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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.
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GeneralExamTakingPhilosophy Youshouldprepareandworktogetthe
correctanswerwheneverpossible.
However,onthoseproblemsyougetstuckon,rememberthatwronganswers arenoworsethannoanswer.
Therefore,firsttrytoeliminateunreasonableanswerstoimprovetheoddsofguessing
right. Thenmakeyourbestguess.
Ifyoudonthavetimetoeliminatewronganswers,pickyourfavoriteletterandguess.
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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.
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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
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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
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SaturationTable
Problem2(cont.)
Temperature
columnonlygoesto374oC,whichisthe
criticalpointfor
water(apexofthevapordome).
Therefore,you
mustusethe
superheatedvapor
tableforProblem2
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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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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
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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
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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
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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
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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
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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
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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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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
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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
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PsychrometricChart
Problem 15 It is desired to condition 35oC 85% relative humidity
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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 )
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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