您好,欢迎访问三七文档
当前位置:首页 > 临时分类 > 类卡诺热机最大功率对应的效率问题
类卡诺热机最大功率对应的效率问题的几个最新研究进展涂展春北京师范大学物理学系Email:tuzc@bnu.edu.cnHomepage:(NO.11075015)中国物理学会2011年秋季学术会议——2011.9.15-18浙江大学Outline●Introduction●Recentadvances–EMP(最大功率对应的效率)ofstochasticheatengine–EMPofFeynman'sratchet*–UniversalityinEMPproblems–BoundsofEMPforlow-dissipationCarnotengines–EMPproblemsviewedfrommin{ΔSir}*●ProspectsIntroductionIsothermalexpansionAdiabaticexpansion(absorb)(release)IsothermalcompressionAdiabaticcomp.●CarnotefficiencyC=1−T2T1Carnotefficiency:=∣W∣/Q1CEfficiency:IdealCarnotheatengineEnginesworkingbetween2reservoirsHowever,theenginesattheCarnotefficiencycannotproduceapoweroutputbecausetheCarnotcyclerequiresaninfinitelyslowprocessToobtainfinitepoweroutput,heatenginesneedspeedingupQuestion:Ifwetrytoachievethemaximumpoweroutputofaheatengine,whatisthecorrespondingefficiency?●MacroscopicCarnot-likeheatengine(1975)22CA=1−T2T1=1−1−C[Remark:Novikov(1957)andChambadal(1957)foratomicpowerstations]!有限时间热力学AbsorbHeat:Q1=T1−T1wt1(1):isothermalexpansion(3):isothermalcompressionReleaseHeat:Q2=T2w−T2t2EndoreversibleassumptionTotaltimeassumption:ttotal=t1t2=Poweroutput:P=Q1−Q2/t1t2⇒Q1T1w=Q2T2w∂P∂T1w=∂P∂T2w=0⇒T2wT1w=T2T1max{P}:Efficiencyatmaximumpower:CA=Q1−Q2Q1=1−T2wT1w=1−T2T1121w2wT1T2T1wT2w(1)(2)(3)(4)Assumptiononheatexchanges:AbsorbHeat:Q1=T1w−1−T1−1t1ReleaseHeat:Q2=T2−1−T2w−1t2CY=C2−CYCCY=2C4−CInparticular,whenα=β,dependonαandβCY=11/我国学者严子俊、陈丽璇、陈金灿、陈林根等的研究也极大地推动了有限时间热力学的发展Recentadvances(1)IsothermaltransitionattemperatureT1duringtimeinterval0tt1(2)Adiabatictransition(quickly)fromtemperatureT1toT2attimet=t1(3)IsothermaltransitionattemperatureT2duringtimeintervalt1tt1+t3(4)Adiabatictransition(quickly)fromtemperatureT2toT1attimet=t1+t3Vx,t=tx22SS=2C4−CT1T2(high)(low)●EMPofstochasticheatengineQuestion:CanthesmallloadbeliftedifT1=T2=T?Answer:Ataglance,itisquitepossible.However,the2ndLawofThermodynamics(Heatcannotbeconvertedtoworkspontaneously)prohibitsthishappening.Uratchetφ●EMPofFeynman'sratchetφUtot=UratchetMMφφ[Chapter46,Feynmanlecture]23RatchetsystemasareversibleengineifT1T2Forward:accumulateenergyMbyvaneAbsorbheatfromthehotbathOutputworkQ1=MM⇒rateF=r0e−M/kBT1Backward:accumulateenergybypawl⇒rateB=r0e−/kBT2TakefromthecoldbathInputworkHeatreleasedtohotbathviathevaneQ2=MReversible=detailedbalance:rateF=rateB⇒Q1T1=Q2T2Efficiency:=MQ1=Q1−Q2Q1=1−T2T1≡Cmaximum!φUtotMφθ2θ[Chapter46,Feynmanlecture]HeatreleasedtocoldbathviathepawlQ2=Q1=MBreakdetailedbalance:rateF≠rateBRatchetsystemasanirreversibleengineQuestion:theefficiencyatmaximumpowerforthissystem?Let'sgoahead---rateF=r0e−M/kBT1rateB=r0e−/kBT2=rateF−rateB=r0[e−M/kBT1−e−/kBT2]P=M=r0e−/kBT1M[e−M/kBT1−e−/kB1/T2−1/T1]Rotationalpower:Introducingtwodimensionlessvariables:m≡M/kBT1,≡/kBT2⇒Pm,=kBT1r0e−1−Cme−m−e−CAverageangularvelocity:Forwardandbackwardrates:Optimizingthepower∂P/∂m=0⇒1−m∗e−m∗=e−∗C∂P/∂=0⇒1−Ce−m∗=e−∗Cm∗=C∗=1−ln1−ccEfficiencyatmaxpower:∗=m∗m∗∗1−C=C[CC−1−Cln1−C]=MQ1=MM⇒(1)slightlyhigherthanthevalueofC-A(2)slightlyhigherthanthevalueofS-Satasmallrelativetemperaturedifference;evidentdeviationfromS-Satalargerelativetemperaturedifference.T1T2∗=C2C287C396OC4CA=1−T2T1=C2C286C396OC4SS=2C4−C=C2C283C396OC4“universalefficiencyatmaximumpower,C/2C2/8,shouldexistatsmallrelativetemperaturedifferences”Conjecture[Tu,JPA(2008)]φ●UniversalityinEMPproblems●BoundsofEMPforlow-dissipationCarnotenginesLow-dissipationansatz:SirT∝1tTimeforcompletingan“isothermal”process∗=C2−ECwith0γE1[SeealsoChen-Yan(1989),Schmiedl-Seifert(2008)]●EMPproblemsviewedfrommin{ΔSir}Carnot-likecycle1.``Isothermalexpansionprocess.2.Adiabaticexpansionprocess.3.``Isothermalcompressionprocess.4.Adiabaticcompressionprocess.absorbQ1,bathT1,timet1S1=Q1/T1S1irS2=0,Q2=0timet2releaseQ3,bathT3,timet3S3=−Q3/T3S3irS4=0,Q4=0timet4Nonetenergyandentropyproduction[arXiv:1108.2873]KeyassumptionsTotaltimeassumption:ttotal=t1t3SjirTj=j−1∫0tj[qj]2dj=1,3Quadratic-formansatz:Heattransferratebetweenworkingsubstanceandthermalbath(note:linearirreversiblethermodynamics)max{P}&min{ΔSir}Thermodynamics+Totaltimeassumption:=P∝=max{P}≡min{Sjir}forgiventj(j=1,3)min{ΔSir}&timemin{Sjir}withconstraint∫0tjqjd=Qjmin{Sjir}=Qj2jtjS1=Q1/T1S1irS3=−Q3/T3S3ir=S2Tj2jtj,tj∞S,tj00qj=Qj/tj=const.duringtjC2≡−maxPC2−C≡Low-dissipationansatzorquadratic-formansatzisasufficientbutnon-necessaryconditionfortheexistenceoftwobounds.Theseboundsmightexistinamoregeneralconditionthanlow-dissipationansatzorquadratic-formansatz.Efficiencyatmax{P}anditsboundsSubstitutingtheexpressionsofmin{ΔSjir}intotheexpressionofP∂P∂t1=∂P∂t3=0Q3maxPST3=21/322T3/T111/3≡MQ1maxPST1=4MT3/T14−MMT3/T1maxP=1−Q3maxP/Q1maxP=C2−C,=1/13/1max{P}andendoreversibleS1=Q1/T1S1irS3=−Q3/T3S3irqj=Qj/tj=const.duringtjmin{Sjir}=Qj2jtjSimilarly,Recall,⇒Q1T1e=Q2T2eEnginesw
本文标题:类卡诺热机最大功率对应的效率问题
链接地址:https://www.777doc.com/doc-3297351 .html