Presentation on theme: "The Second Law of Thermodynamics"— Presentation transcript:
1The Second Law of Thermodynamics Consider the following process:A rock spontaneously rises by loweringits temperature such that mCpDT = mghso that DU = 0.Since energy is conserved, this typeof process is not forbidden by thefirst law, but we know this neverhappens! There is a natural flow ofthings or direction for spontaneousprocesses to occur.We also know that various forms ofwork can be completely convertedinto heat, e.g., rubbing of two rockstogether in a heat reservoir such thatthey undergo no temperature changeso that W = Q and DU = 0.mghIn general work of any kind can be done on a system in contactwith a reservoir giving rise to a flow of heat without alteringthe state of a system, W = Q. Work can be converted entirely into heat by a suitable dissipative process.
2Heat Engines - The conversion of heat into work In order to convert heat into work we require a machine that willconsume heat and produce work. The machine itself must not suffer anypermanent change; it must play a passive role in that following the processit must return to its initial state. The machine must pass through a cycle.def.Thermal efficiency,Applying the first law to the operation of the machine or engine, W = Q1- Q2where Q2 corresponds to any heat rejected from the engine,Q1WHeat enginesystemQ2
3Carnot CycleThe Carnot cycle is a reversible cycle operating between two temperatures.A B: Isothermal expansion adsorbing heat Q1.B C: Adiabatic expansion decreasing T from T1 to T2.C D: Isothermal compression rejecting heat Q2.D A: Adiabatic compression increasing T from T2 to T1.*Note that if the cycle is operated in reverse refrigerator.pVABCDT1T2Q1Q2WQ1Q2T1T2WCarnot cycle for a gasSince all the steps are reversible DU = 0, W = Q1 - Q2 and
4The 2nd LawKelvin Statement - No process is possible whose sole result is the completeconversion of heat into work. This addresses the efficiencyof conversion.Clausius Statement - No process is possible whose sole result is the transfer of heatfrom a colder to a hotter body. Spontaneity of processes and theirreversibility of nature.Kelvin Statement Clausius StatementCarnot’s Theorem:No engine operating between two given reservoirs can be more efficient than aCarnot engine operating between the same two reservoirs.
5Proof of Carnot’s Theorem WC = QC1 - QC2WH = QH1 - QH2T1T2QH1QH2QC2QC1Assume the existence of a Hypothetical engine such that,Since the Carnot engine is reversible we can drive it backwards using the mechanical energy from H. The Carnot cycle can be adjusted (adiabats) so that in one cycle it uses exactly as much work as H produces.
6Proof of Carnot’s Theorem WC = QC1 - QC2WH = QH1 - QH2T1T2QH1QH2QC2QC1Composite EngineNow consider C and H asa Composite Engine.This composite engineproduces no net work butsimply extracts heat from acold reservoir and deliversan amount of heat,to a hotter reservoir.This is a violation of the Clausius statementof the second law!Carnot’s Theorem:
7are equally efficient. For any reversible engine, Corollary: All reversible engines operating between the same temperature reservoirs are equally efficient.Thus the efficiency of any reversible engine operating between the same reservoirsare equally efficient. For any reversible engine,Qn+2Tn+2Wn+2Tn+1TnTn-1Cn+2Cn+1CnQn+1Qn-1QnWn+1WnConsider a series of Carnot Engines
8to the ratios of heat exchanged by a reversible For the composite engine,Qn+2Tn+2Wn+2Tn+1TnTn-1Cn+2Cn+1CnQn+1Qn-1QnWn+1WnThen,This can only be true if the f’s factorize such thatTherefore: The ratio of the temperatures of the reservoirs is equalto the ratios of heat exchanged by a reversibleengine operating between the same reservoirs.
9According to Carnot’s theorem and its corollary we can make the following statements: Therefore Taking the heat entering the system as positive, we can sayFor any closed cycle, , where the equality necessarily holdsfor a reversible cycle.
10Entropydef:We can now define a new variable, the entropy S, by the relationdS = for an infinitesimal reversible change. This definitionholds for reversible changes only. For a finite reversible change of state, the change in entropy is given by,QrevT
11Entropy in Irreversible Processes Since entropy is a state function, the change in entropy accompanying a statechange must always be the same regardless of how the state change occurs. Onlywhen the state change occurs reversibly is the entropy change related to theheat transfer by the equationBXConsider an irreversible change AB. Constructany reversible path R thus forming an irreversiblecycle ABRA. For the irreversible cycle theClausius theorem says,RAxDetermination of the change inentropy for an irreversible change
12Taking the integral in two parts, i.e.,Butby definition of entropy. ThusThus we have this general result for a differentialirreversible change.orFor a thermally isolated system Q = 0 and we have thegeneral result known as the law of increase of entropy.
13Some Interesting Examples Isothermal expansion of an Ideal GasAdiabatic Free expansion of an Ideal Gas(Joule expansion - no Q or W exchangedwith surroundings)This is an irreversible process, but we canalways use the combined Law and integratefrom the initial to final state by a convenientreversible path.Combined 1st & 2nd Law
14Isothermal dissipation of Work Electrical work is dissipated isothermally by heat flow intoa reservoir. There is no entropy change of the system becauseit’s thermodynamic coordinates do not change. Thereservoir adsorbs Q = W units of heat at temperature T so itsentropy change isDiathermal WallsP1, V1Reservoir (T)Adiabatic dissipation of WorkElectrical work is dissipated in a thermally isolatedsystem maintained at fixed pressure. The T of the system increases irreversibly. The coordinates of the system change from P,T1 to P,T2 . The entropy changecan be calculated byComposite systemAdiabatic WallsP, T1
15Examples No good! An inventor claims to have developed a power cycle capable of delivering anet work output of 410 kJ for an energyinput by heat transfer of 1000 kJ. Thesystem undergoing the cycle receivesheat transfer from hot gases atT = 500 K and discharges energy byheat transfer to the atmosphere atT = 300 K. Evaluate this claim.The thermal efficiency isThe maximum efficiency of any power cycle isNo good!
16Different Forms of the Combined 1st and 2nd Law Using the definition of entropyWe can use these forms to determinethe entropy change of an ideal gassubjected to changes in p, v, T.Since enthalpy is defined as,H = U + pVRearranging these equations andwriting them on a unit mass basis,We already know that for an ideal gas,
17Entropy Production B X R A x Determination of the change in It is convenient to define a quantity such thatAxDetermination of the change inentropy for an irreversible changeRecall the determination of entropy change for an irreversible process. is necessarily a positive quantity called entropy production.Def:
18Entropy Production If the end states are fixed the entropy change on the left hand side of this equation can bedetermined.Rewriting the expressionThe 2 terms on the RHS of theequation are path dependant.we obtain,The 1st term on the RHS of theequation is the entropy transfer associatedwith heat transfer. The direction or sign ofthe entropy transfer is the same as heattransfer.The 2nd term is the entropy production term.
19Entropy Production can not be less than zero The entropy balance can be expressedin various forms. If heat transfer takesplace along several locations on theboundary of the system where the temperaturesdo not vary with position or time,Here Qj/Tj is the amount of entropytransferred to the portion of theboundary at temperature Tj . can not be less than zeroBy contrast the change in the entropyof the system can be positive, negativeor zero:On a time rate basis for a closedsystem
20ExamplesWater initially a saturated liquid at 100 ºC is contained in a piston cylinder assembly.The water undergoes a process to the corresponding saturated vapor during which thepiston moves freely in the cylinder. If the change of state is brought about by heatingthe water as it undergoes an internally reversible process at constant pressure andtemperature determine the work and the heat transfer per unit of mass in kJ/kgAt constant pressure the work is simply,Table A-2
21ExamplesSince the process is reversible and occurs at constant temperatureThis could also have been calculated our old way
22Examples The figure shows a system receiving heat Q from a reservoir. By definition the reservoiris free of irreversibilities, but the system is not,fluid friction, etc. Let’s determine theentropy change of the system and that of thereservoir.For the system,For the reservoir
23ExamplesWater initially a saturated liquid at 100 ºC is contained in a piston cylinder assembly.The water undergoes a process to the corresponding saturated vapor during which thepiston moves freely in the cylinder. There is no heat transfer with the surroundings.If the change in state is brought about by the action of a paddle wheel, determine thenet work per unit mass in kJ/kg and the amount of entropy produced per unit mass inkJ/kg-K.As the volume of the system increases during this process, there is an energy transferby work from the system during the expansion as well as an energy transfer by workto the system done by the paddle wheel. The net work is evaluate from the changein internal energy.
24Examples From the 1st Law, U = -W. On a unit mass basis we have, The minus sign indicates that the work input by stirring is greater in magnitude than thework done by the water as it expands.The entropy produced is evaluated by applying the entropy balance,On a unit mass basis,
26Entropy Diagrams T S Carnot cycle on a T – S Area representation of heat transferfor an internally reversible processof a closed system.STCarnot cycle on a T – Sdiagram.CCW – refrigeration cycleCW – power cycle
27Thermodynamic Potentials Combined 1st and 2nd LawdU = TdS - pdVEnthalpy is a function of S and pPotential Function in terms of S and p, EnthalpyLengendre Transform subtract a -d(pV) termfrom dUdU + d(pV) = TdS - pdV + d(pV)d(U + pV) = TdS + Vdpwhere H = (U + pV) is the Enthalpyand H = H (S, p)Potential Function in terms of T and p,Gibbs Free EnergyLengendre Transform add a -d(TS) termto dHd ( H-TS ) = Vdp - SdTwhere G = (H-TS) is the Gibbs Free Energyand G = G (p, T)
28Four Fundamental Thermodynamic Potentials Potential Function in terms of T and V,Helmholtz Free EnergyFour Fundamental Thermodynamic PotentialsdU = TdS - pdVdH = TdS + VdpdG = Vdp - SdTdA = -pdV - SdTThe appropriate thermodynamic potentialto use is determined by the constraintsimposed on the system. For example,since entropy is hard to control (adiabaticconditions are difficult to impose) G and Aare more useful. Also in the case of solidsp is a lot easier to control than V so G isthe most useful of all potentials for solids.Lengendre Transform subtract a -d(TS) term from dUd(U-TS) = -pdV - SdT = dAwhere A = A(V, T) is theHelmholtz Free EnergyThe Maxwell relations are useful in thatthe relate quantities that are difficult orimpossible to measure to quantities thatcan be measured.
29Some important bits of information For a mechanically isolated system kept at constant temperature and volumethe A = A(V, T) never increases. Equilibrium is determined by the state ofminimum A and defined by the condition, dA = 0.For a mechanically isolated system kept at constant temperature and pressurethe G = G(p, T) never increases. Equilibrium is determined by the state ofminimum G and defined by the condition, dG = 0.Consider a system maintained at constant p. Then