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127942_3FirstLawofThermodynamics_ClosedSystems.pdf

M.BahramiENSC388(F09)1

LawofThermodynamics:ClosedSystems1

TheFirstLawofThermodynamics:ClosedSystems

Thefirstlawofthermodynamicscanbesimplystatedasfollows: duringaninteraction betweenasystemanditssurroundings,theamountofenergygainedbythesystemmust beexactlyequaltotheamountofenergylostbythesurroundings. Aclosedsystemcanexchangeenergywithitssurroundingsthroughheatandwork transfer.Inotherwords,workandheataretheformsthatenergycanbetransferred acrossthesystemboundary. Basedonkinetictheory,heatisdefinedastheenergyassociatedwiththerandom motionsofatomsandmolecules.HeatTransfer Heatisdefinedastheformofenergythatistransferredbetweentwosystemsbyvirtueof atemperaturedifference.

Note:therecannotbeanyheattransferbetweentwo

systemsthatareatthesame temperature. Note:Itisthethermal(internal)energythatcanbestoredinasystem.Heatisaformof energyintransitionandasaresultcanonlybeidentifiedatthe systemboundary. HeathasenergyunitskJ(orBTU).Rateofheattransferistheamountofheattransferred perunittime. Heatisadirectional(orvector)quantity;thus,ithasmagnitude,directionandpointof action.

Notation:

-Q(kJ)amountofheattransfer -Q°(kW)rateofheattransfer(power) -q(kJ/kg)Ͳheattransferperunitmass -q°(kW/kg)Ͳpowerperunitmass

Signconvention

:HeatTransfertoasystemispositive,andheattransferfromasystemis negative.Itmeansanyheattransferthatincreasestheenergyofasystemispositive,and heattransferthatdecreasestheenergyofasystemisnegative.

M.BahramiENSC388(F09)1

LawofThermodynamics:ClosedSystems2

Fig.1:Signconvention:positiveiftothesystem,negativeiffromthesystem.

ModesofHeatTransfer

Heatcanbetransferredinthreedifferentmodesconduction,convection,andradiation. Allmodesofheattransferrequiretheexistenceofatemperaturedifference.

Conduction

:isthetransferofenergyfromthemoreenergeticparticlestotheadjacent lessenergeticparticlesasaresultofinteractionsbetweenparticles. Insolids,conductionisduetothecombinationofvibrationsofthemoleculesinalattice andtheenergytransportbyfreeelectrons.

Convection

:isthemodeofenergytransferbetweenasolidsurfaceandtheadjacent liquidorgaswhichisinmotion,anditinvolvesthecombinedeffectsofconductionand fluidmotion(advection). Convectioniscalledforcedifthefluidisforcedtoflowbyexternalmeanssuchasa fanor apump.Itiscalledfreeornaturalifthefluidmotioniscausedbybuoyancyforcesthatare inducedbydensitydifferencesduetothetemperaturevariationinafluid.

Radiation

:istheenergyemittedbymatterintheformofelectromagneticwaves(or photons)asaresultofthechangesintheelectronicconfigurationsoftheatomsor molecules. Work Workistheenergyinteractionbetweenasystemanditssurroundings.Morespecifically, workistheenergytransferassociatedwithforceactingthroughadistance.

Notation:

-W(kJ)amountofworktransfer -W°(kW)power

Q=5kJQ=Ͳ5kJ

HeatinHeatout

System

M.BahramiENSC388(F09)1

LawofThermodynamics:ClosedSystems3

-w(kJ/kg)Ͳworkperunitmass -w°(kW/kg)Ͳpowerperunitmass

Signconvention

:workdonebyasystemispositive,andtheworkdoneonasystemis negative.

Fig.2:Signconventionforheatandwork.

Similaritiesbetweenworkandheattransfer:

Botharerecognizedattheboundariesofthesystemastheycrossthem(boundary phenomena). Systems possesenergy,butnotheatorwork(transferphenomena). Bothareassociatedwithaprocess,notastate.Heatorworkhasnomeaningata state. Botharepathfunctions,theirmagnitudesdependonthepathfollowedduringa processaswellastheendstates.

Pathfunctions

:haveinexactdifferentialsdesignatedbysymbolɷ.Properties,onthe otherhand,arepointfunctions whichdependonthestateonly(notonhowasystem reachesthatstate),andtheyhaveexactdifferentials. )W-nor WW not function,Path (Wfunction)(Point 12122
1122
1

WVVVdV

ElectricalWork

Theworkthatisdoneonasystembyelectrons.WhenNcoulombsofelectronsmove throughapotentialdifferenceV,theelectricalworkdoneis:

System

W(+)(Ͳ)

(+) (Ͳ)

M.BahramiENSC388(F09)1

LawofThermodynamics:ClosedSystems4

)(Was form rate in the explained becan Which )( e kWVIkJVNW e

Example1:Electricalwork

AwellͲinsulatedelectricalovenisbeingheatedthroughitsheatingelement.Determine whetheritisworkorheatinteraction.Considertwosystems:a)theentireoven(including theheater),andb)onlytheairintheoven(withouttheheater)seeFig3Ͳ3.

Solution:

Theenergycontentoftheovenisincreasedduringthisprocess. a) Theenergytransfertotheovenisnotcausedbyatemperaturedifference betweentheovenandair.Instead,itiscausedbyelectricalenergycrossingthe systemboundaryandthus:thisisaworktransferprocess. b) Thistime,thesystemboundaryincludestheoutersurfaceoftheheaterandwill notcutthroughit.Therefore,noelectronswillbecrossingthesystemboundary. Instead,theenergytransferisaresultofatemperaturedifferencebetweenthe electricalheaterandair,thus:thisisaheattransferprocess.

Fig.3:Schematicforexample1.

Mechanicalwork

Thereareseveralwaysofdoingwork,eachinsomewayrelatedtoaforceactingthrough adistance. )(.kJsFWElectric oven air

Heater

HeaterElectric

oven air

Systemboundary

M.BahramiENSC388(F09)1

LawofThermodynamics:ClosedSystems5

Iftheforceisnotconstant,weneedtointegrate: 2 1 )(.kJdsFW

Therearetworequirementsforaworkinteraction:

theremustbeaforceactingontheboundary theboundarymustmove Therefore,thedisplacementoftheboundarywithoutanyforcetoopposeordrivethis motion(suchasexpansionofagasintoevacuatedspace)isnota workinteraction,W=0. Also,iftherearenodisplacementsoftheboundary,evenifanactingforceexists,there willbenoworktransferW=0(suchasincreasinggaspressureinarigidtank).

MovingBoundaryWork

Theexpansionandcompressionworkisoftencalledmovingboundarywork,orsimply boundarywork. WeanalyzethemovingboundaryworkforaquasiͲequilibriumprocess.Considerthegas enclosedinapistonͲcylinderatinitialPandV.Ifthepistonisallowedtomoveadistance dsin aquasiͲequilibriummanner,thedifferentialworkis:

PdVPAdsdsFW

b . ThequasiͲequilibriumexpansionprocessisshowninFig.4.Onthisdiagram,the differentialareadAundertheprocesscurveinPͲVdiagramisequaltoPdV,whichisthe differentialwork. Note:agascanfollowseveraldifferentpathsfromstate1to2,andeachpath willhavea differentareaunderneathit(workispathdependent). ThenetworkorcycleworkisshowninFig.5.Inacycle,thenetchangeforanyproperties (pointfunctionsorexactdifferentials)iszero.However,thenetworkandheattransfer dependonthecyclepath.

ȴU=ȴP=ȴT=ȴ(anyproperty)=0foracycle

M.BahramiENSC388(F09)1

LawofThermodynamics:ClosedSystems6

Fig.4:theareaunderPͲVdiagramrepresentstheboundarywork.

Fig.5:networkdoneduringacycle.

PolytropicProcess

Duringexpansionandcompressionprocessesofrealgases,pressureandvolumeare oftenrelatedbyPV n =C,wherenandCareconstants.Themovingworkforapolytropic processcanbefound: nVPVPdVCVPdVW n polytopic 1 11222
12 1 Since CVPVP nn 2211
.Foranidealgas(PV=mRT)itbecomes: VP W net 12 V 1 V 2

M.BahramiENSC388(F09)1

LawofThermodynamics:ClosedSystems7

)(1,1 12 kJnnTTmRW polytropic

Thespecialcasen=1istheisothermalexpansionP

1 V 1 =P 2 V 2 =mRT 0 =C,whichcanbe foundfrom: )(1,ln 12 112
12 1 , kJnVVVPdVVCPdVW isothermalb

Sinceforanidealgas,PV=mRT

0 atconstanttemperatureT 0 ,orP=C/V.

Example2:Polytropicwork

AgasinpistonͲcylinderassemblyundergoesapolytropicexpansion.Theinitialpressureis

3bar,theinitialvolumeis0.1m

3 ,andthefinalvolumeis0.2m 3 .Determinetheworkfor theprocess,inkJ,ifa)n=1.5,b)n=1.0,andc)n=0.

Solution:

Assumethati)thegasisaclosedsystem,ii)themovingboundaryisonlyworkmode,and iii)theexpansionispolytropic. a)n=1.5 2 1 1122
1 V V nVPVPPdVW

WeneedP

2 thatcanbefoundfrom nn VPVP 2211
: barbarVVPP n

06.12.01.03

5.1 2 1 12 kJmNkJ barmNmbarW6.17.101 1/10

5.111.032.006.1

3253
b)n=1,thepressurevolumerelationshipisPV=constant.Theworkis: kJmNkJ barmNmbarWVVVPPdVW

79.201.02.0ln.101

1/101.03ln

325
312
112
1 c)Forn=0,thepressureͲvolumerelationreducestoP=constant(isobaricprocess)andthe integralbecomeW=P(V 2 ͲV 1 ). Substitutingvaluesandconvertingunitsasabove,W=30kJ.

M.BahramiENSC388(F09)1

LawofThermodynamics:ClosedSystems8

Springwork

Forlinearelasticsprings,thedisplacementxisproportionaltotheforceapplied: xkF s wherek s isthespringconstantandhastheunitkN/m.Thedisplacementxismeasured fromtheundisturbedpositionofthespring.Thespringworkis: )(21 2 12 2 kJxxkW sspring Note:theworkdoneonaspringequalstheenergystoredinthespring.

NonͲmechanicalformsofwork

NonͲmechanicalformsofworkcanbetreatedinasimilarmannertomechanicalwork. SpecifyageneralizedforceFactinginthedirectionofageneralizeddisplacementx,the worktransferassociatedwiththedisplacementdxis: dxFW.

Example3:Mechanicalwork

Calculatetheworktransferinthefollowingprocess:

Fig.6:SchematicPͲVdiagramforExample3.

Solution:

Process1Ͳ2isanexpansion(V

2 >V 1 )andthesystemisdoingwork(W 12 >0),thus: W 12 =P 1 (V 2 ͲV 1 )+[0.5(P 1 +P 2 )-P 1 ](V 2 -V 1 ) =(V 2 -V 1 )(P 1 +P 2 )/2

Process2Ͳ3isanisometricprocess(constantvolumeV

3 =V 2 ),soW 23
=0

Process3Ͳ1isacompression(V

3 >V 1 ),workisdoneonthesystem,(W 31
<0) 12 3 V V 2 V 1 P 1 P 2 P

M.BahramiENSC388(F09)1

LawofThermodynamics:ClosedSystems9

W 31
=ͲP 1 (V 2 -V 1 ) W cycle =W net =W 12 +W 23
+W 31
=(V 2 -V 1 )(P 2 -P 1 )/2 NotethatinacycleȴU=ȴP=ȴT=ȴ(anyproperty)=0

FirstLawofThermodynamics

Firstlaw,ortheconservationofenergyprinciple,statesthatenergycanbeneithercreated nordestroyed;itcanonlychangeforms.

Thefirstlawcannotbeprovedmathematically,

itisbasedonexperimentalobservations, i.e.,therearenoprocessinthenaturethatviolatesthefirstlaw. Thefirstlawforaclosedsystemorafixedmassmaybeexpressedas: netenergytransferto(orfrom) thesystemasheatandwork=netincrease(or decrease)inthetotal energyofthesystem

Q-W=ȴE(kJ)

where

Q=netheattransfer(=ɇQ

in -ɇQ out )

W=networkdoneinallforms(=ɇW

in -ɇW out )

ȴE=netchangeintotalenergy(=E

2 -E 1 ) Thechangeintotalenergyofasystemduringaprocesscanbeexpressedasthesumof thechangesinitsinternal,kinetic,andpotentialenergies:

ȴE=ȴU+ȴKE+ȴPE(kJ)

122
12 212
21
zzmgPEVVmKEuumU ' Note:forstationarysystemsȴPE=ȴKE=0,thefirstlawreducesto

Q-W=ȴU

ThefirstlawcanbewrittenonaunitͲmassbasis:

q-w=ȴe(kJ/kg) orindifferentialform:

ɷQ-ɷW=dU(kJ)

ɷq-ɷW=du

(kJ/kg) orintherateform:

M.BahramiENSC388(F09)1

LawofThermodynamics:ClosedSystems10

Q°-W°=dE/dt(kW) Foracyclicprocess,theinitialandfinalstatesareidentical,thusȴE=0.Thefirstlaw becomes:

Q-W=0(kJ)

Note:fromthefirstlawpointofview,thereisnodifferencebetweenheattransferand work,theyare bothenergyinteractions.Butfromthesecondlawpointofview,heatand workareverydifferent.

Example4:Fistlaw

AiriscontainedinaverticalpistonͲcylinderassemblyfittedwithanelectricalresistor.The atmosphericpressureis100kPaandpistonhasamassof50kg andafaceareaof0.1m 2 . Electriccurrentpassesthroughtheresistor,andthevolumeofairslowlyincreasesby

0.045m

3 .Themassoftheairis0.3kganditsspecificenergyincreasesby42.2kJ/kg. Assumetheassembly(includingthepiston)isinsulatedandneglectthefrictionbetween thecylinderandpiston,g=9.8m/s 2 .Determinetheheattransferfromtheresistortoair forasystemconsistinga)theairalone,b)theairandthepiston.

Fig.7:Schematicforproblem4.

Assumptions:

Twoclosedsystemsareunderconsideration,asshowninschematic. Theonlyheattransferisfromthe resistortotheair.ȴKE=ȴPE=0(forair) Theinternalenergyofthepistonisnotaffectedbytheheattransfer. a)Takingtheairasthesystem, (ȴKE+ȴPE+ȴU) air =Q-W

AirPistonAirPiston

System

boundary partb

System

boundary parta

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LawofThermodynamics:ClosedSystems11

Q=W+ȴU air Forthissystemworkisdoneatthebottomofthepiston.Theworkdonebythesystemis (atconstantpressure): 2 1 12V V

VVPPdVW

Thepressureactingontheaircanbefoundfrom:

PA piston =m piston g+P atm A piston kPakPaPakPa mNPa msmkgPP AgmP atm

91.10410010001

/11

1.0/81.950

222pistonpiston

Thus,theworkis

W=(104.91kPa)(0.045m

3 )=4.721kJ

WithȴU

air =m air ȴu air ,theheattransferis Q=W+m air ȴu air =4.721kJ+(0.3kg)(42.2kJ/kg)=17.38kJ b)systemconsistingtheairandthepiston.Thefirstlawbecomes: (ȴKE+ȴPE+ȴU) air +(ȴKE+ȴPE+ȴU) piston =Q-W where(ȴKE=ȴPE) air =0and(ȴKE=ȴU) piston =0.Thus,itsimplifiesto: (ȴU) air +(ȴPE) piston =Q-W Forthissystem,workisdoneatthetopofthepistonandpressureistheatmospheric pressure.Theworkbecomes W=P atm

ȴV=(100kPa)(0.045m

3 )=4.5kJ Theelevationchangerequiredtoevaluatethepotentialenergychangeofthepistoncan befoundfromthevolumechange:

ȴz=ȴV/A

piston =0.045m 3 /0.1m 2 =0.45m (ȴPE) piston =m piston gȴz=(50kg)(9.81m/s 2 )(0.45m)=220.73J=0.221kJ

Q=W+(ȴPE)

piston +m air ȴu air

Q=4.5kJ+0.221kJ+(0.3kg)(42.2kJ/kg)=17.38kJ

Notethattheheattransferisidenticalinbothsystems.

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LawofThermodynamics:ClosedSystems12

SpecificHeats

Thespecificheatisdefinedastheenergyrequiredtoraisethetemperatureofaunitmass ofasubstancebyonedegree.Therearetwokindsofspecificheats: specificheatatconstantvolume,Cv(theenergyrequiredwhenthevolumeis maintainedconstant) specificheatatconstant pressure,C p (theenergyrequiredwhenthepressureis maintainedconstant)

ThespecificheatatconstantpressureC

p isalwayshigherthanC v becauseatconstant pressurethesystemisallowedtoexpandandenergyforthisexpansionmustalsobe suppliedtothesystem. Let'sconsiderastationaryclosedsystemundergoingaconstantͲvolumeprocess(w b =0).

Applyingthefirstlawinthedifferentialform:

ɷq-ɷw=du

atconstantvolume(nowork)andbyusingthedefinitionofC v ,onecanwrite: vvv

TuCordudTC

Similarly,anexpressionforthespecificheatatconstantpressureCpcanbefound.From thefirstlaw,foraconstantpressureprocess(w b +ȴu=ȴh).Ityields: pp ThC specificheats(bothC v andC p )arepropertiesandthereforeindependentofthe typeofprocesses. C v isrelatedtothechangesininternalenergyu,andC p tothechangesinenthalpy, h.

Itwouldbemoreappropriatetodefine:C

v isthechangeinspecificinternalenergyper unitchangeintemperatureatconstantvolume.C p isthechangeinspecificenthalpyper unitchangeintemperatureatconstantpressure.

Specificheatsforidealgases

Ithasbeenshownmathematicallyandexperimentallythattheinternalenergyisa functionoftemperatureonly. u=u(T)

M.BahramiENSC388(F09)1

LawofThermodynamics:ClosedSystems13

Usingthedefinitionofenthalpy(h=u+Pv)andtheidealgasequationofstate(Pv=RT), wehave: h=u+RT

SinceRisaconstantanduisafunctionofTonly:

h=h(T)

Therefore,atagiventemperature,

u,h,C v andC p ofanidealgaswillhavefixedvalues regardlessofthespecificvolumeorpressure.Foranidealgas,wehave: dTTCdhdTTCdu pv Thechangesininternalenergyorenthalpyforanidealgasduringaprocessare determinedbyintegrating: )/()/( 2 1 122
1 12 kgkJdTTChhhkgkJdTTCuuu pv Aslowpressures,allrealgasesapproachidealͲgasbehavior,andthereforetheirspecific heatsdependontemperatureonly.Thespecificheatsofrealgasesatlowpressuresare calledidealͲgasspecificheats(orzeroͲpressurespecificheats)andareoftendenotedby C p0 andC v0 .Tocarryouttheaboveintegrations,weneedtoknowC v (T)andC p (T).These areavailablefromavarietyofsources: TableAͲ2a:forvariousmaterialsatafixedtemperatureofT=300K TableAͲ2b:variousgasesoverarangeoftemperatures250чTч1000K TableAͲ2c:variouscommongasesintheform ofathirdorderpolynomial

Foranidealgas,wecanwrite:

CCRdTdu

dTdhRTuThRT Theratioofspecificheatsiscalledthespecificheatratiok=C p /C v : varieswithtemperature,butthisvariationisverymild. formonatomicgases,itsvalueisessentiallyconstantat1.67. Manydiatomicgases,includingair,haveaspecificheatratioofabout1.4atroom temperature.

M.BahramiENSC388(F09)1

LawofThermodynamics:ClosedSystems14

Specificheatsforsolidsandliquids

Asubstancewhosespecificvolume(ordensity)isconstantiscalledincompressible substance.Thespecificvolumesofsolidsandliquids(whichcanbeassumedas incompressiblesubstances)essentiallyremainconstantduringaprocess. Theconstantvolumeassumptionmeansthatthevolumework(boundarywork)is negligiblecomparedwithotherforms ofenergy.Asaresult,itcanbeshownthatthe constantͲvolumeandconstantͲpressurespecificheatsareidenticalforincompressible substances: C p =C v =C Specificheatsofincompressiblesubstancesareonlyafunctionoftemperature,

C=C(T)

Thechangeofinternalenergybetweenstate1and2canbeobtainedbyintegration: )/( 2 1 12 kgkJdTTCuuu Forsmalltemperatureintervals,aCataveragedtemperaturecanbeusedandtreatedas aconstant,yielding: 12 TTCu ave Theenthalpychangeofincompressiblesubstancecanbedeterminedfromthedefinition ofenthalpy(h=u+Pv) h 2 -h 1 =(u 2 -u 1 )+v(P 2 -P 1 )

ȴh=ȴu+vȴP(kJ/kg)

ThetermvȴPisoftensmallandcanbeneglected,soȴh=ȴu.

Example5:Specificheatandfirstlaw

Twotanksareconnectedbyavalve.Onetankcontains2kgofCO 2 at77°Cand0.7bar. Theothertankhas8kgofthesamegasat27°Cand1.2bar.Thevalveisopenedand gasesareallowedtomixwhilereceivingenergybyheattransferfromthesurroundings. Thefinalequilibriumtemperatureis42°C.Usingidealgasmodel,determine a)thefinal equilibriumpressureb)theheattransferfortheprocess.

Assumptions:

ThetotalamountofCO 2 remainsconstant(closedsystem). IdealgaswithconstantC v . Theinitialandfinalstatesinthetanksareequilibrium.Noworktransfer.

M.BahramiENSC388(F09)1

LawofThermodynamics:ClosedSystems15

Thefinalpressurecanbefoundfromidealgasequationofstate: 2121
21

VVRTmm

VVRTmP

fft f

Fortank1and2,wecanwrite:V

1 =m 1 RT 1 /P 1 andV 2 =m 2 RT 2 /P 2 .Thus,thefinalpressure, P f becomes: bar barKkg barKkgKkgPPTm

PTmTmm

P RTm

PRTmRTmmP

fff f 05.1

2.13008

7.0350231510

222
11121
2 22
11121
b)Theheattransfercanbefoundfromanenergybalance:

ȴU=Q-W

WithW=0,

Q=U f -U i whereinitialinternalenergyis:U i =m 1 u(T 1 )+m 2 u(T 2 )

Thefinalinternalenergyis:U

f =(m 1 +m 2 )u(T f )

Theenergybalancebecomes:

Q=m 1 [u(T f )-u(T 1 )]+m 2 [u(T f )-u(T 2 )]

SincethespecificheatC

v isconstant Q=m 1 C v [T f -T 1 ]+m 2 C v [T f -T 2 ] kJKKKkgkJkgKKKkgkJkgQ25.37300315.745.08350315.745.02 Theplussignindicatesthattheheattransferisintothesystem. CO 2

8kg,27C,1.2barCO

2kg,77C,0.7bar

Valve

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