UnitOperationsofChemicalEngineering化工单元操作.docx
- 文档编号:7913778
- 上传时间:2023-01-27
- 格式:DOCX
- 页数:35
- 大小:442.47KB
UnitOperationsofChemicalEngineering化工单元操作.docx
《UnitOperationsofChemicalEngineering化工单元操作.docx》由会员分享,可在线阅读,更多相关《UnitOperationsofChemicalEngineering化工单元操作.docx(35页珍藏版)》请在冰豆网上搜索。
UnitOperationsofChemicalEngineering化工单元操作
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:
Thevariablesofmainpipearedenotedbyasubscript1,andbranchpipebysubscript2.Thefrictionlossforp
- 配套讲稿:
如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
- 特殊限制:
部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。
- 关 键 词:
- UnitOperationsofChemicalEngineering 化工 单元 操作
![提示](https://static.bdocx.com/images/bang_tan.gif)