Chapter 7: The Second Law of Thermodynamics

Transcription

1 Chapter 7: he Second Law of hermodynamics he second law of thermodynamics asserts that processes occur in a certain direction and that the energy has quality as well as quantity he first law places no restriction on the direction of a process, and satisfying the first law does not guarantee that the process will occur hus, we need another general principle (second law) to identify whether a process can occur or not Q (Heat transfer) Hot container Possible Cold surroundings Impossible Fig 7-: Heat transfer from a hot container to the cold surroundings is possible; howeer, the reeres process (although satisfying the first law) is impossible A process can occur when and only when it satisfies both the first and the second laws of thermodynamics he second law also asserts that energy has a quality Presering the quality of energy is a major concern of engineers In the aboe example, the energy stored in a hot container (higher temperature) has higher quality (ability to work) in comparison with the energy contained (at lower temperature) in the surroundings he second law is also used in determining the theoretical limits for the performance of commonly used engineering systems, such as heat engines and refrigerators etc hermal Energy Reseroirs hermal energy reseroirs are hypothetical bodies with a relatiely large thermal energy capacity (mass x specific heat) that can supply or absorb finite amounts of heat without undergoing any change in temperature Lakes, riers, atmosphere, oceans are example of thermal reseroirs A two-phase system can be modeled as a reseroir since it can absorb and release large quantities of heat while remaining at constant temperature A reseroir that supplies energy in the form of heat is called a source and one that absorbs energy in the form of heat is called a sink Chapter 7, ECE 309, Spring 06

2 Heat Engines Heat engines conert heat to work here are seeral types of heat engines, but they are characterized by the following: - hey all receie heat from a high-temperature source (oil furnace, nuclear reactor, etc) - hey conert part of this heat to work 3- hey reject the remaining waste heat to a low-temperature sink 4- hey operate in a cycle Energy source (furnace) Source ( H ) Q in Boiler urbine Q in W in W net W out - W in W out Heat engine W net Pump Q out Condenser Sink ( L ) Q out Energy sink (rier, lake) W net Q in + Q out Fig 7-: Steam power plant is a heat engine hermal efficiency: is the fraction of the heat input that is conerted to the net work output (efficiency benefit / cost) Chapter 7, ECE 309, Spring 06 W ηth Qin Q ηth Q net, out out in and W net, out Q in Q out

3 he thermal efficiencies of work-producing deices are low Ordinary spark-ignition automobile engines hae a thermal efficiency of about 0%, diesel engines about 30%, and power plants in the order of 40% Is it possible to sae the rejected heat Q out in a power cycle? he answer is NO, because without the cooling in condenser the cycle cannot be completed Eery heat engine must waste some energy by transferring it to a low-temperature reseroir in order to complete the cycle, een in idealized cycle he Second Law: Kelin-Planck Statement It is impossible for any deice that operates on a cycle to receie heat from a single reseroir and produce a net amount of work In other words, no heat engine can hae a thermal efficiency of 00% Source ( H ) Q in W net Q in Heat engine hermal efficiency of 00% Chapter 7, ECE 309, Spring 06 3 Fig7-3: A heat engine that iolates the Kelin-Planck statement of the second law cannot be built Refrigerators and Heat Pumps Q out 0 In nature, heat flows from high-temperature regions to low-temperature ones he reerse process, howeer, cannot occur by itself he transfer of heat from a lowtemperature region to a high-temperature one requires special deices called refrigerators Refrigerators are cyclic deices, and the working fluids used in the cycles are called refrigerant Heat pumps transfer heat from a low-temperature medium to a high-temperature one Refrigerators and heat pumps are essentially the same deices; they differ in their objecties only Refrigerator is to maintain the refrigerated space at a low temperature On the other hand, a heat pump absorbs heat from a lowtemperature source and supplies the heat to a warmer medium

4 WARM enironment WARM house Q H 3 Condenser Q H Q H desired output Compressor R W in HP W in Expansion Vale W c Q L desired output Q L 4 Eaporator COLD refrigerated space COLD enironment QL Refrigerator Heat pump Fig7-4: Objecties of refrigerator and heat pump Coefficient of Performance (COP) he performance of refrigerators and heat pumps is expressed in terms of the coefficient of performance (COP) which is defined as Benefit q COP R Cost w It can be seen that L c COP HP COPR + Benefit q COP HP Cost w Air conditioners are basically refrigerators whose refrigerated space is a room or a building he Energy Efficiency Rating (EER): is the amount of heat remoed from the cooled space in BU s for Wh (watt-hour) EER 34 COP R Most air conditioners hae an EER between 8 to (COP of 3 to 35) he Second Law of hermodynamics: Clausius Statement It is impossible to construct a deice that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to highertemperature body In other words, a refrigerator will not operate unless its compressor is drien by an external power source H c Chapter 7, ECE 309, Spring 06 4

5 Kelin-Planck and Clausius statements of the second law are negatie statements, and a negatie statement cannot be proed So, the second law, like the first law, is based on experimental obserations he two statements of the second law are equialent In other words, any deice iolates the Kelin-Planck statement also iolates the Clausius statement and ice ersa Source ( H ) Source ( H ) Q H W net Q H Heat engine η 00% Refriger ator Q H + Q L W net 0 Q H Refriger ator Q L 0 Equialent Q L Q L Source ( L ) Source ( L ) Fig 7-5: he iolation of the Kelin-Planck statement leads to iolation of Clausius Any deice that iolates the first law of thermodynamics (by creating energy) is called a perpetual-motion machine of the first kind (PMM), and the deice that iolates the second law is called a perpetual-motion machine of the second kind (PMM) Reersible and Irreersible Process A reersible process is defined as a process that can be reersed without leaing any trace on the surroundings It means both system and surroundings are returned to their initial states at the end of the reerse process Processes that are not reersible are called irreersible Chapter 7, ECE 309, Spring 06 5

6 Reersible processes do not occur and they are only idealizations of actual processes We use reersible process concept because, a) they are easy to analyze (since system passes through a series of equilibrium states); b) they sere as limits (idealized models) to which the actual processes can be compared Some factors that cause a process to become irreersible: Friction Unrestrained expansion and compression mixing Heat transfer (finite ) Inelastic deformation Chemical reactions In reersible process things happen ery slowly, without any resisting force, without any space limitation eerything happens in a highly organized way (it is not physically possible - it is an idealization) Internally reersible process: if no irreersibilities occur within the boundaries of the system In these processes a system undergoes through a series of equilibrium states, and when the process is reersed, the system passes through exactly the same equilibrium states while returning to its initial state Externally reersible process: if no irreersibilities occur outside the system boundaries during the process Heat transfer between a reseroir and a system is an externally reersible process if the surface of contact between the system and reseroir is at the same temperature otally reersible (reersible): both externally and internally reersible processes Boundary at 0 C 0 C 0 C otally reersible Source at Heat Heat Source at Internally reersible H 000 H 30 Chapter 7, ECE 309, Spring 06 6

7 he Carnot Cycle he efficiency of a heat-engine cycle greatly depends on how the indiidual processes that make up the cycle are executed he net work (or efficiency) can be maximized by using reersible processes he best known reersible cycle is the Carnot cycle Note that the reersible cycles cannot be achieed in practice because of irreersibilities associated with real processes But, the reersible cycles proide upper limits on the performance of real cycles Consider a gas in a cylinder-piston (closed system) he Carnot cycle has four processes: - Reersible isothermal expansion: he gas expands slowly, doing work on the surroundings Reersible heat transfer from the heat source at H to the gas which is also at H -3 Reersible adiabatic expansion: he cylinder-piston is now insulated (adiabatic) and gas continues to expand reersibly (slowly) So, the gas is doing work on the surroundings, and as a result of expansion the gas temperature reduces from H to L 3-4: Reersible isothermal compression: he gas is allowed to exchange heat with a sink at temperature L as the gas is being slowly compressed So, the surroundings is doing work (reersibly) on the system and heat is transferred from the system to the surroundings (reersibly) such that the gas temperature remains constant at L 4-: Reersible adiabatic compression: he gas temperature is increasing from L to H as a result of compression Carnot cycle is the most efficient cycle operating between two specified temperature limits he efficiency of all reersible heat engines operating between the two same reseroirs are the same he thermal efficiency of a heat engine (reersible or irreersible) is: η th Q Q L H For the Carnot cycle, it can be shown: η th, Carnot L H Chapter 7, ECE 309, Spring 06 7

8 P Q H H Const W net 4 Q L 3 L Const Fig 7-6: P- diagram for the Carnot cycle he efficiency of an irreersible (real) cycle is always less than the efficiency of the Carnot cycle operating between the same two reseroirs Chapter 7, ECE 309, Spring 06 8 < ηth η th ηth, > ηth, re re, re irreersible heat engine reersible heat engine impossible heat engine! Consider a Carnot heat engine working between two thermal reseroirs L 300 K and H he thermal efficiency of the heat engine increases as the heat source temperature H is increased H K η th % he thermal efficiency of actual heat engine can be maximized by supplying heat to the engine at the highest possible temperature (limited by material strength) and rejecting heat to lowest possible temperature (limited by the cooling medium temperature such as atmosphere, lake, rier temperature)

9 From the aboe table, it can also be seen that the energy has a quality More of the high-temperature thermal energy can be conerted to work herefore, the higher the temperature, the higher the quality of the energy will be he Carnot Refrigeration and Heat Pump Cycle A refrigerator or heat pump that operates on the reerse Carnot cycle is called a Carnot Refrigerator, or a Carnot heat pump he Coefficient of performance of any refrigerator or heat pump (reersible or irreersible) is gien by: COP R Q H / Q L and COP HP Q / Q COP of all reersible refrigerators or heat pumps can be determined from: COP R, re H / L and Also, similar to heat engine, one can conclude: COP R < COP COPth > COP R, re, re th, re COP HP, re L L H / irreersible refrigerator reersible refrigerator impossible refrigerator! Example 7-: Refrigerator Performance A refrigerator maintains the temperature of the freezer compartment at -5 C when the air surrounding the refrigerator is at C he rate of heat transfer from the freezer compartment to the refrigerant (the working fluid) is 8000 kj/h and the power input required to operate the refrigerator is 300 kj/h Determine the coefficient of performance of the refrigerator and compare with the coefficient of performance of a reersible refrigeration cycle operating between reseroirs at the same temperatures H Assumptions: Steady-state operation of the refrigerator he freezer compartment and the surrounding air play the roles of the cold and hot reseroirs, respectiely Chapter 7, ECE 309, Spring 06 9

10 Source ( H ) C 95 K W net 300 kj/h Refrigerator Source ( L ) - 5 C 68 K he coefficient pf performance of the refrigerator is: COP R Q C / W cycle COP R 8000 (kj/h) / 300 (kj/h) 5 he coefficient of performance of a Carnot refrigerator working between the same two reseroirs is: COP 99, / 95/ 68 R Carnot H C Chapter 7, ECE 309, Spring 06 0

11 Chapter 8: Entropy he second law leads to the definition of a new property called entropy he Clausius Inequality he first law is simply an energy balance Howeer, the second law leads to an inequality; an irreersible process is less efficient than a reersible process Another important inequality in thermodynamics is the Clausius inequality: δq 0 hat is, the cyclic integral of δq / is always less than or equal to zero his is alid for all cycles, reersible or irreersible For internally reersible cycles, it can be shown that: δq int, re Entropy he Clausius inequality forms the basis for the definition of a new property called entropy As can be seen in the equation aboe, for an internally reersible process the cyclic integral of δq / is zero A quantity whose cyclic integral is zero depends on the state only and not the process path, and thus it is a property Clausius in 865 realized that he discoered a new property and he called it entropy: Q ds δ int,re 0 (kj/k) Entropy per unit mass is designated by s (kj/kgk) he entropy change of a system during a process can be calculated: S S δq S int, re (kj/k) o perform this integral, one needs to know the relation between Q and during the process Note that the cyclic integral of δq / will gie us the entropy change only if the integration carried out along an internally reersible path between two states For irreersible processes, we may imagine a reersible process between the two states (initial and final) and calculate the entropy change (since entropy is a property) Chapter 8, ECE 309, Spring 06

12 he Increase of Entropy Principle Entropy change of a closed system during an irreersible process is greater that the integral of δq / ealuated for the process In the limiting case of a reersible process, they become equal δq ds he entropy generated during a process is called entropy generation, and is denoted by S gen, Chapter 8, ECE 309, Spring 06 S S S δq + S gen Note that the entropy generation S gen is always a positie quantity or zero (reersible process) Its alue depends on the process, thus it is not a property of a system he entropy of an isolated system during a process always increases, or in the limiting case of a reersible process remains constant (it neer decreases) his is known as the increase of entropy principle he entropy change of a system or its surroundings can be negatie; but entropy generation cannot S gen > 0 0 < 0 irreersible process reersible process impossible process - A process must proceeds in the direction that complies with the increase of entropy principle, S gen > 0 A process that iolates this principle is impossible - Entropy is a non-consered property, and there is no such thing as the conseration of entropy herefore, the entropy of unierse is continuously increasing 3- he performance of engineering systems is degraded by the presence of irreersibility he entropy generation is a measure of the magnitudes of the irreersibilities present during the process Entropy Balance Entropy is a measure of molecular disorder or randomness of a system, and the second law states that entropy can be created but it cannot be destroyed he increase of entropy principle is expressed as Entropy change Entropy transfer + Entropy generation S his is called the entropy balance system S transfer S gen

13 Entropy Change he entropy balance is easier to apply that energy balance, since unlike energy (which has many forms such as heat and work) entropy has only one form he entropy change for a system during a process is: Entropy change Entropy at final state - Entropy at initial state S S + S system herefore, the entropy change of a system is zero if the state of the system does not change during the process For example entropy change of steady flow deices such as nozzles, compressors, turbines, pumps, and heat exchangers is zero during steady operation Mechanisms of Entropy ransfer Entropy can be transferred to or from a system in two forms: heat transfer and mass flow hus, the entropy transfer for an adiabatic closed system is zero Heat ransfer: heat is a form of disorganized energy and some disorganization (entropy) will flow with heat Heat rejection is the only way that the entropy of a fixed mass can be decreased he ratio of the heat transfer Q/ (absolute temperature) at a location is called entropy flow or entropy transfer final Entropy tr ansfer with heat initial Q S heat Since (in Kelin) is always positie, the direction of entropy transfer is the same of the direction of heat transfer When two systems are in contact, the entropy transfer from warmer system is equal to the entropy transfer to the colder system since the boundary has no thickness and occupies no olume Note that work is entropy-free, and no entropy is transferred with work Mass Flow: mass contains entropy as well as energy, both entropy and energy contents of a system are proportional to the mass When a mass in the amount of m enters or leaes a system, entropy in the amount of ms (s is the specific entropy) accompanies it Entropy Balance for a Closed System A closed system includes no mass flow across its boundaries, and the entropy change is simply the difference between the initial and final entropies of the system he entropy change of a close system is due to the entropy transfer accompanying heat transfer and the entropy generation within the system boundaries: Entropy change of the system Entropy transfer with heat + Entropy generation Chapter 8, ECE 309, Spring 06 3

14 Q k S S + k herefore, for an adiabatic closed system, we hae: S S adiabatic S gen For an internally reersible adiabatic process S 0, because S gen 0 he total entropy generated during a process can be determined by applying the entropy balance to an extended system that includes both the system and its immediate surroundings where external irreersibility might be occurring Example 8-: Entropy balance for a closed system Saturated liquid water at 00 C is contained in a piston-cylinder assembly he water undergoes a process to the corresponding saturated apor state, during which the piston moes freely in the cylinder here is no heat transfer with the surroundings If the change of state is brought about by the action of a paddle wheel, determine the network per unit mass, in Kj/kg, and the amount of entropy produced per unit mass, in kj/kgk gen Water Insulated Paddle wheel Assumptions: - he water in the piston-cylinder assembly is a closed system - here is no heat transfer with the surroundings 3- he system is at an equilibrium state initially and finally PE KE 0 Solution he network can be calculated by using the law: U + KE + PE Q W hat is simplifies to: U - W Chapter 8, ECE 309, Spring 06 4

15 On a unit mass basis, the energy balance becomes: From able A-4, W / m - (u g u f ) W / m kj/kg he negatie sign indicates that the work input by the stirring is greater than the work done by the water as it expands Using an entropy balance, the amount of entropy produced can be found Since there is no heat transfer, On a unit mass basis, this becomes: δq S + S gen S gen 3 0 Using able A-4 S gen / m s g - s f S gen / m 6048 kj / kgk Entropy Balance for a Control Volume In addition to methods discussed for closed system, the entropy can be exchanged through mass flows across the boundaries of the control olume m i s i Control olume Q m o s e he entropy balance in the rate form for a control olume becomes: ds dt CV Qk + m i si k m e se + For a steady-state steady-flow process, it simplifies to: S gen, CV m e se m i si Q S gen, CV k k Chapter 8, ECE 309, Spring 06 5

16 Example 8-: Entropy balance for a CV Steam enters a turbine with a pressure of 3 MPa, a temperature of 400 C, and a elocity of 60 m/s Saturated apor at 00 C exits with a elocity of 00 m/s At steady-state, the turbine deelops work equal to 540 kj/kg Heat transfer between the turbine and its surroundings occur at an aerage outer surface temperature of 350 K Determine the rate at which entropy is produced within the turbine per kg of steam flowing, in kj/kgk Neglect the change in potential energy between inlet and exit Assumptions: - Steady state operation in CV PE 0 - urbine outer surface is at a specified aerage temperature P 3 MPa 400 C V 60 m/s W / m 540 kj/kg urbine b 350 K 00 C V 00 m/s Sat apor From the mass balance, we know that m m m Since the process is steady-state, one can write: Chapter 8, ECE 309, Spring Qk + m ( si se ) + S gen, CV k he heat transfer occurs at b 350 K, the first term of the right hand side of the entropy balance reduces to Q / b S m gen, CV Qk m k + ( s s) We need to calculate the rate of heat transfer he first law (energy balance) can be used to find the heat transfer rate Combining the mass balance and the first law, one finds: Q W CV CV m m + ( h ) + h V V

17 From able A-6, h 3309 kj/kg, and From A-4 h 676 kj/kg After substitution, and conerting the units, one finds: Q CV 6 kj / kg m From able A-4, s kj/kgk and from able A-6, s 69 kj/kgk Inserting alues into the expression for entropy production: S gen, CV m Qk + ( s s) m k kj / kg K Entropy Entropy can be iewed as a measure of molecular disorder, or molecular randomness As a system becomes more disordered, the positions of the molecules become less predictable and the entropy increases Entropy kj/(kgk) Gas Liquid Solid Fig 8-: Entropy of a substance (leel of disorder) increases when it melts from solid phase to liquid S solid < S liquid < S gas Some remarks: Work is an organized form of energy, free of disorder or randomness, thus free of entropy herefore, there is no entropy associated with energy transfer as work he quantity of energy is always presered during an actual process, based on the first law, but the quality is bound to decrease (the second law) Processes can occur only in the direction of increased oerall entropy or molecular disorder hus, the entire unierse is getting more and more chaotic eery day At absolute zero (0 K), molecules become completely motionless, this represents a state of ultimate molecular order (and minimum energy) herefore, the entropy of Chapter 8, ECE 309, Spring 06 7

18 a pure crystalline substance at zero temperature is zero hat is because; there is no uncertainty about the state of the molecules at that instant his statement is the third law of thermodynamics Since there is a reference for entropy (absolute zero), entropy is an absolute property he entropy measured with respect to absolute zero is called absolute entropy he two diagrams used most extensiely in the second-law analysis are the -s and h-s diagrams For an internally reersible process, one can write: δ Q re ds int, (kj) Internally reersible process Q ds Fig 8-: On a -s diagram, the area under an internally reersible process presents the heat transfer for the process For an internally reersible isothermal process, we hae: Q int,re 0 ds In a -s diagram, an isentropic process is represented by a ertical line An isentropic process is a process in which entropy remains constant As a result an isentropic process inoles no heat transfer herefore: Isentropic process (s s ) Reersible + Adiabatic Ealuation of Entropy Change he differential form of the conseration of energy for a closed system (fixed mass) for an internally reersible process is: δq int,re - δw int,re du where, δq int,re ds δw int,re PdV hus, Chapter 8, ECE 309, Spring 06 8 s

19 or, per unit mass his is called the first Gibbs equation ds du + PdV ds du + Pd From the definition of enthalpy, h u + P, one can find: h u + P dh du + Pd + dp Eliminating du from the first Gibbs equation, one finds the second Gibbs equation: ds dh dp Explicit relations for differential changes in entropy can be obtained from Gibbs equations: du Pd ds + dh dp ds o calculate the entropy change, we must know the relationship between du or dh and temperature Calculation of the Entropy for Saturated Mixture Use ables A-4 and A-5 to find s f, s g and/or s fg for the following: s ( x)s f + x s g or s s f + x s fg Calculation of the Entropy for Superheated Vapor Gien two properties or the state, such as temperature and pressure, use able A- 6 Calculation of the Entropy for Compressed Liquid In the absence of compressed liquid data for a property s s f@ Entropy Change of Solids and Liquids Solids and liquids can be assumed as incompressible substances since their olumes remains essentially constant during a process hus, the first Gibbs equation becomes: ds du d s s C( ) cd Assuming an aeraged alue for specific heat, one obtains: Chapter 8, ECE 309, Spring 06 9

20 Chapter 8, ECE 309, Spring 06 0 ln C s s ae Note that the entropy change of an incompressible substance is only a function of temperature herefore, for an isentropic process where s s, one can find: Entropy Change of Ideal Gas he entropy change of an ideal gas can be obtained, by substituting du C d and P R / into Gibbs equation ln ) ( R d C s s d R d C ds + + Assuming aeraged alues for specific heats, one obtains: K kg kj P P R C s s K kg kj R C s s ae p ae ln ln ln ln,, + For isentropic processes of ideal gases, the following relationships can be found by setting ds 0, isentropic process ln ln ln ln k C R or C R Since R C p C, k C p / C, and thus R / C k In a similar manner, one finds: isentropic process isentropic process / ) ( k k k P P P P hese equations can be expressed in the following compact forms: k - constant P ( k) / k constant

21 P k constant he specific ratio k, aries with temperature, and in isentropic relations aboe an aerage k alue should be used Example 8-3: Isentropic process of ideal gas A rigid, well-insulated tank is filled initially with 5 kg of air at pressure 500 kpa and a temperature 500 K A leak deelops, and air slowly escapes until the pressure of the air remaining in the tank is 00 kpa Using the ideal gas model, determine the amount of mass remaining in the tank and its temperature Mass initially in the tank that remains in the tank (m ) Slow leak Insulated tank Mass initially in the tank that escapes (m ) Assumptions: - As shown in the figure, the closed system is the mass initially in the tank that remains in the tank - here is no significant heat transfer between the system and its surroundings 3- Irreersibilities within the tank can be ignored as the air slowly escapes Solutions: Using the ideal gas equation of state, the mass initially in the tank that remains in the tank at the end of process is: m m PV R m PV R P P Since the olume of the tank V remains constant during the process We need to find the final temperature For the closed system under consideration (m ), there are no irreersibilities, and no heat transfer Accordingly, it is an isentropic process, and thus the isentropic relationships can be used: m Chapter 8, ECE 309, Spring 06

22 P P ( k ) P P / k ( k ) / k With a constant k 4 for air, after substituting alues, one finds: 3555 K Finally, inserting alues into the expression for system mass m (00/500) (500/3555) (5 kg) 58 kg Reersible Steady-Flow Work he conseration of energy equation for a steady-flow deice undergoing an internally reersible process can be expressed in differential form as δqre δwre dh + dke + dpe But δqre ds δqre dh dp ds dh dp Substituting into the relation aboe, after canceling dh, it yields, Integrating, we find w -δw re dp + dke + dpe dp ke pe With negligible changes in potential and kinetic energies, w re re dp ( kj / kg) ( kj / kg) From the aboe equation can be seen that, the larger the specific olume the larger the reersible produced or consumed work by the steady-flow deice hus, eery effort should be made to keep the specific olume of the flow as small as possible during a compression process to minimize the input work When the fluid is incompressible, the specific olume remains constant during the process, thus the aboe equation becomes: δw re (P P) ke pe (kj/kg) For a steady-state flow of a liquid through a deice that inoles no work interactions (such as nozzle or a pipe section), the work term is zero, Chapter 8, ECE 309, Spring 06

23 V V ( P P ) + + g( z z ) 0 his is known as Bernoulli equation in fluid mechanics Chapter 8, ECE 309, Spring 06 3