1Chapter 6: Using EntropyCombining the 1stand the 2ndLaws of ThermodynamicsPhoto courtesy of U.S. Military Academy photo archives.Introducing Entropy• The Clausius and Kelvin-Plank Statements constitute the 2ndlaw of thermodynamics• Carnot corollaries apply to systems communicating with a hot and cold reservoir only• They are used to establish the max efficiency and COP for various thermodynamic cycles• Another corollary, known as Clausius Inequality, will be introduced that can be applied to any cycle independent of the bodies it interacts with.• This corollary will lead to the definition of entropy.2Clausius Inequality0 no irreversibilities present within the system0 irreversibilities present within the system0 impossiblecyclecyclecycleσσσ=><cyclebTQσ−=⎟⎠⎞⎜⎝⎛δ∫• Inequality:• : cyclic integral over the boundary and cycle• Unlike energy, entropy in the presence of irreversibilities, is always produced.Tb δQ∫Sketch of ProofbrbrTTQQ=δδ0≤δ∫rQTemperature Scale:By the 2ndLaw:0≤⎟⎠⎞⎜⎝⎛δ∫bTQδQrδQbReversible cycleδWReservoir, Trsystemsystem3Defining Entropy• The Clausius inequality states that for an internally reversible cycle• This implies that is independent of the path from state 1 to 2.• Therefore can be considered a property change• This integral is defined to be the entropy at state 2 relative to 1:0=⎟⎠⎞⎜⎝⎛δ∫bTQ∫δ21TQ∫⎟⎠⎞⎜⎝⎛δ21revintTQ∫⎟⎠⎞⎜⎝⎛δ=−21revint12TQSSFinding Entropy DataFor Water and Refrigerants:Motivation: Application of the 2ndLaw of Thermodynamics frequently requires finding specific entropy data.Using Tables A-2 – A-18For example:P1 = 3 MPaT1= 500 oCUse Table A-4;s1 = 7.2338 kJ/kg K(1 ) ( )fgf gfsxs xs s xs s=−+=+−Saturation Data:Liquid Data:In the absence of tables, use the Compressed Liquid Rule4Graphical Entropy DataTemperature-Entropy DiagramEnthalpy-Entropy (Mollier) DiagramT dS EquationsTdS dU PdV=+TdS dH VdP=−Tds du Pdv=+Per Unit Mass:Per Unit Mass:Tds dh vdP=−5Finding Entropy DataFor Ideal Gases:Motivation: Application of the 2ndLaw of Thermodynamics frequently requires finding specific entropy data.For use with Table A-22()()vpdu C Tdh C TPvRT===For an Ideal Gas:Using these relations with the Tds equations yields:()()222 11 2 11,,()()lnPsTP sTP sT sT RP⎛⎞−=−−⎜⎟⎝⎠oo()()222 11 2 11,,()()lnPsTP sTP sT sT RP⎛⎞−=−−⎜⎟⎝⎠ooFor use with Table A-23Finding Entropy Data Internally Reversible ProcessesInternally reversible processes are idealizations, but are found in all Carnot cycles. intrev2intrev1QdSTQTdSδ⎛⎞=⎜⎟⎝⎠=∫For internally reversible processes, the area under the process curve represents the process Heat Transfer.• Isothermal IR Process: Qir=T(S2-S1)• Adiabatic IR Process: Tds=0⇒S2=S16Carnot CycleReversible power cycle: Two adiabatic processes alternated with two isothermal processesCarnot power cycles operated in reverse may be regarded as a reversible refrigeration or heat pump cycle, with maximum coefficient of performanceExample: Problem 6.367Ideal Gaseswith Constant Specific Heats()()2222 1111, , ln lnpTPsT P sTP C RTP⎛⎞ ⎛⎞−= −⎜⎟ ⎜⎟⎝⎠ ⎝⎠()()2222 1111, , ln lnvTvsT v sTv C RTv⎛⎞ ⎛⎞−= +⎜⎟ ⎜⎟⎝⎠ ⎝⎠Use Specific heat data in Tables A-20 and A-21For example:Air (Ideal Gas), Constant Specific HeatsP1 = 3 MPa, P2= 1 MPaT1= 500 K, T1= 300 KUse Table A-20, at Tavg= 400 K;Cp = 1.013 kJ/kg KΔs= -0.202 kJ/kg KFinding Entropy DataFor Incompressible SubstancesIncompressible substance model is generally used with liquids and solids only, and assumes constant specific volume.2211lnTssCT−=C = Specific Heat or Heat CapacityClipArt courtesy of M.S. Office 20028Entropy Balance: Closed Systems2211bQSSTδσ⎛⎞−= +⎜⎟⎝⎠∫Entropy ChangeEntropy TransferEntropy ProductionSince σ measures the effect of irreversibilities present within the system during a process, its value depends on the nature of the process, and thus is NOT a propertyOther common forms of the entropy balance:21bQSSTσ−=+Uniform Boundary TemperaturejjjQdSdt Tσ••=+∑Rate basisIncrease in Entropy Principle:Closed SystemsThe summation of the entropy changes of surroundings AND system must always increase or remain the same]]totalsystem surrSSσΔ+Δ=0 ideal0 actual0 impossibletotaltotaltotalσσσ=><9Example Problem 6.59Example Problem 6.7010Example Problem 6.77The Reynold Transport Theoremwhere β is the specific property corresponding to B. ThennVVnrr•=Where is the velocity normal to the control surface and dBCV/dt is the rate of change of B within the control volume:Let B be an extensive property of a system: ∫βρ=CVdVtB )(sys∫βρ+=CSnCVdAVdtdBdtdBsys()dVtdtdBCVCV∫∂βρ∂=nrCVCSΩCS: Control SurfaceCV: Control VolumeSys=CV at time t11Entropy Balance• Letting β=s (specific entropy) and B=S in the RTT,• By the 2ndlaw• Combing the 2 equations give∫ρ+=CSnCVsysdAVsdtdSdtdSσ+⎟⎟⎠⎞⎜⎜⎝⎛=∫&&CSbsysTQdtdSnrCVCSσ+ρ−⎟⎟⎠⎞⎜⎜⎝⎛=∫∫&&CSnCSbCVdAVsTQdtdSEntropy Rate Balance:Control Volumesjcvii ee CVji ejQdSms msdt Tσ••••=+ − +∑∑ ∑Rate of entropy changeRates of entropy transferRate of entropy productionIntegral Form:()()nnVA A AieiebdqsdV dA s V dA s V dAdt Tρρρ•⎛⎞⎜⎟=+ −⎜⎟⎝⎠∑∑∫∫ ∫ ∫At Steady-State:0CVjii eeji ejQms msTσ••••=+ − +∑∑ ∑12Special Case: SISO• The entropy balance for a Single Inlet Single Outlet:• If the system is adiabatic (no heat transfer), then∑∑σ+⎟⎟⎠⎞⎜⎜⎝⎛=−⇒=σ+−+⎟⎟⎠⎞⎜⎜⎝⎛jcvjjiecveijjjmTQmssssmTQ&&&&&&&10)(msscvie&&σ=−Example: Problem 6.141 (a)13Isentropic EfficienciesComparing actual (adiabatic) and isentropic devices with- same inlet states- same exit pressureTurbinesThis helicopter gas turbine engine photo is courtesy of the U.S. Military Academy.1stLaw:2ndLaw:12cvWhhm••=−WTurbine2 Turbine1 WWTurbine2 Turbine1 210cvssmσ••=−≥Isentropic Turbine Efficiency:1212cvTscvsWmhhhhWmη••••⎛⎞⎜⎟−⎜⎟−⎝⎠==−⎛⎞⎜⎟−⎜⎟⎝⎠Example: Problem 6.141 (b)14Isentropic EfficiencyNozzlesV1NozzleV2NozzleV2Clipart courtesy of M.S. Office 20022222/2/2NsVVη=Isentropic Efficiencies: Compressors and