine. If the boundaries are not at the same temperature as the fluid, and they radiate to the thermocouple, there can be an error in the temperature that the thermocouple reads. The discussion was about different possible sources of this type of error. HT.22 Still muddy about thermocouples. I didn’t mean to strew confusion about these. As on page 59 of the notes, if we take a pair of dissimilar wires (say copper and constantan (an alloy of tin and several other metals) or platinum and rhodium) which are joined at both ends and subject the two junctions to a temperature difference, it is found that a voltage difference will be created. If we know the temperature of one junction (say by use of an ice bath) and we know the relation between voltage and temperature difference (these have been measured in detail) we can find the temperature of the other junction from measurement of the voltage. The assumption is that the temperature of the junction is the temperature that is of interest; the possible error in this thinking is the subject of muddy point 1. HT.23 Why does increasing the local flow velocity decrease the temperature error for the thermocouple? The heat transferred by convection is Q& convection = h(Tthermocouple − T fluid ). If we neglect conduction from the wire junction (in other words assume the thermocouple wires are thin and long), the heat balance is between convection heat transfer and radiation heat transfer. For a fixed Q& convection if the heat transfer increases the temperature difference between the thermocouple and the fluid decreases. We have seen, however, that the heat transfer may be estimated using the Reynolds analogy (Section 3.1). For fixed skin friction coefficient the higher the velocity the higher the heat transfer coefficient. Index absorptance HT-54 absorption HT-57 adiabatic 0-5, 0-9 adiabatic efficiency see efficiency adiabatic flame temperature 2C-7 Biot number HT-30, HT-36 black body HT-56, HT-63 see also radiation blade see turbine blade Brayton cycle 1A-5, 1C-5, 2A-5 efficiency 1A-7, 1C-7, 2A-5 maximum work 1A-10 net work 2A-5 Breguet Range equation 2A-9 Carnot cycle 1A-3, 1C-1, 1C-4 efficiency 1A-4, 1C-2 two-phase 2B-8 coefficients of absorption see absorptance reflection see reflectance transmission see transmittance coefficient of performance 1A-12 combined First and Second Law 1B-5 composite slab see slab compressor adiabatic efficiency 2A-15 conduction HT-5, HT-26 heat flux HT-6 one-dimensional HT-7, HT-8, HT-14, HT-17,HT- 26 continuum 0-1 control volume 0-12 see also first law for a control volume convection HT-5, HT-19, HT-26 heat transfer coefficient HT-20, HT-23 near a wall HT-26 convective heat transfer coefficient see convection COP see coefficient of performance critical point 2B-3 critical radius of insulation HT-30 cycle see thermodynamic cycle cylindrical geometry see non-planar geometry Diesel cycle 2A-4 diffusivity HT-22 dissipation 1C-10 drag HT-24 efficiency adiabatic efficiency 1C-17, 2A-14, 2A- 15 overall efficiency 2A-10 propulsive efficiency 2A-10 thermal efficiency 1A-2, 1A-4, 1A-7, 1C-2, 1C-7, 2A-2, 2A-5, 2A-8, 2B-14 emittance HT-57 energy 0-5 internal 0-5, 0-6, 0-7 kinetic 0-5 potential 0-5 enthalpy 0-6, 0-7 of formation 2C-4 entropy 1B-2, 1C-24 generated 1C-10, 1C-20 microscopic 1D entropy changes 1B-5, 1B-6, 1C-9 equilibrium 0-2, 0-3 extensive properties 0-1 fin HT-36 First Law of thermodynamics 1B-3 control mass 0-5 control volume 0-13, 0-14 steady flow 0-14 Fourier's law HT-7 free expansion 1A-14, 1B-9 fuel-air ratio 2A-12, 2C-2 Gibbs equation 1B-5 gray body HT-57, HT-59 heat 0-5 two-phase systems 2B-6 heat exchanger HT-24, HT-45 efficiency HT-52 heat flux HT-7 heat pumps 1A-11 heat reservoir heat transfer from 1B-7 heat transfer between two 1B-7 single 1B-8 heat sources HT-33 heat transfer coefficient see convection h-s diagram 1C-6 ideal gases 0-7 enthalpy 0-6, 0-7 entropy changes 1B-6 equation of state 0-7 internal energy 0-6, 0-7 specific heats 0-7, 0-8 see also specific heats insulated wall HT-12 intensive properties 0-1 internal energy see energy irreversibility 1A-12, 1C-8, 1C-24 isentropic efficiency see adiabatic efficiency isothermal process 1A-1, 1A-14, 1B-9, 1C-18 Kirchoff's law HT-57 lost work 1C-9, 1C-10, 1C-13 maximum work 1A-10, 1C-2 microscopic see entropy Newton's Law of cooling HT-20 non-planar geometry HT-14 cylindrical HT-14, HT-27, HT-30, HT- 47, HT-65 spherical HT-17, HT-65 one-dimensional problem HT-34 see also conduction Otto cycle 2A-1 efficiency 2A-2 work produced 2A-3 overall efficiency see efficiency phase 2B-1 planar geometry HT-7, HT-26, HT-59 Prandtl number HT-22 process 0-2 adiabatic 0-5 isothermal 1A-1, 1A-14 reversible 0-4, 1B-4 see also reversibility reversible and adiabatic 0-9 propulsive efficiency see efficiency quality 2B-5 quasi-equilibrium 0-3 quasi-steady heat transfer HT-14 radiation HT-5, HT-54 of a black body HT-54 ramjet 2A-6 efficiency 2A-8 fuel-air ratio 2A-12 specific impulse 2A-8, 2A-12 thrust 2A-7 Rankine cycle 2B-13 efficiency 2B-14 superheated 2B-16 reflectance HT-54 reflection HT-54, HT-57 refrigerators 1A-11 reversibility 0-4, 0-9, 1A-12, 1A-15, 1A-16, 1B-4, 1C-2, 1C-7, 1C-24 Reynolds analogy HT-20, HT-23, HT-25 saturated liquid 2B-3 saturated vapor 2B-3 saturation state 2B-1 Second Law of thermodynamics 1B-1, 1B-4 Combined with First Law 1B-5 shape factor HT-63 reciprocity relation HT-64 shear stress HT-20, HT-24 skin friction coefficient HT-23 slab HT-9, HT-16 composite HT-11 with heat sources HT-33 sources see heat sources specific heats 0-6, 0-7, 0-8, 0-9 specific impulse 2A-8, 2A-12 specific properties 0-1 spherical geometry see non-planar geometry stagnation state 1C-15 Stanton number HT-23 state see thermodynamic state state equation 0-3, 0-7 steady flow energy equation 0-14 steady heat transfer HT-8 see also quasi- steady Stephan-Boltzmann constant HT-54 stoichiometric 2C-1 superheated Rankine cycle see Rankine cycle superheated vapor 2B-3 temperature distribution through a cylinder section HT-15 through a spherical geometry HT-18 temperature scale 1C-3 theoretical air 2C-1 thermal conductivity HT-7 thermal efficiency see efficiency thermal resistance circuits series HT-11, HT-12, HT-26 parallel HT-12 thermocouples HT-60 thermodynamic cycle 1A-2, 1C-19 Brayton cycle see Brayton cycle Carnot cycle see Carnot cycle combined cycle 2B-19 Diesel cycle see Diesel cycle Otto cycle see Otto cycle non ideal 2A-14 Rankine cycle see Rankine cycle thermodynamic properties 0-1, 0-2 extensive properties 0-1 intensive properties 0-1 specific properties 0-1 thermodynamic state 0-2, 0-3 thermodynamic system 0-2 closed system 0-1 simple system 0-2 throttle 0-9, 1C-14 thrust 2A-7 transient heat transfer HT-42 transmittance HT-54 T-s diagrams 1C-4 turbine adiabatic efficiency 1C-17, 2A-14 turbine blade HT-19 turbulent boundary layer HT-20 two-phase systems 2B heat transfer 2B-6 internal energy 2B-6 P-T diagram 2B-1 P-v diagram 2B-2 T-v diagram 2B-3 work 2B-6 unavailable energy 1C-11 vapor dome 2B-2, 2B-4 view factor see shape factor viscosity dynamic HT-20 kinematic HT-22 work 0-4, 1A-1 from a single heat reservoir 1B-8 isothermal process 1A-2, 1A-14 net work 2A-5, 2A-17 two-phase system 2B-6 Zeroth Law of Thermodynamics 1B-3