Unit-Operations-of-Chemical-Engineering(化工单元操作)

abab.1.1Whatwillbethegaugeand(b)theabsolutepressureofwaterdepth12mbelowthesurface?ρwater=1000andPatmosphere101kN/m2.Solution:Rearrangingtheequation1.1-4ppba

ghpressureoftobezero,thendepth12mbelowsurfaceispp10009.81kPabpressurewaterat12mppb

gh1010001000218720PakPa1.3Aasshownmeasurepressurethereadingtheinequal.fluidBmethane（烷）,thatCingravity=andthatliquidAinUwater.The51mm,IfthereadingofmanometerthepressuretheinstrumentInofwhenthechangeinthethereservoirsis(b)whenthechangeintheintakenintoaccount?inanswertothe(a)?：=1000kg/m

=815kg/m3

3

D/d=8R=0.145mthepressuretworeservoirsisincreased,thechangesinreservoirsandUtubes4

D

4

d

so

hydrostaticfollowingrelationshippgpg12sop)12

theequation(2)forxintogivesppR()gA

（）intheintheis123123ppRR(

)(Ac

.)0.145815263PaAc（bthechangeintheintakenaccountpp

R(

gcR(

g9.810.145815281.851error=

281.8281.8

6.7％1.4twoU-tubefluidasinthereadingsoftwoU-tube=400mm=50mm,Theindicatingliquidisofwithpreventfromthediffusingintoandheight=50mm.TrycalculatepressurepointandFigureforproblem1.4Thereagaseousmixtureinmanometermeter.Thedensitiesofareby

,

2O

,

ThepressurepointAisbyhydrostatic12111211.p

O

gR()Hg2g2

g

isnegligibleinwith

Hg

ρH2O

equationbepA

c

=

2O

gR3

2=1000×9.81×0.05+13600×9.81×0.05=7161N/m²ppDA

=7161+13600×9.81×0.4=60527N/m1.5Waterdischargesfromthethewhichdiameterisd.ratiooftodequals1.25.Theverticalbetweentheofdrainpipe2m.heightHthecenterlineofdrainpipewaterinreservoirrequiredforthewaterthetankAtotheofthethatfluidflowpotentialThetankandtheexitof

p

allopenair.H

D

dp

hFigureforproblem1.5ASolution:Bernoulliequationwrittenbetweenand2-2,withbeingreferenceplane:pp122Whereppusimplificationofequationu2Hg22oD44111221211212oD44111221211212.Therelationshipvelocityoutletandvelocitycanthe2u

u

Bernoulliequationwrittenbetweenthestationpu200

Combining1,2,andgivesu2112Hg＝＝＝21.2510002.44forHH=1.39m1.6AwithaρhorizontalpipeofAm2

isflowingunknownthroughat,andthenittosectionoftheinreduced2

pressureisAssumingfrictioncalculateVandifthepressuredifference(-p)is:Fig1.6,flowdiagramisshownwithpressuretapstopp.Fromthemass-balancecontinuity,constantρρ=ρ,2222.A12theitemstheBernoulliequationahorizontalpipe,z=1ThenBernoulliequationbecomes,after2212p2p011012Rearranging,

A12

V2p12

2(1

1

V＝

PerformingsamederivationtermsofV,2p＝1

1.7Aofviscosityisflowscriticalforlaminarflowincircularpipeofdandwithmeanthatthepressureinlengthofpipe

32isL

OilofPasflowsofdiameter0.1mwithaof0.6m/s.RRABABBARRABABBA.Calculatelossofinalengthof:heforasectionfoundallcrosssectionanddividingbythecross-sectionalareaV

A0

2

profileforlaminarL

equationforintoequationandintegratingV

p0LD

2

332L2320.05d

PaInverticalwater,pressureareinsertedAandBwherethepipeare0.15mThebelowAandwhentheflowratedown0.02/s,atBN/m2

greaterthatattheintheAandB2expressedwhereisthevelocityatA,findthe2

Figureforproblem1.8ofk.IftheABtubesfilledwithwatertoaU-tubecontainingmercuryofgivesketchtheintheofthediffercalculatevalueofthisinSolution:=0.15m;z-z=lQ/s,-pN/m2BAhfcd1BAhfcd1.

4

d

d4

0.020.15

m/

4

d

d4

0.020.7850.075

mthefluidflowsdown,writingbalancep

g

B2

VA222.5

1.132147154.5321.1321000

214.7150.638k

makingthestaticequilibriump

gHgR

pHg

147159.81126009.81

mm．liquidflowsdownthroughthefromtheathethetubefromthecstationd,infigure.Twosegmentsofthecdthesametheand（1the

ab

fab

cd

h

Figureforproblem1.9（2therelationshipRRtheUtube.Solution:h

fab

lV2d2ab1d21ab1d21.hso

lVdfabFluidflowsfromstationatostationconservationppabhence

fabpa

fab

tostationpphence

p

d

static=R（-）g-lρg=R（-）gSubstitutingequationinequation2，then

1

fabh

fab

1

g

Substitutingequationinequation3，thenh

2

g

ThusR=R22.Waterpassesapipediameteri=0.004withtheinWhatthepressureflowsthroughlengthinHOFindmaximumvelocityandpointrwhichitFindtherwhichtheaveragevelocity

Lequalsifkerosenethroughthispipevariables？（theandWaterarePas

rkg/m

，；theviscositydensityofarePasand

Figureforproblemkg/m3

，respectively）solution:）

ud0.0040.001

Hagen-Poiseuilleequation

32d2

0.40.0010.0042

h

160010009.81

m）occursatthe0.5uso=0.4×2=0.8m）u=V=0.4m/su

umax

rr

r10.004

Vu

0.5r0.0040.004kerosene:

ud0.003

0.00316004800ee.h

48009.81

0.611Asshowninthekeepsconstant.Asteel(withtheofisU-tubeistothethefromtheofthereservoir,andtheUisfilledwithmercuryandtheleft-sidearmoftheUtheiswithwater.distancebetweentheandthepipelineis20m.a)Whenvalveisgatevalveispartly,h=1400mm.Thefrictioncoefficientλis0.025,losscoefficientofentranceis0.5.Calculaterateofwatergatevalveispartly.(in/h)thegateiswidelyopen,attap(inpressure,N/m²).l/≈15valveopen,thefrictioncoefficientλis0.025.Figureforproblem：Whenthevalveisopenedtheequationbetweensurfaceofreservoir1—

’

thesectionofpoint2，andthecenter’then

2pu12

f—2

（）Intheequation

(thepressure)pgR2Hg

O

ghN/

2.0thevalvefullyclosed,theheightofwaterlevelinthereservoircanrelateddistancebetweenthemeniscusofleftU

HO

g(Z)gR1Hg

（bh=1.5mubstitutetheknownvariablesintoequationb

136001000

h

f_2

lV215K)d0.12

2.13V

thea9.81×6.66=

V396302.1321000

2theisVtheflowrateofwaterisV3600

4

d

4

0.1288.5m

/）thepressureofpointwhereismeasuredvalvemechanicalenergystations—1

，then1

p

3

32

p3

h

f3

（c）

6.66Z3u01p1

3

f3

l

)

2[0.025(

V24.81

2inputintoequationc，

6.66

V22

22c2c.theis:Vmechanicalenergybetween—’and——2’,forthesamesituationwater

pV2p11122h226.66

f,1—

（dZ21m/20(page）1

f,1_2

l2K)(0.025J/2inputintoequationd9.81×6.66=theis:

3.51p2100032970

26.2℃passesthroughawithanof300mmandlong.Thereisattached-pipe(Φ60whichwiththepipe.lengthincludingequivalentlengthofformoftheattached-pipe10m.Arotameterinbranchpipe.readingoftheis3

/h,calculateflowratethepipeandtheThecoefficientoftheisand0.03,：ofmainpipedenotedbyasubscript1,bysubscript2.Thefrictionforparallel

h

f

f2V1

S2Theenergyinthebranchpipeis

h

f

ld

2

u2Intheequation

0.0312f2112f21.ll2

2

d2u23600

4

2

/sinputintoequationc

h

f2

100.34320.03/kg0.053Theenergyintheis

f1

f

l21

u1

0.3330.30.018

2.36/ThewaterisVh

4

2.36601

/Totalwaterdischargeism3/hhAVenturimeterisusedforflowofwaterpipe.TheoftheVenturithroatistwofifthsdiameteroftheinletarefilledtubesamanometer.Thevelocityofflowalongfoundtobe2.5R

manometerreadinginmetresoflossofheadinletandofVenturiwhenR(RelativeisSolution:WritingequationbetweeninletandoforVenturip1

pog2

above,z)=p1

2o2

f

continuityequation

Figureforproblemo219.032.52o219.032.52f1300.V1

dd

V6.25

equationforVinto2gives39.06111fRf

xg

f

thehydrostaticequilibriumforpp(1

Hg

x

equationforpressuredifferenceinto4obtains(

118.94

f

6hf

(

123.61118.94RR2.288/kg1.17.Sulphuricacidgravityisflowingapipeof50internaldiameter.Athin-lippedintheanddifferentialbymercuryisAssumingthattheleadstothemanometerarewithacid,(a)theweightofflowingsecond,and(b)frictionincausedbyorifice.Thecoefficientorificemaytakenas0.61,specificgravityofmercuryasdensityofwater10003Solution:a)

100501p(1

Hg

g0.1(136001300)2

o101

4

1

2

20.1(136001300)140.614.31/s22.m

4

D

2

4

0.01

2.631300kgsapproximatedropp(1

Hg

g0.1(136001300)

12066.3Padueofvelocityinthroughorificep2

2o212

1300

(14)2

Padropby4488.8Paf2.1totestforofThepressurethereadingatkPaasflowrate26m3/h.Theshaftiswhilethecentrifugaloperatesthespeedof2900r/min.Ifthedistancebetweenthesuctionand0.4m,thebothlineareCalculateefficiencyofandlisttheoftheunderthisoperatingSolution:themechanicalequationconnection2p2pH2g

H

f_2

pressure）1.522H0f,12totalheadsof

H

51000

5

18.41efficiencyofis

Ne

e

QH10003600

1.3oof1oof1.N=2.45kWefficiency

45

100%Theperformanceofpumpisrate，m³/hTotal，power，，%2.2Wateristransportedbyammvacuum,thetank,inwhichtheis0.5asintotal

2equivalentofpipealllocal

loss.Thepipelineis

1×mmtheorificecoefficientofCandorificediameterd0.62and25mm,coefficientDevelopedHofinm(theofUpressuregaugeinorificeisSolution:Equation(1.6-9)V

（

0.620.1689.81(13600

0.62

6.444.12m/sratemS

3.144

10002.02kg/sFluidflowpipethetotheisforV=VH

p2

H

f2200ovgf2200ovgf.

200760

g=7.7mThebetweenthevelocityofpipeVmsDFrictionlossff

lu20010.025d2g29.81

mso2.3.Acentrifugalistousedtoextractwaterinthemmofmercury,asinfigure.Atratedpositivesuctionbe3mthecavitationvaporofIfinthepipeaccountedforheadof1.5m.Whatmustbeleastheightofliquidlevelintheabove:From

H

gHg

po

HNPSHfWhereUseofwillgivetheminimumheightHasH

ppvNPSH1360010009.81

m2.4Sulphuricatkg/sa60m25mmdropinpressure.Ifthedropfallsbyhalf,whatwilltheflowratebe•

of1840kg/m3•

Viscosityacid×10

PasSolution:Velocityinpipe:dffdff.u

flowratecrosstionalareaof0.78518400.02524

2

/sReynolds

d0.0251840

Fig.1.22forwhendropcalculated1.4-9h4

l2603.32d22

2

/kg1840frictionisfromh4f

llu2603.32446109d20.025

2

426J/kg426783.84kPaifpressuredrop

lu23919200.046Re＝40.0462d

d

l1.81.20.0461840

1840

600.0252

sou

391920

2.27m/

ρ=0.785×0.0252×2.27×1840=2.05kg/s2.4Sulphuricatkg/sa60m25mmdropinpressure.Ifthedropfallsbyhalfoffrictionisnegligible,willthenewbeofViscosityacid×10

Friction

f

0.32

forsmoothpipeSolution:12ffof12ffof.energyequation:pup11H22lu22

f4

d

uu

32

121840

3.32msRe

184025

61150.0056

0.00560.00870.32l60h422Δ×1840×9.81=847.0kpa

46.922.6fluidisfromAtoBwithφ

diameterlengthofmeters,figure.orifice16.4mmusedmeasureflowOrificecoefficient＝0.63.losspressureis×102,frictionfind:WhatispressurealongAB?Whatisofobliteratedintototalpowerfluidshaftis500W,60%efficiency?(Thefluidis3solution：

)zg

p

u2

pgAh

fpB

lhd

022.u

1

2gR

0.630.68700.97870

/∴u(16.4/33)×8.5=2.1m/s∴

pp

h0.024870f

302.10.033

4m

2Wm

4

d2u768550.7850.033

2

sotheratiopowerinfrictionlossesinABpowerfluid1385000.6

％％3.2Aquartzose（颗粒）withdensityof2650（淀）freelyinthe℃trytocalculatethemaximumdiameterStockslawandtheminimumdiameterobeyinglaw.olution:hegravitysettlingfollowedStocks’maximumsettledbeRethatis1Ret

dut

thent

fortheterminalc

(cSsolvingcriticaldiameterdc

(

Thedensityof20airρkg/m³viscosityµ=1.81×10N·2pttptt.d1.2243

)(26501.205)1.205m

m≥lawandterminalvelocitycanbecalculatedbygdu1.75criticalReynoldsnumber

Ret

d

c

ut

，

givest

combinationofwith3d1.74cdcsolvingcriticaldiameterd

3c

d

c

)21.205)1.2051.5121512um

m3.3Itisdesiredremoveparticles50infrom226.5m

ofusingasettlingforThetemperatureare21

C1atm.Theparticledensityis3.Whatminimumdimensionsofthechamberareconsistentwith(themaximumoftheairsolution:calculateterminalvelocityfromtheequation3.2-16.Thedensityof21airρkg/m³viscosityµ=1.81×10N·2ut

d

18

2

m/s

t

soBL

Q20.86mu60t

2

LHuutthepermissibleoftheairis3m/sLH30.181

setbemthenequationLmH3.4Acycloneusedtoseparatedustofof2300kg/m³fromTheflowgas/h,ofthe·s/m²,kg/3

Ifdiameterofcycloneis400mm,totheSolution:D=0.4mBDDui

Q1000/shB0.2Accordingtothefor：d

(

99(3.6m0)i3.6Apressof

filteringareaisusedforfilteringaslurry.Thefiltrationcarriedoutconstantpressurewith500mmHg.Theoffiltratecollectedinthefirstwasoneliterafurther5min,an0.6literwasHowmuchfiltratewillbefiltrationoutfor15minassumingcakebeincompressible?mfm0.00930.0093mfm0.00930.0093.heforconstant-pressureKAt5min

l.

1

2

VK0.1

2

10min

l

2

V0.1

2

solvingequationsaboveVandK

KFor

min

V

22

forV=2.073l3.7Thefollowingareforafilterpressofm

filteringareaintestPressuredifference/kg2

filteringtime/s50×-39.10-32.27×-39.10×-3filtrationconstantKatthepressure1.05iftheofthefilterfilledwiththecake660s,whatisthefinaloffiltrationdt

Eandconstantofn?①fromq

Kt

For

p1.05

㎏/㎝2.27

2

0.0093

2

K

solvingequationsandgivesqm

mm

32

K

m

2

/0.00930.00930.00930.0093.②KAdt2(E

=

m3/For

3.5

㎏㎝2

2.270.0093

q

m

m

solvingequationsandgivesq

m

mm

32

K

m

2

/4.37

1

1

lnsolvingn3.8Aslurryiffilteredbyfilter2aconstantfiltration2t

filteringareaconstantequationforq=filtratefilteringarea,inl/m2,filteringinminfiltratewillafter249.6min?Ifpressuredifferencebothresistancesoffiltrationmediumandcakearehowmuchfiltratewillbeafter249.6min?solution:q

2

250(250(249.6

22V=24lq11222mq11222m.thedifferenceisKK

1.41250l

l3.9Filtrationoutplatefilterwith20frames0.3and50mmthick.Atapressuredifferenceof248.7kN/m

one-quarterofthetotalfiltratecycleisobtainedforfirst300s.Filtrationiscontinuedataconstantforfurther1800s,theframesfull.Thetotalfiltrate0.7

dismantlingandtakes500sItdecidedusearotaryfilter,1.5mindiameter,placeoffilterpress.thattheresistanceofthesamethatthefilterisspeedofrotationofdrumwhichwillinsameratefiltrationaswaswithfilterpress.Thefiltrationinthefilterisoutconstantof2

with25%ofthesubmergedSolution:filtration:A=2×0.32×20=3.6mΔp2tV×0.7=0.175m3tt3tV0.175

2

V

2

t3.6

2

300KV3.622100KV=0.2627m3K

0.723.62

Vm210012.96

Q

2100500

2.692

m

/fortherotarydrumfiltermmmmmm.filtration:AdL×2.2×1.52pressure:=70kN/m2thedrumsubmerged:ForrotarydrumfiltrationkeepVKchangeswithchanginginK

703.1517248.7

thecapacityofdrumfilteristooffilter

10.3622

0.2627

0.2627

fortheequationtrialand

0.0002381nVK20.0690.2627mm转/min)Afilterof0.093m2

filteringwasusedsuspensioncontainingcarbonate,operatinginpressure.Thefiltratecollected2.27×103

duringthe50volumeofcollected3.35×10

3

duringthe100s.Howwillfiltratebefilteringfor：filteringaream

;filtrate-3

3

fort=50sandV=3.35×10

3

fort=100sforpressurefiltrationKAtvariablesgivengives

2

2.27

0.093

2

2

2

2

combining12V=3.85×10-4

3

-5

2

/sSubstitutionofandKforconstant-pressureequationVVA2tsolvingvolumeofgainedfor200smm.V-3

3Aslurrywithcakeis

filterconstantpressure.IfthefilteringconstantK2

/min,operatingpressuredifference

2atmtheresistancecanbeFindthefiltratevolume,in

filtering10ifthebeandVis1m

whatfiltratevolumeV,in3

filtering10?whatfiltrationdV/dtwhenthe10forcase(1)?Solution:forconstant-pressure

2

2

tfiltersetnegligibleV2KA2VAKt

②Thefilteringmediumtakenaccount

2

2

tV2V

2

9.05m

③theequationforconstant-pressurefilteringisderivativeasmediumresistancedoesntbeaccountdV210dt22Aflatwallconstructedofalayerofbrick,withaconductivityof0.138W/(m·°C)bycommonbrick,conductivityW/(m·°C).temperatureofinnerofthewallisC,andof76.6°C.()Whatheatthroughthewall?(Whattemperatureoftheinterfacebrickcommonbrick?(c)Supposingbetweenthetwobricklayerspoorthata“contactresistance”of2/Wpresent,theloss?Solution:iioooiiooo.heatloss:

qA

i

BB2kk2

7600.1140.1381.38

688.89/m

temperatureinterface:/AB1k0.1381℃

2(c)is0.088°C·m2

/W,heatloss:qA

i

760BB0.1140.2290.088kk1.38

W/m

(dbyk=0.615C).IfthetemperatureofoutersurfaceofpipeCtemperaturethelayeris38C,whataretheheatlosstemperaturewithinlayer?：Similartoinseriesthroughtheflatwall,thetotalresistanceacrossinsulating

B0.615

0.426)/ln(1.278/0.426)LLHeatpertubewallq177589.7WLtemperatureqLR

177

r0.213

589.7/mt2277.5

routerdiameterofasteel150mm.Thetubebackedtwolayersheatofconductivityoftwoinsulatingmaterials1i1oiioop1i1oiioop/21

.bothinsulatingmaterialsthesamethicknessIfthedifferencebetweenwallandsurfaceinsulatingconstant,insulatingmaterialshouldbepackedinsideheatsolution：k（lower）thewithlowerconductivity

i

bk1

bk

A

2

L(r)3r3r2

m

L(r)21r2r1

thicknessoftwolayersareequalb=b=b,

/22

k11

12

1AAm2contrarily,materialwiththermalconductivityinside,kQbbb1112AkA2A2A1mm2m11mm1m1QA2Amm2

for

12

sowithheatobtainedasthematerialwiththermalconductivityisAiratnormalpressurethrough20andiso

C.Whatfilmheattransfercoefficienthpipewalliftheaveragevelocityofairis10m/s?propertiesat60CAir:densitykg/m,viscosityµ=0.02conductivityW/(mC),=1kJ/(kgSolution:piodipiodi.

du

0.02

0.0289

0.023

di

0.8

106000.02

0.8

0.4

0.03324Methylalcoholflowinginthepipeofadouble-pipecooledwithwaterflowinginthejacket.Theinnerfrom25-mmTheconductivityofsteelisW/(m·oandfoulingareinthefollowingTable.Whatoverallcoefficient,ontheoutsideofinnerpipe?Datafor8W/(m

hWatercoefficientInsidefoulinghfoulingfactorSolution:equation

do

U

dbd1hhkhhidiiiooappendixthewallthickness3.25mmandoutsidediameterisfornominaldiameter25mmSchedulestandardsteelpiped

2

30.25mmoiokmpoiokmp.U

2.1841.2167.9983.521

/

o

CBenzenewithmass1800kgthroughthedoublepipeexchanger.fromo

Cto

C.TheofinnerisandtheouterWhatisindividualtransfercoefficientbenzene?Solution:diameterinner=19mm,insidediameteroftubeTheequivalentdiameterofannulartube

(2)D)i

32mmioequationh0.023

k

u

80t2

thebenzene:-3Pas，，℃ρ=850kg/m

attemperatureof50℃velocityofbenzeneflowingthroughthespaceofisas

4

(2)uiu

42)36000.0322io

1.13m/Substitutingtheintoequation1givespko12ppoipko12ppoi.h

k

e

0.4＝

0.0130.45

3

0.40.24427748

2

4.10.Air2atmand20Cinthepipeoftubular80ofis57×3.5mmitslengthislong.Whattransfercoefficientofside?olution:ofairatinletandoot2080t122

o

Cofair

Cμ=0.019×10Pas,densityatonideallaw,

229273)

kg/m311,air

2

0.028

W/mKApp.13heatofair=1KJ/kgKThevelocityofairflowingthroughtubeu

4603.140.05

8.49

m/s

du

0.019

48884.5

10.028

0.68h

k0.0280.0230.8Pr0.4dd0.05

W/m2

K4.12.Ahotfluidwithaflowrate2250through25x2.5mmtube.physicalofareasfollows:C),c

=4kJ/(kgNdensityHeatfilminW/(miio1/30.81/3p11iio1/30.81/3p11.If1125kg/hotherconditionswhatishIfof(inside10mm,andthethesameasofcalculatethetemperatureoffluidandquantityofheatflowtubeareW/m,temperatureofwallfora?thisinordertoheattransferfilmcoefficientenhanceheatwhatkindsofmethodscanuseandisHint:forforturbulentflowPrsolution:

u

4236000.02

2

1.99

m/s

du

1.99100010

4

kNuh0.0230.8Prd0.02

2.05503.1/m

2

K

hu

coefficienthasratedecreases1125kg/hhh

0.8

1125

0.8

0.574´=5503.1×0.574=3158.8W/m

Kkh0.023Red

0.8

13hh

0.2

0.2

1.15´2

KffLffL.qAh(t)f

dlht)fwtf

q40040dlh3.140.02

o

Cthevelocityofflowingthesaturatedat100o

Cissaturatedliquidsurfaceofasingleverticalwithanddiametertemperatureofwall92C.WhatisthevaporIfhorizontally,thequantityofvaporperhour?Solution:From6,ofsaturatedat100

oTheaveragetemperatureof℃Thepropertiesofwaterat96

oareasfollows:

Pas(fromappendix=958kg/m

;fromequation

3hof

2

/(

C)m

hA()

3

kg/20.8kg/the

md

0.038

)440.282

flow)Ifisplacedfrom(4.5-14)h

0.729

ffdof

0.773poooioo0.773poooioo.h0.729h

Ld

2.50.7730.0382mhm4.14.Inadoublepipecold(ckJ/(kgflowkg/h)throughthepipehotfluidthroughtheTheinletoffluidareand80C,inletandoutlettemperaturesoffluidare150and90ChinsideaW/(mo

heattransfercoefficientU(basedsurfaceofpipe)300W/(mo

C),Ifheatisandthewall(steel)takenas45W/(mfind:heatpipe?thepipelengthforinispipeifheatingchangestosaturatedC)whichsaturatedliquidconditionstheusedforitisthatitmeetproduction(thetemperatureofcoldfluidcannot80C),explainSolution:(1)FromequationUo

1bd1oodhhiimoo

o1222m112212o1222m112212.0.0017294.868

o

℃writeenergybalanceequation(t)UAUp21

dLo

forcountercurrentflowToto

C

T=90o

∆t1so∆t

=150-80=70C;∆to=70oCtoofdoubleL

(t)21U

50070

T

C

∆to;∆tCt

C

tC

sot22L

mc(t)Uo

mct)1mtU36003000.023U12oolndepositingonthesurfacesofthegiveadditionalthermaloooiiiiioiooooooiiiiioiooo.15.WaterturbulentlyinthethevelocityofwaterVistransfercoefficientU(basedoutersurfacepipe)W/(m2

C).Ifthe1.5m/sUis2660W/(m

CWhatisfilmhoutside(Heatofpipewallandscaleignored)Solution:equation(4.3-37),neglectingthermalresistanceoftubewallU

111thevelocityofis1m/s,coefficientoftowalloftubesiscoefficientUo

2115

oftheindividualwaterwallofh,tohby

0.8

1.5

0.8

coefficientisUo

i

hi

solving1and2forhandh=2858W/(m

);hm2

4.17.Waterpassparallellythroughexchangerwhich1mlong.inletandoutletofwater15CandC,andthoseareand100Iftheofdecreasestotheflowandphysicalp222oop222oo.inlettemperatureswaterwhatlengthofexchanger?(Heatwallare：energyforforwater

q100)1qc15)2

qc(150100)c80)UA1p2differentoperatinginparallel℃15010060℃13560℃135ln60

foroil

c(1501p1

（）forwater

qmct22

（5qcc(t111p

2

qmcmctqm100)cUAp2solvingfort℃temperaturedifference℃5030℃135℃135ln30equation

mm

q150Lq150100

solving

L

=1.86L4.18.Airwhichpipeinturbulentflowisheatedfrom20Cto80C.ThesaturatedatCsaturatedthepipe.Ifrateoforiginandinlettemperaturesofairwhatkindof.employintodothat?(Heatresistancewallscalebe：temperatureofair:

t

202

=50Ctheairat℃

1.093

kg/m3

c1.017p

kJ/kg.k

k

thevaporpressureofsteamtomeettheforrisingintherateofair20%,temperatureofdenotedbyTq(8020)p

（1q

20)p

（2mm∵coefficientwithchangeisverycomparedwithcoefficientofair,theofsideinwithresistanceandUh1.2Uh

0.8

1.157

ln

80

61.5

℃

temperatureof