FE Thermodynamics Review 2011

8/12/2019 FE Thermodynamics Review 2011

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Definitions

System

collection

of

matter

which

we

are

interestedinstudying.

Surroundings everythingoutsideofthesystem.

Universe system+surroundings.

Puresubstance itschemicalcomposition

remainsthesameregardlessofphase. Intensiveproperty thermodynamicpropertyindependentofthesystemmass. Independentproperty astandaloneproperty

thatcanbeusedtoidentifythestateofasystem.


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Definitions(cont.)

Statepostulateforapuresubstance takestwoindependentintensivepropertiestofixitsstate. Closedsystem massdoesnotcrossthesystem

boundaries,butheatandworkcancrossthesystemboundaries

Opensystem masscrossesthesystem

boundaries,heatandworkcanalsocrossthesystemboundaries

Steadyflow opensystemwithnomass

accumulation Unsteadyflow opensystemwith massaccumulation

Signconvention:Heatintosystemispositive.Workoutofsystemispositive.


4/31

GeneralExamTakingPhilosophy Youshouldprepareandworktogetthe

correctanswerwheneverpossible.

However,onthoseproblemsyougetstuckon,rememberthatwronganswers arenoworsethannoanswer.

Therefore,firsttrytoeliminateunreasonableanswerstoimprovetheoddsofguessing

right. Thenmakeyourbestguess.

Ifyoudonthavetimetoeliminatewronganswers,pickyourfavoriteletterandguess.


5/31

Problem1.Whatisthemassofaircontainedinaroom

20mx40m

x3m

atstandard

conditions?

Standardconditionsmayvarydependingonwhat

organizationisdefining

them.

Itisgenerally,

P=1atm =101kPa atsealevel, T=25oC=298K Assume: airisanidealgas

Youaregiventheuniversalgasconstantas

Ru

=8.314kJ/kmol K,whereR=Ru /M,M=molecularweightfortheparticulargasof

interest.


6/31

Problem1(cont.)

IdealgasLaw:PV=mRT

m=PV/RT

R=[8.314kJ/kmol K]/[29kg/kmol]=0.287kJ/kgKV=[20x40x3]m3

=2400m3

Therefore,

m=[(101)kPa(2400)m3]/[(0.287)kJ/kgK(298)K]m=2834kg


7/31

Problem2 Findthevolumeoccupiedby20kg

ofsteamat0.4MPa,400o

C

UsesteamtablesprovidedinyourFEbooklet

Atthispointyoudonotknowwhethertousethesaturatedmixturetableorthe

superheatedsteam

table,

sothat

must

first

be

determined.

Recallthevapordomeassociatedwiththe

saturationconditionsforwater(seenext

slide)


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Temperaturevs.volumeand

Pressurevs.volumediagrams

P


9/31

SaturationTable

Problem2(cont.)

Temperature

columnonlygoesto374oC,whichisthe

criticalpointfor

water(apexofthevapordome).

Therefore,you

mustusethe

superheatedvapor

tableforProblem2


10/31

Superheated

VaporTable

Problem2(cont.)

At0.4MPa,400oCv=0.7726m3

/kg

V=mv

=(20)kg(0.7726)m3/kg

V=15.45

m3


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12/31

LinearInterpolationforProblem3

Sinceittakestwoindependentintensivepropertiestofixthestateofapuresubstance,inthecaseofproblem3,thestate2isfixedsinceweknowthe

pressureandspecific

volume.However,sincethetablesarenottabulatedinroundnumbersofspecific

volume,wemustinterpolatebetweentwovaluesthatbracketthespecificvolume

given.

Lookingatthesuperheatedwatertable,wenoticethatforapressureof0.4MPa,the

specificvolumeisbracketedbetweenthetemperaturesof800oCand900oC.Wenote

theinternalenergiesforthesetemperatures.Thenextstepistosetupalinear

relationshipbetweenthesevaluesandsolveforthedesiredinternalenergycorrespondingtotheknownspecificvolume.

u2

=3793.2

kJ/kg

P bl 4 H h h t i d d t l t l


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Problem4 Howmuchheatisneededtocompletelyvaporize100kgofwaterfrom temperatureT1

=20 oCifthe

pressureismaintainedataconstantP=200kPaabsolute?

T

v

P=200kPa

1

2

Solution:

TheprocessisshownintheTvdiagramdrawnonthe

left.State1beginsasacompressedliquidandstate2isshownasasaturatedvapor.Theprocessfollowsa

lineofconstantpressure(asindicatedintheproblem

statement).Sincewedonothaveaccesstoa

compressedliquidtableforthisexam,youhaveto

makeanassumptionregardingthepropertyselection

forstate1.Sincethewaterisinitiallyintheliquidstate,

weconsiderittobelargelyincompressible.Therefore,

wecannotaddmuchinternalenergytothewaterby

compressingit.However,asmalltemperatureincreaseinthewaterdoescontributetothewaterinternal

energy.Therefore,wewillassumethatduetoliquid

incompressibility,wecanusethepropertyofthe

saturatedliquid,evaluatedatthisinitialtemperature,torepresentstate1oftheliquid.


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P bl 5 C l l h k d b i i d


15/31

Problem5 Calculatetheworkdonebyapistoncontainedwithinacylinderwithairif2m3

istripledwhilethe

temperature

is

maintained

at

a

constant

T=30oC.TheinitialpressureisP1

=400kPaabsolute.

1

2

Solution:

Assumeair

isanideal

gas.

PV

=mRT

V1

=2m3

;V2

=3V1

=6m3

P=mRT/V

Work,

but Therefore,

P bl 6 H h k i i i


16/31

Problem6 Howmuchworkisnecessarytocompressairinan

insulatedcylinderfrom0.2m3

to0.01m3?

UseT1

=20oCandP1

=100kPaabsolute.

Solution:

Assume:

Idealgas

Useabsolutetemperature

Isentropicprocess

T1

=20oC=293K

ConstantcvoAdiabaticprocess(insulatedcontainer)

1st

Law(closedsystem)

Q=U+W= m(u)+W

W=mcvo

(T2

T

1

); W/m=w=cvo

(T2

T

1

)

Isentropicrelationships

T1

v1k1

=T2

v2k1

wherek=1.4(forair)

T2

=T1

{v1

/v2

}k1

=T1

{V1

/V2

}k1

=(293)K{(0.2)/(0.01)}1.41

=971K

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


17/31

Problem7 A10cmthickslabofwood(kwood

=0.2W/mK)is3mhigh

and10mlong.Calculatetheheattransferrateifthetemperatureis

25oContheinsideand 20oContheoutside.Neglectconvection.

q

Tin

=25oCTout

=-20oC

woodSolution:

Assume:conductionheattransferonly

A=LW=(10)m(3)m=30m2

q=

kA(dT/dx)~

kA(T/x)

q=(0.2)W/mK(30)m2

(20 25)

oC/(0.1)m

q=270W

Problem 8 The surface of the glass in a 1 2 m x 0 8 m


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Problem8

Thesurfaceoftheglassina1.2mx 0.8m

skylightismaintainedat20oCandtheoutsideairtemperature

is 20oC.Estimatetheheatlossratebyconvectionwhenthe

convectivecoefficientish=12W/m2 K.skylight

Solution:

Assumeconvectiveheattransferonly

q=hA(Ts

T

)

q=(12)W/m2

K(1.2)m(0.8)m[20 (20)]oC

q=460.8W

Problem 9 A 2 cm diameter heating element located within an


19/31

Problem9

A2cmdiameterheatingelementlocatedwithinan

ovenismaintainedat1000oCandtheovenwallsareat500oC.Ifthe

elementemissivityis0.85,estimatetherateofheatlossfromthe

2mlongelement.

qSolution:

Assumeradiationheattransferonly

Useabsolutetemperatures,

Treattheovenwallsaslargesurroundings.

Ts

=1000oC=1273K;Tsur

=500oC=773K

q=

A(Ts

4

Tsur

4

); where A=dL(circumferencial

area)

q=(0.85)(5.67x108

)W/m2

K4

()(0.02)m(2)m[(1273)4

(773)4]K4

q = 13,742W=13.7kW


20/31

Problem10

RefrigerantR134aexpandsthroughavalve

fromapressureof800kPatoapressureof100kPa.Whatis

thefinalquality?

1

2

Expansionvalve

Solution:

Thereisnotableofthermodynamicdatafor134a

containedwithinyourFEbook.Thereforeyoumust

usethePressure Enthalpydiagramprovided(see

subsequentpages).Thisdiagramismuchlikethe

Temperature volumediagramforwaterinthatthe

apexofthedomeisthecriticalpoint(CP).Thesolid

linecorrespondingtosaturatedliquidisontheleftandthatassociatedwiththesaturatedvaporisonthe

right.

1st

Law(SteadyFlow,expansionvalve),

h1

=h2

Wecommonlyuseexpansionvalvesinvaporcompressionrefrigeratorstomovefrom

thehighpressureandtemperatureportionofthecycletothelowpressureand

temperatureportionofthecycle.State1istypicallyconsideredtobeinasaturated

liquidstate.PleaseseeFEbookforlistofcommonlyusedsteadyflowdevicesandtheir

associatedmodelingequations.


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Problem10(cont.)

Solution(cont.)

h1

=hf

=244kJ/kg=h2

FromthePvs.hplot(nextpage),findthequality(thesearethecurvesemanatingfromthecriticalpoint)

x=0.37

P E th l di f R


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Pressure EnthalpydiagramforR 134a


23/31

Problem11 Whatistheminimumpumppowerrequiredto

increasethepressureofwaterfrom2kPato6MPawithamass

flowrateof10kg/s?

pump

1

2Solution:

1st

Law(pumps,isentropicorreversiblesteadyflow)

Use

the

saturation

table

for

water

to

find

vf

.

vf

=0.001002m3/kg

w= (0.001002)m3/kg(6000 2)kPa

w=

6.0kJ/kg;


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25/31

Problem13 Aheatpumpdelivers20,000kJ/hrwithq

1.39

kW

electrical

input.

Calculate

the

COP.Solution:

Aheatpumpislikearefrigerator.Itdifferentinthatincludesaspecial4wayvalvethat

allows

the

user

to

switch

the

functions

of

the

heat

exchangers

for

the

heating

season

(winter)andthecoolingseason(summer).Unlessotherwisestatedintheproblem

statement,theheatpumpisusuallyconsideredtobeintheheatingmodeforthese

analyses.Intheheatingmode,thatwhichwedesireisQH

.Therefore,theCOPisdefined

asfollows.

Problem 14 A thermometer with a wet cloth attached to its bulb


26/31

Problem14

Athermometerwithawetclothattachedtoitsbulb

reads20oCwhenairisblowingaroundit.Iftheairhasadrybulb

temperatureof33oC,whatistherelativehumidityanddewpoint

temperature?Howmuchwatercouldbecondensedoutofa100m3

volume?

Solution:

YouwillneedtousethepsychrometricchartprovidedinyourFE

book.Ithasbeencopied

onthefollowingpages.

Awetbulbtemperatureof20oCisprovidedintheproblemstatement.Thelines

correspondingtothewetbulbtemperatureonthepsychrometricchartaredashedand

movefromtheupperlefttothelowerright(locatethemonthechart).Thedrybulbtemperatureistheabscissaofthepsychrometricchart.Theintersectionbetweenthe

20oCwetbulbandthe33oCdrybulbtemperaturecorrespondstoarelativehumidityof

30%(therelativehumiditycurvesmovefromthelowerlefttotheupperrightonthe

chart).

Thedewpointtemperatureisthattemperaturewhenwaterbeginstocondense.Since

theordinateofthepsychrometricchartisthehumidityratio(gramsmoisture/kgdryair),

whichisproportionaltotheabsolutehumidity.Thedewpointtemperatureisthedry

bulbtemperaturethatcorrespondstotheintersectionofahorizontallinedrawnthrough

state1andthesaturationline(correspondingto100%relativehumidity).Thisisadew

pointtemperatureof~13oC.


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Problem14(cont.)

Solution(cont.)

Condensateistakenfromtheairwhenitiscooledtobelowthedewpointtemperature.Theamountofcondensatethatcanberemovedcorrespondstothat

associatedwiththe

differenceinthemoisturecontentatthedewpointtemperatureandthatat0oC.Below

0oC,thephasewillbefrostorice,correspondingtoacrystallization(vapor

solid)phase

changeprocess.

Therefore,

MassofCondensate=m(2

1

),

wheremisthemassofdryairand

isthehumidityratio(takenfrompsychrometricchart)

Assumingdryairisanidealgas

m=PV/RT=[(101)kPa(100)m3

]/[(287)kJ/kgK(33+273)K]=115kg

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

P h t i Ch t


28/31

PsychrometricChart

Problem 15 It is desired to condition 35oC 85% relative humidity


29/31

Problem15

Itisdesiredtocondition35 C,85%relativehumidity

airto24oC,50%relativehumidity.If100m3/minairistobe

conditioned,howmuchenergyisrequiredinthecoolingprocessand

howmuchintheheatingprocess?

Solution:

Onewaytoremovethemoistureistofirstcooltheairtothesaturationcondition(100%

relativehumidityline)andcontinuethecoolingprocesstoremovemoistureuntilthedesiredhumidityratioisachieved.Therefore,theengineermust

locatetheinitial(1)and

final(4)statepointsfirstandthenworkwiththepsychrometricchart(asshownonthe

followingpages)todeterminetheenergyrequirements.

Frompsychrometriccharth1

=110kJ/kgdryair,h2

=106kJ/kgdryair(seeenthalpyon

upperleft).Thisisthesensible(nophasechange)coolingpart.

P bl 15 ( t )


30/31

Problem15(cont.)

Solution(cont.)

Sensibleheating

Fromthecharth3

=37kJ/kgdryair,h4

=48kJ/kgdryair

Latentandsensiblecooling

TotalCooling=7.62131.4kW=139kW;TotalHeating=21kW

P h t i Ch t 2


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PsychrometricChart2

FE Thermodynamics Review 2011

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