[PDF] FE Thermodynamics Review

102736_3FEThermodynamicsReview2011.pdf

FEThermodynamicsReview

Dr.DavidR.Muñoz

Definitions

• System-collectionofmatterwhichweare interestedinstudying. • Surroundings-everythingoutsideofthesystem. • Universe-system+surroundings. • Puresubstance-itschemicalcomposition remainsthesameregardlessofphase. • Intensiveproperty- thermodynamicproperty independentofthesystemmass. • Independentproperty-astandaloneproperty thatcanbeusedtoidentifythestateofasystem.

Definitions(cont.)

• Statepostulateforapuresubstance- takestwo independentintensivepropertiestofixitsstate. • Closedsystem-massdoesnotcrossthesystem boundaries,butheatandworkcancrossthe systemboundaries • Opensystem-masscrossesthesystem boundaries,heatandworkcanalsocrossthe systemboundaries Ͳ Steadyflow-opensystemwithnomass accumulation Ͳ UnsteadyflowͲ opensystemwith mass accumulation • Signconvention:Heatintosystemispositive. Workoutofsystemispositive.

GeneralExamTakingPhilosophy

• Youshouldprepareandworktogetthe correct answerwheneverpossible. • However,onthoseproblemsyougetstuckon, rememberthatwronganswersarenoworse thannoanswer. • Therefore,firsttrytoeliminateunreasonable answerstoimprovetheoddsofguessing right. • Thenmakeyourbestguess. • Ifyoudon'thavetimetoeliminatewrong answers, pickyourfavoriteletterandguess.

Problem1.Whatisthemassofaircontainedinaroom

20mx40mx3matstandardconditions?

• Standardconditionsmayvarydependingonwhat organizationisdefiningthem.Itisgenerally, • P=1atm =101kPa atsealevel, • T=25 o

C=298K

• Assume:airisanidealgas • Youaregiventheuniversalgasconstantas R u =

8.314kJ/kmol

K,whereR=R u /M,

M=molecularweightfortheparticulargasof

interest.

Problem1(cont.)

IdealgasLaw:PV=mRT

m =PV/RT

R=[8.314kJ/kmol

K]/[29kg/kmol]=0.287kJ/kgK V =[20x40x3]m 3 =2400m 3

Therefore,

m =[(101)kPa(2400)m 3 ]/[(0.287)kJ/kgK(298)K] m=2834kg

Problem2-Findthevolumeoccupiedby20kg

ofsteamat0.4MPa,400 o C • UsesteamtablesprovidedinyourFEbooklet • Atthispointyoudonotknowwhethertouse thesaturatedmixturetableorthe superheatedsteamtable,sothatmustfirstbe determined. • Recallthevapordomeassociatedwiththe saturation conditionsforwater(seenext slide)

Temperaturevs.volumeand Pressurevs.volumediagrams

P

SaturationTable

Problem2(cont.)ͲTemperature

column onlygoesto 374
o

C,whichisthe

criticalpointfor water (apexofthe vapordome). Ͳ

Therefore,you

mustusethe superheated vapor tableforProblem2

Superheated VaporTable

Problem2(cont.)

At0.4MPa,400

o C v=0.7726m 3 /kg V=mv =(20)kg(0.7726)m 3 /kg

V=15.45m

3

Problem3-

Howmuchheatmustbeaddedto2kgofsteam initiallyatT 1 =300 o

C,containedwithinarigidvolumeto

increasethepressurefrom0.2MPa to0.4MPa?

Solution:

Assume:

ClosedSystem[NomasscrossestheControlSurface (CS)butheatorworkcancrosstheCS.]

Rigidtankmeansthevolumedoesnotchange:V

1 =V 2 , v 1 =v 2 Rigid Tank Q 1 st

Law(closedsystem)

Q =ȴU=m(ȴu) AtT 1 =300 o C,P 1 =0.2MPa(usingthesuperheattableforwater) v 1 =1.3162m 3 /kg;u 1 =2808.6kJ/kg

Sincev

1 =v 2 ,atP 2 =0.4MPa,v 2 =1.3162m 3 /kg u 2 =3793.2kJ/kg(byinterpolation)

Therefore:Q=(2)kg(3793.2-

2808.6)kJ/kg

Q=1969kJ

LinearInterpolationforProblem3

Sinceittakestwoindependentintensivepropertiestofixthestateofapuresubstance, inthecaseofproblem3,thestate2isfixedsinceweknowthe pressureandspecific volume.However,sincethetablesarenottabulatedinroundnumbersofspecific volume, wemustinterpolatebetweentwovaluesthatbracketthespecificvolume given. Lookingatthesuperheatedwatertable,wenoticethatforapressureof0.4MPa,the specific volumeisbracketedbetweenthetemperaturesof800 o

Cand900

o

C.Wenote

theinternalenergiesforthesetemperatures.Thenextstepistosetupalinear relationshipbetweenthesevaluesandsolveforthedesiredinternalenergy corresponding totheknownspecificvolume. u 2 =3793.2kJ/kg

Problem4-

Howmuchheatisneededtocompletely vaporize100kgofwaterfromtemperatureT 1 =20 o C ifthe pressureismaintainedataconstantP=200kPaabsolute? T v

P=200kPa

1 2

Solution:

TheprocessisshownintheTͲvdiagramdrawnonthe

left.State1beginsasacompressedliquidandstate2 isshownasasaturatedvapor.Theprocessfollowsa lineofconstantpressure(asindicatedintheproblem statement).Sincewedonothaveaccesstoa compressedliquidtableforthisexam,youhaveto makeanassumptionregardingthepropertyselection forstate1.Sincethewaterisinitiallyintheliquidstate, weconsiderittobelargelyincompressible.Therefore, wecannotaddmuchinternalenergytothewaterby compressingit.However,asmalltemperatureincrease inthewaterdoescontributetothewaterinternal energy.Therefore,wewillassumethatduetoliquid incompressibility,wecanusethepropertyofthe saturatedliquid,evaluatedatthisinitialtemperature, torepresentstate1oftheliquid.

Problem4(cont.)

Solution(cont.)

1 st

Law(closedsystem)

Q =ȴU=m(ȴu) Where u 1 =u f (at20 o C);u 1 =83.95kJ/kg(saturationtable) u 2 =u g (at200kPa);u 2 =2529.5kJ/kg(superheattable) Q =(100)kg(2529.5- 83.95)kJ/kg

Q=244,555kJ

Problem5-

Calculatetheworkdonebyapistoncontained withinacylinderwithairif2m 3 is tripledwhilethe temperatureismaintainedataconstant T=30 o

C.TheinitialpressureisP

1 =400kPaabsolute. 12

Solution:

Assumeairisanidealgas.PV=mRT

V 1 = 2m 3 ;V 2 = 3V 1 = 6m 3

P=mRT/V

Work, butTherefore,

Problem6-

Howmuchworkisnecessarytocompressairinan insulatedcylinderfrom0.2m 3 to0.01m 3 ? UseT 1 =20 o CandP 1 =100kPaabsolute.

Solution:

Assume:ͲIdealgas

Useabsolutetemperature

ͲIsentropicprocess

T 1 =20 o C =293K

ͲConstantc

vo

ͲAdiabaticprocess(insulatedcontainer)

1 st

Law(closedsystem)

Q =ȴU+W=m(ȴu)+W W =Ͳmc vo (T 2 Ͳ T 1 );W/m=w=c vo (T 2 Ͳ T 1 )

Isentropicrelationships

T 1 v

1kͲ1

=T 2 v

2kͲ1

where k=1.4(forair) T 2 =T 1 {v 1 /v 2 } k Ͳ1 =T 1 {V 1 /V 2 } k Ͳ1 =(293)K{(0.2)/(0.01)}

1.4Ͳ1

=971K w=Ͳ(0.719)kJ/kgK(971- 293)K=Ͳ
487kJ/kg

Problem7-A10cmthickslabofwood(k

wood =0.2W/mK)is3mhigh and10mlong.Calculatetheheattransferrateifthetemperatureis 25
o

ContheinsideandͲ20

o

Contheoutside.Neglectconvection.

q T in =25 o C T out =-20 o C wood

Solution:

Assume:conductionheattransferonly

A=LW=(10)m(3)m=30m

2 q = ͲkA(dT/dx)~Ͳ kA(ȴT/ȴx) q =Ͳ(0.2)W/mK(30)m 2 (

Ͳ20Ͳ25)

o

C/(0.1)m

q =270W

Problem8-

Thesurfaceoftheglassina1.2mx0.8m skylightismaintainedat20 o

Candtheoutsideairtemperature

is

Ͳ20

o

C.Estimatetheheatlossratebyconvectionwhenthe

convectivecoefficientish=12W/m 2 K. skylight

Solution:

Assumeconvectiveheattransferonly

q =hA(T s Ͳ T ь ) q =(12)W/m 2 K (1.2)m(0.8)m[20-(Ͳ20)] o C q =460.8W

Problem9-

A2cmdiameterheatingelementlocatedwithinan ovenismaintainedat1000 o

Candtheovenwallsareat500

o

C.Ifthe

element emissivityis0.85,estimatetherateofheatlossfromthe

2mlongelement.

q

Solution:

Assumeradiationheattransferonly

Ͳ

Useabsolutetemperatures,

rTreattheovenwallsaslargesurroundings. T s = 1000
o

C=1273K;T

sur = 500
o

C=773K

q =ɸ ʍ A(T s4 -T sur4 ); where A=ʋdL("circumferencial" area) q =(0.85)(5.67x10 Ͳ8 ) W/m 2 K 4 ( ʋ ) (0.02)m(2)m[(1273) 4 - (773) 4 ]K 4 q =

13,742W=13.7kW

Problem10-

RefrigerantR134aexpandsthroughavalve from apressureof800kPatoapressureof100kPa.Whatis thefinalquality? 1 2

ExpansionvalveSolution:

Thereisnotableofthermodynamicdatafor134a

containedwithinyourFEbook.Thereforeyoumust usethePressure-Enthalpydiagramprovided(see subsequentpages).Thisdiagramismuchlikethe

Temperature-volumediagramforwaterinthatthe

apexofthedomeisthecriticalpoint(CP).Thesolid linecorrespondingtosaturatedliquidisontheleftand thatassociatedwiththesaturatedvaporisonthe right. 1 st

Law(SteadyFlow,expansionvalve)

, h 1 =h 2 Wecommonlyuseexpansionvalvesinvaporcompressionrefrigeratorstomovefrom thehighpressureandtemperatureportionofthecycletothelowpressureand temperatureportionofthecycle.State1istypicallyconsideredtobeinasaturated liquidstate.PleaseseeFEbookforlistofcommonlyusedsteadyflowdevicesandtheir associatedmodelingequations.

Problem10(cont.)

Solution(cont.)

h 1 =h f =244kJ/kg=h 2 FromthePvs.hplot(nextpage),findthequality(thesearethecurvesemanating fromthecriticalpoint) x=0.37

Pressure- EnthalpydiagramforRͲ 134a

Problem11-Whatistheminimumpumppowerrequiredto

increasethepressureofwaterfrom2kPato6MPawithamass flowrateof10kg/s?pump 1 2

Solution:

1 st

Law(pumps,isentropicorreversiblesteadyflow)

Use thesaturationtableforwatertofindv f . v f =0.001002m 3 /kg w=Ͳ(0.001002)m 3 /kg(6000-2)kPa w=Ͳ6.0kJ/kg; Problem12-ArefrigerationsystemusingRͲ134aoperatesbetween

Ͳ20

o

Cand40

o

C.Whatisthemaximumcoefficientofperformance

(COP)?

Refrigerator

T H T L W Q H Q L

Solution:

ThemaximumCOPisachievedinaCarnotcycle.

ForaCarnotcycle,

Remembertouseabsolutetemperatures.

Problem13-Aheatpumpdelivers20,000kJ/hrwithq

1.39kWelectricalinput.CalculatetheCOP.

Solution:

Aheatpumpislikearefrigerator.Itdifferentinthatincludesaspecial4Ͳwayvalvethat allowstheusertoswitchthefunctionsoftheheatexchangersfortheheatingseason (winter)andthecoolingseason(summer).Unlessotherwisestatedintheproblem statement,theheatpumpisusuallyconsideredtobeintheheatingmodeforthese analyses.Intheheatingmode,thatwhichwedesireisQ H .Therefore,theCOPisdefined asfollows.

Problem14-

Athermometerwithawetclothattachedtoitsbulb reads20 o

Cwhenairisblowingaroundit.Iftheairhasadrybulb

temperatureof33 o C, whatistherelativehumidityanddewpoint temperature?

Howmuchwatercouldbecondensedoutofa100m

3 volume?

Solution:

YouwillneedtousethepsychrometricchartprovidedinyourFE book.Ithasbeencopied onthefollowingpages.

Awetbulbtemperatureof20

o C isprovidedintheproblemstatement.Thelines corresponding tothewetbulbtemperatureonthepsychrometricchartaredashedand movefromtheupperlefttothelowerright(locatethemonthechart).Thedrybulb temperatureistheabscissaofthepsychrometricchart.Theintersectionbetweenthe 20 o

Cwetbulbandthe33

o Cdrybulbtemperaturecorrespondstoarelativehumidityof

30%(therelativehumiditycurvesmovefromthelowerlefttotheupperrightonthe

chart). Thedewpointtemperatureisthattemperaturewhenwaterbeginstocondense.Since theordinateofthepsychrometricchartisthehumidityratio(gramsmoisture/kgdryair), whichisproportionaltotheabsolutehumidity.Thedewpointtemperatureisthedry bulbtemperaturethatcorrespondstotheintersectionofahorizontallinedrawnthrough state1andthesaturationline(correspondingto100%relativehumidity).Thisisadew point temperatureof~13 o C.

Problem14(cont.)

Solution(cont.)

Condensateistakenfromtheairwhenitiscooledtobelowthedewpointtemperature. Theamountofcondensatethatcanberemovedcorrespondstothat associatedwiththe differenceinthemoisturecontentatthedewpointtemperatureandthatat0 o

C.Below

0 o C,thephasewillbefrostorice,correspondingtoacrystallization(vapor- solid)phase changeprocess.

Therefore,

MassofCondensate=m(ʘ

2 - ʘ 1 ), wheremisthemassofdryairandʘ isthehumidityratio(takenfrompsychrometricchart)

Assumingdryairisanidealgas

m=PV/RT=[(101)kPa(100)m 3 ]/[(287)kJ/kg

K(33+273)K]=115kg

Massofcondensate=(115)kgdryair(9.3-3.7)gmmoisture/kgdryair=644gmmoisture

PsychrometricChart

Problem15-

Itisdesiredtocondition35 o

C,85%relativehumidity

airto24 o C,

50%relativehumidity.If100m

3 /min airistobe conditioned,howmuchenergyisrequiredinthecoolingprocessand howmuchintheheatingprocess?

Solution:

Onewaytoremovethemoistureistofirstcooltheairtothesaturationcondition(100% relativehumidityline)andcontinuethecoolingprocesstoremovemoistureuntilthe desiredhumidityratioisachieved.Therefore,theengineermust locatetheinitial(1)and final(4)statepointsfirstandthenworkwiththepsychrometricchart(asshownonthe followingpages)todeterminetheenergyrequirements.

Frompsychrometriccharth

1 =110kJ/kgdryair,h 2 =106kJ/kgdryair(seeenthalpyon upperleft).Thisisthesensible(nophasechange)coolingpart.

Problem15(cont.)

Solution(cont.)

Sensibleheating

Fromthecharth

3 =37kJ/kgdryair,h 4 =48kJ/kgdryair

Latent

andsensiblecooling

TotalCooling=Ͳ7.62Ͳ

131.4kW=Ͳ139kW;TotalHeating=21kW

PsychrometricChart2