Engineering Science Second Year - Second Law Fundamentals Lecture Notes PDF

Engineering Science Second Year Second Law of Thermodynamics Dr Tobias Hermann Notes developed by P.D.McFadden (Last edit: Dr Tobias Hermann – September 2024) Michaelmas Term 2024 CONTENTS...

Engineering Science Second Year Second Law of Thermodynamics Dr Tobias Hermann Notes developed by P.D.McFadden (Last edit: Dr Tobias Hermann – September 2024) Michaelmas Term 2024 CONTENTS 1 Contents 1 Introduction 4 1.1 The Second Law............................. 4 1.2 Books................................... 4 1.3 Syllabus.................................. 5 1.4 Lectures and Example Sheets...................... 5 1.5 Learning Outcomes............................ 6 1.6 Acknowledgements............................ 7 2 Revision 8 2.1 Introduction................................ 8 2.2 First Law for a Closed System...................... 8 2.3 Steady Flow Energy Equation...................... 9 2.4 Reversible Work.............................. 10 2.5 Reversibility................................ 10 2.6 Heat Engines............................... 12 2.7 Heat Pumps................................ 13 3 Second Law 14 3.1 Introduction................................ 14 3.2 Clausius’ Statement........................... 14 3.3 Corollary 1: Planck’s Statement..................... 15 3.4 Equivalence of Clausius and Planck Statements............ 15 3.5 Corollary 2: Reversible Engine is the Most Efficient.......... 17 3.6 Corollary 3: Reversible Engines have Same Efficiency......... 18 3.7 Corollary 4: Thermodynamic Temperature Scale............ 18 3.8 Corollary 5: Clausius Inequality..................... 20 3.9 Corollary 6: (dQ/T )rev = 0 for a Reversible Engine......... 21 H 3.10 Relationship between Corollaries 5 and 6................ 22 3.11 Corollary 7: Definition of Entropy.................... 22 2 CONTENTS 3.12 Entropy as Criterion of Reversibility................... 24 3.13 What is Entropy?............................. 25 3.14 Corollary 8: Adiabatic Reversible Processes are Isentropic...... 26 3.15 Corollary 9: Entropy Tends to Increase................. 27 3.16 Example 1................................. 28 3.17 Example 2................................. 28 3.18 Example 3................................. 29 3.19 Example 4................................. 30 3.20 Entropy Change in a Perfect Gas.................... 30 3.21 Example 5................................. 33 3.22 Example 6................................. 33 4 Thermodynamic Relations 36 4.1 Introduction................................ 36 4.2 First and Second General Thermodynamic Relations......... 36 4.3 Helmholtz and Gibbs Functions..................... 37 4.4 Maxwell’s Relations............................ 37 4.5 Internal Energy of an Ideal Gas..................... 38 4.6 Enthalpy of an Ideal Gas......................... 39 4.7 Relationships between Specific Heats.................. 39 4.8 Change of State – Clausius-Clapeyron Equation............ 42 4.9 Fluid Flow and the SFEE......................... 44 4.10 Example.................................. 45 5 Cycles 46 5.1 Introduction................................ 46 5.2 Carnot Cycle............................... 46 5.3 Carnot Cycle with a Perfect Gas..................... 48 5.4 Heat and Work for an Perfect Gas in an Isothermal Process...... 49 5.5 Work for a Perfect Gas in an Isentropic Process............ 51 5.6 Summary of Heat and Work in Carnot Cycle for Perfect Gas..... 51 5.7 Stirling Cycle............................... 52 6 Temperature 55 6.1 Introduction................................ 55 CONTENTS 3 6.2 Temperature Scales............................ 55 6.2.1 Internal Energy of an Ideal Gas................. 55 6.2.2 Charles’ Law........................... 56 6.2.3 Kelvin Temperature Scale.................... 56 6.2.4 Equivalence of Kelvin and Ideal Gas Temperatures...... 57 6.2.5 Temperature Scale and Absolute Zero............. 59 6.3 Practical Temperature Scales...................... 59 7 Property Diagrams 60 7.1 Introduction................................ 60 7.2 Constant T and s Lines on a p –v Diagram............... 60 7.3 Isobars on a T – s Diagram for an Ideal Gas.............. 61 7.4 Isobars on a T – s Diagram for a Two-Phase System.......... 62 7.5 Example 1................................. 62 7.6 Example 2................................. 63 7.7 Example 3................................. 66 7.8 Example 4................................. 67 7.9 The h– s Diagram or Mollier Chart.................... 68 4 PART 1. INTRODUCTION Part 1 Introduction 1.1 The Second Law We have already met the first law of thermodynamics in the first year of the course, and seen that it is a statement of the principle of conservation of energy in terms of the thermodynamic quantities heat and work. We have also seen how the work done by an engine can be expressed as the difference between the heat energy taken in by the engine from a high-temperature source and the heat energy rejected by the engine to a low-temperature sink, and how the engine efficiency can be calculated as the work done divided by the heat taken in. The first law, however, gave no indication of there being an upper limit to the efficiency of an engine, and might even suggest that we could convert all the heat we take from the source into useful work, thus achieving an efficiency of 100%. It is the second law of thermodynamics that defines the upper limit of the efficiency with which some particular engine can convert heat into work, and similarly the upper limit of the coefficient of performance of a particular heat pump or refrigerator. An understanding of the second law therefore helps us to improve the efficiency of our engine, or the coefficient of performance of our heat pump, which is of great importance given the growing concerns about global warming and dwindling fossil fuel supplies. In addition, the second law throws new light on temperature and how we can measure it, on the concept of reversibility that we touched on briefly in first year, and a new property we shall meet called entropy which enables us to determine whether or not certain processes are possible. 1.2 Books Among the most useful texts for this course are: Rogers, G.F.C. and Mayhew, Y.R., Engineering Thermodynamics Work and Heat Transfer 4th edition (Longman). The most complete. Beware opposite 1.3. SYLLABUS 5 sign convention for work. Eastop, T.D. and McConkey, A., Applied Thermodynamics for Engineering Technologists 5th edition (Pearson Prentice Hall). An easy introduction. Beware opposite sign convention for work. Whalley, P.B., Basic Engineering Thermodynamics (OUP). Well suited to the course. Other possibilities are: Spalding, D.B. and Cole, E.H., Engineering Thermodynamics (Arnold). An easy introduction. Sears, F.W. and Salinger, G., Thermodynamics Kinetic Theory and Statistical Thermodynamics (Addison-Wesley). Good for kinetic theory. Wark, K., Thermodynamics (McGraw-Hill). Beware mixture of SI and imperial units. 1.3 Syllabus Thermodynamics (10 lectures): Introduction to the second law Second law and corollaries, reversible processes, thermodynamic temperature scale and entropy. Equivalence of thermodynamic and ideal gas temperature scales. Second law and cycles. Definition of isentropic efficiency and its application to steam and gas turbine cycles. Applications of the second law Heat exchanger analysis and design. Use of temperature vs. enthalpy plots in heat exchanger and cycle design. Ideal and practical refrigeration systems. Concept of thermodynamic equilibrium; general conditions for equilibrium. Gibbs free energy; equilibrium of mixtures. The Clausius-Clapeyron equation; phase (liquid-vapour) equilibrium of ideal binary mixtures. Chemical reaction equilibrium and the equilibrium constant. 1.4 Lectures and Example Sheets There are ten lectures, five given by Dr T. Hermann and five by Prof D. Gillespie, and 2.5 example sheets. The first four lectures concentrate on the fundamental 6 PART 1. INTRODUCTION aspects of the second law, and topics covered include: The alternative definitions of the second law. The corollaries of the second law. Temperature scales and the concept of an absolute zero of temperature. The Carnot and Stirling cycles. A colour version of these lecture notes, in PDF format with hyperlinks, will be placed on Canvas near the end of the lecture series. Students with special visual needs are welcome to raise these with the lecturer. 1.5 Learning Outcomes After the completion of this course of ten lectures and the accompanying example sheets, you should: Have a good understanding of the second law of thermodynamics and its corollaries. Understand the Carnot cycle, and be able to show the equivalence of the thermodynamic and ideal gas temperature scales. Be able to analyse heat engines (and reversed heat engines) to decide if they are irreversible, reversible or impossible. Understand the derivation of entropy and be able to calculate entropy changes for ideal gases and use tabulated data to determine entropy values for real fluids. Be familiar with the derivation of the general thermodynamic relations, and be able to use them. Understand the Joule cycle for a gas turbine, be able to analyse systems with irreversibilities. Be familiar with the operation of a turbo-jet engine and its performance analysis. Understand the Rankine cycle for steam power plants, and be able to analyse systems with irreversibilities using data from charts and tables. Be able to evaluate the performance of more complex steam cycles. Have an appreciation of the means for improving steam cycle performance. 1.6. ACKNOWLEDGEMENTS 7 Be able to analyse refrigeration cycles, including those with irreversibilities. Be familiar with the pressure-enthalpy chart for refrigerants and be able to use tabulated data for system performance calculations. Be able to analyse combined cycle gas turbine power plant and other combinations of thermodynamic systems. Appreciate the trade-offs between subsystem performance and the need to optimise total system performance. Understand phase (liquid-vapour) equilibrium of ideal binary mixtures. Evaluate chemical reaction equilibrium using the equilibrium constant. 1.6 Acknowledgements These lecture notes and the accompanying example sheets are almost entirely the work of the late Dr Peter McFadden, who spent much of his career creating high quality teaching materials for undergraduates. The quality of the presentation is entirely down to his extensive efforts. The course material draws extensively on past work for this and related courses by Dr Nick Hankins, Dr Colin Wood, Prof Peter Ireland, Prof Richard Stone and the late Dr Peter Whalley (whose book largely covers this course and 1st year thermodynamics work at Oxford). Their contributions are also most gratefully acknowledged. If you have comments on these notes e.g. what goes too fast, what is unclear, what is not worth having, then please e-mail me at [email protected]. 8 PART 2. REVISION Part 2 Revision 2.1 Introduction In this chapter we shall look briefly at the first law of thermodynamics that we met last year, and at the concept of reversibility that we glossed over first time around. 2.2 First Law for a Closed System The first law of thermodynamics for a closed system may be defined in several ways. One possible verbal definition is: If a closed system is taken through a cycle, then the net work delivered to the surroundings is equal to the net heat taken from the surroundings. This can be written mathematically as X X △Q = △W cycle cycle X where means the summation around the cycle. cycle A more useful definition, applicable to individual processes in a closed system, relates the heat in and work out to the change in internal energy of the system by Q − W = △U or q − w = △u For a process from state 1 to state 2, the first law can be written as Q − W = U2 − U1 or q − w = u2 − u1 The first law can also be written in differential form as dQ − dW = dU or dq − dw = du 2.3. STEADY FLOW ENERGY EQUATION 9 2.3 Steady Flow Energy Equation Consider an open system with a steady mass flow ṁ kg s−1 into and out of the system boundary, as illustrated in figure 2.1. The heat flow rate into the system is Q̇ J s−1 (or W) (or kW) and the shaft work out of the system is Ẇs J s−1 (or W) (or kW)... Q Ws c2. m. c1 p2 , v 2 m p1 , v 1 z1 z2 Figure 2.1: Steady flow system Therefore the first law equation can be written as the steady flow energy equation, or SFEE 2 Q̇ − Ẇs = ṁ△ u + pv + g z + 21 c 2 1  or more simply as Q̇ − Ẇs = ṁ△ u + pv + g z + 12 c 2
where u is the internal energy, pv can be thought of as a flow work , g z is the gravitational potential, and 21 c 2 the kinetic energy, all per unit mass. If we divide by the mass flow rate ṁ, we get the specific form of the SFEE q − ws = △ u + pv + g z + 12 c 2
Note that the work terms in the SFEE are shaft work ws and Ws. The group of terms u + pv often occurs in thermodynamic problems, so we define a new thermodynamic property called the enthalpy H J (or kJ) or specific enthalpy h J kg−1 (or kJ kg−1 ) such that H = U + pV or h = u + pv The enthalpy, like the internal energy, is a property of state, because u , p and v are properties of state. 10 PART 2. REVISION We can then rewrite the SFEE in even more compact and useful form as Q̇ − Ẇs = ṁ△ h + g z + 12 c 2
or q − ws = △ h + g z + 12 c 2
2.4 Reversible Work (a) p (b) p 1 closed 1 open system system work work 2 2 0 v 0 v Figure 2.2: Reversible work on p –v diagram For a closed system, the reversible work is given by Z2 w= p dv 1 which on a p –v diagram is equal to the area between the line and the v axis as illustrated in figure 2.2(a). For an open system, the reversible shaft work is given by Z2 ws = − v dp 1 which is equal in magnitude to the area on a p –v diagram between the line and the p axis, as shown in figure 2.2(b). Some care is required when interpreting the sign. In figure 2.2(b), the pressure decreases as the system changes from state 1 to state 2, so the area shown is in fact negative and the resulting shaft work is positive. 2.5 Reversibility Last year we defined a reversible process as follows: 2.5. REVERSIBILITY 11 A reversible process is one which can be reversed so that the system and the surroundings are unchanged at the end of the process and the reverse process. In practical terms, this means a reversible process is one where there is no mechanical friction or fluid friction, and which happens slowly or via a number of quasi-equilibrium states. The reasons for the absence of friction are obvious – friction always opposes the motion, so if we attempt to reverse a process in which friction is present, we will not recover the energy that was dissipated during the forward process but will instead dissipate more energy! Less obvious is the reason why the process must happen slowly – if we want to convert heat into useful work that we can do on the surroundings, we do not want to give unnecessary kinetic energy to the internal components of our system. One aspect we completely ignored last year was the transfer of heat between bodies of different temperatures. For such a process to be reversible, the temperature difference must be zero (or infinitesimally small). We shall learn how to deal with this mathematically when we meet the second law, but we can get a qualitative feel for why it is so by realising that if we bring together two bodies of different temperature and allow heat to transfer between them, we are wasting the opportunity to generate useful work by connecting them by a heat engine. Experience tells us that heat flows from a hotter body to a cooler body. Thus in figure 2.3(a), where TA > TB , heat will flow from A to B and the process will be irreversible because there is a finite temperature difference. Similarly in figure 2.3(b), where TA < TB , the heat is transferred from B to A and the process is again irreversible. However, if TA = TB as in figure 2.3(c), we can argue that the heat flow will be zero, but that in the limiting case, when the difference in temperatures is negligible, the direction of the heat flow has become reversible. Thus we can (a) TA>TB (b) (c) TA=TB TA 1. 14 PART 3. SECOND LAW Part 3 Second Law 3.1 Introduction The second law actually comprises TEN associated statements. The second law as a whole cannot be proved (nor can the first law!) but if we take any one of the ten statements as a starting point, then the remaining nine can be proved from the first. Different texts choose the fundamental statement of the second law in different ways, but in these notes we shall take Clausius’ statement as the starting point because it makes obvious sense, and the other nine we shall refer to as the corollaries, or alternative statements, of the second law. 3.2 Clausius’ Statement It is impossible to construct a system operating in a cycle which transfers heat from a cooler body to a hotter body without work being done on the system by the surroundings. TH TH > TC Q W=0 impossible Q TC Figure 3.1: Clausius’ statement of the second law This is illustrated schematically in figure 3.1. We cannot actually prove Clausius’ statement, yet it is consistent with our experience that heat always flows from a hotter body to a cooler body. 3.3. COROLLARY 1: PLANCK’S STATEMENT 15 Rudolf Clausius (1822–1888) was a German physicist who reconciled the results of Joule with the theories of Carnot. He stated formally the equivalence of heat and work (the first law of thermodynamics) and developed the concept of entropy (which he named in 1865) to explain the directionality of physical processes. He discovered the fact that the entropy of the universe can never decrease in a real process and can only remain constant in a reversible process (the second law of thermodynamics). [scienceworld.wolfram.com/biography/Clausius.html] 3.3 Corollary 1: Planck’s Statement It is impossible to construct a system operating in a cycle which extracts heat from a reservoir and does an equivalent amount of work on the surroundings. TH impossible Q W(=Q) Figure 3.2: Corollary 1: Planck’s statement This is illustrated schematically in figure 3.2 3.4 Equivalence of Clausius and Planck Statements We can demonstrate that the Clausius and Planck statements of the second law are equivalent: First suppose that Planck’s statement is not true, so that by removing heat QH from the hot reservoir with a heat engine E we can provide work W = QH , as shown in figure 3.3(a). With this work we can drive a heat pump P to remove heat QC from the cold reservoir and deliver QH + QC to the hot reservoir, as in figure 3.3(b). By combining the heat engine E and the heat pump P and the heat and work transfers between them, as contained within the system boundary designated by the dotted line in figure 3.3(b), into a new heat pump P′ as shown in figure 3.3(c), we see that we can use a heat pump to remove QC from the cold reservoir and deliver it to the hot reservoir without having to 16 PART 3. SECOND LAW Max Planck was born in Kiel, Germany, on April 23, 1858. He studied at the Universities of Munich and Berlin, where his teachers included Kirchhoff and Helmholtz, and received his doctorate of philosophy at Munich in 1879. Planck’s earliest work was on the subject of thermodynamics, an interest he acquired from his studies under Kirchhoff, whom he greatly admired, and very considerably from reading Clausius’ publications. He published papers on entropy, on thermoelectricity and on the theory of dilute solutions. [nobelprize.org/nobel_prizes/physics/laureates/1918/planck-bio.html] do any work on the pump. This means Clausius’ statement is not true. Now suppose that Clausius’ statement is not true, so that we can remove heat QC from the cold reservoir with a heat pump P and deliver it to the hot reservoir without doing any work, as shown in figure 3.4(a). With a heat engine E we can then remove heat QH + QC from the hot reservoir to produce work W = QH and reject heat QC to the cold reservoir, as in figure 3.4(b). By combining the heat pump P and heat engine E and the heat and work transfers between them, as contained within the system boundary designated by the dotted line in figure 3.4(b), into a new heat engine E′ as shown in figure 3.4(c), we see that we can use a heat engine to remove QH from the hot reservoir and produce equal work W = QH. This means Planck’s statement is not true. The only way out of this logical puzzle is to conclude that Clausius’ statement and Planck’s statement are equivalent statements, although we cannot prove that either of them is true. (a) TH (b) TH (c) TH QH QH QH + QC QC W=0 E E P P’ W = QH W = QH QC QC TC TC Figure 3.3: If Planck’s statement is not true then Clausius’ statement is not true 3.5. COROLLARY 2: REVERSIBLE ENGINE IS THE MOST EFFICIENT 17 (a) TH (b) TH (c) TH QC QC QH + QC QH W=0 P P E E’ W=0 W = QH W = QH QC QC QC TC TC Figure 3.4: If Clausius’ statement is not true then Planck’s statement is not true 3.5 Corollary 2: Reversible Engine is the Most Efficient The reversible heat engine is the most efficient heat engine operating between two given reservoirs. (a) TH (b) TH (c) QH QH QH QH WX – WR WR WX – WR WX X R WR X R X’ QH – WX QH – WR QH – WX QH – WR WX – W R TC TC TC Figure 3.5: Proof of corollary 2 To prove this corollary, we first suppose that the converse is true, and that engine X is more efficient than the reversible engine R. Engine X takes heat QH from a hot reservoir and rejects heat QH − WX to a cold reservoir while producing work WX , as shown in figure 3.5(a). If engine X is indeed more efficient than a reversible engine R operating between the same reservoirs, then WX > WR. Next we run the reversible engine R backwards R to pump heat QH − WR from the cold reservoir and deliver heat QH to the hot reservoir for a work input of WR , as shown in figure 3.5(b). The work produced by engine X can be be used to drive engine R, and because WX > WR there will still be WX − WR net work left. Finally, we can combine engines X and R and the heat and work transfers between them, as contained within the system boundary designated by the dotted line in figure 3.5(b), into a new engine X′, as shown in figure 3.5(c). The net result is that 18 PART 3. SECOND LAW we can take heat WX − WR from the cold reservoir and do an equal amount of work WX − WR , thus converting heat entirely into work. This violates Planck’s statement of the second law and so our original assumption that engine X was more efficient than engine R cannot be true. Thus the reversible heat engine must be the most efficient. Hint: Use the same approach for question 2 of example sheet 2A4G. 3.6 Corollary 3: Reversible Engines have Same Efficiency All reversible engines operating between the same two heat reservoirs have the same efficiency. The proof for this corollary follows closely that given in the previous section. First, we assume that reversible engine X is more efficient than reversible engine R when running between the same two reservoirs. Engine X will therefore produce work WX which is more than the work WR produced by engine R. We can reverse engine R and drive it using engine X and still have net work WX − WR available. By combining engines X and R into a new engine X′, we see that we are taking heat from the cold reservoir to produce an equal amount of work, which violates Planck’s statement. Therefore our original assumption that engine X is more efficient than engine R cannot be true, nor can engine R be more efficient than engine X. That is, both engines must have the same efficiency. 3.7 Corollary 4: Thermodynamic Temperature Scale A temperature scale can be defined which is independent of any particular thermometric substance and which provides an absolute zero of temperature. Consider a reversible heat engine R operating between two reservoirs taking heat QH from a hot reservoir at temperature TH and rejecting heat QC to a cold reservoir at temperature TC , as shown on the left-hand side of figure 3.6. Corollary 3 showed us that all reversible engines operating between two reservoirs have the same efficiency, regardless of the amount of heat being transferred, therefore the efficiency of a reversible engine must depend on the temperatures of the reservoirs. Thus we can write the efficiency of the engine as QC η =1− = 1 − f (TC , TH ) QH where f (TC , TH ) = QC /QH is some yet to be defined function of TC and TH. 3.7. COROLLARY 4: THERMODYNAMIC TEMPERATURE SCALE 19 TH QH QH R W = QH – QI QI R TI W = QH – QC QI QC R W = QI – QC QC TC Figure 3.6: Proof of corollary 4 What do we know about the form of the function f (TC , TH )? Consider two more heat engines, identical to the first, running between the hot and cold reservoirs and an intermediate heat reservoir at temperature TI , as shown on the right-hand side of figure 3.6. As before we find QC QC QI QH = f (TC , TH ) = f (TC , TI ) = f (TI , TH ) or = f (TH , TI ) QH QI QH QI from which QC QC QI = × QH QI QH so f (TC , TI ) g (TC ) f (TC , TH ) = f (TC , TI ) × f (TI , TH ) = =. f (TH , TI ) g (TH ) That is f (TC , TH ) is independent of the intermediate temperature TI , and must be of the form g (TC )/g (TH ) where g (T ) is some yet to be determined function of temperature. Hence we can see that the ratio of the two heat flows is given by QC g (TC ) =. QH g (TH ) There are no further restrictions on the choice of function g (T ). If we choose the simplest possible function, g (T ) = T , we get QC TC = QH TH although other choices are possible. (Kelvin originally proposed a logarithmic scale!) This is the thermodynamic definition of temperature. 20 PART 3. SECOND LAW Planck’s statement of the second law says that QC > 0 and so TC > 0. Hence temperature must be greater than zero, the absolute zero of temperature, and absolute zero is unobtainable. 3.8 Corollary 5: Clausius Inequality I dQ When any system undergoes a cycle ≤ 0. T T0 dWA R dWE T A dQA Figure 3.7: Proof of corollary 5 Here means the integral around the cycle. To prove this corollary, consider a H system A undergoing a cycle. During part of this cycle, the system receives heat dQA in a part of the system where the temperature is T, as shown in figure 3.7. The system does work dWA on the surroundings. Imagine that the heat is supplied from an external reservoir of heat at a temperature T0 via a reversible heat engine R which does work dWE on the surroundings. From the thermodynamic definition of temperature presented in corollary 4 we know that QH TH = QC TC and hence for this system we get dWE + dQA T0 = dQA T or T0 dWE = dQA − dQA T Integrating over a whole cycle of the system gives I I I dQA dWE = T0 − dQA T  21 H 3.9. COROLLARY 6: (DQ/T )REV = 0 FOR A REVERSIBLE ENGINE Applying the first law to the system gives net heat in = net work out or I I dQA = dWA and so I I I dQA dWE = T0 − dWA T giving I I dQA (dWE + dWA ) = T0. T This equation is for the system A and the reversible heat engine R. This combined system gets heat only from the reservoir at T0 and rejects no heat. Thus, from the second law, the work it does on its surroundings must be zero or negative, hence I (dWE + dWA ) ≤ 0 which implies that I dQA T0 ≤0 T or I dQA ≤0 T which is the Clausius inequality applied to system A. H 3.9 Corollary 6: (dQ /T )rev = 0 for a Reversible Engine I   dQ When any system undergoes a reversible cycle = 0. T rev To prove this corollary, we first assume that for a reversible cycle the inequality of corollary 5 is possible, so that when the cycle is run in the forward direction we get I   dQ ≤ 0. T rev forward If we reverse that cycle, we would then have I   dQ ≥0 T rev reverse but this contravenes corollary 5. Thus neither of the above inequalities is possible. The only solution is therefore I   dQ = 0. T rev 22 PART 3. SECOND LAW 3.10 Relationship between Corollaries 5 and 6 Corollaries 5 and 6 can be used together to determine whether cycles are reversible, irreversible or impossible:  I    = 0 if the cycle is reversible dQ  < 0 if the cycle is irreversible T   > 0 is impossible.  If each of the heat transfers occurs at constant temperature, we can write this in discrete form:  = 0 if the cycle is reversible  X Qi   < 0 if the cycle is irreversible Ti  i  > 0 is impossible.  TH hot reservoir QH heat engine W QC TC cold reservoir Figure 3.8: Schematic diagram for heat engine Thus for the heat engine shown in figure 3.8, remembering that heat transfer into the system is positive, we find X Qi QH QC = − Ti TH TC i and so    = 0 if the engine is reversible QH QC  − < 0 if the engine is irreversible TH TC   > 0 is impossible.  Hint: The above equations are useful for question 3 of example sheet 2A4G. 3.11 Corollary 7: Definition of Entropy dQrev There exists a thermodynamic property entropy S such that dS =. T 3.11. COROLLARY 7: DEFINITION OF ENTROPY 23 Recall that the changes in the value of a thermodynamic property when we take the system from one state to another depend only on the initial and final states, not on the path by which the system is taken. Thus if S is a property, and we take a system from state 1 to state 2, then the change in entropy for a reversible process Z2   dQ S2 − S1 = T rev 1 does not depend on the path from 1 to 2. To prove this, consider a cycle 1–2–1 as shown in figure 3.9. There are many reversible cycles possible between states 1 and 2, for example path A + path B, and path A + path C. For these Z Z Z Z dQ dQ dQ dQ + = + =0 T T T T A B A C because of the Clausius inequality so Z Z dQ dQ = T T B C which is independent of path as long as it is a reversible path. So we can write Z2 dQ = S2 − S1 T 1 where S is a property, a function of two other properties such as pressure and volume. For a reversible process, dS = dQ/T, but what about an irreversible process? Consider the two cycles above, but suppose that one of the return paths B and C is reversible and the other is irreversible. Path A is still reversible, so for path A + p C 2 B A 1 v Figure 3.9: Proof of corollary 7 24 PART 3. SECOND LAW reversible return path, dQ/T = 0, hence H Z Z dQ dQ + = 0. T T A rev For path A + irreversible return path, dQ/T < 0 from the Clausius inequality, H hence Z Z dQ dQ + < 0. T T A irrev From these equations we get Z Z dQ dQ < T T irrev rev or in differential form   dQ < dS. T irrev In general dQ dS ≥ T or  dQ > for irreversible process   dS T dQ for reversible process.  =  T Note that the units of entropy S are heat transfer ≡ J K−1 (or kJ K−1 ) absolute temperature and the units of specific entropy s are J kg−1 K−1 (or kJ kg−1 K−1 ) where dq ds ≥. T 3.12 Entropy as Criterion of Reversibility dQ surroundings system Figure 3.10: Heat transfer between surroundings and system 3.13. WHAT IS ENTROPY? 25 Consider now not just the system but also the surroundings, as illustrated in figure 3.10. Suppose the system undergoes a differential reversible change caused by the transfer of heat dQ at temperature T from the surroundings, so that dQ dSsystem =. T But this heat is exchanged with the surroundings and if the surroundings undergo a reversible change then dQ dSsurroundings = − T so that for the universe, comprising system plus surroundings, we get dSuniverse = 0. Thus the entropy of the universe, not the system, is constant if all processes are reversible. Suppose now that the system process is irreversible and the surroundings process is still reversible, so that dQ dSsystem > T and dQ dSsurroundings = − T so that dSuniverse > 0. This applies whether dQ > 0 or dQ < 0. Thus the entropy of the universe, not the system, increases if any process is irreversible. Since all actual processes are to some extent irreversible, then the entropy of the universe is continually increasing. 3.13 What is Entropy? Ultimately it can be shown that entropy is related to order and disorder. If ω is a measure of the probability that a given state will occur, then it can be shown that S = κ ln ω where κ = Boltzmann’s constant = R0 /N , and N = Avogadro’s number = number of molecules in 1 kmol = 6.023 × 1026 kmol−1. Left to its own devices, a system, for example molecules in a box as shown in figure 3.11, will assume a state with a high probability rather than a low probability. 26 PART 3. SECOND LAW So for state A, which has a low probability, to change to state B, which has a high probability, ω increases and so entropy s increases. In macroscopic terms, the expansion of a gas at constant temperature increases the entropy. A B Figure 3.11: Molecules in box 3.14 Corollary 8: Adiabatic Reversible Processes are Isentropic The entropy of any closed system which is thermally isolated from its surroundings remains constant under a reversible process To prove this corollary, consider a reversible process from state 1 to state 2. From corollary 7, the entropy change will be Z2   dQ = S2 − S1. T rev 1 If the process is adiabatic, dQ = 0 therefore S2 − S1 = 0 and the process has constant entropy, and is described as isentropic. Isentropic processes are very important for engineers as benchmarks, because as we saw in corollary 2 the reversible engine is the most efficient engine. Gases and two-phase systems are often used in thermodynamic machines because their expansion is much greater than for solids and liquids. The flow through a component in a system can be so fast that sometimes little heat transfer takes place, and we can therefore model the process as adiabatic. Components of thermodynamic machines (e.g. the fan, compressor and turbine aerofoils in a jet engine) are designed to minimise irreversibilities. An isentropic process is thus often the best process that could be achievable in theory. Since all real processes are to some extent irreversible, the entropy of the universe is continually increasing. 3.15. COROLLARY 9: ENTROPY TENDS TO INCREASE 27 An important consequence of this corollary is that      reversible and adiabatic     isentropic    if a process is reversible and isentropic it is adiabatic.     adiabatic and isentropic reversible     3.15 Corollary 9: Entropy Tends to Increase The entropy of any closed system which is thermally isolated from its surroundings must increase under a non-reversible process. Proof: For process A (non-reversible, adiabatic) from state 1 to state 2 shown in figure 3.12, dQ = 0 hence Z2   dQ =0 T A 1 and for process B (non-adiabatic, reversible) from state 2 to state 1 Z1   dQ = S1 − S2. T B 2 Because the first process is not reversible, the whole cycle 1–2–1 is not reversible, hence Z2   Z1   dQ dQ + S1 and the entropy increases during the non-reversible, adiabatic process A. p non-reversible 2 adiabatic A B non-adiabatic reversible 1 v Figure 3.12: Proof of corollary 9 28 PART 3. SECOND LAW 3.16 Example 1 dQ T Figure 3.13: Example 1 Problem: Calculate the entropy change of a block of metal of mass m and specific heat c when it is heated from T1 to T2 as shown in figure 3.13. Solution: When a solid or a liquid is heated, the change in volume is extremely small so cp ≈ cv = c and therefore a single specific heat c is sufficient. When heat is added reversibly, dS = dQ/T and this heat alters the temperature as dQ = mc dT , so that dT dS = mc T which after integrating gives T2 S2 − S1 = mc ln. T1 Note that the right-hand side contains only properties, so the entropy change does not depend on the path. Even if the heat is added irreversibly , then this is still the entropy change for the block, as it is only a function of the beginning and end points of the process. This equation will hold for a solid or a liquid, where the volume change on heating is small. 3.17 Example 2 Problem: Saturated water at 1 bar flows steadily through a boiler and emerges as dry saturated steam at 1 bar as shown in figure 3.14. Calculate the change in entropy of the steam. Solution: From the SFEE, with the shaft work ws = 0 and with negligible kinetic and potential energy terms, we get q = h2 − h1 = hg − hf = hfg = 2257.9 kJ kg−1 3.18. EXAMPLE 3 29 q 1 2 water steam Figure 3.14: Example 2 from the saturated table in HLT on pages 58–59 for 1 bar. The saturation temperature at 1 bar is Ts = 99.63o C = 372.78 K. Since heat is added at this constant temperature we get Z2   dq q 2257.9 s2 − s1 = = = = 6.057 kJ kg−1 K−1. T rev T 372.78 1 The entropy change when going from saturated liquid to saturated vapour at constant pressure or temperature is called the entropy of vaporisation, denoted by sfg , and is tabulated in the saturated tables in HLT on pages 54–55 and 56–65. 3.18 Example 3 Problem: An adiabatic steam turbine, shown in figure 3.15, takes in steam at 500o C, 100 bar and exhausts it at 20o C, dryness 0.8. Show that the turbine is irreversible. 1 ws turbine 2 Figure 3.15: Example 3 Solution: The initial state is 500o C, 100 bar and final state 20o C, dryness 0.8. The turbine is adiabatic, so there is no heat exchange with the surroundings, and hence △Ssurroundings = 0 or △ssurroundings = 0. From the superheat table in HLT on page 70, at 500o C, 100 bar we get s1 = 6.5994 kJ kg−1 K−1. From the saturated table in HLT on pages 54–55, at 20o C, x2 = 0.8, we get s2 = sf2 + x2 sfg2 = 0.2963 + 0.8 × 8.3721 = 6.9940 kJ kg−1 K−1. 30 PART 3. SECOND LAW Note now that △ssystem = s2 − s1 = + 0.395 kJ kg−1 K−1 and so △suniverse = △ssystem + △ssurroundings = + 0.395 kJ kg−1 K−1. The overall entropy change of the universe is positive, and so the turbine is irreversible. Note the per kg part of the units of △suniverse refers to per kg of the system, not the universe! 3.19 Example 4 Problem: Calculate the shaft work required for the isentropic compression of a liquid, as in the Rankine cycle feed pump, and the change in the enthalpy of the liquid. Solution: We shall see soon from the general thermodynamic relations in section 4.2 that dh = T ds + v dp so if s = constant, then dh = v dp , and as v ≈ constant for a liquid then the feed pump work is Z2 h2 − h1 = v dp ≈ v (p2 − p1 ). 1 For the SFEE when kinetic energy and gravity terms are negligible, we get in differential form dq − dws = dh. Substituting dh = T ds + v dp again gives dq − dws = T ds + v dp but if the process is reversible and adiabatic dq = T ds so Z2 dws = − v dp or ws = − v dp. 1 3.20 Entropy Change in a Perfect Gas We shall see soon from the general thermodynamic relations in section 4.2 that for a reversible, closed system du = T ds − p dv. 3.20. ENTROPY CHANGE IN A PERFECT GAS 31 Recalling from first year that for an ideal gas du = cv dT , we find cv dT = T ds − p dv or dT p ds = cv + dv. T T By substituting the ideal gas equation pv = RT we find dT dv ds = cv +R. T v which after integration this becomes for a perfect gas (cv = constant) T2 v2 s2 − s1 = cv ln + R ln. (3.1) T1 v1 Although the general thermodynamic relation from which we started was applied to a reversible process in a closed system, this final result contains only properties at the initial and final states and is therefore independent of the path which the process followed between states. It is thus true for any process for a perfect gas. Note that by choosing values of T1 , T2 , v1 and v2 then either s1 > s2 or s1 < s2 or s1 = s2 but these inequalities tell nothing about the reversibility of the process because they are entropy changes only of the system, not of the universe. An alternative equation can be found using the second general thermodynamic relation from section 4.2 dh = T ds + v dp. For an ideal gas dh = cp dT dT v ds = cp − dp T T which after substituting the ideal gas equation gives dT dp ds = cp −R (3.2) T p which after integrating for a perfect gas (cp = constant) becomes T2 p2 s2 − s1 = cp ln − R ln. (3.3) T1 p1 32 PART 3. SECOND LAW Again this is true for any process for a perfect gas because it contains only properties at the initial and final states and is therefore independent of the path which the process followed between those states. Equations 3.1 and 3.3 can both be found in HLT on page 86. For an isentropic process in a perfect gas, ds = 0 so equation 3.2 becomes dT dp cp =R. T p For an ideal gas pv = RT. Taking logs and differentiating gives dp dv dT + = p v T so we can eliminate dT /T in both the above to get dp dv dp cp + cp =R p v p hence dp cp dv + =0 p cp − R v which may be integrated for a perfect gas to give pv cp /(cp − R) = constant which is a polytropic process with a polytropic index cp n=. cp − R In section 4.7 we will see that for an ideal gas cp − cv = R and we shall define the ratio of the principal specific heats of a gas to be cp γ= cv where γ > 1. That is, for a perfect gas, an isentropic process follows a polytropic equation with index equal to the ratio of specific heats pv γ = constant. 3.21. EXAMPLE 5 33 insulated container 1 kmol of ideal vacuum gas Figure 3.16: Example 5 3.21 Example 5 Problem: An insulated, rigid box is divided in two equal parts as shown in figure 3.16. One half contains 1 kmol of a perfect gas and the other half is evacuated. The dividing wall is then punctured. What is the change in entropy? Solution: The box is insulated and rigid, so Q = 0 and W = 0. The first law states Q − W = △U therefore U1 = U2. For an ideal gas △u = cv △T and hence T2 = T1. For a perfect gas we have seen the change in entropy is T2 v2 s2 − s1 = cv ln + R ln T1 v1 and here T2 /T1 = 1, v2 /v1 = 2 so s2 − s1 = R ln 2. Given n = 1 kmol, the mass of gas is m = nM kg where M kg kmol−1 is the molar mass. Hence S2 − S1 = m(s2 − s1 ) = nMR ln 2 = nR0 ln 2 = 1 × 8.315 ln 2 = 5.764 kJ K−1. where R0 kJ kmol−1 K−1 is the molar gas constant. 3.22 Example 6 Problem: A copper block with mass 0.2 kg and temperature 95o C is lowered slowly into an insulated bath containing 0.5 kg of water at 5o C, open to the atmosphere. Find the final equilibrium temperature and the changes in entropy of the block and the water. Solution: Consider the first law for a closed system comprising both the metal block and the liquid, written as Q − W = U2 − U1. 34 PART 3. SECOND LAW The bath is insulated and there is no other information suggesting that heat is being transferred to the surroundings, therefore Q = 0 for the block and water combined. Strictly speaking, there will be very small changes of volume of the block as it cools and of the water as it heats, and the net volume change would be doing work against the atmosphere. However, these are so small that we can safely neglect them here, hence W = 0 and U2 = U1 , △Ub = −△Uw , and Qb = −Qw where subscripts b and w refer to the block and the water. Thus mb cb (T2 − Tb1 ) = −mw cw (T2 − Tw1 ) and therefore mb cb Tb1 + mw cw Tw1 0.2 × 380 × 95 + 0.5 × 4180 × 5 T2 = = = 8.158o C. mb cb + mw cw 0.2 × 380 + 0.5 × 4180 Corollary 7 in section 3.11 defined the entropy for a reversible process as dQrev dS =. (3.4) T For the block dQb = mb cb dTb hence dTb dS = mb cb Tb which after integrating from state 1 to state 2 gives T2 273.16 + 8.158 Sb2 − Sb1 = mb cb ln = 0.2 × 380 × ln = −186.6 J K−1. (3.5) Tb1 273.16 + 95 Similarly we could show that T2 273.16 + 8.158 Sw2 − Sw1 = mw cw ln = 0.5 × 4180 × ln = 1023.2 J K−1. (3.6) Tw1 273.16 + 5 Note that the overall entropy change for the system S2 − S1 = −186.6 + 1023.2 = +836.6 J K−1 is positive. As there was no heat transfer to the surroundings, this indicates that the process was irreversible. Why? Because there was heat transfer occurring between the block and the water at a finite temperature difference. But wasn’t there a flaw in our reasoning above? Equation 3.4 was for a reversible process, and we have just shown that our process is irreversible! However, our 3.22. EXAMPLE 6 35 process has an initial state 1 defined by two known properties, temperature and pressure, and a final state 2, also defined by two known properties. From the phase rule, if any two properties are defined then all other properties are defined, including entropy. These properties exist independently of the actual path followed by the process that takes us from state 1 to state 2. Even though our path was irreversible, it is always possible to find a reversible path between two states, by doing or extracting work as necessary until the temperatures of the block and the water change from their initial to their final values, and therefore our result holds. 36 PART 4. THERMODYNAMIC RELATIONS Part 4 Thermodynamic Relations 4.1 Introduction In this chapter we shall examine the four general thermodynamics relations, meet the Helmholtz and Gibbs functions, and touch briefly on Maxwell’s relations. We shall also discover some of the useful conclusions that follow from these relations. Note that here we are making no distinction between the thermodynamic temperature and the ideal gas temperature. We shall explore this subtle detail later and see that they are in fact equivalent. 4.2 First and Second General Thermodynamic Relations The first law in differential form is dq − dw = du (4.1) and for a closed system, the reversible work in differential form is dw = p dv. (4.2) We saw from the definition of entropy in section 3.11 that for a reversible process dq = T ds. (4.3) If we substitute equations 4.2 and 4.3 into equation 4.1, we get the first general thermodynamic relation du = T ds − p dv. (4.4) Note that equations 4.2 and 4.3 are restricted to a reversible process, and equation 4.2 applies only to a closed system. However, after combining, equation 4.4 only contains properties of state. It is thus true for any change, whether reversible or irreversible, whether closed system or steady flow. A similar expression can be obtained by remembering that h = u + pv or in differential form dh = du + p dv + v dp. Substituting for du gives the second 4.3. HELMHOLTZ AND GIBBS FUNCTIONS 37 general thermodynamic relation dh = T ds + v dp. (4.5) These can be found, along with the third and fourth general relations, in HLT on page 86. 4.3 Helmholtz and Gibbs Functions In first year, we noted that the group u + pv often appeared in thermodynamic equations, suggesting that this group has some underlying importance, and we defined the new property enthalpy h as h = u + pv. Similarly, it is found that the group u − T s occurs in the analysis of chemical reactions and in statistical thermodynamics, and the group h − T s occurs in the analysis of chemical reactions and during changes of state. We therefore define two new properties, known as the Helmholtz and Gibbs functions. The Helmholtz function is defined as f = u − Ts and gives an indication of the maximum possible work output from a system when heat is transferred at constant temperature. Note that some texts use a for the Helmholtz function instead of f. The Gibbs function is defined as g = h − Ts and gives an indication of the maximum possible work output from a system when it changes between states. The Gibbs function is sometimes referred to as the free energy function. These definitions can be found in HLT on page 86. Our involvement with these functions is limited, but you may meet them again next year. 4.4 Maxwell’s Relations Maxwell’s relations are not on the second year syllabus but we shall list them here because they are useful for deriving several other thermodynamic relations. If you are revising for exams using papers from the old course, you will encounter several problems that use them. They are given in HLT on page 86. 38 PART 4. THERMODYNAMIC RELATIONS     ∂T ∂p =− (M1) ∂v s ∂s v     ∂T ∂v = (M2) ∂p s ∂s p     ∂p ∂s = (M3) ∂T v ∂v T     ∂v ∂s =− (M4) ∂T p ∂p T 4.5 Internal Energy of an Ideal Gas The internal energy of an ideal gas is a function of temperature only. Proof: The phase rule states that F +P =C +2 therefore for a system with C = 1 component and P = 1 phase, a gas, there will be F = 2 degrees of freedom, hence we must define two properties in order to define completely the state of the system. Thus, for example, we can express the internal energy u of such a system as a function of temperature T and specific volume v using u = u(T , v ) which can be differentiated to gives     ∂u ∂u du = dT + dv. (4.6) ∂T v ∂v T The first general thermodynamic relation du = T ds − p dv can be differentiated with respect to v at constant T to give     ∂u ∂s =T −p ∂v T ∂v T and using Maxwell relation M3 we get     ∂u ∂p =T − p. (4.7) ∂v T ∂T v For an ideal gas RT p= v therefore   ∂p R p = = ∂T v v T 4.6. ENTHALPY OF AN IDEAL GAS 39 so   ∂p T − p = 0. (4.8) ∂T v Combining equations 4.7 and 4.8 gives   ∂u =0 ∂v T which can be substituted into equation 4.6 to show that   ∂u du = dT (4.9) ∂T v proving that u is a function of temperature T only, written as u = u(T ). The specific heat cv is defined as   ∂u cv = ∂T v hence equation 4.9 can be written du = cv dT. Thus for an ideal gas, du = cv dT, and u = u(T ) only. This only applies for an ideal gas because we used pv = RT. We used this result without proof in first year. 4.6 Enthalpy of an Ideal Gas The enthalpy of an ideal gas is a function of temperature only. Proof: By definition h = u + pv and for an ideal gas pv = RT so h = u + RT so if u = u(T ) only then h = h(T ) only and dh = cp dT. Alternatively, proceed as for the previous proof, but use h = h(T , p) and dh = T ds + v dp. 4.7 Relationships between Specific Heats In this section we shall find a general expression for cp − cv. For a system containing a single phase of a single component, we can write any property as a function of any two other properties. Thus   ∂u u = u(T , v ) so du = cv dT + dv (4.10) ∂v T 40 PART 4. THERMODYNAMIC RELATIONS and   ∂h h = h(T , p) so dh = cp dT + dp. (4.11) ∂p T We can combine the first two general thermodynamic relations du = T ds − p dv and dh = T ds + v dp by eliminating T ds to give du + p dv = dh − v dp. (4.12) Substituting equations 4.10 and 4.11 into equation 4.12 gives         ∂u ∂h cv dT + p + dv = cp dT − v − dp. ∂v T ∂p T For the particular case of a constant pressure process, dp = 0 so     ∂u cv dT + p + dv = cp dT ∂v T and by differentiating with respect to T at constant p      ∂u ∂v cp − cv = p +. ∂v T ∂T p From du = T ds − p dv by differentiating with respect to v at constant T we get     ∂u ∂s =T −p ∂v T ∂v T and using Maxwell relation M3     ∂u ∂p =T −p ∂v T ∂T v so finally substituting gives    ∂p ∂v cp − cv = T. ∂T v ∂T p If we use the definitions in HLT on page 86 for the bulk coefficient of thermal expansion   1 ∂v β= v ∂T p and the compressibility coefficient   1 ∂v κ=− v ∂p T 4.7. RELATIONSHIPS BETWEEN SPECIFIC HEATS 41 where the minus sign occurs because v decreases as p increases, we get β 2T v       ∂v ∂v ∂p = −T. κ ∂T p ∂T p ∂v T The “chain rule” for partial derivatives is       ∂x ∂y ∂z = −1 ∂y z ∂z x ∂x y so in terms of p , v and T we can write       ∂v ∂p ∂T = −1 ∂T p ∂v T ∂p v therefore       ∂v ∂p ∂p =− ∂T p ∂v T ∂T v hence β 2T v     ∂v ∂p =T = cp − cv κ ∂T p ∂T v and finally     ∂v ∂p cp − cv = T ∂T p ∂T v giving β 2T v cp − cv =. (4.13) κ For an ideal gas pv = RT so     ∂p R ∂v R = and = ∂T v v ∂T p p therefore R R RT cp − cv = T × × =R× =R v p pv so that for an ideal gas cp − cv = R. We can obtain this last result more simply, because we have already shown that for an ideal gas u = u(T ) only and h = h(T ) only therefore du dh cv = and cp = dT dT and using h = u + pv = u + RT 42 PART 4. THERMODYNAMIC RELATIONS gives dh du = +R or cp = cv + R. dT dT For liquids and solids, cp − cv is small. For example, for copper at T = 300 K, using β = 51 × 10−6 K−1, κ = 7.7 × 10−9 m2 N−1 and v = 1.12 × 10−4 m3 kg−1, equation 4.13 gives cp − cv = 0.01 J kg−1 K−1 so cp ≈ cv ≈ 380 J kg−1 K−1. Hence for all practical purposes, solids and liquids only have one specific heat c. 4.8 Change of State – Clausius-Clapeyron Equation vapour liquid Figure 4.1: System of liquid and vapour Consider a liquid in equilibrium with its vapour, that is, both are saturated, as illustrated in figure 4.1. If a small amount of liquid vaporises, the temperature and pressure are constant so that dp = 0 and dT = 0. From the definition of the Gibbs function given in section 4.3 we get g =h−T s dg = dh − T ds − s dT dg = (v dp + T ds) − T ds − s dT dg = v dp − s dT so that during the change of state dg = 0 hence gf = gg. This can be demonstrated using the steam tables in HLT. From pages 58–59, we find that for liquid water at 1 bar, 372.8 K gf = hf − T sf = 417.5 − 372.8 × 1.3027 = −68.2 kJ kg−1 4.8. CHANGE OF STATE – CLAUSIUS-CLAPEYRON EQUATION 43 and for steam at 1 bar, 372.8 K gg = hg − T sg = 2675.4 − 372.8 × 7.3598 = −68.2 kJ kg−1. If the system pressure is increased by a small amount dp , the saturation temperature increases by dT and the Gibbs free energy g of each phase increases. However gf + dgf = gg + dgg but gf = gg or dgf = dgg so that vf dp − sf dT = vg dp − sg dT or dp sf − sg =. dT vf − vg The change from liquid to vapour occurs at constant pressure dp = 0 so from dh = T ds + v dp we get Zg △h = T ds = T △s because T = constant f hence hf − hg = T (sf − sg ) where T K is the saturation temperature. Hence dp hf − hg = dT T (vf − vg ) which is known as the Clausius-Clapeyron equation. Note that it is an ordinary differential equation, not a partial differential, because on the saturation line p is a unique function of T only. Now suppose that vg >> vf , which it usually is, and that the vapour behaves like an ideal gas, so that RT vg =. p 44 PART 4. THERMODYNAMIC RELATIONS ln p slope = - h /R fg 1/T Figure 4.2: Graph of ln p against 1/T With these assumptions, and putting hfg = hg − hf , we get dp hfg p = dT RT 2 but   dp 1 dT d(ln p) = and d =− 2 p T T or d(ln p) hfg =− d(1/T ) R which is the gradient of a graph of ln p versus 1/T. If hfg is constant, a graph of ln p against 1/T as shown in figure 4.2 will be a straight line. 4.9 Fluid Flow and the SFEE The SFEE was q − ws = △(h + 21 c 2 + g z) and in differential form dq − dws = dh + c dc + g dz. If the change is reversible, that is, there is no friction, then dq = T ds , and if no shaft work is done then T ds = dh + c dc + g dz but dh = T ds + v dp so v dp + c dc + g dz = 0. If the specific volume is constant then, using v = 1/ρ where ρ is the density, integrating gives p 1 2 + c + g z = constant. ρ 2 4.10. EXAMPLE 45 This is Bernoulli’s equation for fluid flow. It can also be derived from force = rate of change of momentum as covered in the fluid mechanics lectures. Conditions for the equation to apply are steady flow reversible flow so no friction no shaft work incompressible flow so ρ = constant. It is not necessary to specify that heat transfer is zero. 4.10 Example Water from a large reservoir flows through a smooth pipe to a point 10 m below the reservoir where it discharges to the atmosphere. Ignoring fluid friction, calculate the velocity of the water leaving the pipe. Bernoulli’s equation gives p1 1 2 p2 1 2 + 2 c1 + g z1 = + 2 c2 + g z2. ρ ρ The reservoir is large, so the initial velocity is effectively c1 = 0. Both the reservoir and the discharge pipe are open to the atmosphere so p1 = p2 = p0. If we take the height of the reservoir to be z1 = 0, then z2 = −10 m. The density is ρ = 1000 kg m3. Hence 105 1 2 105 1 2 + (0) + g (0) = 3 + 2 c2 + 9.8(−10) 103 2 10 √ ⇒ c2 = 2 × 9.8 × 10 = 14.0 m s−1. 46 PART 5. CYCLES Part 5 Cycles 5.1 Introduction In this chapter we shall look at some thermodynamic cycles, including the Carnot cycle. We shall also see how cycles can be represented on the T – s diagram, to complement the p –v diagram we met in first year. 5.2 Carnot Cycle In discussing the second law, we have become familiar with the symbolic language of the reversible heat engine depicted in figure 5.1, but we have not thought how a reversible engine might be made to work. For that implementation we now design a sequence of processes making a cycle in which heat and work are conveyed to and from a working fluid. The cycle is known as the Carnot cycle. In order to mimic a reversible engine, the cycle must include a process in which heat is received reversibly at a constant high temperature TH (process 1–2). There must also be a reversible process where heat is rejected at a lower temperature TC (process 3–4). If we draw a T – S diagram, these isothermal processes are horizontal straight lines. In the link processes 2–3 and 4–1 the temperature varies so we must design Sadi Carnot (1796–1832) was a French physicist and military engineer who gave the first theoretical account of heat engines, now known as the Carnot cycle, thereby laying the foundations of the second law of thermodynamics. He is often described as the ‘father of thermodynamics’, being responsible for such concepts as Carnot efficiency, Carnot theorem, Carnot heat engine, and others. [en.wikipedia.org/wiki/Nicolas_Leonard_Sadi_Carnot] 5.2. CARNOT CYCLE 47 TH hot reservoir QH heat engine W QC TC cold reservoir Figure 5.1: Schematic diagram for heat engine them to have no heat exchange. These processes are therefore adiabatic as well as reversible. By corollary 7 this means that they are isentropic and therefore vertical on the T – S diagram in figure 5.2. T W12 QH TH 1 2 isothermal isentropic W41 isentropic W23 isothermal TC 4 W34 QC 3 S 0 S1 S2 Figure 5.2: T – S diagram for heat engine Let’s compare our theoretical reversible engine and the Carnot cycle. For the theoretical reversible engine, from corollary 4 in section 3.7 we find QC QH = TC TH therefore QC TC = QH TH and from the first law W = QH − QC we can calculate the efficiency W QC TC η= =1− =1−. QH QH TH 48 PART 5. CYCLES For the Carnot cycle, from corollary 7 in section 3.11 we see dQ = T dS but if we integrate over the isothermal processes at constant temperatures TH and TC we get QH = TH (S2 − S1 ) and QC = TC (S2 − S1 ) hence QC TC = QH TH and from the first law W12 + W23 − W34 − W41 = QH − QC we can calculate the efficiency QH − QC QC TC η= =1− =1−. QH QH TH That is, they have identical efficiency! 5.3 Carnot Cycle with a Perfect Gas So far, we have a cycle comprising two isothermal and two isentropic processes which represents the heat exchange and efficiency of the notional reversible engine, but no working fluid was proposed. We just sketched a rectangle on a T – S diagram. Now we take a further step towards the design of a practical machine by defining the four reversible processes in terms of the properties of an ideal gas enclosed in a cylinder with a piston In order to retain a good physical awareness of the processes in terms of a piston moving in and out of a cylinder, it is desirable to sketch the Carnot cycle not only on a T – s diagram, but also on a p –v diagram, as in figure 5.3. Drawing p –v (or p –V ) diagrams is an excellent habit to develop anyway. The following small piece of analysis tells us how to draw the slopes of isothermal and isentropic processes on a p –v diagram. For an isothermal process on a p –v diagram for an ideal gas, we can differentiate the ideal gas equation pv = RT to give dp dv + =0 p v 5.4. HEAT AND WORK FOR AN PERFECT GAS IN AN ISOTHERMAL PROCESS49 T w12 p 1 1 q12 2 isothermal isentropic T1=T2 w12 q12 isothermal isentropic w41 2 isentropic isentropic w23 w23 w41 4 isothermal T3 =T4 3 4 w34 q34 3 w34 q isothermal 34 s v 0 s1 s2 0 Figure 5.3: T – s and p –v diagrams for Carnot cycle because T is constant, hence the slope is   ∂p p =−. ∂v T v For an isentropic process on a p –v diagram for a perfect gas, we can differentiate the polytropic equation pv γ = constant for a perfect gas to give dp dv +γ =0 p v hence the slope is   ∂p p = −γ. ∂v s v That is, both isothermal and isentropic lines on a p –v diagram are hyperbolas, but the isentropic lines have steeper negative slope. In thermodynamics we often use T – s diagrams to represent cycles. However, for closed systems, such as internal combustion engines, p –V diagrams are generally more useful. In certain circumstances other properties give valuable insight. For example, the Mollier diagram for steam that we shall meet in section 7.9 is a graph of enthalpy h against entropy s. Refrigerators are elegantly described using p –h diagrams. It is fortunate that two properties are required to define state otherwise we would need three-dimensional graph paper! 5.4 Heat and Work for an Perfect Gas in an Isothermal Process Processes 1–2 and 3–4 in the T – s diagram for a Carnot cycle, shown in figure 5.3, are isothermal. For an ideal gas, we know that the internal energy is related to the 50 PART 5. CYCLES temperature by △u = cv △T hence for an isothermal process 1–2, u1 = u2. Thus from the first law q12 − w12 = 0. (5.1) If the work is reversible, we can write Z2 w12 = p dv. (5.2) 1 For an isothermal process at temperature T1 in an ideal gas RT1 p= (5.3) v and by substituting equation 5.3 into equa