Chapter 5: The Second Law of Thermodynamics PDF

Chapter 5 The Second Law of Thermodynamics Thermodynamics treats the principles of energy transformations, and the laws of thermodynamics establish the bounds within which these transformations are observed to occur. The first law states the principle of energy conservation, leading to energy balances in which work and heat are included as simple additive terms. Yet work and heat are otherwise quite different. Work is directly useful in ways that heat is not, e.g., for the elevation of a weight or for the acceleration of a mass. Evidently, work is a form of energy intrinsically more valuable than an equal quantity of heat. This difference is reflected in a second fundamental law which, together with the first law, makes up the foundation upon which the edifice of thermodynamics is built. The purpose of this chapter is to: ∙ Introduce the concept of entropy, an essential thermodynamic property ∙ Present the second law of thermodynamics, which reflects the observation that limits exist to what can be accomplished even by reversible processes ∙ Apply the second law to some familiar processes ∙ Relate changes in entropy to T and P for substances in the ideal-gas state ∙ Present entropy balances for open systems ∙ Demonstrate the calculation of ideal work and lost work for flow processes ∙ Relate entropy to the microscopic world of molecules 5.1 AXIOMATIC STATEMENTS OF THE SECOND LAW The two axioms presented in Chapter 2 in relation to the first law have counterparts with respect to the second law. They are: Axiom 4: There exists a property called entropy1 S, which for systems at internal equilibrium is an intrinsic property, functionally related to the 1Pronounced en’-tro-py to distinguish it clearly from en-thal’-py. 173 174 CHAPTER 5. The Second Law of Thermodynamics measurable state variables that characterize the system. Differential changes in this property are given by the equation:  dS  t= dQ rev / T (5.1) where St is the system (rather than the molar) entropy. Axiom 5: (The Second Law of Thermodynamics) The entropy change of any system and its surroundings, considered together, and resulting from any real process, is positive, approaching zero when the process approaches reversibility. Mathematically,  ΔS total ≥ 0  (5.2) The second law affirms that every process proceeds in such a direction that the total entropy change associated with it is positive, the limiting value of zero being attained only by a reversible process. No process is possible for which the total entropy decreases. The practical utility of the second law is illustrated by application to two very common processes. The first shows its consistency with our everyday experience that heat flows from hot to cold. The second shows how it establishes limits to the conversion of heat to work by any device. Application of the Second Law to Simple Heat Transfer First, consider direct heat transfer between two heat reservoirs, bodies imagined capable of absorbing or rejecting unlimited quantities of heat without temperature change.2 The equation for the entropy change of a heat reservoir follows from Eq. (5.1). Because T is constant, integration gives: Q ΔS = __     T A quantity of heat Q is transferred to or from a reservoir at temperature T. From the res- ervoir’s point of view the transfer is reversible, because its effect on the reservoir is the same regardless of source or sink of the heat. Let the temperatures of the reservoirs be TH and TC with TH > TC. Heat quantity Q, transferred from one reservoir to the other, is the same for both reservoirs. However, QH and QC have opposite signs: positive for the heat added to one reservoir and negative for the heat extracted from the other. Therefore QH = −QC, and the entropy changes of the reservoirs at TH and at TC are: t QH   − Q C t Q C ΔS  H   = ___     = _____    and ΔS  C   = ___     T H T H T C These two entropy changes are added to give: ( T HT C ) t t − Q C ___ Q C T H − T C ΔS total = ΔS  H   + ΔS  C   = _____     +     = QC _______     T H T C 2The firebox of a furnace is in effect a hot reservoir, and the surrounding atmosphere, a cold reservoir. 5.1. Axiomatic Statements of the Second Law 175 Because the heat-transfer process is irreversible, Eq. (5.2) requires a positive value for ΔStotal, and therefore Q C(T H − T C) > 0 With the temperature difference positive, QC must also be positive, which means that heat flows into the reservoir at TC, i.e., from the higher to the lower temperature. This result conforms to universal experience that heat flows from higher to lower temperature. A formal statement conveys this result: No process is possible which consists solely of the transfer of heat from one temperature level to a higher one. Note also that ΔStotal becomes smaller as the temperature difference decreases. When TH is only infinitesimally higher than TC, the heat transfer is reversible, and ΔStotal approaches zero. Application of the Second Law to Heat Engines Heat can be used far more usefully than by simple transfer from one temperature level to a lower one. Indeed, useful work is produced by countless engines that employ the flow of heat as their energy source. The most common examples are the internal-combustion engine and the steam power plant. Collectively, these are heat engines. They all rely on a high-temperature source of heat, and all discard heat to the environment. The second law imposes restrictions on how much of their heat intake can be converted into work, and our object now is to establish quantitatively this relationship. We imagine that the engine receives heat from a higher-temperature heat reservoir at TH and discards heat to a lower-temperature reservoir TC. The engine is taken as the system and the two heat reservoirs comprise the surroundings. The work and heat quantities in relation to both the engine and the heat reservoirs are shown in Fig. 5.1(a). With respect to the engine, the first law as given by Eq. (2.3) becomes: ΔU = Q + W = QH + QC+ W Because the engine inevitably operates in cycles, its properties over a cycle do not change. Therefore ΔU = 0, and W = −QH − QC. The entropy change of the surroundings equals the sum of the entropy changes of the reservoirs. Because the entropy change of the engine over a cycle is zero, the total entropy change is that of the heat reservoirs. Therefore QH QC ΔStotal= − ___ − ___ TH TC Note that QC with respect to the engine is a negative number, whereas QH is positive. Combining this equation with the equation for W to eliminate QH yields: ( TC ) TH− TC W = THΔStotal+ QC ______ 176 CHAPTER 5. The Second Law of Thermodynamics Hot Reservoir Hot Reservoir QH QH W W QC QC Cold Reservoir Cold Reservoir (a) (b) Figure 5.1: Schematic diagrams: (a) Carnot engine; (b) Carnot heat pump or refrigerator. This result gives the work output of a heat engine within two limits. If the engine is totally ineffective, W = 0; the equation then reduces to the result obtained for simple heat transfer between the two heat reservoirs, i.e.: ( THTC ) TH− TC ΔStotal= − QC ______ The difference in sign here simply reflects the fact that QC is with respect to the engine, whereas previously it was with respect to the lower-temperature reservoir. If the process is reversible in all respects, then ΔStotal = 0, and the equation reduces to: ( T C ) T H W = QC ___     − 1 (5.3) A heat engine operating as described in a completely reversible manner is very special and is called a Carnot engine. The characteristics of such an ideal engine were first described by N. L. S. Carnot3 in 1824. Note again that QC is a negative number, as it represents heat transferred from the engine. This makes W negative, in accord with the fact that work is not added to, but is produced by, the engine. Clearly, for any finite value of W, QC is also finite. This means that a portion of the heat transferred from the higher-temperature reservoir must inevitably be exhausted to the lower-temperature reservoir. This observation can be given formal statement: It is impossible to construct an engine that, operating in a cycle, produces no effect (in system and surroundings) other than the extraction of heat from a reservoir and the performance of an equivalent amount of work. 3Nicolas Leonard Sadi Carnot (1796−1832), French engineer. See http://en.wikipedia.org/wiki/Nicolas_Leonard_ Sadi_Carnot. 5.1. Axiomatic Statements of the Second Law 177 The second law does not prohibit the continuous production of work from heat, but it does place a limit on how much of the heat taken into a cyclic process can be converted into work. Combining this equation with W = −QH − QC to eliminate first W and then QC leads to Carnot’s equations: − QC ___ ____ QH = (5.4) TC TH W ___ TC = ___ − 1 (5.5) QH TH Note that in application to a Carnot engine QH, representing heat transferred to the engine, is a positive number, making the work produced (W) negative. In Eq. (5.4) the small- est possible value of QC is zero; the corresponding value of TC is absolute zero on the Kelvin scale, which corresponds to −273.15°C. The thermal efficiency of a heat engine is defined as the ratio of the work produced to the heat supplied to the engine. With respect to the engine, the work W is negative. Thus, −W η ≡ ___ (5.6) QH In view of Eq. (5.5) the thermal efficiency of a Carnot engine is: TC η  Carnot = 1 − ___ (5.7) TH Although a Carnot engine operates reversibly in all respects, and cannot be improved, its efficiency approaches unity only when TH approaches infinity or TC approaches zero. Neither condition exists on earth; all terrestrial heat engines therefore operate with thermal efficiencies less than unity. The cold reservoirs available on earth are the atmosphere, lakes, rivers, and oceans, for which TC ≃ 300 K. Hot reservoirs are objects such as furnaces where the temperature is maintained by combustion of fossil fuels or by fission of radioactive elements, and for which TH ≃ 600 K. With these values, η = 1 − 300/600 = 0.5, an approximate realistic limit for the thermal efficiency of a Carnot engine. Actual heat engines are irreversible, and η rarely exceeds 0.35. Example 5.1 A central power plant, rated at 800,000 kW, generates steam at 585 K and discards heat to a river at 295 K. If the thermal efficiency of the plant is 70% of the maximum possible value, how much heat is discarded to the river at rated power? Solution 5.1 The maximum possible thermal efficiency is given by Eq. (5.7). With TH as the steam-generation temperature and TC as the river temperature: 295 η  Carnot = 1 − ____ = 0.4957 and η = (0.7)(0.4957) = 0.3470 585 178 CHAPTER 5. The Second Law of Thermodynamics where η is the actual thermal efficiency. Combining Eq. (5.6) with the first law, written W = −QH − QC, to eliminate QH, yields: ( η ) ( 0.347 ) 1−η 1 − 0.347 QC= ___ W = _______ (− 800,000) = −1,505,475 kW This rate of heat transfer to a modest river would cause a temperature rise of several °C. 5.2 HEAT ENGINES AND HEAT PUMPS The following steps make up the cycle of any Carnot engine: ∙ Step 1: A system at an initial temperature of a cold reservoir at TC undergoes a reversible adiabatic process that causes its temperature to rise to that of a hot reservoir at TH. ∙ Step 2: The system maintains contact with the hot reservoir at TH and undergoes a reversible isothermal process during which heat QH is absorbed from the hot reservoir. ∙ Step 3: The system undergoes a reversible adiabatic process in the opposite direction of Step 1 that brings its temperature back to that of the cold reservoir at TC. ∙ Step 4: The system maintains contact with the reservoir at TC, and undergoes a reversible isothermal process in the opposite direction of Step 2 that returns it to its initial state with rejection of heat QC to the cold reservoir. This set of processes can in principle be performed on any kind of system, but only a few, yet to be described, are of practical interest. A Carnot engine operates between two heat reservoirs in such a way that all heat absorbed is transferred at the constant temperature of the hot reservoir and all heat rejected is transferred at the constant temperature of the cold reservoir. Any reversible engine operating between two heat reservoirs is a Carnot engine; an engine operating on a different cycle must necessarily transfer heat across finite temperature differences and therefore cannot be reversi- ble. Two important conclusions inevitably follow from the nature of the Carnot engine: ∙ Its efficiency depends only on the temperature levels and not upon the working substance of the engine. ∙ For two given heat reservoirs no engine can have a thermal efficiency higher than that of a Carnot engine. We provide further treatment of practical heat engines in Chapter 8. Because a Carnot engine is reversible, it may be operated in reverse; the Carnot cycle is then traversed in the opposite direction, and it becomes a reversible heat pump operating between the same temperature levels and with the same quantities QH, QC, and W as for the engine but reversed in direction, as illustrated in Fig. 5.1(b). Here, work is required, and it is used to “pump” heat from the lower-temperature heat reservoir to the higher-temperature heat reservoir. Refrigerators are heat pumps with the “cold box” as the lower-temperature reservoir and some portion of the environment as the higher-temperature reservoir. The usual measure 5.3. Carnot Engine with Ideal-Gas-State Working Fluid 179 of quality of a heat pump is the coefficient of performance, defined as the heat extracted at the lower temperature divided by the work required, both of which are positive quantities with respect to the heat pump: QC ω ≡ ___ (5.8) W For a Carnot heat pump this coefficient can be obtained by combining Eq. (5.4) and Eq. (5.5) to eliminate QH: TC ωCarnot≡ ______ (5.9) TH− TC For a refrigerator at 4°C and heat transfer to an environment at 24°C, Eq. (5.9) yields: 4 + 273.15 ωCarnot= ________ = 13.86 24 − 4 Any actual refrigerator would operate irreversibly with a lower value of ω. The practical aspects of refrigeration are treated in Chapter 9. 5.3 CARNOT ENGINE WITH IDEAL-GAS-STATE WORKING FLUID The cycle traversed by a working fluid in its ideal-gas state in a Carnot engine is shown on a PV diagram in Fig. 5.2. It consists of four reversible processes corresponding to Steps 1 through 4 of the general Carnot cycle described in the preceding section: ∙ a → b Adiabatic compression with temperature rising from TC to TH. ∙ b → c Isothermal expansion to arbitrary point c with absorption of heat QH. ∙ c → d Adiabatic expansion with temperature decreasing to TC. ∙ d → a Isothermal compression to the initial state with rejection of heat QC. In this analysis the ideal-gas-state working fluid is regarded as the system. For the iso- thermal steps b → c and d → a, Eq. (3.20) yields: V  cig  V  aig  Q H = RT H ln ___   ig  and Q C = RT C ln ___   ig  V  b  V  d  Dividing the first equation by the second gives: QH ___ H ( c b) T ln Vig / Vig = ____________    QC TCln (Vaig/ Vdig) For an adiabatic process Eq. (3.16) with dQ = 0 becomes, C  ig  dT dVig − ____   V ___ = ____ ig R T V 180 CHAPTER 5. The Second Law of Thermodynamics TH b QH Figure 5.2: PV diagram showing a P Carnot cycle for a working fluid in the ideal-gas state. c TC TH a d QC TC V For steps a → b and c → d, integration gives: T H T H ig C  ig  dT V  aig  C    dT V  ig  ∫ T C R T ∫ T C R T   ___   V   ___  = ln ___   ig  and   ___   V   ___  = ln ___   dig  V  b  V  c  Because the left sides of these two equations are the same, the adiabatic steps are related by: Vig Vaig ig Vc Va ig ln ___ dig = ln ___ ig or ln ___ = − ln ___ Vc Vb Vbig Vdig Combining the second expression with the equation relating the two isothermal steps gives: QH ___ TH − QC QH = − ___ or ____ = ___ QC TC TC TH This last equation is identical with Eq. (5.4). 5.4 ENTROPY Points A and B on the PVt diagram of Fig. 5.3 represent two equilibrium states of a particular fluid, and paths ACB and ADB represent two arbitrary reversible processes connecting these points. Integration of Eq. (5.1) for each path gives: ∫ ACB T dQrev ∫ ADB T _____ dQrev _____ ΔSt=     and ΔSt=     Rev. Confirming Pages 5.4. Entropy 181 D B P Figure 5.3: Two reversible paths A joining equilibrium states A and B. C Vt t t Because ΔSt is a property change, it is independent of path and is given by SB− SA. If the fluid is changed from state A to state B by an irreversible process, the entropy t t change is again ΔSt = SB − SA, but experiment shows that this result is not given by ∫dQ / Tevaluated for the irreversible process itself, because the calculation of entropy changes by this integral must, in general, be along reversible paths. The entropy change of a heat reservoir, however, is always given by Q/T, where Q is the quantity of heat transferred to or from the reservoir at temperature T, whether the transfer is reversible or irreversible. As noted earlier, the effect of heat transfer on a heat reservoir is the same regardless of the temperature of the source or sink of the heat. If a process is reversible and adiabatic, dQrev = 0; then by Eq. (5.1), dSt = 0. Thus the entropy of a system is constant during a reversible adiabatic process, and the process is said to be isentropic. The characteristics of entropy may be summarized as follows: ∙ Entropy relates to the second law in much the same way that internal energy relates to the first law. Equation (5.1) is the ultimate source of all equations that connect entropy to measurable quantities. It does not represent a definition of entropy; there is none in the context of classical thermodynamics. What it provides is the means for calculating changes in this property. ∙ The change in entropy of any system undergoing a finite reversible process is given by the integral form of Eq. (5.1). When a system undergoes an irreversible process between two equilibrium states, the entropy change of the system Δ S  tis evaluated by application of Eq. (5.1) to an arbitrarily chosen reversible process that accomplishes the same change of state as the actual process. Integration is not carried out for the irreversible path. Because entropy is a state function, the entropy changes of the irreversible and reversible processes are identical. ∙ In the special case of a mechanically reversible process (Sec. 2.8), the entropy change of the system is correctly evaluated from ∫dQ / Tapplied to the actual process, even though the heat transfer between system and surroundings represents an external irreversibility. The reason is that it is immaterial, as far as the system is concerned, smi96529_ch05_173-209.indd 181 06/19/17 12:21 PM 182 CHAPTER 5. The Second Law of Thermodynamics whether the temperature difference causing the heat transfer is infinitesimal (making the process externally reversible) or finite. The entropy change of a system result- ing only from the transfer of heat can always be calculated by ∫ dQ / T,whether the heat transfer is accomplished reversibly or irreversibly. However, when a process is irreversible on account of finite differences in other driving forces, such as pressure, the entropy change is not caused solely by the heat transfer, and for its calculation one must devise a mechanically reversible means of accomplishing the same change of state. 5.5 ENTROPY CHANGES FOR THE IDEAL-GAS STATE For one mole or a unit mass of fluid undergoing a mechanically reversible process in a closed system, the first law, Eq. (2.7), becomes: dU = dQrev − PdV Differentiation of the defining equation for enthalpy, H = U + PV, yields: dH = dU + PdV + V dP Eliminating dU gives: dH = dQrev − PdV + PdV + VdP or dQrev = dH − VdP For the ideal-gas state, dHig = CPigdT and Vig = RT / P. With these substitutions and division by T, dQrev _____ dT dP = CPig___ − R ___ T T P As a result of Eq. (5.1), this becomes: dT dP dSig CPig___ dT dSig= CPig___ − R ___ or ____= ___ − d ln P T P R R T where Sig is the molar entropy for the ideal-gas state. Integration from an initial state at condi- tions T0 and P0 to a final state at conditions T and P gives: ΔS  ig T C  ig  dT ∫ T 0 R T P  ____  =   ___   P   ___   − ln ___     (5.10) R P 0 Although derived for a mechanically reversible process, this equation relates properties only, independent of the process causing the change of state, and is therefore a general equation for the calculation of entropy changes in the ideal-gas state. 5.5. Entropy Changes for the Ideal-Gas State 183 Example 5.2 For the ideal-gas state and constant heat capacities, Eq. (3.23b) for a reversible adia- batic (and therefore isentropic) process can be written: T  1 ( P  1 ) (γ−1)/γ ___T  2 P  2     = ___       Show that this same equation results from application of Eq. (5.10) with ΔSig = 0. Solution 5.2 Because CPigis constant, Eq. (5.10) becomes: Cig T2 P2 T2 R P2 0 = ___ P ln ___ − ln ___ = ln ___ − ___ ig ln ___ R T1 P1 T1 CP P1 By Eq. (3.12) for the ideal-gas state, with γ = CPig/ CVig: R γ−1 CPig= CVig+ R or ___ = ___ ig CP γ Whence, T2 γ − 1 ___ P2 ln ___= ____ ln T1 γ P1 Exponentiating both sides of this equation leads to the given equation. Equation (4.5) for the temperature dependence of the molar heat capacity CPig allows integration of the first term on the right of Eq. (5.10). The result can be written as: + B + (C + 2 2 )( 2 ) (T − T0) T0 [ ] T CigdT T D _____ T + T0 _____ ∫ ___ P ___ = A ln ___ (5.11) T0 R T T 0 T As with the integral ∫(CP/R)dT of Eq. (4.8), this integral is often evaluated; for computational purposes we define the right side of Eq. (5.11) as the function ICPS(T0, T; A, B, C, D) and presume the existence of a computer routine for its evaluation.4 Equation (5.11) then becomes: T CigdT ∫ T 0 R T   ___ P ___   = ICPS(T0, T; A, B, C, D) Also useful is a mean heat capacity, here defined as: T ∫ T    C  Pig  dT /T  ⟨CP⟩  = __________ ig   0   (5.12) S ln (T /T 0) In accord with this equation, division of Eq. (5.11) by ln (T/T0) yields: ⟨ P ⟩S = A + B + (C +   2 2 )(  Cig   [ ) ] ln (T / T0) _____ T + T 0 ______ D T − T0  ______     _______ (5.13) R T  0  T  2 4Examples of these defined functions implemented in Microsoft Excel, Matlab, Maple, Mathematica, and Mathcad are provided in the online learning center at http://highered.mheducation.com:80/sites/1259696529. 184 CHAPTER 5. The Second Law of Thermodynamics The right side of this equation is defined as another function, MCPS (T0, T; A, B, C, D). Equation (5.13) is then written: ⟨CPig⟩   ______   S = MCPS(T0, T; A, B, C, D) R The subscript S denotes a mean value specific to entropy calculations. Comparison of this mean value with the mean value specific to enthalpy calculations, as defined by Eq. (4.9), shows the two means to be quite different. This is inevitable because they are defined for the purpose of evaluating entirely different integrals. Solving for the integral in Eq. (5.12) gives: T   = ⟨CPig⟩ ln ___ ∫ T 0 dT T   CPig___ T S T0 and Eq. (5.10) becomes: ΔSig ⟨ P ⟩S _ T P Cig    _=  ______   ln − ln _   (5.14) R R T0 P0 This form of the equation for entropy changes for the ideal-gas state is often useful in iterative calculations where the final temperature is unknown. Example 5.3 Methane gas at 550 K and 5 bar undergoes a reversible adiabatic expansion to 1 bar. Assuming ideal-gas-state methane at these conditions, find its final temperature. Solution 5.3 For this process ΔSig = 0, and Eq. (5.14) becomes: ⟨ P ⟩S T2 2 P 1 Cig    ______   ln ___ = ln ___ = ln __ = − 1.6094 R T1 P1 5 Because ⟨CPig⟩ depends on T2, we rearrange this equation for iterative solution: S − 1.6094 ( ⟨CP⟩S  / R ) T2 − 1.6094 ________ ln ___ = ________   ig   or T2= T1exp   ig   T1 ⟨CP⟩  / R S With constants from Table C.1 of App. C, ⟨CPig⟩  / Ris evaluated by Eq. (5.13) S written in its functional form: ⟨CPig⟩   ______ ( ) S   = 𝖬𝖢𝖯𝖲 550, T ; 1.702, 9.081 × 10−3, − 2.164 × 10−6, 0.0 2 R 5.6. Entropy Balance for Open Systems 185 For an initial value of T2 < 550, compute a value of ⟨ CPig⟩  / Rfor substitution into the equation for T2. This new value of T2 allows recalculation of ⟨ CPig⟩  / R, and the S process continues to convergence on a final value of T2 = 411.34 K. S As with Ex. 4.3 a trial procedure is an alternative approach, with Microsoft Excel’s Goal Seek providing a prototypical automated version. Example 5.4 A 40 kg steel casting (CP = 0.5 kJ⋅kg−1⋅K−1) at a temperature of 450°C is quenched in 150 kg of oil (CP = 2.5 kJ⋅kg−1⋅K−1) at 25°C. If there are no heat losses, what is the change in entropy of (a) the casting, (b) the oil, and (c) both considered together? Solution 5.4 The final temperature t of the oil and the steel casting is found by an energy bal- ance. Because the change in energy of the oil and steel together must be zero, (40)(0.5)(t− 450) + (150)(2.5)(t − 25) = 0 Solution yields t = 46.52°C. (a) Change in entropy of the casting: C P dT T 2 ∫ ΔS  t = m   _____      = mCP  ln ___     T T 1        273.15 + 46.52 = (40)(0.5) ln  _____________     = −16.33 kJ⋅K  −1    273.15 + 450 (b) Change in entropy of the oil: 273.15 + 46.52 ΔSt= (150)(2.5)ln ____________   = 26.13 kJ⋅K−1    273.15 + 25 (c) Total entropy change: ΔStotal= −16.33 + 26.13 = 9.80 kJ⋅K−1 Note that although the total entropy change is positive, the entropy of the casting has decreased. 5.6 ENTROPY BALANCE FOR OPEN SYSTEMS Just as an energy balance is written for processes in which fluid enters, exits, or flows through a control volume (Sec. 2.9), so too may an entropy balance be written. There is, however, an important difference: Entropy is not conserved. The second law states that the total entropy change associated with any process must be positive, with a limiting value of 186 CHAPTER 5. The Second Law of Thermodynamics zero for a reversible process. This requirement is taken into account by writing the entropy balance for both the system and its surroundings, considered together, and by including an entropy-generation term to account for the irreversibilities of the process. This term is the sum of three others: one for the difference in entropy between exit and entrance streams, one for entropy change within the control volume, and one for entropy change in the surroundings. If the process is reversible, these three terms sum to zero, making ΔStotal = 0. If the process is irreversible, they sum to a positive quantity, the entropy-generation term. The statement of balance, expressed as rates, is therefore: ⎧Time rate of⎫ ⎧ Net rate of ⎫ ⎧ Time rate of ⎫ ⎪ change of ⎪ ⎪ change in ⎪ ⎪ change of ⎪ Total rate entropy of ⎪ ⎪ entropy in ⎪ { generation} ⎨ entropy  ⎬+ ⎨     ⎬+ ⎨  ⎬=   of entropy ⎪ in control ⎪ ⎪ ⎩flowing streams⎭ ⎩surroundings⎭ ⎩ volume ⎭ The equivalent equation of entropy balance is t d(mS)cv _______ dSsurr ∙ + Δ(Sm)  fs + _____ ∙ = S  G ≥ 0 (5.15) dt dt ∙ where S  G is the rate of entropy generation. This equation is the general rate form of the entropy balance, applicable at any instant. Each term can vary with time. The first term is the time rate of change of the total entropy of the fluid contained within the control volume. The second term is the net rate of gain in entropy of the flowing streams, i.e., the difference between the total entropy transported out by exit streams and the total entropy transported in by entrance streams. The third term is the time rate of change of the entropy of the surroundings, resulting from heat transfer between system ∙ and surroundings. Let rate of heat transfer Q   jwith respect to a particular part of the control surface be associated with Tσ,j where subscript σ,j denotes a temperature in the surroundings. ∙ The rate of entropy change ∙ in the surroundings as a result of this transfer is then − Q   j / T   σ, j. The minus sign converts Q  j, defined with respect to the system, to a heat transfer rate with respect to the surroundings. The third term in Eq. (5.15) is therefore the sum of all such quantities: ∙ t dSsurr j   Q  _____  = −∑  ____  dt j T σ, j Equation (5.15) is now written: ∙ d(mS)  cv j ∙   Q  _______   + Δ(Sm∙ )    fs – ∑ ____     = S  G ≥ 0 (5.16) dt j T σ, j ∙ The final term, representing the rate of entropy generation S    G, reflects the second-law requirement that it be positive for irreversible processes. There are two sources of irreversibility: (a) those within the control volume, i.e., internal irreversibilities, and (b) those resulting from heat transfer across finite temperature differences between system ∙ and surroundings, i.e., external thermal irreversibilities. The limiting case for which S  G = 0 applies when the process is completely reversible, implying: ∙ The process is internally reversible within the control volume ∙ Heat transfer between the control volume and its surroundings is reversible Rev. Confirming Pages 5.6. Entropy Balance for Open Systems 187 The second item means either that heat reservoirs are included in the surroundings with temperatures equal to those of the control surface or that work-producing Carnot engines are interposed in the surroundings between the control-surface temperatures and the heat-reservoir temperatures. For a steady-state flow process the mass and entropy of the fluid in the control volume are constant, and d(mS)cv/dt is zero. Equation (5.16) then becomes: ∙ j ∙ ∙   Q   )  fs – ∑  ____  = S  G ≥ 0   Δ(Sm (5.17) j T σ, j If in addition there is but one entrance and one exit, m ∙ is the same for both streams, and divid- ing through by m ∙  yields: ∙ Q    j ΔS − ∑ _ = SG≥ 0 (5.18) j Tσ, j Each term in Eq. (5.18) is based on a unit mass of fluid flowing through the control volume. Example 5.5 In a steady-state flow process carried out at atmospheric pressure, 1 mol⋅s−1 of air at 600 K is continuously mixed with 2 mol⋅s−1 of air at 450 K. The product stream is at 400 K and 1 atm. A schematic representation of the process is shown in Fig. 5.4. Determine the rate of heat transfer and the rate of entropy generation for the process. Assume the ideal-gas state for air with Cig P= (7/ 2)R, that the surroundings are at 300 K, and that kinetic- and potential-energy changes of the streams are negligible. n. 3 mol s 1 T 400 K Figure 5.4:. 1. 1 Process n 1 mol s n 2 mol s A Control Volume B described in TA 600 K TB 450 K Ex. 5.5.. Q smi96529_ch05_173-209.indd 187 06/19/17 12:54 PM 188 CHAPTER 5. The Second Law of Thermodynamics Solution 5.5 We start by applying an energy balance to determine the rate of heat transfer, which we will need to know in order to compute the rate of entropy generation. Writing the energy balance, Eq. (2.29), with m ∙ replaced by n ∙ , and then replacing ∙ n ∙ with n ∙ A  + n B ,   ig− n∙   AHAig− n  B HBig= n  A(Hig− HAig)+ n  B(Hig− HBig) ∙ ∙ ∙ ∙ Q  = n∙ H ∙ = n  AC  Pig  (T − T A) + n  BC  Pig  (T − T B) = C  Pig  [n  A (T − T A) + n  B (T − T B)] ∙ ∙ ∙ ∙ Q                = (7 / 2)(8.314)[(1)(400 − 600) + (2)(400 − 450)] = − 8729.7 J⋅s  −1 ∙ The steady-state entropy balance, Eq. (5.17), with m ∙ again replaced by n, can similarly be written as ∙ ∙ Q ∙ Q     = n  A( S  ig− S  Aig  )+ n  B( S  ig− S  Big  ) − ___ ∙ S   G = nS  ig− n  AS  Aig  − n  BS  Big  − ___ ∙ ∙ ∙ ∙     Tσ  Tσ  ( T B ) Tσ  ∙ ∙ T T Q T T Q   =  n∙   AC  Pig  ln  ___                   + n  BC  Pig  ln ___ ∙     − ___     = C  pig  n∙   A ln ___     − ___     + n  B ln ___ ∙     T A T B Tσ  T A [ ] 400 400 8729.7 = (7 / 2)(8.314) (1) ln ____     + (2) ln ____      + ______     = 10.446 J⋅K  −1⋅s  −1 600 450 300 The rate of entropy generation is positive, as it must be for any real process. Example 5.6 An inventor claims to have devised a process which takes in only saturated steam at 100°C and which by a complicated series of steps makes heat continuously available at a temperature level of 200°C, with 2000 kJ of energy available at 200°C for every kilogram of steam taken into the process. Show whether or not this process is possible. To give this process the most favorable conditions, assume cooling water available in unlimited quantity at a temperature of 0°C. Solution 5.6 For any process to be even theoretically possible, it must meet the requirements of the first and second laws of thermodynamics. The detailed mechanism need not be known to determine whether this is true; only the overall result is required. If the claims of the inventor satisfy the laws of thermodynamics, fulfilling the claims is theoretically possible. The determination of a mechanism is then a matter of ingenuity. Otherwise, the process is impossible, and no mechanism for carrying it out can be devised. In the present instance, a continuous process takes in saturated steam at 100°C, and makes heat Q continuously available at a temperature level of 200°C. Because cooling water is available at 0°C, the maximum possible use is made of the steam if it is condensed and cooled (without freezing) to this exit temperature and discharged 5.6. Entropy Balance for Open Systems 189 at atmospheric pressure. This is a limiting case (most favorable to the inventor), and requires a heat-exchange surface of infinite area. It is not possible in this process for heat to be liberated only at the 200°C tem- perature level, because as we have shown no process is possible that does nothing except transfer heat from one temperature level to a higher one. We must therefore suppose that some heat Qσ is transferred to the cooling water at 0°C. Moreover, the process must satisfy the first law; thus by Eq. (2.32): ΔH = Q + Qσ + Ws where ΔH is the enthalpy change of the steam as it flows through the system. Because no shaft work accompanies the process, Ws = 0. The surroundings con- sist of cooling water, which acts as a heat reservoir at the constant temperature of Tσ = 273.15 K, and a heat reservoir at T = 473.15 K to which heat in the amount of 2000 kJ is transferred for each kilogram of steam entering the system. The diagram of Fig. 5.5 indicates overall results of the process based on the inventor’s claim. Heat reservoir at 200° C Q 2000 kJ Saturated steam at 100° C Liquid water at 0° C Figure 5.5: Process Apparatus H1 2676.0 kJ kg 1 H2 0.0 described in Ex. 5.6. S1 7.3554 kJ kg 1 K 1 S2 0.0 Q Heat reservoir at 0 ° C (cooling water) The values of H and S shown on Fig. 5.5 for liquid water at 0°C and for satu- rated steam at 100°C are taken from the steam tables (App. E). Note that the val- ues for liquid water at 0°C are for saturated liquid (Psat = 0.61 kPa), but the effect of an increase in pressure to atmospheric pressure is insignificant. On the basis of 1 kg of entering steam, the first law becomes: ΔH = H2− H1= 0.0 − 2676.0 = −2000 + Qσ and Qσ = −676.0 kJ The negative value for Qσ indicates heat transfer is from the system to the cooling water. We now examine this result in the light of the second law to determine whether the entropy generation is greater than or less than zero for the process. Equation (5.18) is here written: Qσ Q ΔS − ___ − ______ = SG Tσ 473.15 190 CHAPTER 5. The Second Law of Thermodynamics For 1 kg of steam, ΔS = S2− S1= 0.0000 − 7.3554 = −7.3554 kJ⋅K−1 The entropy generation is then: −676.0 ______ − 2000 S G = − 7.3554 −  ______     −               273.15 473.15 = − 7.3554 + 4.2270 + 2.4748 = − 0.6536 kJ⋅K  −1 This negative result means that the process as described is impossible; the second law in the form of Eq. (5.18) requires SG ≥ 0. The negative result of the preceding example does not mean that all processes of this general nature are impossible; only that the inventor claimed too much. Indeed, the maximum amount of heat which can be transferred to the hot reservoir at 200°C is readily calculated. The energy balance is: Q + Qσ = ΔH (A) The maximum heat transfer to the hot reservoir occurs when the process is completely revers- ible, in which case SG = 0, and Eq. (5.18) becomes Q Qσ __ + ___ = ΔS (B) T Tσ Combination of Eqs. (A) and (B) and solution for Q yields: T Q = ____ (ΔH − Tσ ΔS) T − Tσ With respect to the preceding example, [ ] = − 1577.7 kJ⋅ kg−1 473.15 − 2676.0 − (273.15)(− 7.3554) Q = ______ 200 2000 kJ⋅ kg−1claimed. This value is smaller in magnitude than the − 5.7 CALCULATION OF IDEAL WORK In any steady-state flow process requiring work, an absolute minimum amount of work is required to produce the desired change of state in the fluid flowing through the control vol- ume. In a process producing work, an absolute maximum amount of work can be realized from a given change of state of the fluid flowing through the control volume. In either case, the limiting value results when the change of state associated with the process is accomplished completely reversibly. For such a process, the entropy generation is zero, and Eq. (5.17), writ- ten for a uniform surroundings temperature Tσ, becomes: ∙ Q  ∙ Δ(Sm∙ )    fs − ___     = 0 or Q  = Tσ Δ(Sm∙ )    fs Tσ  5.7. Calculation of Ideal Work 191 ∙ Substituting this expression for Q in the energy balance, Eq. (2.29): Δ[ (H + _  12 u  2 + zg)m∙    ]   = Tσ  Δ(Sm∙ )    fs + W  s(rev) ∙ fs ∙ ∙ The shaft work, W  s  (rev), ∙ is here the work of a completely reversible process. If W  s  (rev) is given the name ideal work, W    ideal, the preceding equation may be rewritten: W   ideal = Δ[ (H + _  12 u  2 + zg)m∙    ]   − Tσ  Δ(Sm∙ )    fs ∙ (5.19) fs In most applications to chemical processes, the kinetic- and potential-energy terms are negligible compared with the others; in this event Eq. (5.19) reduces to: ∙ ∙  W  ideal  = Δ(Hm∙ )    fs − Tσ Δ(Sm)f  s  (5.20) For the special case of a single stream flowing through the control volume, m ∙ may be factored. The resulting equation may then be divided by m ∙ to express it on the basis of a unit amount of fluid flowing through the control volume. Thus, ∙ W  ideal = m∙ (  ΔH − Tσ ΔS) (5.21) Wideal= ΔH − Tσ ΔS (5.22) A completely reversible process is hypothetical, devised here solely for determination of the ideal work associated with a given change of state. The sole connection between an actual process and an imagined hypothetical reversible process employed for determining ideal work is that they both apply to the same changes of state. ∙ Our objective is to compare the actual work of a process W s  (or Ws) as given by an energy balance to the ideal work as given by Eqs. (5.19) through (5.22) for a hypothetical reversible process that produces the same property changes. No description of the hypothetical process is required, as it ∙may always be imagined (see Ex. 5.7). When W    ideal (or Wideal) is positive, it is the minimum work required to ∙ bring about a given change in the properties of the flowing streams, and is smaller than Ws  . In this case a thermodynamic efficiency ηt is defined as the ratio of the ideal work to the actual work: ∙ W  ideal ηt(work required) = _____   ∙   (5.23) Ws  | | ∙ ∙ When W  ideal (or Wideal) is negative,  W  ideal  is the maximum work∙ obtainable from a given change in the properties of the flowing streams, and is larger than  Ws . In this case, the | | thermodynamic efficiency is defined as the ratio of the actual work to the ideal work: ∙ Ws  η(work produced) = _____   ∙   t (5.24) W  ideal 192 CHAPTER 5. The Second Law of Thermodynamics Example 5.7 What is the maximum work that can be obtained in a steady-state flow process from 1 mol of nitrogen in its ideal-gas state at 800 K and 50 bar? Take the temperature and pressure of the surroundings as 300 K and 1.0133 bar. Solution 5.7 The maximum possible work is obtained from any completely reversible process that reduces the nitrogen to the temperature and pressure of the surroundings, i.e., to 300 K and 1.0133 bar. Required here is the calculation of Wideal by Eq. (5.22), in which ΔS and ΔH are the molar entropy and enthalpy changes of the nitrogen as its state changes from 800 K and 50 bar to 300 K and 1.0133 bar. For the ideal-gas state, enthalpy is independent of pressure, and its change is given by: T 2 ∫ T 1 ΔHig=   CPigdT  The value of this integral is found from Eq. (4.8), and is represented by: 8.314 × ICPH(800, 300; 3.280, 0.593 ×𝟣𝟢−3, 0.0, 0.040 × 𝟣𝟢−5) = −15,060 J⋅mol−1 where the parameters for nitrogen come from Table C.1 of App. C. Similarly, the entropy change is found from Eq. (5.10), here written: T 2 P2 ∫ T 1 dT ΔSig=   CPig___  − R ln ___ T P1 The value of the integral, found from Eq. (5.11), is represented by: 8.314 × ICPS(800, 300; 3.280, 0.593 ×𝟣𝟢−3, 0.0, 0.040 × 𝟣𝟢−5) = −29.373 J⋅mol−1⋅K−1 1.0133 ΔSig= −29.373 − 8.314 ln ______ Whence, = 3.042 J⋅mol−1⋅K−1 50 With these values of ΔHig and ΔSig, Eq. (5.22) becomes: Wideal= −15,060 − (300)(3.042) = −15,973 J⋅mol−1 One can easily devise a specific reversible process to bring about the given change of state of the preceding example. Suppose the nitrogen is continuously changed to its final state at 1.0133 bar and T2 = Tσ = 300 K by the following two-step steady-flow process: ∙ Step 1: Reversible, adiabatic expansion (as in a turbine) from initial state P1, T1, H1 to 1.0133 bar. Let T′ denote the discharge temperature. ∙ Step 2: Cooling (or heating, if T′ is less than T2) to the final temperature T2 at a constant pressure of 1.0133 bar. For step 1, with Q = 0, the energy balance is Ws = ΔH = (H′ − H1), where H′ is the enthalpy at the intermediate state of T′ and 1.0133 bar. 5.7. Calculation of Ideal Work 193 For maximum work production, step 2 must also be reversible, with heat exchanged reversibly with the surroundings at Tσ. This requirement is satisfied when Carnot engines take heat from the nitrogen, produce work WCarnot, and reject heat to the surroundings at Tσ. Because the temperature of the nitrogen decreases from T′ to T2, Eq. (5.5) for the work of a Carnot engine is written in differential form: (T ) ( T) Tσ Tσ dWCarnot= __ − 1 (−dQ)= 1 − __ dQ Here dQ refers to the nitrogen, which is taken as the system, rather than to the engine. Integration yields: T 2 ∫ T’ T dQ WCarnot= Q − Tσ   ___   The first term on the right is the heat transferred with respect to the nitrogen, given by Q = H2 − H′. The integral is the change in entropy of the nitrogen as it is cooled by the Carnot engines. Because step 1 occurs at constant entropy, the integral also represents ΔS for both steps. Hence, WCarnot= (H2− H′)− Tσ ΔS The sum of Ws and WCarnot gives the ideal work; thus, Wideal=(H′− H1)+ (H2− H′)− Tσ ΔS = (H2− H1)− Tσ ΔS or Wideal= ΔH − Tσ ΔS which is the same as Eq. (5.22). This derivation makes clear the difference between Ws, the reversible adiabatic shaft work of the turbine, and Wideal. The ideal work includes not only Ws, but also all work obtain- able from Carnot engines for the reversible exchange of heat with the surroundings at Tσ. In practice, work produced by a turbine can be as much as 80% of the reversible adiabatic work, but usually no mechanism is present for extraction of WCarnot.5 Example 5.8 Rework Ex. 5.6, making use of the equation for ideal work. Solution 5.8 The procedure here is to calculate the maximum possible work Wideal which can be obtained from 1 kg of steam in a flow process as it undergoes a change in state from saturated steam at 100°C to liquid water at 0°C. The problem then reduces to the question of whether this amount of work is sufficient to operate a Carnot 5A cogeneration plant produces power both from a gas turbine and from a steam turbine operating on steam gen- erated by heat from the gas-turbine exhaust. 194 CHAPTER 5. The Second Law of Thermodynamics refrigerator rejecting 2000 kJ as heat at 200°C and taking heat from the unlimited supply of cooling water at 0°C. From Ex. 5.6, we have ΔH = −2676.0 kJ⋅kg−1 and ΔS = −7.3554 kJ⋅K−1⋅kg−1 With negligible kinetic- and potential-energy terms, Eq. (5.22) yields: Wideal= ΔH − Tσ ΔS = −2676.0 − (273.15)(−7.3554) = −666.9 kJ⋅kg−1 If this amount of work, numerically the maximum obtainable from the steam, is used to drive the Carnot refrigerator operating between the temperatures of 0°C and 200°C, the heat rejected is found from Eq. (5.5), solved for QH: ( Tσ − T ) ( 200 − 0 ) T 200 + 273.15 = Wideal ____ QH = 666.9 __________    = 1577.7 kJ As calculated in Ex. 5.6, this is the maximum possible heat release at 200°C; it is less than the claimed value of 2000 kJ. As in Ex. 5.6, we conclude that the process as described is not possible. 5.8 LOST WORK Work that is wasted as the result of irreversibilities in a process is called lost work, Wlost, and is defined as the difference between the actual work of a change of state and the ideal work for the same change of state. Thus by definition, Wlost≡ Ws− Wideal (5.25) In terms of rates, ∙ ∙ ∙ W  lost ≡ W  s − W  ideal (5.26) The actual work rate comes from the energy balance, Eq. (2.29), and the ideal work rate is obtained using Eq. (5.19): [( ) ] ∙ 1 ∙ W  s = Δ H +  __ u  2+ zg m    − Q ∙ 2 fs            [( ) ] ∙ 1 W  ideal = Δ H + __    u  2+ zg m    − Tσ  Δ(Sm)  fs ∙ ∙ 2 fs By difference, as given by Eq. (5.26), ∙ ∙  W  lost = Tσ Δ(Sm)  fs − Q  ∙ (5.27) For the case of a single surroundings temperature Tσ, the steady-state entropy balance, Eq. (5.17), becomes: ∙ ∙ Q S  G = Δ(Sm)  fs − ___ ∙     (5.28) Tσ  5.8. Lost Work 195 ∙ ∙ Multiplication by Tσ gives: Tσ S  G = Tσ Δ(Sm)  fs − Q ∙ The right sides of this equation and Eq. (5.27) are identical; therefore, ∙ ∙  W  lost = Tσ S  G  (5.29) ∙ ∙ As a consequence of the second law, S  G ≥∙ 0;it follows that W  lost ≥ 0.When a process is completely reversible, ∙ the equality holds, and W   lost = 0.For irreversible processes the inequal- ity holds, and W   lost,i.e., the energy that becomes unavailable for work, is positive. The engineering significance of this result is clear: The greater the irre- versibility of a process, the greater the rate of entropy production and the greater the amount of energy made unavailable for work. Thus every irreversibility carries with it a price. Minimizing entropy production is essential for conservation of the earth’s resources. ∙ For the special case of a single stream flowing through the control volume, mfactors and ∙ becomes a multiplier of the entropy difference in Eqs. (5.27) and (5.28). Division by m con- verts all terms to the basis of a unit amount of fluid flowing through the control volume. Thus, ∙ ∙ ∙ W  lost = mTσ ΔS − Q (5.30) Wlost= Tσ ΔS − Q (5.31) Q (5.32) Q S∙   G = m∙ ΔS − ___     (5.33) SG= ΔS − ___ Tσ  Tσ ∙ Recall here that Q and Q represent heat exchange with the surroundings, but with the sign referred to the system. Equations (5.31) and (5.33) combine for a unit amount of fluid to give: Wlost= Tσ SG (5.34) Again, because SG ≥ 0, it follows that Wlost ≥ 0. Example 5.9 The two basic types of steady-flow heat exchanger are characterized by their flow pat- terns: cocurrent and countercurrent. Temperature profiles for the two types are indi- cated in Fig. 5.6. In cocurrent flow, heat is transferred from a hot stream, flowing from left to right, to a cold stream flowing in the same direction, as indicated by arrows. In countercurrent flow, the cold stream, again flowing from left to right, receives heat from the hot stream flowing in the opposite direction. The lines ∙ relate the temperatures of the hot and cold streams, TH and TC respec- tively, to Q  C, the accumulated rate of heat addition to the cold stream as it progresses through the exchanger from the left end to an arbitrary downstream location. The fol- lowing specifications apply to both cases: ∙ TH1= 400 K TH2= 350 K TC1= 300 K n  H = 1 mol⋅s−1 196 CHAPTER 5. The Second Law of Thermodynamics TH1 400 K 400 K TH1 390 K TC2 TH TH TC T 350 K TH2 350 K T 340 K TC TC1 300 K TC1 300 K 0. 0. QC QC (a) (b) Figure 5.6: Heat exchangers. (a) Case I, cocurrent. (b) Case II, countercurrent. The minimum temperature difference between the flowing streams is 10 K. Assume the ideal-gas state for both streams with CP = (7/2)R. Find the lost work for both cases. Take Tσ = 300 K. Solution 5.9 With∙ the assumption ∙ of negligible kinetic- and potential-energy changes, with Ws = 0,and with Q = 0(there is no heat exchange with the surroundings) the energy balance [Eq. (2.29)] can be written: n  H(ΔHig)H+ n  C(ΔHig)C= 0 ∙ ∙ With constant molar heat capacity, this becomes: n  H CPig(TH2 − TH1) + n  CCPig(TC2 − TC1) = 0 ∙ ∙ (A) The total rate of entropy change for the flowing streams is: Δ (Sign)  fs = n  H(ΔSig)H+ n  C(ΔSig)C ∙ ∙ ∙ By Eq. (5.10), with the assumption of negligible pressure change in the flowing streams, ( TH1 n∙   H C1) TH n∙   C TC Δ(Sign)  fs = n  HCPig ln ___ 2 + ___    ln ___ 2 ∙ ∙ (B) T 5.9. The Third Law of Thermodynamics 197 ∙ Finally, by Eq. (5.27), with Q = 0, W  lost = Tσ Δ (Sign)  fs ∙ ∙ (C) These equations apply to both cases. ∙ Case I: Cocurrent flow. By Eqs. (A), (B), and (C), respectively: ∙ n  C 400 − 350 ___   ∙   = ________ = 1.25 n  H 340 − 300 ( 400 300 )                 350 340 Δ(S  ign)f  s = (1)(7 / 2)(8.314) ln ____ + 1.25 ln ____ = 0.667 J·K−1·s−1 ∙ ∙ W  lost = ( 300)(0.667)= 200.1 J⋅s−1 ∙ Case II: Countercurrent flow. By Eqs. (A), (B), and (C), respectively: ∙ n  C 400 − 350 ___   ∙   = ________ = 0.5556 n  H 390 − 300 ( 400 300 )                 350 390 Δ (Sign)f  s = (1)(7 / 2)(8.314) ln ____ + 0.5556 ln ____ = 0.356 J·K−1·s−1 ∙ ∙ W  lost = (300)(0.356) = 106.7 J⋅s−1 Although the total rate of heat transfer is the same for both exchangers, the temperature rise of the cold stream in countercurrent flow is more than twice that for cocurrent flow. Thus, its entropy increase per unit mass is larger than for the cocurrent case. However, its flow rate is less than half that of the cold stream in cocurrent flow, so that the total entropy increase of the cold stream is less for countercurrent flow. From a thermodynamic point ∙ of view, the ∙ countercurrent case is much more efficient. Because Δ(S  ign)fs = SG , both the rate of entropy generation and the lost work for the countercurrent case are only about half the value for the cocurrent case. Greater efficiency in the countercurrent case would be anticipated based on Fig. 5.6, which shows that heat is transferred across a smaller temperature difference (less irreversibly) in the countercurrent case. 5.9 THE THIRD LAW OF THERMODYNAMICS Measurements of heat capacities at very low temperatures provide data for the calculation from Eq. (5.1) of entropy changes down to 0 K. When these calculations are made for different crystal- line forms of the same chemical species, the entropy at 0 K appears to be the same for all forms. When the form is noncrystalline, e.g., amorphous or glassy, calculations show that the entropy of the disordered form is greater than that of the crystalline form. Such calculations, which are sum- marized elsewhere,6 lead to the postulate that the absolute entropy is zero for all perfect crystalline substances at absolute zero temperature. While this essential idea was advanced by Nernst and 6K. S. Pitzer, Thermodynamics, 3d ed., chap. 6, McGraw-Hill, New York, 1995. Rev. Confirming Pages 198 CHAPTER 5. The Second Law of Thermodynamics Planck at the beginning of the twentieth century, more recent studies at very low temperatures have increased confidence in this postulate, which is now accepted as the third law of thermodynamics. If the entropy is zero at T = 0 K, then Eq. (5.1) lends itself to the calculation of absolute entropies. With T = 0 as the lower limit of integration, the absolute entropy of a gas at temper- ature T based on calorimetric data is: T f ΔHf Tυ  T (C ) (CP )s (CP )l ΔHυ ∫0 Tf ∫ T f T Tυ ∫ Tυ  T P g S =   _____  dT + ____ +   ____  dT + ____ +   _____  dT (5.35) T This equation7 is based on the supposition that no solid-state transitions take place and thus no heats of transition need appear. The only constant-temperature heat effects are those of melt- ing at Tf and vaporization at Tυ. When a solid-phase transition occurs, a term ΔHt /Tt is added. Note that although the third law implies that absolute values of entropy are obtainable, for most thermodynamic analyses, only relative values are needed. As a result, reference states other than the perfect crystal at 0 K are commonly used. For example, in the steam tables of App. E, saturated liquid water at 273.16 K is taken as the reference state and assigned zero entropy. However, the absolute or “third law” entropy of saturated liquid water at 273.16 K is 3.515 kJ⋅kg−1⋅K−1. 5.10 ENTROPY FROM THE MICROSCOPIC VIEWPOINT Because molecules in the ideal-gas state do not interact, their internal energy resides with individual molecules. This is not true of the entropy, which is inherently a collective property of a large number of molecules or other entities. The microscopic interpretation of entropy is suggested by the following example. Suppose an insulated container, partitioned into two equal volumes, contains Avoga- dro’s number NA of molecules in one section and no molecules in the other. When the partition is withdrawn, the molecules quickly distribute themselves uniformly throughout the total vol- ume. The process is an adiabatic expansion that accomplishes no work. Therefore, ig ΔU = CVΔ T = 0 and the temperature does not change. However, the pressure of the gas decreases by half, and the entropy change, as given by Eq. (5.10), is: P2 ΔSig= −R ln ___ = R ln 2 P1 Because this is the total entropy change, the process is clearly irreversible. At the instant when the partition is removed, the molecules occupy only half the space available to them. In this momentary initial state, the molecules are not randomly distributed over the total volume to which they have access but are crowded into just half the total vol- ume. In this sense they are more ordered than they are in the final state of uniform distribution throughout the entire volume. Thus, the final state can be regarded as a more random, or more disordered, state than the initial state. Generalizing from this example, and from many other 7Evaluation of the first term on the right is not a problem for crystalline substances, because CP  /T remains finite as T → 0. smi96529_ch05_173-209.indd 198 06/19/17 12:32 PM 5.10. Entropy from the Microscopic Viewpoint 199 similar observations, one is led to the notion that increasing disorder (or decreasing structure) at the molecular level corresponds to increasing entropy. The means for expressing disorder in a quantitative way was developed by L. Boltzmann and J. W. Gibbs through a quantity Ω, defined as the number of different ways that micro- scopic particles can be distributed among the “states” accessible to them. It is given by the general formula: N! Ω = _____________    (5.36) (N1!)(N2!)(N3!)... where

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