Second Law of Thermodynamics and Entropy

Second Law of Thermodynamics and Entropy
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Published Date:02-08-2017
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Second Law of Thermodynamics and Entropy Transport and Production Mechanisms 7.1 INTRODUCTION In this chapter, we introduce the second law of thermodynamics and an important new thermodynamic property called entropy. The theory is presented first then applied to a variety of closed and open systems of engineering interest in the following chapters. The details of the second law of thermodynamics and its associated entropy balance are presented, along with a detailed discussion of the entropy transport mechanisms associated with the energy transports of heat and work. Unlike mass, energy, and momentum, entropy is not conserved. Consequently, the mechanisms of entropy production must be well understood to produce an effective entropy balance equation. In Chapter 8, the focus is on applying the theory presented in this chapter to the same steady state closed systems considered in Chapter 5. The second law of thermodynamics is expanded in Chapter 9 to cover open Modern Engineering Thermodynamics. DOI: 10.1016/B978-0-12-374996-3.00007-5 © 2011 Elsevier Inc. All rights reserved. 205206 CHAPTER 7: SecondLawofThermodynamicsandEntropyTransportandProductionMechanisms systems, as in Chapter 6. Then appropriate applications are presented, dealing with a variety of common open systems of engineering interest, such as nozzles, diffusers, throttling devices, heat exchangers, and mixing. Chapter 9 ends with a brief discussion of shaft work machines and unsteady state processes. 7.2 WHAT IS ENTROPY? When we discussed the first law of thermodynamics in Chapter 4, it was fairly easy to apply the general balance equations to the energy concept and to invoke the conservation of energy principle to obtain a workable energy balance equation. Energy is a common English word, and it is also a well-accepted technical term. Everyone has a basic understanding of what the word means, though we would all have a difficult time defining it precisely. Thesamecanbesaidforthewords force and momentum. They are such familiar words that we easily accept mathematical formulae and logical arguments structured around them. Most people are intrigued by seeing a movie run backward because it produces images of things never observed in the real world. What they do not realize is that they are seeing the effects of the second law of thermo- dynamics in action. The second law dictates the direction of the arrow of time. That is, things occur only in a certain way in the real world; and by applying the second law to an observation (like the screening of a movie), we can determine whether the event is running forwardorbackwardintime.Thesecondlawofthermo- dynamics is what prohibits us from actually traveling backward in time. Curiously, it is the only law of nature that has such a restriction. All the other laws of mechanics and thermodynamics are valid regardless of whether time is moving forward or backward. Only the second law of thermodynamics is violated when time is reversed. At this point we need to introduce a new thermodynamic property that is simply a measure of the amount of molecular disorder within a system. The name of this new property is entropy. The meaning of this particular name is explained later, but note that a system that has a high degree of molecular disorder (such as a high- temperature gas) has a very high entropy value and, conversely, a system that has a very low degree of molecular disorder (such as a crystalline solid) has a very low entropy value. This new property is very important because entropy is at the core of the second law of thermodynamics. A system that has all its atoms arranged in some perfectly ordered manner has an entropy value of zero. This is the substance of the third law of thermodynamics. This law, introduced in 1906 by Walter H. Nernst (1864–1941), states that the entropy of a pure substance is a constant at absolute zero temperature. In 1911, Max Planck modi- fied this law by setting the entropies of all pure substances equal to zero at absolute zero temperature. This had the effect of normalizing entropy values and thus creating a uniform absolute entropy scale for all substances. Therefore, we can write the following simple mathematical statement of the third law of thermodynamics: The third law of thermodynamics: The entropy of a pure substance is zero at absolute zero temperature, or limðÞ Entropyof apuresubstance = 0 (7.1) T0 With this simple entropy-disorder concept in mind, we would expect that the entropy of solid water (ice), with its highly ordered molecular structure, would be less than that of liquid water with its amorphous molecular structure, which in turn would be less than that of water vapor with its highly random molecular order. This is in fact true, for at the triple point of water (the only point where the solid, liquid, and vapor phases coexist in equilibrium), the values of the specific entropies of these phases are . . Specificentropyof iceatthetriplepoint = −1:221kJ/ðÞ kg K = −0:292Btu/ðÞ lbm R . . Specificentropyof liquidwateratthetriplepoint = 0:0kJ/ðÞ kg K = 0:0Btu/ðÞ lbm R . . Specificentropyof watervaporatthetriplepoint = 9:157kJ/ðÞ kg K = 2:187Btu/ðÞ lbm R So clearly the entropy of a solid the entropy of a liquid the entropy of a vapor or gas. IS ENTROPY LIMITED TO MOLECULES? Even though the physical concept of entropy is based on the behavior of molecular systems, order and disorder phenom- ena exist at all levels. For example, if entropy is a measure of disorder, then what is the entropy of your bedroom? If you have a messy room it is very disordered, so its entropy is high, but if you are a neat person and keep things picked up and put away, then it has a low entropy. Things seem to get messy easily, and to maintain cleanliness and order requires con- stant effort. This is a fundamental characteristic of the second law of thermodynamics. The natural progression of things is from ordered to disorganized and to keep it organized requires the input of energy. If entropy had been named disorder, perhaps it would not be so difficult to understand.7.3 The Second Law of Thermodynamics 207 FOR WHAT IS THE THIRD LAW USED? This law is used to define an absolute measurement scale for entropy, but it does not otherwise contribute to a thermo- dynamic analysis of an engineering system. Numerical values for specific entropy are listed in thermodynamic tables along with values for specific volume, specific internal energy, and specific enthalpy. For convenience, most thermodynamic tables are developed around a “relative” measurement scale, where the values of entropy and internal energy are arbitrarily set equal to zero at a point other than at absolute zero temperature. For example, in the steam tables, the specific internal energy and specific entropy of saturated liquid water are arbitrarily set equal to zero at the triple point of water (0.01°C, 0.6113 kPa or 32.018°F, 0.0887 psia). Thus, the specific internal energies and specific entropies of the less-disordered mole- cular states of water (like ice) have negative values on this relative scale. THE TRUTH ABOUT ENTROPY 1. Entropy is a measure of the amount of molecular disorder within a system. 2. Entropy can only be produced (but not destroyed) within a system. 3. The entropy of a system can be increased or decreased by entropy transport across the system boundary. Since it always takes an input of energy to create order within a system, it seems reasonable to postulate that a relation exists between the energy transports of a system and its order, or entropy value. Thus, we arrive at the three basic elements of the second law of thermodynamics: We begin this chapter by assuming the existence of a disorder-measuring thermodynamic property that we call entropy.Weusethesymbol S to represent the total entropy (an extensive property), and use s = S/m for the specific entropy (an intensive property). 7.3 THE SECOND LAW OF THERMODYNAMICS We can use the general balance equation of Chapter 2 to analyze any concept whatsoever. Introducing the total entropy S into balance Eq. (2.11) provides the following total entropy balance (SB): S = S +S (7.2) G T P where S is the gain or loss of total entropy of the system due to the transport of total entropy S into or out of G T the system and the production or destruction of total entropy S by the system. A total entropy rate balance (SRB) P is easily obtained from Eq. 7.2 by differentiating it with respect to time to give _ _ _ S = S +S (7.3) G T P _ where the overdot indicates material time differentiation (i.e., S = dS/dt). Unlike energy, mass, and momentum, entropy is not conserved in any real process. Processes that have zero entropy production are called reversible and are characterized by the fact that they can occur equally well in either the forward or backward direction of time. The thing that makes entropy a unique concept worthy of a thermodynamic law of its own is that entropy is never destroyed in any real process. Now, it happens that some processes have very small amounts of entropy production, and it is a useful approximation for these processes to set their entropy production equal to zero. This can be stated in a very succinct mathematical form as The second law of thermodynamics Theentropyproduction,S ≥0 (7.4a) P and _ Theentropyproductionrate,S ≥0 (7.4b) P where the equality sign applies only to a reversible process. Equations (7.2) and (7.3), as modified by Eqs. (7.4a) and (7.4b), form the mathematicalbasis for a workingformof theentropy balance and the entropy ratebalance.208 CHAPTER 7: SecondLawofThermodynamicsandEntropyTransportandProductionMechanisms IS ENTROPY CONSERVED LIKE ENERGY AND MOMENTUM? No, but it turns out that, if you “assume” the entropy production equal to zero, then the second law becomes a conserva- tion law. This makes entropy problems easier to solve, because it eliminates the entropy production term in the entropy balance equation. But, since entropy is never actually conserved in any real process, processes that “assume” zero entropy production are only approximations of real world behavior. We have a special word that we use to tell you when you are to assume the entropy production is zero. The word is reversi- ble. When you see that word in a problem statement, you know that the entropy production term is assumed to be zero. However, remember that reversible processes are only approximations of real (irreversible) processes. At this point we must develop the auxiliary formulae for the entropy transport and production terms before Eqs. (7.2) and (7.3) can be put to any practical use. Unfortunately, this is not an easy task. To understand the concepts of entropy transport and production, we must go back to the original 19th century classical ideas of Carnot, Clausius, and Thomson (Lord Kelvin), the early developers of this field. When this is completed, we bring the subject forward to a modern formulation. 7.4 CARNOT’S HEAT ENGINE AND THE SECOND LAW OF THERMODYNAMICS The origins of the second law of thermodynamics lie in the work of a young 19th century French military 1 engineer named Nicolas Leonard Sadi Carnot (1796–1832). Sadi was the son of one of Napoleon’smost successful generals, Lazare Carnot, and was educated at the famous Ecole Polytechnique in Paris. This institution was established in 1794 as an army engineering school and provided a rigorous program of study in chemistry, physics, and mathematics. Between 1794 and 1830, the Ecole Polytechnique had such famous instructors as Lagrange, Fourier, Laplace, Ampere, Cauchy, Coriolis, Poisson, Guy-Lussac, and Poiseuille. After his formal education, Carnot chose a career as an army officer. At that time, Britain was a powerful military force, primarily as a result of the Industrial Revolution brought about by the British development of the steam engine. French technology was not developing as fast as Britain’s, and in the 1820s, Carnot became convinced that France’s inadequate utilization of steam power had made it militarily inferior. He began to study the fundamentals of steam engine technology, and in 1824, he published the results of his studies in a small book entitled Reflections on the Motive Power of Fire (the French wordfor firewas thena common term for whatwe call heattoday). Sadi Carnot was trained in the basic principles of hydraulics, pumps, and water wheels at the Ecole Polytechni- que. It was clear to him that the power of a steam engine was released as the heat fluid (caloric) fell from the high temperature of the boiler to the lower temperature of the condenser, in much the same way that water falls through a water wheel to produce a mechanical shaft work output. He conjectured that According to established principles at the present time, we can compare with sufficient accuracy the motive power of heat to that of a waterfall. The motive power of a waterfall depends on its height and on the quantity of liquid; the motive power of heat depends also on the quantity of caloric used, and on what may be termed, on what in fact we will call, the height of its fall, that is to say, the difference of temperature of the bodies between which the exchange of caloric is made. WHAT IS HEAT ANYWAY? The essence of the concept of heat was a very actively debated scientific topic at the time. In 1789, the great French chemist Antoine Lavoisier (1743–1794) proposed the caloric theory of heat, in which heat was presumed to be a colorless, odorless, weightlessfluidcalled caloricthat couldbepouredfromoneobject toanother.Whenan objectbecamefull of caloric,it wassaid tobe saturated with it. This was theoriginoftheterms saturated liquid, saturated vapor, and soonthatweuse in thermodynamics today. These terms were introduced into the scientific literature in the early 19th century, when the caloric theory of heat was popular, and they were never removed when it was later proven that heat was not a fluid. Today, they are simply misnomers. Now,whenweusetheword heat inatechnicalsense,wenormallymeanan energy transport arisingfroma temperature difference. 1 Pronounced car-no. Many Frenchwords have a silent t ending. For example, Peugot, Tissot, Monet, ballet, chalet, Chevrolet, and Renault. Sadiwas named after the medievalPersian poetSaadi Musharif edDin,whose poems becamepopular in France in the late 18th century.7.4 Carnot’s Heat Engine and the Second Law of Thermodynamics 209 By the 1820s, a great deal of work had already been done on the efficiency of water wheels, and the water wheel–steam engine analogy must have seemed to Carnot like a good way to approach the problem of improv- ing steam engine efficiency. Two important conclusions came from his work with this analogy. First, he knew that no one could build a water wheel that would produce a continuous work output unless water actually entered and exited the wheel. And, if water with a certain kinetic and potential energy entered the wheel, then the same amount of water with a lower energy level must also exit the wheel. In other words, it is impossible to make a water wheel that converts all the energyoftheinletwaterintooutputshaftwork.There must also be an outflow of water from the wheel, and this outflow must have some energy. These rather obvious statements are illustrated in Figure 7.1. Now, if you extend this idea to a steam engine (or any type of heat engine) by replacing the word water by the hypothetical heat fluid caloric, then it is easy to conclude that when caloric at a certain energy level (tempera- ture) enters a work-producing heat engine, it must also exit the engine at a lower energy level (temperature). This concept was later refined into the following form, known today as the Kelvin-Planck statement of the sec- ond law of thermodynamics: You cannot make a continuously operating heat engine that converts all of its heat input directly into work output (see Figure 7.2). Second, Carnot observed that the maximum efficiency of a water wheel was independent of the type of liquid used and depended only on the inlet and outlet flow energies. This led him to the conclusion that The motive power of heat is independent of the agents employed to realize it; its quantity is fixed solely by the temperatures of the bodies between which is effected, finally, the transfer of caloric. Or, the maximum efficiency of a steam engine (or any type of heat engine) is dependent only on the tempera- tures of the high- and low-temperature thermal reservoirs of the engine (the boiler and condenser temperatures inthecaseofasteamengine)andisindependentoftheworkingfluidoftheengine(waterinthecaseofa steam engine). Of course, to achieve the maximum possible efficiency, the water wheel and the heat engine must be completely reversible; that is, they cannot possess any mechanical friction or other losses of any kind. Water flow Water flow in in No water leaving Water Shaft Shaft the wheel flow out work out work out Common overshot A clearly impossible water wheel type of water wheel FIGURE 7.1 Water wheel operation. High-temperature High-temperature thermal source thermal source Heat in Heat in Cyclic Cyclic Work Work heat heat out out engine engine No heat leaving Heat out the engine An impossible type of Low-temperature continuously operating thermal sink heat engine A common continuously operating heat engine FIGURE 7.2 Heat engine operation.210 CHAPTER 7: SecondLawofThermodynamicsandEntropyTransportandProductionMechanisms CLAUSIUSSTATEMENTOFTHESECONDLAWOFTHERMODYNAMICS It is impossible to build a continuously operating device that will cause heat energy to be transferred from a low-temperature reservoir to a high-temperature reservoir without the input of work energy (Figure 7.3). High-temperature reservoir Heat out No work Cyclic heat engine Impossible Heat in Low-temperature reservoir FIGURE 7.3 Clausius statement. An early nonmathematical version of the second law of thermodynamics, expressed in words, is the Clausius 2 statement of the second law. It is easily understood using the water wheel analogy. Another early nonmathematical verbalized version of the second law is by William Thomson (Lord Kelvin), later 3 modified by Max Planck (1858–1947). Both the Clausius and the Kelvin-Planck “statements” of the second law are really just verbalized consequences of the second law. Its development by recourse to water wheel technology is clearly wrong, since a water wheel is not a heat engine, but the conclusions are nonetheless correct. They have historical explanatory value but can- not be used in engineering design. The power of the second law becomes apparent only when it has a mathema- tical formulation. This comes later in the chapter. The significance of Carnot’s conclusions was not recognized until the 1850s, when Rudolph Clausius (1822–1888) and William Thomson (Lord Kelvin, 1824−1907) worked out a clear formulation of the conservation of energy principle, which was then named the first law of thermodynamics. Carnot’s first conclusion was then named the second law of thermodynamics by Clausius, who also expanded Carnot’s energy transformation concepts into a new property he called entropy. The classical Kelvin-Planck and Clausius statements of the second law also provide a classical means for defining the concept of reversibility. The Kelvin-Planck statement limits the efficiency of a heat engine to something less than 100%, but as yet, we have no idea how much less. To establish a more realistic efficiency limit, we need to define the simplest possible heat engine, one that operates with simple idealized processes. Such an engine would be frictionless and not have any losses. We call such an idealized engine reversible because the energy flow through it could be reversed without leaving any trace on the environment. Since the first law of thermodynamics is a conservative law (i.e., energy cannot be produced or destroyed), it has no effect on the reversibility of a system. But the second law is not a conservative law, since entropy is produced in every real process. Therefore, the second law is what dictates whether or not a system and its surroundings can be returned to their original states, and in general, this is not possible if entropy is produced by the process. Therefore, a reversible process is really synonymous with a zero entropy production process. Processes that are not reversible are called irreversible. Phenomena that cause processes to be irreversible are called process irreversibilities. Some typical process irreversibilities within a system are shown in the following table. 2 The word statement is used here, because it is a verbalized rather than mathematical form of the second law. 3 In the 1890s, Planck added the concept of “continuously operating” to Kelvin’s 1850s verbalized version of the second law, so this statement now has both names associated with it.7.4 Carnot’s Heat Engine and the Second Law of Thermodynamics 211 mechanical friction fluid viscosity electrical resistance shock waves mixing chemical reactions heat transfer plastic deformation hysteresis Thus, it is easy to see that all engineering processes of interest are really irreversible processes, and aside from a few mathematical formalities, we have very little need for the reversibility concept.However,wearenotyet capable of analyzing all the irreversibilities that occur in the complex real world, so we indeed need to model our device as reversible during the design stage, then build and test a prototype of it to determine the effect of KELVIN-PLANCK STATEMENT OF THE SECOND LAW 4 OF THERMODYNAMICS You cannot makea continuously operating heatenginethatconvertsall ofits heat inputdirectly intoworkoutput(Figure 7.4). High-temperature reservoir Heat in Cyclic heat Work out = Heat in engine Impossible No heat out Low-temperature reservoir FIGURE 7.4 Kelvin-Planck statement. 4 The Kelvin-Planck statement has many different written forms. For example, it can be expressed as “It is impossible to build a continuously operating device that produces a work output while absorbing heat from a single thermal reservoir,” and as “No heat engine can be more than 100% efficient.” Another form of the Kelvin-Planck statement is “It is impossible to build a continuously operating device that produces a work output while absorbing heat energy from a single thermal reservoir.” WHAT DOES THE WORD ENTROPY MEAN? Rudolph Clausius was German, and the word he chose for Carnot’s energy transformation concept was verwandlungsinhalt, meaning “transformation content.” Fortunately, in 1865, he chose to rename this concept by choosing the Greek word entropy, meaning simply “to change, or transform.” Later, there was an unsuccessful attempt to name an entropy unit the clausius, Cl, after him. It was defined as 1 Cl = 1 kcal/K = 4.186 J/K, but it was not universally accepted. REVERSIBLE PROCESSES: CLASSICAL DEFINITION A process is called reversible if, at any time, both the system and the environment can be returned to their original states. REVERSIBLE PROCESS: MODERN DEFINITION A reversible process is defined to be any process for which the entropy production or the entropy production rate for the ˙ process is zero (i.e., S = S = 0). P P212 CHAPTER 7: SecondLawofThermodynamicsandEntropyTransportandProductionMechanisms the inherent irreversibilities. We then correct for using the reversibility assumption in the design through an appropriate efficiency calculation. Though reversible processes do not actually exist in nature, they are conceptually necessary for creating perfor- mance limits for heat engine technology. William Thomson used Carnot’s second conclusion regarding maximum (i.e., reversible) engine energy conversion efficiency to develop the concept of an absolute temperature scale. 7.5 THE ABSOLUTE TEMPERATURE SCALE After studying the operation of steam engines for several years, Sadi Carnot concluded in 1824 that the efficiency of a heat engine depended only on the temperatures of the engine’s hot and cold thermal reservoirs and not on the fluid used inside the engine. In 1848, Thomson used Carnot’s conclusion to develop the concept of an absolute temperature scale. Soon afterward, an absolute temperature scale based on the size of the celsius degree (°C) became popular and was given his titled name kelvin (K) by his admirers. By using Eq. (4.70), we can define the thermal energy conversion efficiency (also called the thermal efficiency) η T of a continuously operating closed system heat engine with a net output work or power as _ ðW Þ ðW Þ out out net net η = = (7.5) T _ Q in Q in A closed system heat engine can operate continuously only if it operates in a thermodynamic cycle. A system that undergoes a thermodynamic cycle must end up at the same thermodynamic state at the end of the cycle as it started from at the beginning of the cycle. Because the total system energy E is a point function, the closed system first law of thermodynamics energy balance (EB) applied to a cyclic process yields ðÞ Q−W =ðÞ E −E = 0 (7.6) 2 1 cycle cycle Now, from Figure 7.5, we see that the heat input to a cyclic heat engine is ðÞ Q = Q − jQ j (7.7) in out cycle and High-temperature ðÞ W = W (7.8) out cycle thermal source at temperature T H where jQ j is the absolute value of this energy flow. out Q = Q in H Note thatwe introduce the correct sign with the absolutevalue of the symbol in Eqs. (7.7)and(7.8)toindicatetheproperflowdirection(+forheatinandworkout,and− Cyclic for heat out and work in). Normally we do not introduce the sign convention heat W out directly into the equations themselves. The usual custom is to attach the correctflow engine direction signto the numberand notthe symbol. However, wechange thisnotational Q = Q out L scheme here to help you understand the operation of closed system heat engines. Laterinthischapter,wereverttotheconventionalnotationschemeforalgebraicsigns. Low-temperature thermal sink Combining Eqs. (7.5) through (7.8) and using the simplified notation shown in at temperature T Figure 7.5 yields L ðW Þ ðÞ Q − jQ j jQ j jQ j FIGURE 7.5 out in out out L net η = = = 1− = 1− (7.9) T Q Q Q Q Schematic of a cyclic heat engine. in in in H IS IT K OR ºK? In 1967, the International Bureau of Weights and Measures dropped the prefix degree from the SI absolute temperature scale. So we say “100 degrees celsius is equal to 373.15 kelvin” (or 100.00°C = 373.15 K). Notice that we do not capitalize the terms Celsius and Kelvin, even though they are proper names. Remember that, in Chapter 1, we discussed why (a) we do not capitalize the first letter of a unit whose name is derived from that of a person when the unit’s name is written out and (b) the first letter is capitalized when the unit’s name is abbreviated. Also, in this book, we follow the same scheme of omitting the degree symbol on the Rankine absolute temperature scale, so 100.00°F = 559.67 R (and “559.67 R” is written out in lower case as “559.67 rankine”).7.5 The Absolute Temperature Scale 213 If we now follow Carnot’s lead and presume that the thermal efficiency of a reversible heat engine (η ) T rev depends only on the absolute temperatures of the thermal reservoirs, then we can write   jQ j jQ j out L ðÞ η = 1− = 1− T rev Q Q in H rev rev or   jQ j L T L 1−ðÞ η = = f (7.10) T rev Q T H H rev where f() is an unknown function thateventually is used to define the absolute temperature scale, and the subscripts L and Hrefertothelow-andhigh-temperature reservoirs, respectively. Now consider the two reversible heat engines connected in series shown in Figure 7.6. The thermal efficiency of each of the individual reversible heat engines is determined from an analysis of systems A and B individually. 5 The thermal efficiency of the engine in system A is  jQ j W T 2 A 2 ðÞ η = = 1− = 1−f T A Q Q T 1 1 1 and that in system B is  jQ j T W 3 B 3 ðÞ η = = 1− = 1−f T B Q Q T 2 2 2 Now, if we include both engines inside the system boundary, as in system C of Figure 7.6, then we have W = C W +W , and utilizing the previous results, we can write A B ðÞ Q −jQ j +ðÞ jQ j− jQ j W W +W 1 2 2 3 C A B ðÞ η = = = T C Q Q Q 1 1 1  jQ j jQ j− jQ j jQ j 2 2 3 2 = 1− + Q jQ j Q 1 2 1    (7.11) T T T 2 3 2 = 1−f + 1−f f T T T 1 2 1  T T 2 3 = 1−f f T T 1 2 We can also compute the heat engine thermal efficiency of system C as High-temperature  W Q − jQ j jQ j T 1 3 3 thermal source C 3 η ðÞ = = = 1− = 1−f T C at temperature T Q Q Q T 1 1 1 1 1 (7.12) Q 1 System Comparing Eqs. (7.11) and (7.12), we conclude that the Cyclic heat A following functional relation must hold for the unknown (W ) = W engine A out A (reversible) temperature function, f(): System   C T T T Q at temperature T 3 2 3 2 2 f ≡ f f (7.13) T T T 1 1 2 System Cyclic heat B Many common functions do not satisfy this equation. (W ) = W engine B out B (reversible) For example,  Q 3 T T T 3 2 3 sin ≠ sin sin T T T 1 1 2 Low-temperature  thermal sink T T T 3 2 3 log ≠ log log at temperature T 3 T T T 1 1 2  T T T 3 2 3 FIGURE 7.6 exp ≠ exp exp T T T 1 1 2 Two reversible heat engines connected in series. 5 Since these engines are defined at the outset to be reversible, the rev subscript on the η , Q, and W terms in these equations has T , been dropped for simplicity. This subscript reappears in the equations at the end of this analysis.214 CHAPTER 7: SecondLawofThermodynamicsandEntropyTransportandProductionMechanisms WHAT IS A CARNOT ENGINE? A Carnot engine is a reversible heat engine that operates between a high-temperature heat source at T and a low-temperature H heat sink at T . The thermal efficiency of a Carnot engine is simplyðÞ η = 1−T =T . L L H T Carnot It turns out that no one has ever actually made a running Carnot engine (although Rudolph Diesel thought he did, but he invented the Diesel engine instead). The reversible constant temperature heat transfers T and T require the engine to H L have infinite heat transfer surface areas, and this is not practical. The “concept” of a Carnot engine is important because it turns out that no other heat engine can have a thermal efficiency higher than that of a Carnot engine operating between the same two temperature limits. So the value of the Carnot engine is only as a benchmark with which to compare the thermal efficiencies of other actual operating heat engines. The Carnot cycle has become the universal standard by which the performance of other heat engine cycles can be measured. n and so forth. However, any simple power function of the formfTðÞ /T =ðÞ T /T does satisfy Eq. (7.13), since 3 1 3 1   n n n T T T 3 3 2 = T T T 1 1 2 The simplest such power function is a linear one (n = 1), and this is what Thomson chose to establish his absolute temperature scale. Therefore, if we take    T T T T T T 3 3 2 3 2 3 = f = = f f (7.14) T T T T T T 1 1 1 2 1 2 then Eq. (7.10) becomes   jQ j jQ j out L T L = = (7.15) Q Q T in H H rev rev It should be noted that Eq. (7.14) is not the only function that accurately defines an absolute temperature scale (but it is the simplest). Many other functions also work. However, they produce nonlinear temperature scales in which the size of the temperature unit is not constant but depends on the temperature level. This might be a useful technique to expand or condense a temperature scale in certain temperature regions, but the additional 6 complexity associated with a nonlinear temperature scale makes it generally unsuitable for common usage. Now, clearly, the maximum possible thermal energy conversion efficiency of any real irreversible closed system cyclic heat engine is equal to the thermal energy conversion efficiency that the same heat engine would have if it were somehow made to run reversibly like a Carnot engine. Then, from Eq. (7.9), T L ðÞ η =ðÞ η =ðÞ η = 1− (7.16) T max T rev T Carnot T H EXAMPLE 7.1 If a heat engine burns fuel for its thermal energy source and the combustion flame temperature is 4000.°F, determine the maximum possible thermal efficiency of this engine if it exhausts to the environment at 70.0°F. Solution First, draw a sketch of the system (Figure 7.7). The unknown is the maximum possible thermal energy conversion efficiency of any heat engine. The “maximum” efficiency occurs when an engine operates reversibly (i.e., with no internal losses due to friction, etc.). Since all reversible engines must have the same thermal energy conversion efficiency when operated between the same high- and low-temperature reservoirs, we can apply the results of the reversible Carnot engine analysis to this problem. Equation (7.16) gives the maximum possi- ble thermal efficiency as ð70:0+459:67ÞR T L ðη Þ =ðη Þ = 1− = 1− = 0:881 = 88:1% T max T Carnot T ð4000:+459:67ÞR H 6 It has been suggested that, since many thermal phenomena are inherently nonlinear, the use of a nonlinear (e.g., logarithmic) temperature scale might have some engineering merit.7.5 The Absolute Temperature Scale 215 Burning fuel The results of Example 7.1 are highly unrealistic since no real heat engine can ever be at 4000.°F reversible. The irreversibilities within modern heat engines limit their actual operating thermal energy conversion efficiency to around 30%. Heat from combustion Exercises Heat engine Work 1. Rework Example 7.1 for a flame temperature of 2500.°C and an environmental temperature of 20.0°C. Answer:(η ) = 89.4%. T max Exhaust 2. If the engine described inExample 7.1 has a maximum (reversible orCarnot)thermal v7 = ? efficiency of 60.0% when the environmental temperature is 70.0°F, determine the max Environment flame temperatureofthe combustion process.Answer: T = T = 1324R = 865°F. flame H possible at 70.0°F FIGURE 7.7 Example 7.1. EXAMPLE 7.2 A coal-fired electrical power plant produces 5.00 MW of electrical power while exhausting 8.00 MW of thermal energy to a nearby river at 10.0°C. The power plant requires an input power of 100. kW to drive the boiler feed pump. Determine a. The actual thermal efficiency of the power plant. b. The equivalent heat source temperature if the plant operated on a reversible Carnot cycle. Solution First, draw a sketch of the system (Figure 7.8). Q (from the combustion of coal) H Boiler Electrical Boiler feed W = 5.00 MW Turbine E generator pump W =−1.00 kW P Condenser Q (to river at 10.0°C) = 8.00 MW L FIGURE 7.8 Example 7.2. The unknowns are theactual thermal efficiency of thepower plant and theequivalent heat source temperature if theplant oper- atesona reversible Carnot cycle.If weconstructthe system boundaryasshown in the sketch, the power plant is a closed system. a. The actual thermal efficiency of this system is given by Eq. (7.9) as _ _ _ _ _ ðW Þ out W − jW j Q − jQ j net E P H L η = = = T _ _ _ Q Q Q in H H and the energy rate balance (ERB) for the steady state operation of this system is _ _ _ _ Q − jQ j−ðW − jW jÞ = 0 E P H L or _ _ _ _ Q = jQ j +ðW − jW jÞ H L E P = j−8:00MWj +ð5:00MW−j−0:100MWjÞ = 12:9MW (Continued)216 CHAPTER 7: SecondLawofThermodynamicsandEntropyTransportandProductionMechanisms EXAMPLE 7.2 (Continued) and thus the actual thermal efficiency is 5:00MW−j−0:100MWj 12:9MW−8:00MW η = = = 0:380 = 38:0% T 12:9MW 12:9MW b. From Eq. (7.16), we have T L ðη Þ =ðη Þ = 1− = 0:380 T max T Carnot T H so that T ð10:0+273:15ÞK L T = = H 1−0:380 0:620 = 457K = 184°C The calculations of part a are perfectly valid for this power plant since they deal with actual input and output energy values. The answer to part b, however, is unrealistically lower than the actual coal flame temperature in the boiler due to the many irreversibilities that exist within a real power plant. Exercises 3. If the combustion temperature of the power plant discussed in Example 7.2 were 2000.°C, determine the maximum (reversible or Carnot) thermal efficiency of the facility. Answer:(η ) = 87.5%. T max 7 4. If the heat transfer to the boiler in Example 7.2 were 3.50 × 10 Btu/h, the heat transfer from the condenser were 2.10 × 7 10 Btu/h, and the power into the boiler feed pump were 1.50 hp, determine (a) the power output from the turbine/ generator in MW and (b) the actual thermal efficiency of the power plant. Answers: (a) W = 4:10MW, (b)η = 40.%. act T 7.6 HEAT ENGINES RUNNING BACKWARD When a heat engine is run thermodynamically backward, it becomes a heat pump, a refrigerator, or an air condi- tioner, depending on your point of view. Figure 7.9 shows that, when a heat engine is thermodynamically reversed, the directions of all the energy flows are reversed. Thus, a work input W causes a thermal energy transfer Q from a in L low-temperature reservoir and a thermal energy transfer Q to a high-temperature reservoir. Consequently, the back- H ward running heat engine appears to “pump” heat from a low-temperature reservoir to a high-temperature reservoir. However, since heat is really a thermal energy transport phenomenon and not a fluid, it is somewhat misleading to refer to it as being “pumped.” Yet it is common practice in the heating, ventilating, and air conditioning (HVAC) industry to refer to these devices as heat pumps when they are used to provide a thermal energy transfer to a warm environment (e.g., a house)froma coldenvironment (e.g., the outsideair). High-temperature High-temperature Heat thermal source thermal source engine at temperature T at temperature T H H Heat pump Q H Q H Thermo Thermo W W out in cycle cycle Q L Refrigeration Q L or Low-temperature Low-temperature air conditioning thermal sink thermal sink at temperature T at temperature T L L (a) (b) FIGURE 7.9 (a) Heat engine; (b) thermodynamically reversed heat engine (heat pump, refrigerator, or air conditioner).7.6 Heat Engines Running Backward 217 IS THE EFFICIENCY OF HEAT PUMPS, AIR CONDITIONERS, AND REFRIGERATORS GREATER THAN 100%? If you look closely at the thermal efficiency equations for heat pumps, air conditioners, and refrigerators, you see that their efficiency is going to be more than 100%, because under normal operating conditions, the numerator in their efficiency equa- tion is usually greater than the denominator. Consequently, their energy conversion efficiency is usually greater than 100%. How can that be—nothing should have an energy conversion efficiency greater then 100%. But, it is correct. This is simply due to the way in which the thermal efficiency formula (Eq. (4.70)) is structured: Desiredenergyresult Energyconversionefficiency =η = (4.70) E Requiredenergyinput This makes a heat pump much more attractive for domestic heating than, say, a purely resistive electrical heater. Electrical hea- ters convert all their input electrical energy directly into thermal energy and therefore have energy conversion efficiencies of 100%, whereas most heat pumps have energy conversionefficienciesfar in excess of100%for the same electrical energyinput. Since this could be a problem in public advertising, the industry uses the phrase coefficient of performance (COP) instead of efficiency. The COP is simply the pure efficiency number before it is converted into a percentage. For example, the COP of a heat pump with an energy conversion efficiency of 450% is 4.5. _ _ jQ j jQ j jQ j jQ j H H H H COP =η = = = = (7.17) heat heat _ _ _ jW j jQ j−Q in H L jW j jQ j−Q in H L pump pump The desired energy result in the operation of a heat pump is heat addition to an already warm environment. Therefore, its energy conversion efficiency can be determined from Eq. (4.70) and an energy balance on the device (see Figure 7.9b) as _ _ Desiredenergyresult jQ j jQ j jQ j jQ j H H H H η = = = = = heat _ _ _ Requiredenergyinput jW j jQ j−Q jW j jQ j−Q in H L pump in H L where, as in the previous section, we use the absolute values of certain terms to avoid improper or confusing algebraic signs. If the heat pump is modeled as a backward running Carnot heat engine, then Eqs. (7.15) and (7.17) can be combined to yield the COP for a “reversible” (i.e., frictionless, etc.), or Carnot, heat pump as T H COP = (7.18) Carnot T −T H L heatpump If the removal of heat Q from a space is the desired result of a backward running heat engine, then the engine L is called a refrigerator when food is stored in the cooled space and an air conditioner when people occupy the cooled space. The energy conversion efficiency of a refrigerator or air conditioner can also be obtained from Eq. (4.70). As in the case of a heat pump, these efficiencies are also normally greater than 100% and they too are commonly represented with the pure number coefficient of performance label Desiredenergyresult COP =η = refrig: or refrig: or Requiredenergyinput aircond: aircond: (7.19) _ _ Q Q Q Q L L L L = = = = _ _ _ jW j jQ j−Q in H L jW j jQ j−Q in H L For a backward running Carnot (i.e., reversible) heat engine, Eqs. (7.15) and (7.19) can be combined to give the COP for a reversible refrigerator or air conditioner as T L COP = (7.20) Carnot T −T H L refrig:or aircond:218 CHAPTER 7: SecondLawofThermodynamicsandEntropyTransportandProductionMechanisms CRITICAL THINKING Notice that, the smaller the difference between T and T , the larger the COP defined in Eqs. (7.18) and (7.20) becomes. To H L understand this on a physical basis, note that, since the numerator is a fixed value (Q or Q ) in both equations and the L H denominator represents the net work input (W ) to the device, it stands to reason that, the smaller is the work input for a in given output, the better the efficiency (i.e., COP) is. Comparing Eqs. (7.17) and (7.19), we see that COP = COP +1 heat pump refrig: or air cond: EXAMPLE 7.3 The temperature outside on a hot summer day is 95°F. You would like your room to (COP) = ? be at70.°F,so you go out shopping for an air conditioner. If you are going to buy a Carnot Carnot air conditioner, what should be its coefficient of performance? AC Carnot Solution 70.°F 95.°F air First, draw a sketch of the system (Figure 7.10). conditioner Theunknownisthecoefficientofperformanceofanairconditioneroperatingbetween 70.°F and 95°F. Equation (7.20) gives the coefficient of performance for a Carnot air conditioneras T L 70:+459:67 COP = = = 21 Carnot T −T ð95+459:67Þ−ð70:+459:67Þ H L FIGURE 7.10 refrig:or aircond: Example 7.3. Note that, in the denominator, the temperature difference in this equation can be in eitherrelativeorabsolutetemperature unitsbecause(95+459.67)−(70.+459.67)= 95−70. =25°F orR.Thedegreesizeis thesameforboththeabsoluteand therelative temperaturescales. The coefficient of performance of a real air conditioner is sometimes called its energy efficiency rating (EER) and usually ranges between 3 and 9. The value of 21 calculatedin Example7.3 is unreasonably high, because a Carnot air conditioner is reversible and has no friction or other internal losses. Therefore, it requires less input work than a real air conditioner. Exercises 5. The refrigerator in your kitchen maintains a temperature difference of 2.00°C inside when the outside kitchen temperature is 22.0°C. If it is a Carnot refrigerator, what is its coefficient of performance? Answer: (COP) = 13.8. Carnot ref. 6. If the Carnot air conditioner purchased in Example 7.3 is inserted into the window backward during the winter and operated as a heat pump, determine its coefficient of performance as a heat pump. Answer: (COP) = 22. Carnot HP 7.7 CLAUSIUS’S DEFINITION OF ENTROPY Rudolph Clausius extended Thomson’s absolute temperature scale work by rearranging Eq. (7.15) to read ðÞ Q jQ j H L rev rev = T T H L and since this applies only to a closed system undergoing a thermodynamic cycle, it can also be written as  ðQ Þ jQ j Q H L rev rev ∑ = − = 0 T T T H L cycle If we now take an arbitrary thermodynamic closed system cycle and overlay it with an infinite number of infinitesimal heat engine cycles, as shown in Figure 7.11, then we can extend the finite summation process of the previous equation into a cyclic integral. Also, since each of these infinite number of heat engines is now7.7 Clausius’s Definition of Entropy 219 dT An arbitrary thermodynamic cycle for a An infinite closed system number of T infinitesimal heat engine cycles FIGURE 7.11 An infinite number of infinitesimal heat engine cycles approximating an arbitrary closed system thermodynamic cycle.  operating over an infinitely small temperature difference, T ≈ T = T andðÞ Q − jQ j ≈ dQ .Then,in H L H L rev rev rev the limit, the previous equation becomes "  I Q dQ lim ∑ = = 0 (7.21) n∞ T T n cycle cycles rev The temperature T in this equation is the absolute temperature at the point where the heat transfer dQ occurs. Clausius then noted the remarkable result that, since, by definition, I ðÞ Anythermodynamicpropertydifferential = 0 cycle the argument of the integral in Eq. (7.21) must define a thermodynamic property. That is,  dQ = Differentialof somethermodynamicproperty T rev  But, which property? The term dQ by itself is a path function and thus cannot be a thermodynamic property  rev differential. However, when dQ is divided by T, a property differential results. Clausius realized that he had rev 7 discovered a new thermodynamic property and he chose to name it entropy and represent the total entropy of a system by the symbol S, where  dQ dS = (7.22) T rev or Z 2 dQ S −S = (7.23) 2 1 T 1 rev or  Z 2 dQ S −S 1 2 1 s −s = = (7.24) 2 1 m m T 1 rev Be careful to note that Eqs. (7.22)–(7.24), which define entropy, are for a closed system of fixed mass m only. The effect of mass flow on system entropy is taken up in a separate section of this chapter. The use of a relative temperature scale in a grouping of units can sometimes be confusing. For example, when a temperature unit appears in the denominator of a units grouping, it can be written either as °F or R (or °CorK in SI) because only the degree size there is important. Therefore, Eq. (7.24) indicates that the units of specific entropy can be written correctly in either of the following forms: . . s½inBtu/ðÞ lbm °F≡s½inBtu/ðÞ lbm R 7 Here is a translation of how Clausius, in 1865, described why he chose the word entropy for the name of his new property. “We might call S the transformational content of the body, just as we termed the quantity U the heat and work content of the body. But since I believe it is better to borrow terms for important quantities from the ancient languages so that they may be adopted unchanged in all modern languages, I propose to call the quantity S the entropy of the body, from the Greek word ητροπή, meaning a transformation.” A220 CHAPTER 7: SecondLawofThermodynamicsandEntropyTransportandProductionMechanisms or . . s½inkJ/ðÞ kg °C≡s½inkJ/ðÞ kg K This does not mean that the °FandR(or °C and K) scales are equal but only that their degree sizes are equal. Therefore, when you have units like Btu/(lbm·ºF), you need not use any mathematical formula to convert °Fto R in order to write this grouping as Btu/(lbm·R). This is a simple but often confusing point. However, the temperature unit you choose to place in the denominator of a term’s units grouping may depend on how the term is to be used in relation to other temperature terms in the equation. An example of where this occurs is in the use of specific heats. Equation (3.15) is  ∂u c = (3.15) v ∂T v where c is the constant volume specific heat. v We discussed reversible processes briefly in Chapter 4 and noted that there are few reversible processes in the real world. In fact, every heat transport of energy through a finite temperature difference is irreversible. We are able to write Eqs. (7.21) through (7.24) as reversible heat transfers only because we created a very special situation, in which the heat transport of energy was assumed to take place through an infinitesimal temperature difference. But, in the real world, it would require an infinite amount of time to transport a finite amount of energy by this method. If we try to alter the results of Eq. (7.22) by considering only real irreversible heat transports of energy, we immediately realize that the amount of work done by the cyclic heat engines must be less than in the reversible case. Then, for an actual heat engine, W W actual reversible and using the first law of thermodynamics, we conclude that, since the system total energy E is a point function and therefore independent of whether the process path is reversible or irreversible, dE = dQ − dW = dQ − dW rev rev act act WHEN DO WE USE ºF (OR ºC) AND WHEN DO WE USE R (OR K)? Whether you use relative or absolute temperature units in an equation depends on whether the temperature appears in an equation as a difference or stands alone. For example, assuming c is a constant, integrating Eq. (3.15) gives u − u = c (T − T ), v 2 1 v 2 1 and since the temperature appears as a difference here, we can use either ºF or R (or ºC or K) temperature units, because (T in ºF + 459.67 R) − (T in ºF + 459.67 R) = (T in ºF − T in ºF), as the conversion from ºF to R cancels out. You 1 2 1 2 can use either ºF or R and you get the same answer in each case. Also, the numerical value of c in Btu/(lbm·ºF) has the same value in Btu/(lbm·R). For example, for air, c = 0.172 Btu/ v v (lbm·ºF) = 0.172 Btu/(lbm·R). That is because the temperature unit appears in the denominator as “per degree,” and the Fahrenheit degree is the same size as a Rankine degree (only their zero point is different). Similarly, c = 0.718 kJ/(kg·K) = v 8 0.718 kJ/(kg·ºC) for the same reason. This also applies to numerical values of c , entropy s, and the gas constant R =ℛ/M. p However, in Eqs. (7.18) and (7.20), the temperature stands alone in the numerator, but the denominator has a tempera- ture difference. What do you do now? T H COP = (7.18) Carnot T −T H L heat pump T L COP = (7.20) Carnot T −T H L refrig:or aircond: Theruleisthat,wheneveryouhaveanequationinwhichthetemperature T stands alone (and not as a temperature differ- ence), the temperature must always be in an absolute unit (R or K). So the numerators on Eqs. (7.18) and (7.20) must be in absolute temperature units (R or K), but since the denominator has a temperature difference, the temperatures here can be in either relative or absolute temperature units.Ifyouareunsurewhethertouseabsoluteorrelativetemperatureunitsinan equation, use absolute temperature units, since they always give the correct answer. 8 But be careful if you use a table with c or s in Btu/(lbm·ºF) because you might be tempted to use T in °F to cancel the temperature unit. This would be v incorrect. You have to understand that c or s in Btu/(lbm·ºF) has the same numerical value as c or s in Btu/(lbm·R). v v7.8 Numerical Values for Entropy 221 WHAT ARE PERPETUAL MOTION MACHINES? Devices that supposedly operate using processes that violate either the first or second laws of thermodynamics or that are required to be reversible represent various forms of perpetual motion machines. When the operation of a device depends on the violation of the first law of thermodynamics it is called a perpetual motion machine of the first kind (e.g., a heat engine that produces power but does not absorb heat from the environment). When the operation of a device depends on the viola- tion of the second law, it is called a perpetual motion machine of the second kind (e.g., an adiabatic air compressor in which the air exits at a lower temperature than it entered), and when it requires a reversible process to operate it is called a perpetual motion machine of the third kind (e.g., a wheel on a shaft that, once started, continues to rotate indefinitely). No perpetual motion machines operate as proposed and have for centuries been the source of frauds brought on the unsuspecting public by unscrupulous or naive inventors. and dividing by the appropriate absolute temperature and rearranging gives     dW − dW dQ dQ rev act dS = = + (7.25) T T T rev act   For a work-producing heat engine, dW ≤ dW and both are positive work quantities, since they represent act rev energy leaving the system; Eq. (7.25) can be rearranged to produce  dQ dS (7.26) T act Equation (7.26) is known as the Clausius inequality.ItisClausius’s mathematical form of the second law of thermodynamics for a closed system. Dropping the subscript on the bracketed term and thus allowing it to represent either a reversible or actual process produces the following somewhat more general mathematical second law expression:  dQ dS≥ (7.27) T and I  dQ ≤0 (7.28) T cycle where the equality sign is used for a reversible heat transport of energy. 7.8 NUMERICAL VALUES FOR ENTROPY In Chapter 3, we discussed five methods for finding numerical values for properties: thermodynamic equations of state, thermodynamic tables, thermodynamic charts, direct experimental measurements, and the formulae of statistical thermodynamics. The same five methods can be used to find numerical values for the specific entropy. In this section, we focus on the use of thermodynamic equations of state, tables, and charts. Energy and entropy are thermodynamic properties and therefore mathematical point functions. Consequently, the energy and entropy changes of a system depend only on the beginning and ending states of a process and not on the actual thermodynamic path taken by the process between these states. Therefore, for a closed system, we can write the differential energy and entropy balances as     ðÞ dE =ðÞ dE = dE = dQ − dW = dQ − dW rev act rev rev act act and  dQ ðdSÞ =ðdSÞ = dS = rev act T rev Combining the “reversible” path parts of these two equations, we get   dQ = TdS = dE+ dW (7.29) rev rev222 CHAPTER 7: SecondLawofThermodynamicsandEntropyTransportandProductionMechanisms For a stationary differential closed system at a uniform temperature T containing a pure substance that is subjected to only a mechanical moving boundary work mode, Eq. (7.29) becomes TdS = dU+pdV and on dividing through by the system mass m and the absolute temperature T, p du + ds = dv (7.30) T T Since u = h− pv, this equation can also be written as dh v ds = − dp (7.31) T T InChapter3,wedefinetheconstant volume and constant pressure specific heats for an incompressible substance as du c = c = c = v p dT Since v = constant and dv = 0 for an incompressible material, then Eq. (7.30) becomes  dT ðÞ ds = c incomp: T or Z  T 2 dT ðÞ s −s = c (7.32) 2 1 incomp: T T 1 If the specific heat c is constant over the temperature range from T to T , then this equation can be integrated 1 2 to give ðÞ s −s = clnðÞ T /T (7.33) 2 1 2 1 incompressiblematerial withaconstantc In Chapter 3, we also define the constant volume and constant pressure specific heats for an ideal gas as du c = (3.37) v dT and dh c = (3.40) p dT Consequently, we can now write Eqs. (7.30) and (7.31) as   p dT dT v ðÞ ds = c + dv = c − dp ideal v p T T T T gas Further, for an ideal gas, p/T = R/v and v/T = R/p, so this equation can be integrated to give Z 2 dT v 2 ðÞ s −s = c +R ln (7.34) 2 1 ideal v T v 1 gas 1 Z 2 p dT 2 = c −R ln (7.35) p T p 1 1 and if the specific heats are constant over the temperature range from T to T , then these equations become 1 2 T v 2 2 ðs −s Þ = c ln +R ln (7.36) 2 1 v idealgas T v 1 1 constant c andc p v T p 2 2 = c ln −R ln (7.37) p T p 1 17.8 Numerical Values for Entropy 223 DO ALL ELASTIC MATERIALS HAVE ENTROPIC ELASTICITY? No,most elasticsolidsdonothaveentropicelasticity,buttheelasticitypresent inrubberandpolymersislargelyentropic.When workis doneadiabatically ona materialwith entropic elasticity,thetemperature ofthe materialincreases.You can demonstrate thisbystretchinga rubberbandrapidly thenimmediately touchingittoyourlips(which are verysensitivetotemperature).The rubber band is warmer than it was before it was stretched. Then, if you hold the stretched rubber band long enough for it to returntoroomtemperatureandsuddenlyreleaseitandtouchit toyourlips,itiscolderthanitwasbeforeitwasstretched. When an elastic deformation produces a decrease in the specific entropy of a material, it is said to have entropic elasticity. In the case of an ideal gas, Eq. (7.36) shows that an isothermal (T = T ) compression (v v ) 2 1 2 1 produces a decrease in the specific entropy of the gas. Consequently, ideal gases have entropic elasticity. Any process in which entropy remains constant is called an isentropic process. The term isentropic comes from the Greek words for “constant entropy.” If an ideal gas with constant specific heats undergoes an isentropic process, then s − s and Eqs. (7.36) and 2 1 (7.37) give p T R v R 2 2 2 ln = − ln = ln T c v c p 1 v 1 p 1 or For an isentropic process with an ideal gas,   1−k ðÞ k−1 /k p T v 2 2 2 = = (7.38) T v p 1 1 1 and k k p v = p v = constant (7.39) 1 2 1 2 where k = c /c and R = c −c (7.40) p v p v Consequently, from Eq. (4.27), we see that, in the case of an ideal gas with constant specific heats, an isentropic process is the same as a polytropic process with n = k. These equations for the specific entropy of an incompressible substance and an ideal gas are the only such for- mulae to be introduced at this point. Specific entropy equations for more complex substances are introduced later in the text as they are needed. EXAMPLE 7.4 An insulated apparatus contains 1.5 kg of saturated liquid water at 20.°C. Determine the change in specific entropy of the water as it is pressurized from 0.10 MPa to 10. MPa. Assume the liquid water is an incompressible material. Solution First, draw a sketch of the system (Figure 7.12). The unknown is the change in specific entropy, s − s , for the system. The material is liquid water. 2 1 An energy balance for this process gives Q − W =muðÞ −u 1 2 1 2 2 1 Z and since the apparatus is insulated Q = 0. The only possible work mode here is moving boundary work, so W = pdV. 1 2 1 2 But the water is to taken as incompressible, so V = constant and dV = 0. Also, Eq. (3.33) gives the specific internal energy change of an incompressible material as u −u =cTðÞ −T ,where c is the specific heat of the material. Then, the energy 2 1 2 1 balance equation gives for this process 0−0 = mcðÞ T −T 2 1 (Continued)224 CHAPTER 7: SecondLawofThermodynamicsandEntropyTransportandProductionMechanisms EXAMPLE 7.4 (Continued) Insulation m = 1.5 kg m = 1.5 kg x = 0 1 p = 0.10 MPa 2 T = 20.°C 1 ∀ =∀ 2 1 p = 0.10 MPa 1 s = ? 2 s = ? 1 State 1 State 2 Constant volume process s − s = ? 2 1 FIGURE 7.12 Example 7.4. or, T = T . Therefore, the process must also be isothermal, and Eq. (7.33) gives the specific entropy change as 2 1 s −s = clnðT =T Þ = clnð1Þ = 0 2 1 2 1 Consequently, the entropy of an incompressible material is not altered by changing its pressure. EXAMPLE 7.5 An apparatus contains 0.035 kg of air (an ideal gas). The apparatus is used to compress the air isentropically (i.e., at constant entropy) from a pressure of 0.100 MPa to a pressure of 5.00 MPa. If the initial temperature of the air is 20.0°C, determine the final temperature and specific volume of the air. Solution First, draw a sketch of the system (Figure 7.13). Air Air m = 0.035 kg m = 0.035 kg p = 5.00 MPa 2 p = 0.100 MPa 1 T = ? 2 T = 20.0°C 1 v = ? 2 State 1 State 2 Isentropic process FIGURE 7.13 Example 7.5. The unknowns are the final temperature, T , and specific volume, v , of the air in the system. Since p = 0.100 MPa, 2 2 1 T = 20.0°C, and p = 5.00 MPa, then Eq. (7.38) can be used to find T and v as follows. From Thermodynamic Tables to 1 2 2 2 accompany Modern Engineering Thermodynamics, Table C.13b, we find for air that k = 1.4; and solving Eq. (7.38) for T gives 2 k−1  0:4  p 2 k 5:00MPa 1:4 T = T =ð20:0+273:15KÞ = 896K = 623°C 2 1 p 0:100MPa 1 Using the ideal gas equation of state and Table C.13b for the gas constant of air (R = 0.286 kJ/kg·K), we find the initial air specific volume of the air to be v = mRT /p 1 1 1 2 3 . =ð0:035kgÞð0:286kJ=kg KÞð20:0+273:15KÞ/ð0:100×10 kN=m Þ 3 =kg = 0:02934m Then, solving Eq. (7.38) for v gives 2  1 1   − T 1−k 623+273:15 0:4 2 3 3 v = v =ð0:02934m =kgÞ = 0:00180m =kg 2 1 T 20:0+273:15 1 Exercises 7. Determine the change in specific entropy as a 1.00 kg block of solid incompressible iron is heated from 20.0°C to 100.°C. (See Table 3.6 for specific heat values.) Answer:(s −s ) = 0.108 kJ/kg·K. 2 1 iron 8. Determine the change in specific entropy of air as it is heated from 20.0°C to 100.°C in a constant pressure (isobaric) process. Assume air behaves as an ideal gas. Answer:(s −s ) = 0.242 kJ/kg·K. 2 1 isobaric air 9. Determine the change in specific entropy of air as it is heated from 20.0°C to 100.°C in a constant volume (isochoric) process. Assume air behaves as an ideal gas. Answer:(s −s ) = 0.173 kJ/kg·K. 2 1 isochoric air

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