第三講閉口開口系統(tǒng)能量方程及例題名師優(yōu)質(zhì)課賽課一等獎(jiǎng)市公開課獲獎(jiǎng)?wù)n件

2.3熱力學(xué)第一定律普通表示式

當(dāng)系統(tǒng)由狀態(tài)1經(jīng)歷一系列狀態(tài)改變到達(dá)終狀態(tài)2時(shí)，系統(tǒng)總能量改變?yōu)棣sys＝Ein－Eout＝（minein＋Q）－（mouteout＋W）當(dāng)系統(tǒng)處于宏觀上靜止時(shí)

ΔU＝（minein＋Q）－（mouteout＋W）看問題角度:進(jìn)入系統(tǒng)能量為正,離開為負(fù)

moutminQWΔEsys能量守恒第1頁2.4閉口系統(tǒng)能量方程

AConservationof

EnergyPrincipleforClosedSystemsAclosedsystemundergoingaprocessbetweenstate1andstate2,theenergyequationwillbe:

ΔU＝Q

－W

or

Q＝ΔU＋W

orfor1kgofmass

q＝Δu＋wforadifferentialbasis

δQ＝dU＋δWorδq＝du＋δw

第2頁這里功數(shù)量并不等于能夠利用功Forareversibleprocess,theworkcanbecalculatedas:

Foraclosedsystem,heattransferandworktransferaretheonlymechanismsbywhichenergycanbetransferredacrosstheboundary.Ifweneedtoexpressthegeneralenergybalanceonaratebasisbyafinitetimeinterval.Thisyields:Then,inthelimitasΔtapproacheszero第3頁例題2.1氣缸內(nèi)儲(chǔ)有定量CO2氣體，初態(tài)p1＝300kPa，T1＝200℃，V1＝0.2m3。經(jīng)歷一可逆過程后溫度下降至T2＝100℃。假如過程中壓力和比容間關(guān)系滿足pv1.2＝常數(shù)，試確定該過程中CO2氣體所作功、比功、熱力學(xué)能改變量，氣體與外界之間熱量交換。Solution：Given:initialandendstates,process,CO2inacylinderFind:W,w,ΔU,QModel:closedsystemStrategy:applythebasicclosedsystemenergybalancetosolvefortheQandΔU,applythedefinitionofworktocalculationWandw

State1State2第4頁Analysis:1)work：可逆δw＝pdvW＝mw＝0.671×94500＝63425J

因氣體體積改變，故此功稱為膨脹功。W>0，為氣體對(duì)外做功。在終態(tài)，氣體體積為V2＝0.656m32)系統(tǒng)內(nèi)熱力學(xué)能改變：(ΔE)sys＝ΔU＝mcV(T2－T1)查表取cv＝0.656kJ/kg·K第5頁所以ΔU＝0.671×103×(100－200)＝－44018J

（說明系統(tǒng)熱力學(xué)能是降低）3)Q計(jì)算：Q＝ΔU＋W＝63425＋（－44018）＝19407JComment:該例題中氣體對(duì)外作出功量大于氣體熱力學(xué)能降低許，降低部分由外界對(duì)氣體加熱量所補(bǔ)充。另外此處能量單位用焦耳顯得不方便，用kJ則好些。第6頁2.5開口系統(tǒng)能量方程

ConservationofEnergyPrincipleforAOpenSystemorControlVolumeTwostepsofanalysis1st

ConservationofMassPrincipleforAOpenSystemorControlVolume2ndConservationofEnergyPrincipleforAnOpenSystemorControlVolume1st:ConservationofMassPrincipleforAOpenSystemorControlVolumeIsdefinedasthemassflowratekg/s第7頁Case3forbothsteadystateandone-dimensionalflow:Steadystate:anysysteminsteadystateifthesystempropertiesareconstantwithtimeateverypositionwithinandontheboundariesofthesystems.One-dimensionalflow:

ifthepropertiesataflowboundaryareuniformoverthecross-sectionalarea.Case1Forsteadystate:Case2Forone-dimensionalflow:第8頁2ndConservationofEnergyPrincipleforAOpenSystemorControlVolumeIt’sclearwhenregardlessthekineticandpotentialenergyofasystem,we’llhave:Ifwetaketheassumption-Steadystate:

Ifwetaketheassumption-One-dimensionalflow:

Ifwetakeboththeassumptions-Steadystate

andOne-dimensionalflow:

第9頁Theenergyinoroutofthecontrolvolume：Whenmassgetsinthesystem,itwillcarrytheseveralitemsofenergyintothesystem1)internalenergyofthemass;2)?3)kineticenergyofthemass;4)andpotentialenergyofthemass.1)internalenergyofthemass(ifmkgmassenters)2)howtoletthemassacrosstheboundaryintoasystem?第10頁Accordingtothefig.shownbelow:Forone-dimensionalassumption,Workshouldbedoneas:Wegivethisworkthenameofflowworkorpushwork.As1)and2)discussion,wehave:Whenmkgofmassentersinacontrolvolume

:第11頁Adefinition:orWegivehthenameofenthalpy.Thisisapropertyofmatter.(Combinedproperty)3)kineticenergyofthemass(Whenmkgofmassentersinacontrolvolume

)4)andpotentialenergyofthemass(Whenmkgofmassentersinacontrolvolume)第12頁Fromabovediscussing,foreachmkgmassentersinthesystem,thetotalenergyentersthecontrolvolumeshouldbe:thetotalenergyleavesthecontrolvolumeshouldbe:Thecontrolvolumeenergyequationononedimensionalassumption:Formassrate:

第13頁Ifwetakeanotherassumptionofsteadystate:Therearenumerousapplicationsofthesteadyflowandsteadystateconservationofenergyprincipleforwhichthereisonlyoneinlet(position1)andoneoutlet(position2).Inthiscircumstanceaboveequationwillbeexpressedas:Onaunitmassbasis,itisconvenienttoexpressas:第14頁例題2.2已知汽輪機(jī)進(jìn)口水蒸氣參數(shù)為p1＝9MPa，t1＝500℃，流速cf1＝50m/s；出口水蒸氣參數(shù)為p2＝0.5MPa，t2＝180℃，流速cf2＝120m/s。蒸氣質(zhì)量流量qm＝330t/h。蒸氣在汽輪機(jī)中進(jìn)行穩(wěn)定絕熱流動(dòng)，求汽輪機(jī)功率。解：取系統(tǒng)以下列圖所表示（穩(wěn)態(tài)穩(wěn)流）分析：Q=0ΔU=0ein＝u1＋p1v1＋c12/2＋gz1ein＝u2＋p2v2＋c22/2＋gz2

系統(tǒng)對(duì)外作功：Wt＝m[（u1－u2）＋（p1v1－p2v2）＋（c12/2－c22/2）＋（gz1－gz2）]＝m[（h1－h(huán)2）＋（c12－c22）/2

＋g（z1－z2）]例題2.2附圖moutminws第15頁續(xù)解：對(duì)每kg工質(zhì)，功能夠表示為：wt＝（h1－h(huán)2）＋（c12－c22）/2＋g（z1－z2）略去其中位能差，可計(jì)算得：wt＝（h1－h(huán)2）＋（c12－c22）/2＝(3386.4－2812.1)×103＋(502－1202)/2

＝568.4×104J/kgP＝mwt＝330×103×568.4×104/3600＝52.1×103kW

第16頁特殊情況忽略動(dòng)能差和位能差時(shí)q＝（h2－h(huán)1）＋wtQ＝（H2－H1）＋Wt絕熱時(shí)能量方程Wt＝H1－H2絕熱節(jié)流過程H2＝H1

技術(shù)功Wt＝W－（p2V2－p1V1）

圖2.6技術(shù)功在p-v圖上表示第17頁習(xí)題:閱讀:第二章內(nèi)容思索題:P4.7.習(xí)題:p462-32-4p472-92-10第18頁SummaryMechanicalworkisdefinedas:Expansionorcompressi