流体力学英文版课后习题答案.docx

流体力学英文版课后习题答案.docx

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流体力学英文版课后习题答案

1.1Whatwillbethe（a）thegaugepressureand（b）theabsolutepressureofwateratdepth12mbelowthesurface?

ρwater=1000kg/m3,andPatmosphere=101kN/m2.

Solution:

Rearrangingtheequation1.1-4

Setthepressureofatmospheretobezero,thenthegaugepressureatdepth12mbelowthesurfaceis

Absolutepressureofwateratdepth12m

1.3AdifferentialmanometerasshowninFig.issometimesusedtomeasuresmallpressuredifference.Whenthereadingiszero,thelevelsintworeservoirsareequal.AssumethatfluidBismethane（甲烷）,thatliquidCinthereservoirsiskerosene（specificgravity=0.815）,andthatliquidAintheUtubeiswater.TheinsidediametersofthereservoirsandUtubeare51mmand6.5mm,respectively.Ifthereadingofthemanometeris145mm.,whatisthepressuredifferenceovertheinstrumentInmetersofwater,（a）whenthechangeinthelevelinthereservoirsisneglected,（b）whenthechangeinthelevelsinthereservoirsistakenintoaccount?

Whatisthepercenterrorintheanswertothepart（a）?

Solution：

pa=1000kg/m3pc=815kg/m3pb=0.77kg/m3D/d=8R=0.145m

Whenthepressuredifferencebetweentworeservoirsisincreased,thevolumetricchangesinthereservoirsandUtubes

（1）

so

（2）

andhydrostaticequilibriumgivesfollowingrelationship

（3）

so

（4）

substitutingtheequation

（2）forxintoequation（4）gives

（5）

（a）whenthechangeinthelevelinthereservoirsisneglected,

（b）whenthechangeinthelevelsinthereservoirsistakenintoaccount

error=

1.4TherearetwoU-tubemanometersfixedonthefluidbedreactor,asshowninthefigure.ThereadingsoftwoU-tubemanometersareR1=400mm，R2=50mm,respectively.Theindicatingliquidismercury.Thetopofthemanometerisfilledwiththewatertopreventfromthemercuryvapordiffusingintotheair,andtheheightR3=50mm.TrytocalculatethepressureatpointAandB.

Figureforproblem1.4

Solution:

ThereisagaseousmixtureintheU-tubemanometermeter.Thedensitiesoffluidsaredenotedby

respectively.ThepressureatpointAisgivenbyhydrostaticequilibrium

issmallandnegligibleincomparisonwith

andρH2O,equationabovecanbesimplified

=

=1000×9.81×0.05+13600×9.81×0.05

=7161N/m²

=7161+13600×9.81×0.4=60527N/m

1.5Waterdischargesfromthereservoirthroughthedrainpipe,whichthethroatdiameterisd.TheratioofDtodequals1.25.TheverticaldistancehbetweenthetankAandaxisofthedrainpipeis2m.WhatheightHfromthecenterlineofthedrainpipetothewaterlevelinreservoirisrequiredfordrawingthewaterfromthetankAtothethroatofthepipe?

Assumethatfluidflowisapotentialflow.Thereservoir,tankAandtheexitofdrainpipeareallopentoair.

Solution:

Bernoulliequationiswrittenbetweenstations1-1and2-2,withstation2-2beingreferenceplane:

Wherep1=0,p2=0,andu1=0,simplificationoftheequation

1

Therelationshipbetweenthevelocityatoutletandvelocityuoatthroatcanbederivedbythecontinuityequation:

2

Bernoulliequationiswrittenbetweenthethroatandthestation2-2

3

Combiningequation1,2,and3gives

SolvingforH

H=1.39m

1.6Aliquidwithaconstantdensityρkg/m3isflowingatanunknownvelocityV1m/sthroughahorizontalpipeofcross-sectionalareaA1m2atapressurep1N/m2,andthenitpassestoasectionofthepipeinwhichtheareaisreducedgraduallytoA2m2andthepressureisp2.Assumingnofrictionlosses,calculatethevelocitiesV1andV2ifthepressuredifference（p1-p2）ismeasured.

Solution:

InFig1.6,theflowdiagramisshownwithpressuretapstomeasurep1andp2.Fromthemass-balancecontinuityequation,forconstantρwhereρ1=ρ2=ρ,

FortheitemsintheBernoulliequation,forahorizontalpipe,

z1=z2=0

ThenBernoulliequationbecomes,aftersubstituting

forV2,

Rearranging,

PerformingthesamederivationbutintermsofV2,

1.7AliquidwhosecoefficientofviscosityisµflowsbelowthecriticalvelocityforlaminarflowinacircularpipeofdiameterdandwithmeanvelocityV.Showthatthepressurelossinalengthofpipe

is

.

Oilofviscosity0.05Pasflowsthroughapipeofdiameter0.1mwithaaveragevelocityof0.6m/s.Calculatethelossofpressureinalengthof120m.

Solution:

TheaveragevelocityVforacrosssectionisfoundbysummingupallthevelocitiesoverthecrosssectionanddividingbythecross-sectionalarea

1

Fromvelocityprofileequationforlaminarflow

2

substitutingequation2foruintoequation1andintegrating

3

rearrangingequation3gives

1.8.Inaverticalpipecarryingwater,pressuregaugesareinsertedatpointsAandBwherethepipediametersare0.15mand0.075mrespectively.ThepointBis2.5mbelowAandwhentheflowratedownthepipeis0.02m3/s,thepressureatBis14715N/m2greaterthanthatatA.

AssumingthelossesinthepipebetweenAandBcanbeexpressedas

whereVisthevelocityatA,findthevalueofk.

IfthegaugesatAandBarereplacedbytubesfilledwithwaterandconnectedtoaU-tubecontainingmercuryofrelativedensity13.6,giveasketchshowinghowthelevelsinthetwolimbsoftheU-tubedifferandcalculatethevalueofthisdifferenceinmetres.

Solution:

dA=0.15m;dB=0.075m

zA-zB=l=2.5m

Q=0.02m3/s,

pB-pA=14715N/m2

Whenthefluidflowsdown,writingmechanicalbalanceequation

0.295

makingthestaticequilibrium

1.9．Theliquidverticallyflowsdownthroughthetubefromthestationatothestationb,thenhorizontallythroughthetubefromthestationctothestationd,asshowninfigure.Twosegmentsofthetube,bothabandcd，havethesamelength,thediameterandroughness.

Find:

（1）theexpressionsof

hfab,

andhfcd,respectively.

（2）therelationshipbetweenreadingsR1andR2intheUtube.

Solution:

（1）FromFanningequation

and

so

Fluidflowsfromstationatostationb,mechanicalenergyconservationgives

hence

2

fromstationctostationd

hence

3

Fromstaticequation

pa-pb=R1（ρˊ-ρ）g-lρg4

pc-pd=R2（ρˊ-ρ）g5

Substitutingequation4inequation2，then

therefore

6

Substitutingequation5inequation3，then

7

Thus

R1=R2

1.10Waterpassesthroughapipeofdiameterdi=0.004mwiththeaveragevelocity0.4m/s,asshowninFigure.

1）Whatisthepressuredrop–∆PwhenwaterflowsthroughthepipelengthL=2m,inmH2Ocolumn?

2）Findthemaximumvelocityandpointratwhichitoccurs.

3）Findthepointratwhichtheaveragevelocityequalsthelocalvelocity.

4）ifkeroseneflowsthroughthispipe，howdothevariablesabovechange？

（theviscosityanddensityofWaterare0.001Pasand1000kg/m3，respectively；andtheviscosityanddensityofkeroseneare0.003Pasand800kg/m3，respectively）

solution:

1）

fromHagen-Poiseuilleequation

2）maximumvelocityoccursatthecenterofpipe,fromequation1.4-19

soumax=0.4×2=0.8m

3）whenu=V=0.4m/sEq.1.4-17

4）kerosene:

1.12Asshowninthefigure,thewaterlevelinthereservoirkeepsconstant.Asteeldrainpipe（withtheinsidediameterof100mm）isconnectedtothebottomofthereservoir.OnearmoftheU-tubemanometerisconnectedtothedrainpipeattheposition15mawayfromthebottomofthereservoir,andtheotherisopenedtotheair,theUtubeisfilledwithmercuryandtheleft-sidearmoftheUtubeabovethemercuryisfilledwithwater.Thedistancebetweentheupstreamtapandtheoutletofthepipelineis20m.

a）Whenthegatevalveisclosed,R=600mm,h=1500mm;whenthegatevalveisopenedpartly,R=400mm,h=1400mm.Thefrictioncoefficientλis0.025,andthelosscoefficientoftheentranceis0.5.Calculatetheflowrateofwaterwhenthegatevalveisopenedpartly.（inm³/h）

b）Whenthegatevalveiswidelyopen,calculatethestaticpressureatthetap（ingaugepressure,N/m²）.le/d≈15whenthegatevalveiswidelyopen,andthefrictioncoefficientλisstill0.025.

Figureforproblem1.12

Solution：

（1）Whenthegatevalveisopenedpartially,thewaterdischargeis

SetupBernoulliequationbetweenthesurfaceofreservoir1—1’andthesectionofpressurepoint2—2’，andtakethecenterofsection2—2’asthereferringplane,then

（a）

Intheequation

（thegaugepressure）

Whenthegatevalveisfullyclosed,theheightofwaterlevelinthereservoircanberelatedtoh（thedistancebetweenthecenterofpipeandthemeniscusofleftarmofUtube）.

（b）

whereh=1.5m

R=0.6m

Substitutetheknownvariablesintoequationb

Substitutetheknownvariablesequationa

9.81×6.66=

thevelocityisV=3.13m/s

theflowrateofwateris

2）thepressureofthepointwherepressureismeasuredwhenthegatevalveiswide-open.

Writemechanicalenergybalanceequationbetweenthestations1—1’and3-3´，then

（c）

since

inputtheabovedataintoequationc，

9.81

thevelocityis:

V=3.51m/s

Writemechanicalenergybalanceequationbetweenthestations1—1’and2——2’,forthesamesituationofwaterlevel

（d）

since

inputtheabovedataintoequationd，

9.81×6.66=

thepressureis:

1.14Waterat20℃passesthroughasteelpipewithaninsidediameterof300mmand2mlong.Thereisaattached-pipe（Φ60⨯3.5mm）whichisparallelwiththemainpipe.Thetotallengthincludingtheequivalentlengthofallformlossesoftheattached-pipeis10m.Arotameterisinstalledinthebranchpipe.Whenthereadingoftherotameteris2.72m3/h,trytocalculatetheflowrateinthemainpipeandthetotalflowrate,respectively.Thefrictionalcoefficientofthemainpipeandtheattached-pipeis0.018and0.03,respectively.

Solution：

Thevar