化工原理课程(全英文)教学课件-9

©2015YanweiWangBasicPrinciplesofChemicalEngineeringProcessesLecture9FallSemester,2014YanweiWang(王衍伟)Email:[e-mail]©2015YanweiWang2PreviousLectureDate:March26,2015Pages56-66oftextbookFrictionlossesfromchangesinvelocityordirectionPipeflowsystemsNumericalmethodsforsolvingnon-linearequations(Supplementary)©2015YanweiWang3TypesofFluidFlowProblemsTypeIIandtypeIIIproblemsrequireatrial-and-errortechnique,oraniterativesolutionprocedure(迭代求解过程)©2015YanweiWang4HW-8frompreviouslectureHW8-1:TypeIIpipeflowproblemSolveExample1.7(pp.59-60)Itisrequiredthatyoudrawasystemsketch(示意图!!!)beforesolvingtheproblem.Asamplesketchofpipeflowsystemcanbefoundonpage80(Figureforproblem1.13).©2015YanweiWang5HW-8frompreviouslectureHW8-2:TypeIIIpipeflowproblemWorkExample1.7(pp.59-60)backward,assumingthattheflowrate(𝒒=𝟕𝟓.𝟔𝐦𝟑𝐡)isknown,butthepipediameter(𝑫=𝟎.𝟎𝟕𝟖𝐦)isunknown.Solveforthepipediameter.©2015YanweiWang6Today’stopicDate:March30,2015Finishingchapter2Thethree-reservoirjunctionproblemExercise1.15(page80)Laminarflowthroughaninclinedpipe(HW6-2)Meteringoffluids(pp.66-76)PitottubeVenturimeterOrificemeterRotameter©2015YanweiWang7Thethree-reservoirjunction(1/4)Thethree-reservoirjunctionproblemThreereservoirsatknownelevationsareconnectedtogetherwiththreepipesofknownproperties(lengths,diameters,androughness).Theproblemistodeterminetheflowrateintooroutofeachreservoir.Clearly,fluidflowsoutofreservoirAbecausetheothertwoareatlowerlevels.WhetherthefluidflowsintooroutofreservoirBdepends.Flowrate&directionareTBD!!!©2015YanweiWang8Thethree-reservoirjunction(2/4)Onemayalsowritethecontinuityeq.atthejunctionnode,N𝒊=𝟏𝟑𝒒𝒊=𝒒𝟏+𝒒𝟐+𝒒𝟑=𝟎WhichmeansnoaccumulationatthejunctionnodeN.Inthiscase,positivevaluesof𝒒𝒊meansflowintothejunction,andnegativevalueforflowleavingjunction.AssumingfluidflowsfromreservoirAtoreservoirBandC,theequationofcontinuitybecomes:𝒒𝟏=𝒒𝟐+𝒒𝟑©2015YanweiWang9Thethree-reservoirjunction(3/4)AssumingfluidflowsfromreservoirAtoreservoirBandC,theBernoullieq.foreachbranchingpipefromthejunctiontoreservoircouldbe:𝒉𝒇𝒊isthefrictionlossinbranchingpipe𝒊,specifiedbythesubscript(𝒊=𝟏,𝟐,𝟑).Subscript𝑵referstothejunctionnode.©2015YanweiWang10Thethree-reservoirjunction(4/4)Wecangetridofthemechanicalenergytermforthejunctionnodebysubstitutions.Thisgives𝒑𝑨𝝆+𝒈𝒁𝑨+𝑽𝑨𝟐𝟐=𝒑𝑩𝝆+𝒈𝒁𝑩+𝑽𝑩𝟐𝟐+𝒉𝒇𝟏+𝒉𝒇𝟐𝒑𝑨𝝆+𝒈𝒁𝑨+𝑽𝑨𝟐𝟐=𝒑𝑪𝝆+𝒈𝒁𝑪+𝑽𝑪𝟐𝟐+𝒉𝒇𝟏+𝒉𝒇𝟑Recallthecontinuityequation:𝒒𝟏=𝒒𝟐+𝒒𝟑(AssumingfluidflowsfromreservoirAtoreservoirBandC)Wehavethreeequationsandthreeunknowns.©2015YanweiWang11CasestudyProblem1.15(page80)Threereservoirsareconnectedbythreepipes.Forsimplicityweassumethatallpipeshavethesamediameterof0.305mandthesamefrictioncoefficientof0.02.Becauseofthelargelength-to-diameterratio,minorlossesarenegligible.Determinetheflowrateintooroutofeachreservoir.©2015YanweiWang12SystemsketchC©2015YanweiWang13AssumptionsCAssumingfluidflowsfromreservoirAintoreservoirsBandC©2015YanweiWang14Solutionsteps-1AssumingthatthefluidflowsoutofreservoirAandintoBandC,wehavefromtheeq.ofcontinuity:𝑄1=𝑄2+𝑄3(thecontinuityequation)Sinceallthreepipeshavethesamediameter,wemusthave𝑉1=𝑉2+𝑉3©2015YanweiWang15Solutionsteps-2©2015YanweiWang16Solutionsteps-3•Allreservoirsareopentoatmosphere,𝑝A=𝑝B=𝑝C•Thevelocitiesatallreservoirstations(A,B,&C)arenegligible𝑉A=𝑉B=𝑉C=0,because_____•TakethedatumplaneforelevationsthroughstationC,wehave𝑍C=0,𝑍𝐴=30.5m,𝑍𝐵=6.1m©2015YanweiWang17Solutionsteps-4Knownvariables:𝑓=0.02,𝑍𝐴=30.5m,𝑍𝐵=6.1m,𝐿1=304.8m,𝐿2=152.4m,𝐿3=121.9m,𝐷=0.305m,𝑔=9.80665ms2©2015YanweiWang18Solutionsteps-5𝑉1=𝑉2+𝑉3(Eq.ofcontinuity)Substitutingvaluesfortheknownvariablesleadsto(𝑉1,𝑉2,and𝑉3areinunitofm/s):39.9738𝑉12+19.9869𝑉22−239.282=039.9738𝑉12+15.9869𝑉32−299.103=0©2015YanweiWang19Solutionsteps-6Solvingthethreeequationsgives𝑉1=2.427m/s,𝑉2=0.433m/s,and𝑉3=1.994m/sThecorrespondingflowratesare𝑄1=14𝜋𝐷2𝑉1=0.177m3/s𝑄2=14𝜋𝐷2𝑉2=0.032m3/s𝑄3=14𝜋𝐷2𝑉3=0.145m3/s©2015YanweiWang20Let’stryadifferentassumption??CAssumingfluidflowsoutofreservoirsAandBintoreservoirC©2015YanweiWang21Solutionsteps–1’AssumingthatthefluidflowsoutofreservoirsAandBandintoC,wehavefromtheeq.ofcontinuity:𝑄1+𝑄2=𝑄3(thecontinuityequation)Sinceallthreepipeshavethesamediameter,wemusthave𝑉1+𝑉2=𝑉3©2015YanweiWang22Solutionsteps–2’©2015YanweiWang23Solutionsteps–3’•Allreservoirsareopentoatmosphere,𝑝A=𝑝B=𝑝C•Thevelocitiesatallreservoirstations(A,B,&C)arenegligiblebecauseofthelargediameterofthereservoirincomparisontothepipediameter,𝑉A=𝑉B=𝑉C=0•TakethedatumplaneforelevationsthroughstationC,wehave𝑍C=0,𝑍𝐴=30.5m,𝑍𝐵=6.1m©2015YanweiWang24Solutionsteps–4’Knownvariables:𝑓=0.02,𝑍𝐴=30.5m,𝑍𝐵=6.1m,𝐿1=304.8m,𝐿2=152.4m,𝐿3=121.9m,𝐷=0.305m,𝑔=9.80665ms2©2015YanweiWang25Solutionsteps–5’𝑉1+𝑉2=𝑉3(Eq.ofcontinuity)Substitutingvaluesfortheknownvariablesleadsto(𝑉1,𝑉2,and𝑉3areinunitofm/s):39.9738𝑉12−19.9869𝑉22−239.282=039.9738𝑉12+15.9869𝑉32−299.103=0©2015YanweiWang26Solutionsteps–6’Solvingthosethreeequationsdoesnotyetreal,p