Citation
The Stability of the Pendent Drop and the Enclosed Liquid Bridge

Material Information

Title:
The Stability of the Pendent Drop and the Enclosed Liquid Bridge
Creator:
Lin, Xin
Place of Publication:
[Gainesville, Fla.]
Florida
Publisher:
University of Florida
Publication Date:
Language:
english
Physical Description:
1 online resource (115 p.)

Thesis/Dissertation Information

Degree:
Doctorate ( Ph.D.)
Degree Grantor:
University of Florida
Degree Disciplines:
Chemical Engineering
Committee Chair:
NARAYANAN,RANGANATHA
Committee Co-Chair:
JOHNS,LEWIS E, JR
Committee Members:
CRISALLE,OSCAR DARDO
BALACHANDAR,SIVARAMAKRISHNAN
Graduation Date:
4/30/2016

Subjects

Subjects / Keywords:
Bond number ( jstor )
Critical points ( jstor )
Critical values ( jstor )
Curvature ( jstor )
Eigenfunctions ( jstor )
Eigenvalues ( jstor )
Interfacial tension ( jstor )
Liquid bridges ( jstor )
Liquids ( jstor )
Zero ( jstor )
Chemical Engineering -- Dissertations, Academic -- UF
bridge -- drop -- eigenshape -- rotation -- stability
Genre:
bibliography ( marcgt )
theses ( marcgt )
government publication (state, provincial, terriorial, dependent) ( marcgt )
born-digital ( sobekcm )
Electronic Thesis or Dissertation
Chemical Engineering thesis, Ph.D.

Notes

Abstract:
In this dissertation the stability of two equilibrium configurations are discussed i.e., the pendent drop and the liquid bridge in common terms but with uncommon ideas. The main results that characterize stability in the discussion are obtained from simple ideas in linear algebra. The liquid in the drop is heavier than the surrounding fluid. The only scaled groups that govern the stability are the scaled volume and the Bond number. It can be shown without computations that the volume of the drop has a maximum value beyond which it must break catastrophically and symmetrically. However this upper bound on volume is not the instability limit for the drop for all Bond numbers. There exists a critical Bond number, above which the drop breaks well before the upper bound on volume can be reached. This first critical Bond number can be obtained without a stability analysis and solely from knowing the base shape of the drop. Under the case of zero gravity, the scaled groups for bridge are scaled volumes and the ratio of bridge length to its radius. It can also be proved without computations that for a fixed length to radius, the volume of bridge has a minimum value below which it must break. Meanwhile the ratio of its length to radius exhibits a similar influence on the stability as the Bond number does in pendent drops. There exists a critical ratio, above which the bridge breaks before the lower bound on volume can be obtained. This critical ratio are also be obtained without a stability analysis. A comparison of bridges with drops shows that transverse curvature is of little significance in the latter's stability while it is all important in the former's stability. Several other characteristics of stability are obtained by introducing an auxiliary eigenvalue problem, This problem, termed the diffusion eigenvalue problem is shown to straddle the instability points and to place them in definite hierarchies. The diffusion eigenvalue problem also shows how different controls on the equilibrium configurations affect their relative stabilities. ( en )
General Note:
In the series University of Florida Digital Collections.
General Note:
Includes vita.
Bibliography:
Includes bibliographical references.
Source of Description:
Description based on online resource; title from PDF title page.
Source of Description:
This bibliographic record is available under the Creative Commons CC0 public domain dedication. The University of Florida Libraries, as creator of this bibliographic record, has waived all rights to it worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law.
Thesis:
Thesis (Ph.D.)--University of Florida, 2016.
Local:
Adviser: NARAYANAN,RANGANATHA.
Local:
Co-adviser: JOHNS,LEWIS E, JR.
Electronic Access:
RESTRICTED TO UF STUDENTS, STAFF, FACULTY, AND ON-CAMPUS USE UNTIL 2017-05-31
Statement of Responsibility:
by Xin Lin.

Record Information

Source Institution:
UFRGP
Rights Management:
Copyright Xin Lin. Permission granted to the University of Florida to digitize, archive and distribute this item for non-profit research and educational purposes. Any reuse of this item in excess of fair use or other copyright exemptions requires permission of the copyright holder.
Embargo Date:
5/31/2017
Classification:
LD1780 2016 ( lcc )

Downloads

This item has the following downloads:


Full Text

PAGE 1

THESTABILITYOFTHEPENDENTDROPANDTHEENCLOSEDLIQUID BRIDGE By XINLIN ADISSERTATIONPRESENTEDTOTHEGRADUATESCHOOL OFTHEUNIVERSITYOFFLORIDAINPARTIALFULFILLMENT OFTHEREQUIREMENTSFORTHEDEGREEOF DOCTOROFPHILOSOPHY UNIVERSITYOFFLORIDA 2016

PAGE 2

2016XinLin 2

PAGE 3

Toallchemicalengineers 3

PAGE 4

ACKNOWLEDGMENTS TheauthorissincerelythankfultoRangaNarayananandLewisE.Johns Jr.,professorsintheChemicalEngineeringDepartmentofUniversityofFlorida, forhelpfuladvicethroughoutthevariousstagesofthisworkandtodoctoral candidatesXiaoWang,JasonPicardoandKevinWardforvaluablefeedbackon revisionsofthisthesis.MeanwhilesupportfromNSFPIRE0968313isgratefully acknowledged. 4

PAGE 5

TABLEOFCONTENTS page ACKNOWLEDGEMENTS.............................4 LISTOFTABLES..................................7 LISTOFFIGURES.................................8 ABSTRACT.....................................10 CHAPTER 1INTRODUCTION...............................12 1.1Background...............................12 1.2DescriptionofContents........................13 2LITERATUREREVIEW...........................14 3MODELOFTHEPENDENTDROPINACONSTANTVOLUMEEXPERIMENT..................................21 3.1ArbitraryCross-section,FirstExperiment..............22 3.2TheCaseofthePlaneInterface,FirstExperiment.........26 3.3TheCaseofthe1-DCross-section,FirstExperiment.......28 3.4TheCaseoftheCircularCross-section,FirstExperiment.....38 4MODELOFTHEPENDENTDROPINACONSTANTPRESSURE EXPERIMENT................................47 4.1ArbitraryCross-section,SecondExperiment.............48 4.2TheCaseofthePlaneInterface,1-DCross-section,SecondExperiment.................................51 4.3TheCaseoftheCurvedInterface,1-DCross-section,Second Experiment...............................53 5MODELOFTHECLOSEDLIQUIDBRIDGEANDROTATINGDROP..57 5.1TheCaseoftheNon-rotatingClosedLiquidBridge.........61 5.1.1TheBaseCase,aVerticalLiquidCylinder..........61 5.1.2Decreasing V from V 0 atConstantL < ...........63 5.2TheCaseoftheClosedLiquidBridgeSpinningatConstantAngularVelocity..............................67 5.3TheCaseofthePendentDropSpinningatConstantAngularVelocity...................................75 6CONCLUSIONSANDFUTUREWORK..................84 6.1Conclusions...............................84 5

PAGE 6

6.2FutureWork:ModelofaRotatingPendentDropSpinningatConstantAngularVelocity.........................87 APPENDIX ATHEPATCHEDMETHOD..........................88 A.1PatchMethodforTheFirstExperimentwith1-DCross-section..89 A.2PatchMethodforTheFirstExperimentwithCircularCross-section91 A.3PatchMethodforTheSecondExperimentwith1-DCross-section93 A.4PatchMethodforTheSecondExperimentwithCircularCrosssection..................................94 BTHEDIFFUSIONEIGENVALUEPROBLEM...............97 B.1TheDiffusionEigenvalueProblemforaPendentDrop.......97 B.2TheDiffusionEigenvalueProblemforaClosedLiquidBridge...103 CRESULTSFOR B =0 ............................106 DDOESTHEVELOCITYPERTURBATIONVANISHATCRITICAL?...108 ETHEMETHODOFFROBENIUSTODERIVEESTIMATESOFTHE FIRSTROOTOFJ 0 r ..........................112 REFERENCES...................................113 BIOGRAPHICALSKETCH............................115 6

PAGE 7

LISTOFTABLES Table page 3-1Criticalvaluesof V forgiven B for1-Dcross-section............29 3-2Criticalvaluesof V forgiven B inthecaseofcircularcross-section...41 4-1CriticalvaluesofBondnumbersandappliedpressure P for1-Dcrosssectioninthesecondexperiment......................56 7

PAGE 8

LISTOFFIGURES Figure page 1-1Imageofapendentdropobtainedbycamera[1].............12 2-1ExperimentoftheRayleigh-Taylorinstability................16 2-2Hydrodynamicsimulationofasingle"nger"oftheRayleigh-TaylorinstabilitybyLiandLi[2]............................17 2-3Possibleshapesofdropwhichmayoccurwhenitscross-sectionbecomesnarrow.................................19 3-1Therstexperiment:constantvolumeexperiment.............21 3-2Relationshipbetween P and V foraxedBondnumber B =9,5,3 ...31 3-3Atsmallvaluesof B ,the V vs P curvesas B 0 .............32 3-4Theproleofdropbaseshapewhichcannotbeexpressedbyonly oneindependentvariable..........................33 3-5 P vs V curvesforlargeandsmall B withthestraightline PA + BV =0 whichisthelocusofbifurcationpoints...................35 3-6Dropshapeatthebifurcationpoint.....................36 3-7Dropshapeaswemovecarefullybeyondthebifurcationpoint......37 3-8Relationshipbetween P and V foraxedBondnumber B =14,10,6 .42 3-9 V vs P curveandthelinedenotedbyEq.3.............44 4-1Thesecondexperiment:constantpressureexperiment..........47 4-2 V vs P curvesfortwomodelswithsame B =7 and B =10.8566 ....53 4-3Relationshipbetween P and I = )]TJ/F45 11.9552 Tf 11.291 9.63 Td [( 1 )]TJ/F25 7.9701 Tf 6.587 0 Td [(1 Z d x foraxedBondnumber B =9,3,2 ...................................55 5-1Atypicalshapeofanenclosedliquidbridge................57 5-2 V = V 0 vs P curveswithdifferentvaluesofL.................64 5-3Themethodofpredictingthebifurcationpointwithoutstabilitycalculation.......................................66 5-4Shapesofclosedliquidbridgewithandwithoutspinningforgiven V andL......................................71 5-5Theeffectof 2 ontheliquidbridgebreakupbycomparisonto 2 =0 ..73 8

PAGE 9

5-6Straightlineconstructionforlocatingthebifurcationpointfromthebase calculationinthecaseofspinningliquidbridge...............75 5-7Stableshapeofthespinningdroprotatingatsmall 2 withzerovolumeand B < B crit 2 =0 ..........................80 5-8Unstableshapeofthespinningdroprotatingatlarge 2 withzerovolumeand B < B crit 2 =0 ..........................81 5-9 V vs P curvefor 2 =0 and 2 > 0 ....................82 A-1Theproleofdropbaseshapewhichcannotbeexpressedbyonly oneindependentvariable..........................88 B-1Theproleoftheleft-hand-sideofEq.B,asafunctionof 2 .....99 B-2Theproleoftheleft-hand-sideofEq.B,asafunctionof 2 .....100 B-3Valueof P i =1,3,... I 2 i 2 )]TJ/F28 7.9701 Tf 6.587 0 Td [( 2 i asafunctionof 2 ...................102 B-4Theproleoftheleft-hand-sideofEq.B,asafunctionof 2 ....105 C-1 V vs P curvefor B =0 ............................107 D-1TheFirstExperiment:volumecontrolexperiment.............108 9

PAGE 10

AbstractofaDissertationPresentedtotheGraduateSchool oftheUniversityofFloridainPartialFulllmentofthe RequirementsfortheDegreeofDoctorofPhilosophy THESTABILITYOFTHEPENDENTDROPANDTHEENCLOSEDLIQUID BRIDGE By XinLin May2016 Chair:RangaNarayanan Co-Chair:LewisE.JohnsJr. Major:ChemicalEngineering Inthisdissertationthestabilityoftwoequilibriumcongurationsarediscussedi.e.,thependentdropandtheliquidbridgeincommontermsbutwith uncommonideas.Themainresultsthatcharacterizestabilityinthediscussion areobtainedfromsimpleideasinlinearalgebra. Theliquidinthedropisheavierthanthesurroundinguid.Theonlyscaled groupsthatgovernthestabilityarethescaledvolumeandtheBondnumber.It canbeshownwithoutcomputationsthatthevolumeofthedrophasamaximum valuebeyondwhichitmustbreakcatastrophicallyandsymmetrically.However thisupperboundonvolumeisnottheinstabilitylimitforthedropforallBond numbers.ThereexistsacriticalBondnumber,abovewhichthedropbreaks wellbeforetheupperboundonvolumecanbereached.ThisrstcriticalBond numbercanbeobtainedwithoutastabilityanalysisandsolelyfromknowingthe baseshapeofthedrop. Underthecaseofzerogravity,thescaledgroupsforbridgearescaled volumesandtheratioofbridgelengthtoitsradius.Itcanalsobeproved withoutcomputationsthatforaxedlengthtoradius,thevolumeofbridgehas aminimumvaluebelowwhichitmustbreak.Meanwhiletheratioofitslength 10

PAGE 11

toradiusexhibitsasimilarinuenceonthestabilityastheBondnumberdoes inpendentdrops.Thereexistsacriticalratio,abovewhichthebridgebreaks beforethelowerboundonvolumecanbeobtained.Thiscriticalratioarealsobe obtainedwithoutastabilityanalysis.Acomparisonofbridgeswithdropsshows thattransversecurvatureisoflittlesignicanceinthelatter'sstabilitywhileitisall importantintheformer'sstability. Severalothercharacteristicsofstabilityareobtainedbyintroducingan auxiliaryeigenvalueproblem,Thisproblem,termedthediffusioneigenvalue problemisshowntostraddletheinstabilitypointsandtoplacethemindenite hierarchies.Thediffusioneigenvalueproblemalsoshowshowdifferentcontrols ontheequilibriumcongurationsaffecttheirrelativestabilities. 11

PAGE 12

CHAPTER1 INTRODUCTION ThisChapterisaguidetothetopicscoveredbythisdissertation.The objectivesofthisworkincludeunderstandingtheinstabilityofpendentdrops andclosedliquidbridges.Comparisonofthedifferenttypesofinstabilitiesand patternsofbreakuparepresentedandcompared. Somebasicphysicsassociatedwithapendentdropandaclosedliquid bridgeareintroducedinSection1.1.Theorganizationofthisdissertationis explainedinSection1.2. 1.1Background Astaticpendentdropmaybeformedattheedgeofatubewhenliquidhangsat thelowerendofit,thegravitationalpotentialenergybeinginequilibriumwiththe surfacepotentialenergy.AtypicalpendentdrophastheshapeshowninFigure 1-1.Inthegure,AandBaretwodifferenttypesofimmiscibleuids,and.gis thegravitationalacceleration.Thedropmayhavevariousshapesdependingon itsvolume,thedensitydifferencebetweenuidAandB,theinterfacialtension betweenA,Bandthediameteroftheupperhole. Figure1-1.Imageofapendentdropobtainedbycamera[1]. Therearetypicallytwowaystoformapendentdrop.Onewayistosetthe volumeofthedrop;heretheheightoftheuidinthetubeabovethedropis determinedalongwiththeshapeofthedropitself.Theotherwayistouseset 12

PAGE 13

thepressureatthetopofuidabovethedropandinthiscasetheheightofuid mustbespecied.Inthesecondcasethevolumeofthedropisdetermined alongwiththedrop'sshape.Dropsgeneratedbythesetwomethodsmayormay nothavethesamehydrostaticstabilitycharacteristics.Itisouraimtounderstand thedifferingcharacteristicsandmakesomepredictions. 1.2DescriptionofContents Thedissertationisdividedintosixchaptersthataddresstheinstability analysisofapendentdropandaclosedliquidbridge. Chapter1providesanintroductiontothisthesis,includingthebriefdescriptionaboutelementaryphysicsaboutapendentdropandaclosedliquid bridge. Chapter2isconcernedwithaliteraturereview.Theformationofapendent dropandaclosedliquidbridgeisdiscussed.Severalapplicationsofpendent dropsarepresentedanddiscussed.Theinstabilitymechanismthatcausesa pendentdropandaclosedliquidbridgetobreakisproposed. Chapter3through5addressestheinstabilityanalysisofapendentdrop formedbytwodifferentwaysandaclosedliquidbridge.Analyticalderivations andnumericalcalculationsareusedtoverifythetheorydiscussedinthispart. Chapter6presentsconclusionsandrecommendationsforfuturework. Finallyveappendicesassociatedwiththisdissertationarepresented.They arethesupplementarymaterialstotheworkshowninthisdissertation. 13

PAGE 14

CHAPTER2 LITERATUREREVIEW Scienticstudiesontheformationofthependentdrophavebeendoneover severaldecades.Buttheproblemisstillofinterestbecauseofpromising applications.ForexampleLeandHue[3]provideabriefreviewofthe developmentofink-jetprintingtechnologyanddescribethelatestadvanced ink-jetproductswhoadoptthepotentialink-jettechnologycalledthermaland piezoelectricdrop-on-demandink-jetmethod.HeidegerandWright[4]present theexperimentalresultoftheeffectofformationtimeonliquidextractionduring dropformation.Theirworkhelpsusunderstandthetransferofmassbetween liquiddropsandasurroundingimmiscibleliquidphase,whichprovidesthebasis forindustrialextractionoperations. Oneofthemostobviousapplicationsofpendent-dropstudiesismeasurementof asurfacetension-tensiometerforliquid-uidinterfaces.Inthetypical tensiometer,thesurfacecurvatureismeasuredfromphotographsofthesurface shape[5];thesurfacetensionisthendeterminedfromahydrostaticforce balance.ThistechniqueisbasedontheworkofRotenbergetal.[6]buthasthe disadvantageoftheneedtousethebesttbetweenthenumericalsolutionsof theYoung-Laplaceequationandtheexperimentalprole[7].Inothermeansof measuringsurfacetension,thesurfacetensionisdeterminedbynotingthedrop shapejustasitpinchesoff.Hereneitherthedropvolumenorthepressure requiredtopinchoffisofconsideration.Ofcoursesuchatechniqueis hydro-dynamic.Yet,surfacetensioninitselfisanequilibriumpropertyandthus oughttobemeasuredfromhydrostaticsorfromahydrostaticinstability. Dropbreakupdynamicshavebeeninvestigatedbecauseoftherichnessofthe underlyingphysicsasdescribedbyChenetal.[8]andSchulkes[9].Theworkby Schulkesshowstheinstabilitysignicantlydependsonthedropgrowthrateand 14

PAGE 15

diameterofthetube.WorkonstaticdropsstabilitybyPitts[10,11]andPadday [12]usecalculusofvariationstoconcludethatthedropmusthaveamaximum volumeandbreaksymmetrically.MajumdarandMichael[13]comparedtwo differentwaystoformandmaintainastaticplanependentdrop:constantinput volumeandconstantinputpressure.Theconstantinputvolumeconditionhas beenstudiedbyPitts[10]andprovedtohavemaximumvolumeinstabilitiesto two-dimensionalweakdisturbancesymmetricaboutthecenterplanetowhich thedroppinchesoffatthepointofmaximumvolumepoint.Thestudyof MajumdarandMichael[13]consideredtheconditionthatthedropiscontrolled byconstantpressure,whichcanbeviewedasanextensiontotheworkofPitts [10].Theyshoweddropscontrolledbyconstantpressureareunstableto two-dimensionalweakdisturbancesymmetricaboutthecenterplaneat maximuminternalpressure.Theirstudyalsoshowsthatatwo-dimensionaldrop formedwithprescribedcontactangleisunstabletotwo-dimensional disturbanceswhetheritisvolume-controlledorpressure-controlled.Theybelieve thedropcanonlybreakupatitsmaximumvolumeifitisvolume-controlledand theydidnotstudythepatternofdropbreakupatcriticalvolumeorpressure. MichaelandWilliams[14]alsostudiedthestabilityofaxisymmetricpendent dropstoasymmetricdisturbanceswithazimuthwavenumber m =1 mathematically.Theyfoundthedropisneutrallystabletoasymmetric perturbationswhenitsprolebecomeshorizontalatthepointofsupportandthis asymmetricperturbationalwayssetsinattheplaneinterfacewhentheorice radius a reachesthevalue3.812...inunitsofcapillarylength.Theyalsoshowed when a isgreaterthan3.219,thisinstabilitywillprecedetheaxisymmetric instabilityandotherhigherorderinstabilities m =2,3,4 ....Theonsetofthis m =1 instabilityheldatconstantvolumealwaysoccursatthesamepointasfor thoseheldatconstantpressure.InanotherpaperofMichaelandWilliams[15] 15

PAGE 16

thesameconditionforthe m =1 instabilitywasappliedtoconsideritseffectson sessiledropswithaxisymmetricmeniscussurface.Againitisfoundthatthis instabilityoccurswhenthemeniscusbecomestangentialtoahorizontalplaneof supportatitsedge.Ourproposedworkaimstoconrmtheseresultsandshow manymoreresultsandfeatures.Arecentreviewoncapillaryinstabilityofstatic surfacesbyBostickandSteen[16]isofgeneralthoughnotspecicinterestto ourwork. Whenthetopofdropiswide,thedropbreaksduetotheRayleigh-Taylor instability,whichcanbeillustratedbyexperimentaswellasnumericalsimulation givenbyFigure2-1and2-2.TheRayleigh-Taylorproblemisaninstabilityofan interfacebetweentwouidsofdifferentdensities.Itoccurswhentheheavier uidpushesthelighteruidfromaboveandwhenthebalancebetween gravitationalandsurfacepotentialenergyislost. InFigure2-1,theexperimentbeginswithaatinterfaceshownina.Thetube diameterincreasesfromlefttoright.Inbthelefthalfofupperuidmoves downwardswhiletherighthalfmovesupwards.Finallyitbecomesunstableinc andthenswitchesitspositionwiththeuid,exchangingpositionswhenthe interfacebreakswiththepatternofanantisymmetricfunctionpresentedind.It isshownintheexperimentthatthecongurationisunstablewhentheheavyuid sitsontop.Thegrowthofasmallperturbationattheinterfaceisdrivenbythe mechanismofgravitationalpotentialminimization. Figure2-1.ExperimentoftheRayleigh-Taylorinstability. 16

PAGE 17

Figure2-2.Hydrodynamicsimulationofasingle"nger"oftheRayleigh-Taylor instabilitybyLiandLi[2]. Anantisymmetricbreakupisnottheonlywayinwhichadropcanlosestability. Figure2-2showsthis.InFigure2-2thesimulationtimefromthelefttorightis 0.5,1.0,1.5,and1.9,inscaledunitsrespectively.Notetheformationof Kelvin-Helmholtzinstabilities,inthesecondandlatersnapshotsshownstarting initiallyaroundthelevel y =0 ,aswellastheformationofa"mushroomcap"at alaterstageinthethirdandfourthframeinthesequence.Mostofall,noticethat thedroppinchesoffsymmetrically.Thebreakupinthispictureresemblesajet 17

PAGE 18

wherethemechanismofinstabilityisduetoatrade-offbetweentransverseand longitudinalcurvatures. Thestabilityofastaticliquidbridgeisanothersubjectofinterestand hasbeenstudiedbymanyscholars.Thestabilityandfreeenergyanalysisof Martinez[17]showsthebridgepinchesoffaxisymmetricallyatitsminimum volumewhenitsslenderness L = R 0 < 2.12 andasymmetricallyfor L = R 0 > 2.12 beforeminimumvolumeisobtained.Thisworkisbasedonthebifurcationof staticcongurationofaxisymmetricliquidbridges.Espinoetal.[18]didan experimenttostudythebreakingprocessofaxisymmetricliquidbridges.They analyzedhydrodynamicallylargenumberofliquidbridgecongurationsat minimumvolumeinstabilityandshowedatbreakagetheneckradiusvaries as t 1 = 3 where t isthetimetobreakage.Thoughthisisnotthespecicinterest ofourworkbecauseweconcentratemoreonthehydrostaticstudiesofliquid bridgeswithdifferentpatternsofbreakup. Thedifferenceinthenatureofthebreakupisofinteresttousbecauseitmight telluswhattheoperatingmechanismsforinstabilityare.Infact,whenthetopof dropbecomesnarrowmorecomplicatedshapesofthedropmayshowup, showninFigure2-3.NotethedrophasaneckpartinFigure2-3b.Thiswillbe showntomakethestabilityanalysismorecomplicated.Thisneckisduetothe factthatsurfacetensionplaysasignicantroleindeterminingthecapillary shapeoftheinterfacecomparedtothebodyforceonscaleslessthanthe capillarylength l c = g 1 = 2 ,where g isthebuoyancypervolume accordingtothegravitationalacceleration g anddensitydifference ofthe uidsacrosstheinterface.Wecareaboutthedropinstabilityandhowitbreaks inthisgeometrysinceitisunclearwhetherornottheRayleigh-Taylorinstability isstilltheprimaryfactorcontributingtothebreakupofthedrop. 18

PAGE 19

Figure2-3.Possibleshapesofdropwhichmayoccurwhenitscross-section becomesnarrow. Ourmaingoalisthereforetodeterminethecriticalconditionsandpatternsfor thebreakupofthependentdropindifferentgeometries.Consequenttothiswe aimtoexplainthereasonsforthedifferentmodesofbreakup. ThecriticalconditionsdependonthevalueofBondnumber B g D 2 ,where D isthecharacteristiclength.TheBondnumber,whichgetsitsnamefromthe EnglishphysicistWilfridNoelBond[19],isgenerallyknowninuidmechanicsto describethebalancebetweensurfaceandgravitationalforcesandinfactarises fromscalingtheequationfortheforcebalanceattheinterface.Notethatitmight alsobeviewedasagroupshowingthebalancebetweensurfaceand gravitationalenergy.thesurfacepotentialenergyisthesurfacecurvature multipliedbythesurfacetension.Allgeometriesofahangingdrop,savethoseof one-dimensionalcross-sectionhavetwoeffectsofsurfacecurvaturethatoppose oneanother.Oneoftheseisduetotransversecurvaturewhiletheotherisdue tolongitudinalcurvature.Thusiftheshapeoftwo-dimensionaldropissuchthat itresemblestherightgurewithanarrowneck,thentransversecurvaturemay playadestabilizingroleinadditiontothegravityandthedropcanpinchoffatthe neck.Wewillperformcalculationsfortheunphysicalone-dimensionaldropas wellasthepracticalcaseofatwo-dimensionaldroptodiscernthebehaviorwith transversecurvature.Asurprisewillbeinstore. Theoverallprocedureisasfollows.Werstwriteamodelforthestaticdropand thenobtainthebasesolutionandinvestigateitsstabilitytosmalldisturbances 19

PAGE 20

[20].Wedealwiththenumericalstabilitycomputationsthatincludethe conversionoftheperturbedproblemtoaneigenvalueproblemaswellasthe calculationofthedominanteigenmode,usingChebyshevSpectralmethods[21] inboththe1-Dand2-Dcases,respectively.Therearetwophysicalset-upsthat aregenerictotheformationofpendentdrops.Weturntotheanalysisofeach. 20

PAGE 21

CHAPTER3 MODELOFTHEPENDENTDROPINACONSTANTVOLUMEEXPERIMENT Intherstphysicalset-upofapossibleexperimenttocreateapendentdrop,the uidsliebetweeninnitelylongplanewalls-Dnormaltotheplaneofthe gure.Heavyuidisaddedtothetop,lightuidremovedfromthebottom.Ifthe surfacestartsoutplanarandisstable,uponaddingupperuidandremoving bottomuid,atsomepointthesurfaceoughttobreak.Figure3-1presentsa sketchofthisset-up. Figure3-1.Therstexperiment:constantvolumeexperiment. Inthesimplecasewhereourmodelis1-D,wedenotethesurfaceby z = Z x )]TJ/F55 11.9552 Tf 9.298 0 Td [(L < x < L or x = X z z cl < z < 0 21

PAGE 22

iftheshapeofthecross-sectioniscircular,wethendenotethesurfaceby z = Z r 0 < r < R or r = R z z cl < z < 0 Intheexperimentthesurfaceispinnedatasharpedge.Inordertodenethe shapeofthedropwecanchooseeither x r or z asourindependentvariables. 3.1ArbitraryCross-section,FirstExperiment OurmodelistheYoung-Laplaceequation.Sincethebaseshapeofthedropis symmetricto x =0, thehorizontalforcebalanceisautomaticallysatisedand thusthemodelonlyconsiderstheforcebalancealongtheverticaldirection, wherethepressuredifferenceacrossthesurfaceisbalancedbysurfacetension. Hencewehave P + 2 H = P wherethenormalpointsfromtoand P = )]TJ/F27 11.9552 Tf 9.298 0 Td [( g z + C P = )]TJ/F27 11.9552 Tf 9.298 0 Td [( g z + C where C = P Bot + g H C = P Top + g H and P Bot P Top arethebottomandtoppressure,respectively.Thusfor experimentI,where z = Z x y or z = Z r holds,wehave P )]TJ/F24 11.9552 Tf 11.955 0 Td [( g Z = r r Z + r Z r Z 1 = 2 22

PAGE 23

with Z =0 alongtheperimeterofthecross-sectionand )]TJ/F45 11.9552 Tf 11.955 16.272 Td [( A Z d A = V where V denotesthevolumeofthedropbeinganinputvariableandwhere P denotesthepressuredifferencebetweentopandbottombeinganoutput variable.Fortheplanarcasethisequationhasthesolution Z =0, P =0 at V =0. Andifthesolutionis Z C at V > 0 ,itis )]TJ/F48 11.9552 Tf 9.298 0 Td [(Z )]TJ/F48 11.9552 Tf 9.299 0 Td [(C for )]TJ/F48 11.9552 Tf 9.299 0 Td [(V Wecanscaleourmodelintermsofacharacteristiclength,denoted D characteristicofthecross-sectionandinscaledvariableswehave P )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ = r r Z + r Z r Z 1 = 2 where Z =0 attheedge and )]TJ/F45 11.9552 Tf 11.956 16.273 Td [( A Z d x = V Thedimensionlessgroup B = gD 2 where D ishalfthediameter.Thus V and B areinputsand P and Z areoutputs. Assumingwecanndtheshapeoftheinterface,wewanttoknowifthedropis stabletosmallperturbations.Tondoutweask:istheshapeneutrallystable? Nowastabilitycalculationwouldordinarilydemandthatthevelocity perturbationsalsobecomputed.However,itcanbeshowninastraight-forward mannerusingclassicalenergymethods,thatatneutralconditions,allvelocity perturbationsmustvanishandthatthepressureperturbationsmust,atmost,be independentofspatialdirection.ThisisgiveninAppendixD-1. Henceholdingtheinputsxed,weintroduceasmallperturbationvia Z = Z 0 + Z 1 23

PAGE 24

where Z 0 denotesthebaseshapewhosestabilityisinquestionand isasmall amplitudeoftheperturbationandwehavethelinearperturbedequationfor Z 1 P 1 )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ 1 = L Z 1 where Z 1 =0 attheedge A Z 1 d A =0 and L = rf I + r Z 0 r Z 0 1 = 2 )]TJ 35.276 8.088 Td [(r Z 0 r Z 0 + r Z 0 r Z 0 3 = 2 gr isaself-adjointlinearoperator. Ifwehaveasolutionotherthan Z 1 =0 tothesehomogeneousproblems,the shapeofthesurfaceisatthecriticalpoint.Iftheonlysolutionis Z 1 =0 ,the shapeisnotatthecriticalpoint.Ourplanistoadvanceacontrolorinput variable,obtainafamilyofshapesanddecidewhichmemberofthefamilyis critical. Ineachcaseaneigenvalueproblemhelpsusanswerourquestion.Itis C )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 = L where =0 attheedge and A d A =0 Multiplyingby andintegratingoverthecross-sectionweseethatall 2 s are positivebecause r r r Z 0 r Z 0 > r r Z 0 r r Z 0 24

PAGE 25

Thismeansthatdropswithnegative B cannotbelinearlyunstable. Nowtheeigenvaluesdependonthebaseshapeandasweadvanceourcontrol variable,saythevolumeinexperimentI,wecalculatethebaseshapesandsolve theeigenvalueproblemateachbaseshape.Thenourperturbationproblemhas onlythesolution Z 1 =0 solongas B remainslessthanthelowesteigenvalue. Themodelisstatic,hencewearelookingfornonzerono-growthsolutionsviz., Z 1 6 =0 .Wehaveasolution Z 1 6 =0 if B = 2 1 B V andwehavearrivedatthe criticalshape. Havingarrivedataninterface z = Z r B V whereBisatitscriticalvalueatthe presentvalueof V ,weaskifwecanadvance V andcontinueourcalculationof Z .Tosee,wedifferentiatethebaseproblemwithrespectto V holding B xedat 2 1 anddenotethederivativesby .Inthiscaseweget P )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 1 Z = L Z where Z =0 attheedgeofthecross-section and A Z d A = )]TJ/F24 11.9552 Tf 9.298 0 Td [(1 Thenthequestionis:canwesolvethisfor Z and P ?Thecorresponding homogeneousproblemistheeigenvalueproblemandithasanon-zerosolution 1 ,hencemultiplyingthe equationby 1 ,thehomogeneousproblemby Z subtractingandintegratingover A ,wehave P A 1 d A )]TJ/F48 11.9552 Tf 11.955 0 Td [(C 1 A Z d A =0 i.e., C 1 A Z d A =0 andthismustbesatisedifwearetond Z P .Itissatisedifandonlyif C 1 =0 butordinarily C 1 isnotzeroandthecriticalpointisnotabifurcationpointbut 25

PAGE 26

ratheranose.Butifthecross-sectionhasenoughsymmetryitmaybethat C 1 =0 ,solvabilityissatisedandwecancalculatebaseshapes,unstable thoughtheymaybe,beyondcritical.Togetstartedandproceedmethodicallywe turntoaplanarcross-sectioninonedimensionandrecoverawell-knownresult. Ourmodelhasaplanesolutionviz., Z =0, whichcorrespondstothecontrol variable V =0 nomatterthevalueof B TondoutifwecanobservethissolutioninanexperimentweincreasetheBond number,lookingforthevalue B where B = 2 1 B V =0 Atthisvalueof B theperturbationproblemhasasolution Z 1 6 =0 andwehave arrivedatthecriticalpointforaplanesurface. 3.2TheCaseofthePlaneInterface,FirstExperiment Tondouttheneutralstabilitypointina1-Dcross-sectioncase,westudythe correspondingeigenvalueproblemforaplanesurfacewhichis C )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 = d 2 d x 2 where =0 at x = 1 and 1 )]TJ/F25 7.9701 Tf 6.587 0 Td [(1 d x =0 anditssolutionis = sin x sin =0, C =0 = cos x )]TJ/F55 11.9552 Tf 11.955 0 Td [(cos cos )]TJ/F55 11.9552 Tf 13.151 8.088 Td [(sin =0, C 2 = )]TJ/F55 11.9552 Tf 9.299 0 Td [(cos 26

PAGE 27

whereuponwehaveoddandevensolutionsandthecriticalvalueoftheBond numberis 2 correspondingtoanoddorantisymmetricperturbation.Oncewe takethecaseofabentinterfaceinthebasestate,wemightexpecttheinstability tobeintheformofanantisymmetricperturbation,butwecannotguaranteethis andacalculationwillbeneeded. Inthecaseofcircularcross-sectionwehavethesolution Z 0 =0 at V =0 forall valuesof B astheplanesurfacesolution.Theeigenvalueproblemisthen C )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 = 1 r @ @ r r @ @ r + 1 r 2 @ 2 @ 2 =0 at r =1, 2 0 1 0 r d r d =0 WehavesolutionJ m r cos m J m =0, C =0 for m =1,2, correspondingtotheoddsolutiontothe1-Dproblemwherebecause 2 0 1 0 r d r d iszero. At m =0 wehave = J 0 r )]TJ/F24 11.9552 Tf 13.426 8.088 Td [(2 J 1 J 0 r =2 J 1 C = )]TJ/F24 11.9552 Tf 9.298 0 Td [(2 J 1 andwendthatthesolutionsto J 0 =2 J 1 lietotherightoftheleast positiverootofJ 1 =0 .Hencethecriticalvalueof B is 2 1 J 1 1 =0 ,andthe dropiscriticaltoa m =1 perturbation. Whatweknowthusfaristhatneitherthe1-Dplanarcross-sectionnorthe circularplanarcross-sectionbreakduetoaperturbationsymmetricaroundthe origin.Howeverwearenotsurewhetherthisistruefordropswithnon-zero volume.Acalculationforthedropshapewithcurvedinterfaceiscarriedoutin section3.3and3.4. 27

PAGE 28

3.3TheCaseofthe1-DCross-section,FirstExperiment Inthecaseofa1-Dcross-section,whentheinputBondnumberislessthan 2 theplanesurfaceisstable.Todeterminethecritical V 6 =0, weneedtokeep addingheavyuidfromtopandremovinglightuidfromthebottom,whichleads toacurvedinterface.Inthissituationananalyticalrepresentationeigenfunction isnoteasilyavailableandwehavetocalculateitnumerically. Ourviewoftheproblemthenisthis.Setavalueof B < 2 andincrease V causingthesurfacetopassthroughaseriesofshapes z = Z x B V untilwe reachavalueof V where B = 2 1 B V whereupontheperturbationproblemhasasolution Z 1 6 =0 andwehavearrived attheneutralstabilityshapewheretheinput B isnowcriticalatthe corresponding V TheeigenvalueproblemsforthecurvedinterfacelookmuchasEq3but d 2 d x 2 ontheright-hand-sideisreplacedby d d x 1 + Z 2 x 1 = 2 d d x where Z x B V denotes theinterfaceshapeinquestion.Because 0 < 1 + Z 2 x 1 = 2 < 1 andthegreater magnitudeofthederivative Z x thesmallerthevalueof 1 + Z 2 x 1 = 2 weanticipatethe smallerthevalueof B thelargerthevalueofinput V neededtomake B acritical value.Onholding V xed,thelargerthe V thesmaller B crit andourinterface wouldbreakatasmallerBondnumber.Againtheeigenfunctionswillbeeither oddorevenfunctionsof x Thecomputationalresultsfordifferent B 'satthe neutralstabilitystateareshowninTable3-1.Theseresultsareobtainedbyxing Bondnumberandincreasing V untiltheneutralstabilitypointisreached.For Bondnumberslowerthan1itisnotpossibletondthecritical V using z or x as theonlyoneindependentvariable.Inthissituationwehavetorewritethe stabilityproblemusingapatchedmethodtoobtainthecriticaldata.Thisis describedintheAppendixA.For 5 < B < 2 ,theeigenfunctionatcriticalisodd 28

PAGE 29

whilefor 0 < B < 5 itiseven.Itisfoundthatfor B =5 thelowesteigenvalue correspondstotwodifferenteigenfunctions.Oneisoddandtheotheriseven. Table3-1.Criticalvaluesof V forgiven B for1-Dcross-section. B Critical VB Critical V 2 022.12 9.50.204313.1883 90.32440.84.7147 8.50.41880.65.9081 80.50490.56.8268 70.670.48.166 60.85040.310.3205 51.03850.214.4647 41.290.126.2714 31.610.0548.8 Notingthat V isdenedbytheintegralasmentionedabove,wecannotinputit intothenonlinearproblemexplicitly.Thealternativewayofdoingthisistouse P asanmodelinputandthencalculate V fromtheobtainedbaseshape.The relationshipbetween P and V isgiveninFigure3-2. Figure3-2illustratestheevolutionof P asafunctionoftheinputvariable V Noticethat V hasanupperboundthatgetslargeras B decreases.Notefrom ourearlierderivationsthattheoccurrenceofamaximum V canbeprovedby imposingasolvabilityconditiononthederivativeofthebaseequationwith respectto V .Thismeanswecannotndthebasecurvewiththeinput V greaterthanitsmaximum. 29

PAGE 30

Aswemarchupthecurvebyincreasing V wendthattheeigenfunctionfor 2 o = B isoddi.e.,antisymmetricabout x =0 with P 1 =0 whereasthe eigenfunctionfor 2 e = B iseveni.e.,symmetricabout x =0 forall 5 < B < 2 Infact, 2 e = B occursatthemaximumvalueof V .Thereisavalueof B where 2 o = 2 e = B as B decreasesfrom 2 .Thispointoccursaround B =5 .Forlower B ,theeigenvaluecorrespondingtothesymmetrici.e.,eveneigenfunctionisthe dominanteigenvalue. AllofthisindicatesthatforlargeBondnumbersthedropbreaksinamanner similarastheplanarcross-sectionunderRayleigh-Taylorinstability.For 0 < B < 5 theeigenvalue 2 e = B dominatesanditseigenfunctionissymmetric. Thismeansifthedropbreaksatitsmaximumvolume,thepatternforitsbreakup willlookmuchlikethecaseoftheclassicalliquidbridgediscussedinChapter5. Yetthereisnotransversecurvaturesoeventhoughthebreakuplookslikea bridge,thebreakupisnotforthesamereason. Thuswecanndavalueof B atwhichthefundamentaleigenvalueisadouble root,i.e., 2 o = 2 e andatsmallervaluesof B theeveneigenfunctionbecomesthe fundamental.ItcanalsobeseenfromFigure3-2thatforlargevaluesof B ,the entire V vs P curveislocatedinthesecondquadrant. V istheactualscaled volumeofthedropandisboundedforany P becausethelightuidispushedup above z =0 nearthesharpedgeaswedecrease P .Twotextarrowsindicatethe pointwherethelowestandsecondlowesteigenvalue,i.e., 2 o and 2 e ,equalthe valueof B wechoose.Thecritical V is0.3244for B =9 asshownintable3-1, andoccursbeforeitreachesitsmaximum.Themaximum V inthegure correspondstothesecondsmallesteigenvalueequaltotheBondnumber,i.e., B = 2 e B V .Theinstabilitypointoccurswherethelowesteigenvalueequals theBondnumber. 30

PAGE 31

Figure3-2.Relationshipbetween P and V foraxedBondnumber B =9,5,3 The V vs P curvewillgoesintotherstquadrantandturnsbackintothesecond quadrantif B issmallshowninFigure3-3.Theoutput P hasamaximumvalue where 2 1 isfoundtobeequalto B .Meanwhiletheinputvolumehasamaximum aswellwherethedropmustbreakinasymmetricpatterndueto 2 e = B SolvabilityconditionsEqs.3,3and3failatmaximum V as always. 31

PAGE 32

Figure3-3.Atsmallvaluesof B ,the V vs P curvesas B 0 Thevalueof B forthecurvetojustgointotherstquadrantcanbefoundby consideringthecorrespondingdiffusioneigenvalueproblemEqs.Band B.Thelowestvalueof B forthecurvetoremaininthesecondquadrantis 1 4 2 ,thediffusioneigenvalue. Forsmall B withlargeenoughinputvolume V theshapeoftheinterfaceis showninFigure3-4whereitcannotbeexpressedbyneither Z x nor X z .For thistypeofbaseshapeweusedifferentindependentvariablestorepresenteach partoftheshapeandpatchtheseparttogether,whichiswhatwecallthe patchedmethoddiscussedinAppendixA. 32

PAGE 33

Figure3-4.Theproleofdropbaseshapewhichcannotbeexpressedbyonly oneindependentvariable. Oneparticularcaseis B =0 ,whichisapeculiarproblemandisdiscussedinthe Figure.C-1.Alloftheseresultsareguidedbycalculation,butwecanlearnafair amountwithoutacalculation.Thewaytodeterminestabilitywithoutcalculating eigenvaluesisdescribedasfollows. IntegratingEq3inthecaseof1-Dcross-sectionwith )]TJ/F24 11.9552 Tf 9.299 0 Td [(1 < x < 1 we have PA + BV = Z x + Z 2 x 1 = 2 j 1 )]TJ/F25 7.9701 Tf 6.587 0 Td [(1 where A =2 and V = )]TJ/F45 11.9552 Tf 11.291 9.631 Td [( 1 )]TJ/F25 7.9701 Tf 6.586 0 Td [(1 Z d x .Therighthandsideistheverticalforcethat thepinsexertonthesurface,viz., n = k )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z x i + Z 2 x 1 = 2 t = Z x k + i + Z 2 x 1 = 2 and k t = Z x + Z 2 x 1 = 2 Nowgiventheshapeofthedrop,viz., Z x B V and P B V ,we canplotthecurve V vs P andndthecriticalpointcorrespondingtoanodd eigenfunctionbydrawingtheline PA + BV =0, A =2 33

PAGE 34

Itsintersectionwiththe V vs P graphoccursatthebifurcationpointwhich isseentobeapointwheretheverticalforceexertedbythepinnededges vanishes. Theequationfor Z x ,obtainedbydifferentiatingEq3inthe1-Dcrosssectioncase,is )]TJ/F48 11.9552 Tf 11.956 0 Td [(BZ x = d d x Z x x + Z 2 x 3 = 2 where 1 )]TJ/F25 7.9701 Tf 6.587 0 Td [(1 Z x d x =0 Atthecriticalpoint,viz., B = 2 1 ,wehave )]TJ/F27 11.9552 Tf 11.956 0 Td [( 2 1 = d d x x + Z 2 x 3 = 2 C 1 =0 1 =0 at x = 1 and 1 )]TJ/F25 7.9701 Tf 6.587 0 Td [(1 1 d x =0 whereuponwederive 0= 1 )]TJ/F25 7.9701 Tf 6.587 0 Td [(1 f d d x Z x x + Z 2 x 3 = 2 )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z x d d x x + Z 2 x 3 = 2 g d x = Z x x + Z 2 x 3 = 2 j 1 )]TJ/F25 7.9701 Tf 6.586 0 Td [(1 = 2 Z x x =1 [1+ Z 2 x x =1] 3 = 2 dueto Z beingeven, beingoddand x x =1 6 =0 ,andtherefore Z x x = 1=0 atcritical. Thusfromthedropshapealoneatagiven B andaseriesof V s wecan solvetheperturbationproblemasFigure3-5indicates.InFigure3-5the P vs V curve,the' 'positiondenotesthelimitinginputvolumeabovewhichnobase shapecanbeobtained.The' 'pointsarethecomputationalsolutionstobase 34

PAGE 35

Figure3-5. P vs V curvesforlargeandsmall B withthestraightline PA + BV =0 ,whichisthelocusofbifurcationpoints. problemEqs.3,3and3inthesituationthatsolvabilityconditions Eqs.3,3and3fails.Ifthestraightlineintersects P vs V curve beforeitreachesmaximum V ,theeigenfunctionatcriticalwillbeoddand thedropbreakswithanoddeigenfunctionforlarge B wheretheintersection, denotedasthe'o'inthegure,isthebifurcationpoint.Ifthisoccursafter V reachesitsmaximumvalue,thedropbreakssymmetricallyatmaximumvolume withacorrespondingeveneigenfunctionforsmall B .Atthebifurcationpoint wherethedropbreaksasymmetricallywithanoddeigenfunction,wehavea dropshapeshowninFigure3-6wherethevolumeisatcriticalanditwillbreak asymmetrically.Thefeatureofzerotangentattheedge, Z x =0 at x = 1 ,can beusedtopredictinstabilitiesthatoccurbeforethegreatestvolumeisreached. 35

PAGE 36

Figure3-6.Dropshapeatthebifurcationpoint. Movingbeyondcriticaltowardthegreatestvolume,wediscoverthatthelight uidrisesabovethe z =0 planeasisshowninFigure3-7.Itisfoundthatthe lightuidstartsinvadingintotheheavyuidbeyondcritical. 36

PAGE 37

Figure3-7.Dropshapeaswemovecarefullybeyondthebifurcationpoint. Intermsofwhatwemightexpecttoseeinanexperiment:Thereisavalue of B andacorresponding V where 2 1 B V = 2 2 B V If B liesabovethisvaluethedropbreaksinanantisymmetricpattern,ifitlies belowthedropbreaksinasymmetricpattern. Fordropswith1-Dcross-section,wecanalsoadvance z fromthebottom ofthedrop,viz., z cl ,the z coordinateatthedropcenterline x =0 ,totheits pinnededge,viz., z =0 .Thecorrespondingforcebalanceequation X z for advancing z is P )]TJ/F48 11.9552 Tf 11.955 0 Td [(Bz = )]TJ/F55 11.9552 Tf 13.779 8.088 Td [(d d z X z + X 2 z 1 = 2 IntegrateEq3withrespectto z from z cl to z =0 ,wend )]TJ/F48 11.9552 Tf 11.956 0 Td [(Pz cl + B 2 z 2 cl = )]TJ/F48 11.9552 Tf 33.334 8.088 Td [(X z + X 2 z 1 = 2 j 0 z cl 37

PAGE 38

Atcriticalpointcorrespondingtoanoddeigenfunction,therighthandsideofEq 3iszerosowehave P = B 2 z cl Substitute P intoEq3wend z cl = )]TJ/F48 11.9552 Tf 9.298 0 Td [(V ifthepinnededgeofthedropsatises Z x x = 1=0 atcritical.Meanwhile,it isfoundthat z cl goesintoamaximummagnitudewhen V reachesitsmaximum value. 3.4TheCaseoftheCircularCross-section,FirstExperiment Wenowturntoacircularcross-section,workingouttheplanecaserst andthebentsurfacenext.Foracircularcross-sectionourdropshape,weuse z = Z r torewriteEq3inthecaseofrstexperimentas P )]TJ/F48 11.9552 Tf 11.956 0 Td [(BZ = 1 r d d r r + Z 2 r 1 = 2 d Z d r where Z =0 at r =1 and )]TJ/F24 11.9552 Tf 11.955 0 Td [(2 1 0 Zr d r = V where B and V areinputsand Z and P areoutputs.Forsolvabilityconditions andlinearstabilityanalysis,wehavethecorresponding problem Z r and eigenvalueproblem r whichare P )]TJ/F48 11.9552 Tf 11.955 0 Td [(B Z = 1 r d d r r + Z 2 r 3 = 2 d Z d r where Z =0 at r =1 38

PAGE 39

and )]TJ/F24 11.9552 Tf 9.299 0 Td [(2 1 0 Zr d r =1 and C )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 = 1 r @ @ r r + Z 2 r 3 = 2 @ @ r + 1 + Z 2 r 1 = 2 1 r 2 @ 2 @ 2 where =0 at r =1 and 2 0 d 1 0 r d r =0 At V =0 wehave Z =0 representingtheplanarinterface.Theeigenvalue problemsfor Z =0 lookverysimilartothe1-Dcross-sectioncasebuttheyare alittlemoreinvolvedastheright-hand-sideisdifferentforaxisymmetricaland nonaxisymmetricalperturbations,i.e., m =0 and m =1 denotingperturbations as Z 1 r cos m .Thustheright-hand-sidecanbewrittenas m =0: 1 r d d r r + Z 2 r 3 = 2 d Z 1 d r m =1: 1 r d d r r + Z 2 r 3 = 2 d Z 1 d r )]TJ/F49 7.9701 Tf 31.893 4.884 Td [(Z 1 r 2 + Z 2 r 1 = 2 where Z r B V denotesthesurfaceshapeinquestion.andthesolutionsto oureigenvalueproblemare m =0: = J 0 r )]TJ/F22 11.9552 Tf 11.955 0 Td [(J 0 J 0 +2 J 1 =0, C = 2 J 0 m =1: = J 1 r cos J 1 =0, C =0 etc. Fromthesolutionwendtheleast istherstpositivezeroofJ 1 andthus itssquareisthecriticalvalueof B at V =0 andtheinstabilityistoacos perturbation. 39

PAGE 40

Nowweset B < B crit V =0 andadvance V untiltheleast 2 B V is equalto B tondthecritical V forthecaseofcurvedinterface. Forthiscasetheproblemhassolutions m =0: J 0 r )]TJ/F22 11.9552 Tf 11.956 0 Td [(J 0 , J 0 +2 1 0 J 0 r r d r =0, C = 2 J 0 m =1: J 1 r cos J 0 =0, C =0 etc. WecanobtainJ 0 r andJ 1 r bythemethodofFrobeniusinAppendix EasisdenotedbyEqs.E,E,EandEandderiveestimatesof the 2 s ,butwecanalsosimplydothecalculationaswedidbeforeinSection3.3. Themethodforcomputingcritical V forgiven B withcircularcross-section issimilartothecaseof1-Dcross-section.Ourjobistoset B toavaluebelow B crit V =0 ,thenincrease V untilthesurfacebecomescriticalatthe B of interest.Basedontheplanarsurfaceresult,weneedtosetavalueof B < 2 1 J 1 1 =0 andincrease V causingthesurfacetopassthroughaseriesof shapes z = Z r B V untilwereachavalueof V where B = 2 1 B V m whereupontheperturbationproblemhasasolution Z 1 6 =0 andwehavearrived atthevalueof V wheretheinput B isnowcritical. Similarlyweanticipatethesmallerthevalueof B thelargerthevalueof V neededtomake B acriticalvalue.Onholding V xed,thelarger V thesmaller B crit andoursurfacewouldbreakatasmaller B .Againtheeigenfunctionswill beeitherunsymmetricoraxisymmetric.Thecomputationalresultsfordifferent B 'satcriticalareshowninTable3-2.Theseresultsareobtainedbyxing B and increasing V untiltheneutralstabilitypointisreached.For B greaterthan 10 the dropisalwayscriticalatan m =1 perturbationsincethecritical V for m =1 has 40

PAGE 41

smallervaluescomparedtothosein m =0 case.For B lessthan 10 thedrop volumeiscriticaltoan m =0 perturbationasitalwayshasthelowervalue comparedtothe m =1 perturbation.Alsofor B lowerthan5itisnotpossibleto ndthecritical V usingonlyoneindependentvariable.Inthissituationwehave torewritethestabilityproblemusingthepatchedmethodtoobtainthecritical data.ThismethodisdescribedintheAppendixA. Table3-2.Criticalvaluesof V forgiven B inthecaseofcircularcross-section. B Critical V for m =0 Critical V for m =1 140.32790.2041 120.44730.4272 100.59830.5982 80.75620.802 60.86041.0972 50.88031.31 41.6 32.06Alwaysstable 15.2631since 2 1 > B 0.59.8353 Werecordthe V forincreasing P fordifferent B .Theirrelationshipisshownin Figure3-8. V istheactualvolumeofthedropandisboundedforgiven P becausethelightuidispushedupabove z =0 nearthesharpedgeaswe increase P .Thecritical V is0.2041for B =14 asshownintable3-2,whichis criticalatthe m =1 perturbationandoccursbeforeitreachesitsmaximum.The maximum V inthegurecorrespondstothesmallesteigenvaluefor m =0 perturbationbeingequaltothe B ,i.e., B = 2 1 B V m =0 .Infactforall B 's 41

PAGE 42

lessthanabout10thedropisunstabletothe m =0 perturbation.Whenthe Bondnumberdecreasesfrom14,thecritical V for m =1 willapproachits maximum,i.e.,thecritical V for m =0 perturbation.Thetwocriticalvolumes, onefor m =1 andtheotherfor m =0 coincideat B =10 .Ifwecomparethe resultsto1-Dcross-sectioncase,thesmallesteigenvaluefor m =1 actslike 2 1 in1-Dwhilesmallesteigenvaluefor m =0 islike 2 2 .Thestraightlinesareused tolocatethebifurcationpointwithan m =1 instabilityforanantisymmetric patternofdropbreakup,whichareintroducedandexplainedinEq.3. Figure3-8.Relationshipbetween P and V foraxedBondnumber B =14,10,6 Ateach B lessthan B crit V =0 weincrease V fromzerountilwenda criticalpoint.Atrstthecriticalpointcorrespondsto m =1 ,acos pattern. 42

PAGE 43

Thecriticalpointisabifurcationpointbecause Z existsthereandthebase calculationcanpassthrough.Doingthiswendatapointwherethelowest m =0 eigenvaluebecomesequalto B and Z doesnotexistandwehave reachedthegreatestvalueof V Uponlowering B wereachavaluewhereuponincreasing V fromzerowe ndavalueof V where 2 1 m =1= B = 2 1 m =0 .Thisisthegreatest V anditisnotprecededbyabifurcationpoint.Forlowervaluesof B ,thecritical V alwayscorrespondstothegreatest V andthedropbreaksinaaxisymmetric pattern. Thereisnoqualitativedifferencebetweenthecircularcrosssectionand the1-Dcrosssection.Surfacestoidentify m =1 eigenfunctionsareviewed asanalogoustotheoddeigenfunctionsdiscussedinthe1-Dcaseand m =0 eigenfunctionsasanalogoustoeveneigenfunctions. Nowsolongasthedropiscriticalat m =1 ,wecanlocatethebifurcation pointbydrawingtheline P + BV =0 andplotthecurve V vs P forgiven B inFigure3-9.Thestraightlineisthe locusofthedropbreakingasymmetricallyinan m =1 eigenfunction.Thepoint with' 'markerindicatesthemaximumvolumewecanget.The' 'pointsare thecomputationalsolutionstobaseproblemEqs.3,3and312 inthesituationthatsolvabilityconditionsEqs.3,3and324fails. The V vs P curveintersectsthelinedenotedbyEq.3beforeitreaches themaximumvolume,whichmeansthedropwillbreakasymmetricallyatthe intersectionmarkedwith'o'ina m =1 eigenfunction. B issetto13,whichisless than 3.83 2 ,thecritical B fortheplanesurface V =0 .Theintersectionofthese twolinesisthebifurcationpointwearelookingfor. 43

PAGE 44

Figure3-9. V vs P curveandthelinedenotedbyEq.3. Toseethat Z r r =1=0 atcritical,viz.,at 2 1 B V = B m =1, weobtain bydifferentiatingEq.3withrespectto r : )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ r = 1 r @ @ r r + Z 2 r 3 = 2 @ @ r Z r )]TJ/F24 11.9552 Tf 39.147 8.087 Td [(1 + Z 2 r 1 = 2 1 r 2 Z r 1 0 @ @ r Zr d r =0 wheretheradialpartofan m =1 eigenfunctionsatises )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 = 1 r @ @ r r + Z 2 r 3 = 2 @ @ r )]TJ/F24 11.9552 Tf 39.147 8.087 Td [(1 + Z 2 r 1 = 2 1 r 2 1 0 @ @ r r d r =0 Thussettingthevalueof 2 inEq3to B inEq3andusingthesame methodasweobtainEq3,wehavethefollowingequality 0= )]TJ/F48 11.9552 Tf 33.619 8.087 Td [(Z r + Z 2 r 3 = 2 r r j r =1 r =0 whereuponwend Z r r =1=0 dueto r r =1 6 =0 44

PAGE 45

ThusmultiplyingEq3by r andintegratingover 0 6 r 6 1 werecover thefollowingrelationbetween P and V atcritical P + BV =0 Nowthedropbreaksatlowvaluesof B inasymmetricpatternandthedrop likethejethastransversecurvature.Thejetbreaksinasymmetricpatterndue entirelytotransversecurvature.Butwecannotattributethejet-likenatureof thedropatlowvaluesof B totransversecurvatureforthereisnotransverse curvatureinthe1-Dproblemandyettheimaginary1-Ddropalsobreaks symmetricallyatlow B Fordropswithcircularcross-section,canwerecovertherelationship between z cl ,themagnitudeof z coordinateatthecenterline,anddropvolume atcriticalwith m =1 perturbation?Weadvance z cl ,the z coordinateatthedrop centerline x =0 ,totheitspinnededge,viz., z =0 ,whichissimilartothe methodin1-Dcase.Thecorrespondingforcebalanceequation R z forusing z astheindependentvariableis P )]TJ/F48 11.9552 Tf 11.955 0 Td [(Bz = )]TJ/F48 11.9552 Tf 31.126 8.088 Td [(R zz + R 2 z 3 = 2 + 1 R + R 2 z 1 = 2 Sinceweareunabletointegratetheterm 1 R + R 2 z 1 = 2 inEq3from z cl to 0 analytically,thereisnoexplicitrelationshipbetween z cl andscaleddropvolume atthecriticalstatefor m =1 perturbation.Wecanonlynd z cl goesintoa maximummagnitudewhen V reachesitsmaximumvalueinthiscase. Nowwhataboutaring, 6 r 6 1 where istheinnerradiusoftheliquid ring?Canwedrawourstraightline?Is Z r r =1=0 ?Carryingoutour calculationasabovewendatcritical 1 r Z r + Z 2 r 3 = 2 r at r =1= r Z r + Z 2 r 3 = 2 r at r = 45

PAGE 46

Thismeansif Z r =0 atoneenditmustbezeroattheother.However,wecannot proveneither Z r r =1=0 atcriticalnorourconstructionofthestraightline holdsforliquidrings. 46

PAGE 47

CHAPTER4 MODELOFTHEPENDENTDROPINACONSTANTPRESSURE EXPERIMENT Inthesecondexperimenttheuidsliebetweenplanewalls-Dcross-section orinacylinderofaspeciedanduniformcrosssection.Thetotalvolumesof eachuidisxedandthepressureisappliedatthetopofheavyuid.Thisthen determinesthevolumeofdropmeasuredfrom Z =0 .Iftheinterfacestartsout planarandisstable,uponincreasingpressurefromthetop,atsomepointthe interfaceoughttobreak.Figure4-1presentsasketchofthemodelforthe secondexperiment. Figure4-1.Thesecondexperiment:constantpressureexperiment. Incaseourmodelis1-D,forthesecondexperiment,wedenotethesurfaceby z = Z x )]TJ/F48 11.9552 Tf 9.299 0 Td [(L < x < L or x = X z z CL < z < 0 47

PAGE 48

Ifthecross-sectioniscircularwethenwrite z = Z r 0 < r < R or r = R z z cl < z < 0 dependingonwhatindependentvariablewewouldliketochoose.Inallcases thesurfaceispinnedatasharpedgeforthesimplicityofboundaryconditions and R denotestheboundaryatthecross-sectionofthecylinders. 4.1ArbitraryCross-section,SecondExperiment Thesecondphysicalexperimentdiffersfromtherstonementionedearlierin thatthetotaluidvolumeisheldconstantuponperturbation. AgainthemodelequationistheYoung-Laplaceequationwherethevertical pressuredifferenceacrossthesurfaceisbalancedbysurfacetension.Hencewe have P + 2 H = P wherethenormalpointsfromtoand P = )]TJ/F27 11.9552 Tf 9.298 0 Td [( g z + C P = )]TJ/F27 11.9552 Tf 9.298 0 Td [( g z + C Inthesecondexperimenttheinputisthepressureappliedattheuppersurface oftheheavyuid.Thevolumesoftwouidsremainxedandwedenotetheir 48

PAGE 49

cross-sectionalareasby A and A ,respectively.Thenwehave P = )]TJ/F27 11.9552 Tf 9.298 0 Td [( g z + P + g H P = )]TJ/F27 11.9552 Tf 9.298 0 Td [( g z + P atm + g H V = HA )]TJ/F45 11.9552 Tf 11.955 16.272 Td [( Z d x and V = H + D A )]TJ/F48 11.9552 Tf 11.955 0 Td [(HA + A Z d x HencebysubtractingEq.4fromEq.4wehave P )]TJ/F48 11.9552 Tf 11.955 0 Td [(P atm + g H )]TJ/F27 11.9552 Tf 11.955 0 Td [( g H )]TJ/F24 11.9552 Tf 11.956 0 Td [( g Z = r r Z + r Z r Z 1 = 2 whereinthecaseofplanarinterface, P = P 0 H = H 0 and H = H 0 correspond to Z =0 andduetoconstantvolumesconstraintwehave V = V 0 and V = V 0 andthusweobtainthefollowing H )]TJ/F48 11.9552 Tf 11.955 0 Td [(H 0 = 1 A A Z d A and H )]TJ/F48 11.9552 Tf 11.956 0 Td [(H 0 = H )]TJ/F48 11.9552 Tf 11.955 0 Td [(H 0 A A = A A 1 A A Z d A Thusweobtain P )]TJ/F48 11.9552 Tf 11.955 0 Td [(P 0 + g H )]TJ/F48 11.9552 Tf 11.955 0 Td [(H 0 )]TJ/F27 11.9552 Tf 11.955 0 Td [( g H )]TJ/F48 11.9552 Tf 11.955 0 Td [(H 0 )]TJ/F24 11.9552 Tf 11.955 0 Td [( g Z = r r Z + r Z r Z 1 = 2 andassuming A >> A ,ourmodelequationis P + g 1 A A Z d A )]TJ/F24 11.9552 Tf 11.955 0 Td [( g Z = r r Z + r Z r Z 1 = 2 where P ismeasuredabove P 0 .Nowscalingalllengthsbythediameterofthe cross-sectionwehavethefollowingdimensionlessformofthegoverning 49

PAGE 50

equation P + B 1 A A Z d A )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ = r r Z + r Z r Z 1 = 2 where B = g D 2 B > B and P nowdenotes PD andwhere Z =0 alongthe perimeterofthecross-section.Now P C B areinputs, Z istheoutputandfor theplanarinterfacewehave P =0 Z =0 ,andifthesolutionis Z at P > 0 ,it is )]TJ/F48 11.9552 Tf 9.298 0 Td [(Z at )]TJ/F48 11.9552 Tf 9.298 0 Td [(P .Similarly, )]TJ/F45 11.9552 Tf 9.299 9.63 Td [( A Z d A denotesthevolumeofthedrop. Assumingwecanndtheshapeofthesurface,wewanttoknowifthesurfaceis stabletosmallperturbationsinthesecondexperiment.Tondoutweask:isthe shapeneutrallystable?Henceholdingtheinputsxed,weintroduceasmall perturbationvia Z = Z 0 + Z 1 where Z 0 denotestheshapeofdropwhosestabilityisinquestionandwehave theperturbedequationforthelinearperturbationterm Z 1 inthesecond experiment B 1 A A Z 1 d A )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ 1 = L Z 1 where Z 1 =0 attheedge L = rf )778(! )778(! I + r Z 0 r Z 0 1 = 2 )]TJ 35.276 8.088 Td [(r Z 0 r Z 0 + r Z 0 r Z 0 3 = 2 gr isalinearoperatorwhere Z 0 appears. Ifwehaveasolutionotherthan Z 1 =0 tothesehomogeneousproblems,the shapeofthesurfaceiscritical,iftheonlysolutionis Z 1 =0 ,theshapeisnot critical.Inthesecondexperimentourplanistoadvanceacontrolvariable,i.e., P ,obtainafamilyofshapesanddecidewhichmemberofthefamilyiscritical. 50

PAGE 51

Thecorrespondingeigenvalueproblemthathelpsusunderstandthestabilityof thisexperimentis B 1 A A d A )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 = L where =0 attheedge andmultiplyingby andintegratingoverthecrosssectionweseethatall 2 's arepositivebecause r r r Z 0 r Z 0 > r r Z 0 r r Z 0 Nowtheeigenvaluesdependonthebaseshapeandasweadvanceourcontrol variable,saythepressureappliedatthetopoftheheavyuidinthesecond experiment,wecalculatethebaseshapesandsolvetheeigenvalueproblemat eachbaseshape.Thenourperturbationproblemhasonlythesolution Z 1 =0 so longas B remainslessthanthefundamentaleigenvalue.Ifwehaveasolution B = 2 1 B P B inthesecondexperiment,wehavearrivedatthecritical shape. 4.2TheCaseofthePlaneInterface,1-DCross-section,Second Experiment Thesecondexperimenthasazerosolutionviz., Z =0 fortheplaneinterface.It isfoundat P =0 nomatterthevalueof B and B Tondoutifwecanobservethisplanesolutioninanexperimentweincrease theBondnumber,lookingforthevalueBwhere B = 2 1 B P =0, B Atthisvalueof B theperturbationproblemhasasolution Z 1 6 =0 andwehave arrivedattheRayleigh-Taylorcriticalpointforaplanesurface. 51

PAGE 52

Thecorrespondingeigenvalueproblemforthesecondexperimentis B 1 2 1 )]TJ/F25 7.9701 Tf 6.587 0 Td [(1 d x )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 = d 2 d x 2 =0 at x = 1 anditsanalyticalsolutionsare = sin sin =0 and = cos x + B = 2 1 )]TJ/F48 11.9552 Tf 11.955 0 Td [(B = 2 sin tan =1 )]TJ/F27 11.9552 Tf 14.234 8.088 Td [( 2 B whereuponweagainhaveoddandevensolutionsbutwhatwenddepends uponthevalueof B .Thesmallestrootof tan =1 )]TJ/F28 7.9701 Tf 14.158 4.707 Td [( 2 B liesbetween B and 2 if B islessthan 2 orbetween 2 and B if B isgreaterthan 2 If B < 2 wehaveasolution B < 2 < 2 correspondingtoaneven eigenfunctionandasolution 2 = 2 correspondingtoanoddeigenfunctionbutif B > 2 wehaveasolution 2 = 2 correspondingtoanoddeigenfunctionanda solution 2 < 2 < B correspondingtoaneveneigenfunction.Sodependingon B ,thefundamentaliseitherevenoroddand +1 )]TJ/F25 7.9701 Tf 6.586 0 Td [(1 d x isnotzeroorzero correspondingtoevenorodd.Andwemusthave B < B ,henceif B < 2 all allowableBondnumberscorrespondtoastablesurfacewhereasif B > 2 the criticalBondnumberat P =0 isagain 2 .Hencewecansetavalueof B less than 2 andincrease P .Nowthedropshapesandvolumesinthesecond experimentcanbederivedfromthoseintherstexperiment,andweobtainthe curvesshowninthesketchinFigure4-2,aslongas B isnear 2 .Theinputfor modelIisthevolumeofdrop V whereasformodelIIisthepressure difference P P II inmodelIIcanbederivedfrom P I ,whichistheoutputof modelI.Wecanconstructdifferentstraightlinefordifferentmodels,tolocate thebifurcationpointwithanoddeigenfunction. 52

PAGE 53

Figure4-2. V vs P curvesfortwomodelswithsame B =7 and B =10.8566 WhatweobservefromFigure4-2isthatthesecondexperimentalsohasa straightlineconstructionpredictingthebifurcationpointusingonlythebase shape.Itis P II A )]TJ/F24 11.9552 Tf 13.151 8.088 Td [(1 2 B V + BV =0, A =2 ItisfoundfromFigure4-2thatperturbationsatconstant V arestableathigher V 'sthanperturbationsatconstant P .Ofcoursethisrepeatswhatwelearnedfor arbitrarycross-sections. 4.3TheCaseoftheCurvedInterface,1-DCross-section,Second Experiment Themainideatodeterminethecriticalpointistosetavalueof B < 2 and increasePinthesecondexperimentcausingthesurfacetopassthrougha 53

PAGE 54

seriesofshapes z = Z x B P B untilwereachavalueof P where B = 2 1 B P B whereupontheperturbationproblemhasasolution Z 1 6 =0 andwehavearrived atthevaluesof C wheretheinput B isnowthecriticalvalueof B atthe corresponding B Theeigenvalueproblemsfor Z =0 lookmuchasbeforebut d 2 d x 2 ontheRHSis replacedby d d x 1 + Z 2 x 1 = 2 d d x where Z x B P B denotesthesurfaceshapein question.Because 0 < 1 + Z 2 x 1 = 2 < 1 weanticipatethesmallerthevalueof B the largerthevalueof V or P neededtomake B acriticalvalue.Onholding V or P xed,thelarger V or P thesmaller B crit andoursurfacewouldbreakatasmaller Bondnumber.Againtheeigenfunctionswillbeeitheroddorevenfunctionsof x ThecomputationalresultsfordifferentBondnumbersattheneutralstabilitystate areshownintable4-1. Therelationshipbetween P anddropvolumefordifferent B inthesecond experimentisgivenbyFigure4-3.Here I istheactualvolumeofthedropi.e., I = )]TJ/F45 11.9552 Tf 11.291 9.631 Td [( 1 )]TJ/F25 7.9701 Tf 6.586 0 Td [(1 Z d x .Noticethattheappliedpressure P fromthetopuidisbounded andhasamaximumvaluethatdependsonthevalueof B .Theneutralstability pointisdenedaswhereanyoftheeigenvaluesequalstheBondnumber.Two arrowsindicatethepointwherethelowestandsecondlowesteigenvalue,i.e., 2 o and 2 e ,equalthe B chosen.TheresultsareobtainedbyxingtheBondnumber andincreasing P untiltheneutralstabilitypointisreached.Thecritical P is6.9 for B =9 asshownintable4-1,whichoccursbefore P reachesitsmaximum. ForBondnumberslowerthan3itisnotpossibletondthecritical P usingonly oneindependentvariable.Inthissituationwehavetorewritethestability problemusingamultiple-domainaspatchedmethodtoobtainthecriticaldata. ThisisdescribedinAppendixA.Themaximum P inthegurecorrespondstothe 54

PAGE 55

Figure4-3.Relationshipbetween P and I = )]TJ/F45 11.9552 Tf 11.291 9.63 Td [( 1 )]TJ/F25 7.9701 Tf 6.586 0 Td [(1 Z d x foraxedBondnumber B =9,3,2 casewherethesecondsmallesteigenvalueequalstheBondnumber,i.e., 2 e B P B = B 55

PAGE 56

Table4-1.CriticalvaluesofBondnumbersandappliedpressure P for1-D cross-sectioninthesecondexperiment. BondNumberCritical P 96.9 516.8 319.7 142.5 0.660.4 0.1257.3 56

PAGE 57

CHAPTER5 MODELOFTHECLOSEDLIQUIDBRIDGEANDROTATINGDROP Aliquidbridgecanbecreatedbetweenrigidbodieswitharbitraryshapes. ThesketchofatypicalshapeoftheclosedliquidbridgeisshowninFigure5-1. Figure5-1.Atypicalshapeofanenclosedliquidbridge. Thebridgeiscalledclosediftheinterfaceexistsonlybetweentherigid surfacesasshown.Ifanypartofthehorizontalsurfacesisopenthebridge isseenasopen.Openbridgeswillnotbeincludedinthediscussionofthis chapter. Thebridgeinterface,formingbetweentwodifferentimmiscibleuids,can havevariousshapeswhichdependonthebridgevolumeandforcebalance acrosstheinterface.Ifweassumethatthereisnogravity,wecanobtaintwo mainfeaturesofthebridgeweareinterestedin:thebodyforceofbridgeisthen 57

PAGE 58

zeroandtheinterfaceisformedbecauseofthecapillaryeffect.Theupperpart oftheinterfaceissymmetricwiththelowerpart. Thegoverningequationforcomputingtheshapeofinterfaceisasalways theYoung-Laplaceequation,whichlooksalmostthesameaswhatweuseto modelthedropexceptfortheboundaryconditions.Theunscaledformofthe baseequationdescribingtheshapeofclosedliquidbridgeinacirculargeometry is P )]TJ/F24 11.9552 Tf 11.955 0 Td [( g z = )]TJ/F27 11.9552 Tf 9.298 0 Td [( [ d d z R z + R 2 z 1 = 2 )]TJ/F24 11.9552 Tf 14.509 8.088 Td [(1 R 1 + R 2 z 1 = 2 ] where R = R at Z = L, V = L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L R 2 dz Eq.5helpsussolvethebaseproblemofanon-rotationalclosedliquid bridge.Theinputvariableisthebridgevolume V andtheoutputsarethebase shapeoftheinterface R z and P .Forsimplicityweassume:1thatthereisno gravitywhichisequivalenttosetting g =0 or B =0 ;2bothrigidsurfaceare planesurface.WecanthenwriteEq.5indimensionlessformasEq.5 P = )]TJ/F24 11.9552 Tf 9.298 0 Td [([ d d z R z + R 2 z 1 = 2 )]TJ/F24 11.9552 Tf 14.51 8.088 Td [(1 R 1 + R 2 z 1 = 2 ] where R =1 at Z = L and V = L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L R 2 d z HereEqs.5,5and5needtobescaledwithrespectto R and P now denotes P R andLdenotesL = R Wewillneedtheequationsfor R z later.ThiscanbeobtainedbydifferentiatingEq.5.Thusfor R z wehave 0= 1 R [ d d z R + R 2 z 3 = 2 R z z + 1 R 2 1 + R 2 z 1 = 2 R z ] 58

PAGE 59

and 0= L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L R z d z with R z at z = Ltobedetermined.Using R = d R d V ,wehavethefollowing' problem )]TJ/F24 11.9552 Tf 12.343 2.657 Td [( P = d d z [ R + R 2 z 3 = 2 R z ]+ 1 R 2 1 + R 2 z 1 = 2 R where R =0 at z = L and L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L 2 R R d z =1 Weagainnotethattheperturbedproblemattheneutralstabilitypointwith zerogrowthrateishydrostaticandthatthereisnovelocityperturbationinthe perturbedequation.Thiscanbeprovenmuchasinthecaseofthedrop.Using R = R 0 z + R 1 z ,thelinearperturbationequationfor R 1 isgivenby )]TJ/F48 11.9552 Tf 11.956 0 Td [(P 1 = d d z [ R 0 + R 2 0 z 3 = 2 R 1 z ]+ 1 R 2 0 1 + R 2 0 z 1 = 2 R 1 + R 1 where R 1 =0 at z = Land L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L d zR 0 z 2 0 d R 1 z = V 1 =0 Theproblemishomogeneousin R 1 z with R 0 z givenintheequation.Ifit hasanon-zerosolution,thebridgereachestheneutralstabilitypoint.Tond outif R 1 hasasolutionotherthanzeroweagainintroducethecorresponding eigenvalueproblemforthelinearperturbedequationshownasfollows. C = d d z [ R 0 + R 2 0 z 3 = 2 z ]+ 1 R 2 0 1 + R 2 0 z 1 = 2 + where =0 at z = L 59

PAGE 60

and L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L d zR 0 z 2 0 d =0 Ifwedenote = ^ cos m ,wecanwritethesolutionstotheaboveeigenvalue problemas m =0: 2 1 2 2 1 z 2 z C 1 C 2 m =1: 2 1 2 2 ^ 1 z cos ^ 2 z cos C 1 =0, C 2 =0, IfoneoftheeigenvaluesweobtainfromEqs.5,5and514is equaltozero,wehavesuccessfullyfoundanon-zerosolution R 1 ,tothelinear perturbationproblemandthusthebridgeshapeisatcritical. Writing as ^ cos m andsubstitutingitbacktoEqs.5,5and 5,weexpresstheeigenvalueprobleminthefollowingequivalentform C = d d z [ R 0 + R 2 0 z 3 = 2 ^ z ]+ 1 R 2 0 1 + R 2 0 z 1 = 2 ^ )]TJ/F48 11.9552 Tf 11.955 0 Td [(m 2 ^ + 2 ^ =0 where ^ =0 at z = L and L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L d zR 0 z ^ z 2 0 d cos m =0 Forall m 2 1 and C =0 ,multiplyingEq.5by R 0 ^ andintegratingover )]TJ/F22 11.9552 Tf 9.299 0 Td [(L z Lweobtain )]TJ/F45 11.9552 Tf 11.955 16.273 Td [( L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L [ R 0 + R 2 0 z 3 = 2 ^ 2 z d z ] )]TJ/F24 11.9552 Tf 11.956 0 Td [( m 2 )]TJ/F24 11.9552 Tf 11.955 0 Td [(1 L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L 1 R 2 0 R 0 + R 2 0 z 1 = 2 ^ 2 d z + 2 L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L ^ 2 d z =0 60

PAGE 61

whereuponwendthat 2 =0 istheonlypossibleeigenvalueifandonlyif m =0 Henceweassume m =0 inboththeperturbationandeigenvalueproblem. Asaresult, R 1 and onlydependon r .Thenwecanwritetheperturbation problemas )]TJ/F48 11.9552 Tf 11.956 0 Td [(P 1 = d d z [ R 0 + R 2 0 z 3 = 2 R 1 z ]+ 1 R 2 0 1 + R 2 0 z 1 = 2 R 1 where R 1 =0 at z = L and 2 L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L R 0 R 1 d z =0 Theeigenvalueproblemcanbewrittenas C = d d z [ R 0 + R 2 0 z 3 = 2 z ]+ 1 R 2 0 1 + R 2 0 z 1 = 2 + 2 where =0 at z = L and L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L R 0 d z =0 5.1TheCaseoftheNon-rotatingClosedLiquidBridge 5.1.1TheBaseCase,aVerticalLiquidCylinder Iftheclosedbridgeisnon-rotatingandunderzero-gravity,wehavea solutionforEq.5whichisacylindricalshapeandinunscaledformthe 61

PAGE 62

shapeiswrittenas R = R forall z V 0 = R 2 0 2 L P 0 = R 0 Thesolutionrepresentsacylindricalshapethathasauniformcurvature 1 R 0 .Sinceweareinterestedinwhetherthebasesolutionisstabletosmall perturbation,wewriteEqs.5,5and5fortheverticalcylindrical shapeinscaledformas C = d 2 d z 2 ++ 2 where =0 at z = L and L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L d z =0 Theaboveeigenvalueproblemhasanalyticalsolutionswhichcaneitherbeodd orevenin z .Introducinganewvariable andsetting 2 = 2 )]TJ/F24 11.9552 Tf 11.955 0 Td [(1 weobtain = sin z 2 L 2 = n 2 2 C =0 and = cos z )]TJ/F22 11.9552 Tf 11.955 0 Td [(cos L tan L )]TJ/F27 11.9552 Tf 11.955 0 Td [( L =0, C = )]TJ/F27 11.9552 Tf 9.298 0 Td [( 2 cos L Itisfoundthattheleast 2 is 2 L 2 )]TJ/F24 11.9552 Tf 12.858 0 Td [(1 foroddeigenfunctionsin z androughly 9 4 2 L )]TJ/F24 11.9552 Tf 12.064 0 Td [(1 foreveneigenfunctionsin z Atcritical,theleast 2 iszerosothatL = whichmeansliquidcylinderisstableasweincreaseLuntilLreaches OurplanofcalculationistosetL < observethatthebridgeisstableat V = V 0 where V 0 = 2 Lfortheliquidcylindercaseanddecrease V from V 0 untilwendthepointofneutralstability. 62

PAGE 63

5.1.2Decreasing V from V 0 atConstantL < NomatterthevalueofL,aslongasitislessthan ,alltheeigenvalues at V = V 0 arepositiveandtheliquidcylinderisstablewherethefundamental eigenvaluecorrespondstoanoddeigenfunction.Wealsohave d 2 1 d V > 0 and henceifwedecrease V ,atrsttheoddeigenfunctionremainsfundamentaland thefundamental 2 1 willdecreasetowardzeroasdoallthe 2 's. AteachLand V ,thebasesolution R z satisfyingEqs.5,5and 5isanevenfunctionof z .Theeigenvalueproblem,i.e.,Eqs.5,5 and5atLand V hassolutionswitheitheroddoreveneigenfunctions whereforanoddeigenfunction L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L R d z mustbezerosothat C =0 .For eveneigenfunctions C mustbenon-zerotomake L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L R d z =0 .Wewishto knowifwecandecrease V andndanon-zerosolutiontotheperturbation problem,Eqs.5,5and5sincetheperturbationproblemhas onlythesolutionzerosolongasthecorrespondingeigenvalueproblemhasno solution 2 =0 .Atthevalueof V wherethefundamental 2 isequaltozerothe perturbationproblemhasanon-zerosolution R 1 6 =0 andwereachaneutral stabilitypoint.Ifthecorrespondingeigenfunctioniseven,wehavereachedthe lowestvalueof V andthebridgebreaksinasymmetricpattern.Ifitisodd,the neutralstabilitypointisabifurcationpointandwecancontinuetodecrease V untilthesecondlowesteigenvaluecorrespondinganeveneigenfunctionisequal tozerowherewereachtheminimumvalueof V .Howeverthebridgecanbreak inanantisymmetricpatternbeforethelowestvalueof V isreached. IfwerunanexperimentatdifferentvaluesofL < anddrawagraphof V = V 0 vs P wehavethefollowingFigure5-2.InFigure5-2thewayoflocating thecriticalpointistoslowlydecreasethedropvolumefrom V = V 0 =1 ,where V 0 denotesthebaseliquidcylindersolution,untilwendthelowest 2 ofthe eigenvalueproblemEqs.5,5and5equaltozero.Thenweknow 63

PAGE 64

Figure5-2. V = V 0 vs P curveswithdifferentvaluesofL. theclosedliquidbridgewillpinchoffatthiscritical V = V 0 .Foreachcurvethere isaminimum V = V 0 whereuponthebridgemustbreakinasymmetricpattern withthecriticalpointdenotedasanose,showninthegureas V 3 .Howeverfor valuesofLcloseto ,itcanbreakatthebifurcationpointbeforereachingthe lowestvalueof V = V 0 inanunsymmetricalpattern,showninthegureas V 1 .At L = L thesymmetricandunsymmetricaleigenmodescoincideatthelowest valueof V = V 0 ,showninthegureas V 2 .wecanalsopredictthebifurcation pointusingonlytheshapeofbridge,aswecanalwaysdointhedropproblem. 64

PAGE 65

ByintegratingEq.5,settingg =0 andplugginginEq.5wehave )]TJ/F24 11.9552 Tf 11.955 0 Td [(2 L P = R z + R 2 z 1 = 2 j L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L )]TJ/F45 11.9552 Tf 11.955 16.272 Td [( L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L 1 R 1 + R 2 z 1 = 2 d z Atabifurcationpoint,weanticipatethat R z =0 at z = L.Ifthisistrue,wethen have 2 L P )]TJ/F45 11.9552 Tf 11.955 16.273 Td [( L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L 1 R 1 + R 2 z 1 = 2 d z =0 atthebifurcationpoint.Asaresult,inordertondoutthecriticalvolumeatthe neutralstabilitypoint,wedrawtheagureof 2 L P vs L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L 1 R 1 + R 2 z 1 = 2 d z whereuponthestraightline 2 L P = L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L 1 R 1 + R 2 z 1 = 2 d z willintersectourgraphatabifurcationpointiftheliquidbridgewantstobreak beforereachingtheminimumvolumewithanoddeigenfunction.Thusredrawing thepreviousguresforvariousvaluesofLwehavethefollowingFigure5-3.The straightlineisthelocusofthebridgebreakupinanasymmetricpattern,denoted bytheleft-hand-sideofEq.5.Thecurverepresentstheright-hand-sideof Eq.5andthestraightlineweconstructinFigure5-3is 2 L P 65

PAGE 66

Figure5-3.Themethodofpredictingthebifurcationpointwithoutstability calculation. The' 'pointdenotestheminimumdimensionlessvolume V = V 0 ofthe liquidbridge,whichisthenose,whereuponthebridgemustbreakupwithan eveneigenfunction.The' 'marksrepresentthebaseshapeswheresolvability conditionEqs.5,5and5failsandcannotbephysicallyachieved. ForlargeLthestraightlineintersectsthecurvebeforeitreachesthenoseand thustheliquidbridgebreaksasymmetricallywithanoddeigenfunction.Where asforsmallLitdoesnotintersectthecurvebeforethenoseisreached,which meanstheliquidbridgebreakssymmetricallywithaneveneigenfunction. weobservethatthesurfaceareaofthebridgeis 2 L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L R + R 2 z 1 = 2 d z 66

PAGE 67

andthatsmallLherecorrespondstosmall B inthedropproblemwhilelargeL whichislessthan correspondstolarge B < 2 ToproveEq.5wesetLandsolveforthevalueof V suchthatthe fundamental 2 isequaltozeroandcorrespondingtoanoddeigenfunction ThenmultiplyingEq.5by R ,Eq.5by RR z subtractingandintegrating over )]TJ/F22 11.9552 Tf 9.298 0 Td [(L z Lwehave 0= )]TJ/F48 11.9552 Tf 9.298 0 Td [(R z R + R 2 z 3 = 2 z j L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L andbecause R z isoddand z isevenweconcludethat R z =0 at z = L andthusatabifurcationpointEq.5mustbecorrectanditleadstothe constructionbywhichwecanpredictacriticalpointusingonlytheshapeofthe liquidbridge,i.e., R z Bythesamederivationtheright-hand-sideofEq.Biszeroregardless of R z ifthecriticalpointisanose. 5.2TheCaseoftheClosedLiquidBridgeSpinningatConstantAngular Velocity Theliquidbridgeisnowsettorotateaboutitsaxisofsymmetryatangular velocity )778(! = )778(! k .TheideaistostabilizetheliquidbridgeatgivenvaluesofL and V byweakeningthetransversecurvature.Butthatisnotwhathappens. Theshapeofthesurfaceisdenoted r = R z andourmodelishydrostatic. Thesurfaceshape, R z ,satisestheunscaledequation )]TJ/F48 11.9552 Tf 11.955 0 Td [(P )]TJ/F24 11.9552 Tf 13.151 8.088 Td [(1 2 2 R 2 = d d z [ R z + R 2 z 1 = 2 ] )]TJ/F24 11.9552 Tf 14.51 8.088 Td [(1 R 1 + R 2 z 1 = 2 where R =1 at z = L 67

PAGE 68

and L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L R 2 d z = V wheretheintegralofEq.5over )]TJ/F22 11.9552 Tf 9.299 0 Td [(L z Lis )]TJ/F48 11.9552 Tf 11.955 0 Td [(P 2 L )]TJ/F24 11.9552 Tf 13.15 8.088 Td [(1 2 2 V = R z + R 2 z 1 = 2 j L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L )]TJ/F45 11.9552 Tf 11.955 16.273 Td [( L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L 1 R + R 2 z 1 = 2 d z TheinputvariablesareL, V and 2 .Theoutputsare R z and P WeplantosetLand 2 anddecrease V toobtain R z and P alongthe way.Ateachpointalongthepathwewishtoknowifthebridgeisstabletosmall hydrostaticperturbations.Denotingtheperturbationby R 1 z ,atconstantLand 2 wehave )]TJ/F48 11.9552 Tf 11.955 0 Td [(P 1 )]TJ/F24 11.9552 Tf 11.956 0 Td [( 2 RR 1 = 1 R d d z R + R 2 z 3 = 2 R 1 z + 1 R 2 R + R 2 z 1 = 2 R 1 where R 1 =0 at z = L and L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L 2 RR 1 d z =0 Ifthesolutionof R 1 isnotzero,wereachtheneutralstabilitypoint.Tondout non-zerosolutionof R 1 inEq.E,EandE,wehavethecorresponding eigenvalueproblemas C )]TJ/F24 11.9552 Tf 11.955 0 Td [( 2 R = 1 R d d z R + R 2 z 3 = 2 z + 1 R 2 1 + R 2 z 1 = 2 + 2 where =0 at z = L and L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L 2 R d z =0 68

PAGE 69

wedenoteitssolution 2 1 2 2 ,... 1 2 ,... whichdependonthebaseshape R z ,andbecause R z mustbeaneven functionof z ,theeigenvalueproblemhassolutions, ,oddandevenin z .In ordertohaveanon-zerosolution R 1 toEqs.E,EandEwemust haveoneoftheeigenvaluessatisfyingEqs.5,5and5equalto zero. Alongthepathofdecreasing V ,ifwehaveasolution R z atgivenvaluesof L V and 2 ,wewishtoknowifwecandecrease V andndanothersolution. Toanswerthisquestionwemuststudy R = d R d V atthatpointwherethe R problem is )]TJ/F24 11.9552 Tf 12.343 2.657 Td [( P )]TJ/F24 11.9552 Tf 11.955 0 Td [( 2 R R = R + R 2 z 3 = 2 R z + 1 R 2 1 + R 2 z 1 = 2 R where R =0 at z = L and L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L 2 R R d z =1 Nowwehavethesolution R z =1 forall z at V = V 0 = 2 Lwhere P 0 =1 )]TJ/F24 11.9552 Tf 11.997 0 Td [( 2 Thecorrespondingeigenvalueproblemis C = d 2 d z 2 + 2 +1+ 2 where =0 at z = L and L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L d z =0 69

PAGE 70

whereuponthefundamentaleigenvaluecorrespondingtoanoddeigenfunction andhenceto C =0 ,is 2 = 2 L 2 )]TJ/F24 11.9552 Tf 11.955 0 Td [(+ 2 andthisisalwayspositiveaslongas L 2 + 2 < 2 hencetherotatingbridgeisstablefor 0 < L 2 < 2 1+ 2 = L 2 crit V = V 0 2 Wealsoseethatat V = V 0 d 2 d 2 < 0 forall 2 thusthefundamentaleigenvaluemovestowardzeroas 2 increases.Hencewe haveabridgeshapeadjustingitselfasshowninFigure5-4 70

PAGE 71

Figure5-4.Shapesofclosedliquidbridgewithandwithoutspinningforgiven V andL. as 2 increasesatconstant V OurplanistosetL < L crit V 0 andthendecrease V from V 0 untilwendthe bridgeisunstable. NowatinputvaluesofL 2 and V wehave R z andwecanintroduceour diffusioneigenvalueproblem,viz., 1 R d d z R + R 2 z 3 = 2 z + 1 R 2 1 + R 2 z 1 = 2 + 2 R + 2 =0 where =0 at z = L and L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L R 2 d z =1 71

PAGE 72

wedenoteitssolution 2 1 < 2 2 < 2 3 < ... 1 > 0, 2 3 ,... where 1 2 3 ,... areeven,odd,even,...functionsof z .Thenweseethatthe 0 soddin z are 2 4 ,... andthecorresponding 2 'sare 2 2 2 4 ,... and C =0 Expandingthe 0 sevenin z C isnotequaltozero,inthe 's,evenin z ,we have = X c i i c i = L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L R i d z andmultiplyingEq.5by R i ,Eq.5by R ,subtractingandintegrating over )]TJ/F22 11.9552 Tf 9.298 0 Td [(L z Lwehave CI i = 2 )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 i c i where I i = L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L R i d z .Thenusing L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L R i d z =0 and C 6 =0 wehave X I 2 i 2 )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 i =0, i =1,3,... Thisisanequationforthe 2 'scorrespondingto 'sevenin z andwehave 2 1 < 2 e < 2 3 ... Soourtwoleadingeigenvaluessatisfy 2 o = 2 2 andwhether 2 e or 2 o isthefundamentaleigenvalueas V decreasesdependson L 2 and V At V = V 0 thelowestoddeigenvalueisfundamentalandremainsfundamentalfor V new V 0 .Settingthetwovaluesof 2 ,anddependingonL,therst eigenvaluetoreachzerocorrespondstoanoddeigenfunction.Thisisthecase forLnearL 2 crit V = V 0 ,otherwiseforLnearzerothersteigenvaluetoreach 72

PAGE 73

Figure5-5.Theeffectof 2 ontheliquidbridgebreakupbycomparisonto 2 =0 zerocorrespondstoaneveneigenfunction.Theeffectof 2 onthestabilityof liquidbridgesareshownintheFigure.5-5 BysettingL 2 and V wehave R z z and P .ThendifferentiatingEq. 5wehave )]TJ/F24 11.9552 Tf 11.955 0 Td [( 2 RR z = 1 R d d z [ R + R 2 z 3 = 2 R z z ]+ 1 R 2 1 + R 2 z 1 = 2 R z where L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L R z d z =0= L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L R 2 z d z = L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L 2 RR z d z 73

PAGE 74

NowwemultiplyEq.5by RR z ,Eq.5by R ,subtractandintegrate over )]TJ/F22 11.9552 Tf 9.298 0 Td [(L z L.Thuswehave C L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L RR z d z = L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L f R z d d z [ R + R 2 z 3 = 2 z ] )]TJ/F27 11.9552 Tf 9.298 0 Td [( d d z [ R + R 2 z 3 = 2 R z z ] g d z + 2 L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L RR z d z Now L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L RR z d z iszeroandatcriticalvaluesofL 2 and V 2 iszero whereuponweobtain 0= R z R + R 2 z 3 = 2 z j L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L where R z isanoddfunctionof z .Thusif isoddwecanconclude R z =0 at z = Lbutif iseventheRHSiszeronomatterthevalueof R z .Hencewe canndabifurcationpointbyusingonlythebasesolution.Todothisweplotour basecurveas 2 L P vs L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L 1 R 1 + R 2 z 1 = 2 d z )]TJ/F24 11.9552 Tf 13.151 8.088 Td [(1 2 2 V alongapathofdecreasing V xedvaluesofLand 2 .TheninviewofEq.5 58wedrawboththestraightline 2 L P andthecurve L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L 1 R 1 + R 2 z 1 = 2 d z )]TJ/F25 7.9701 Tf 13.384 4.708 Td [(1 2 2 V in Figure.5-6 74

PAGE 75

Figure5-6.Straightlineconstructionforlocatingthebifurcationpointfromthe basecalculationinthecaseofspinningliquidbridge. whereupontheintersectionisabifurcationpoint. 5.3TheCaseofthePendentDropSpinningatConstantAngularVelocity Wesetadrop,pinnedtotheedgeofacircularcross-section,tospinning atangularvelocity aboutitsverticalaxisofsymmetry.Assuminghydrostatic conditions,wecanobtainitsshape, z = Z r bysolving P + 1 2 2 r 2 )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ = 1 r d d r r + Z 2 r 1 = 2 d Z d r where 75

PAGE 76

Z =0 at r =1 and )]TJ/F24 11.9552 Tf 11.955 0 Td [(2 1 0 Zr d r = V Theangularvelocityisscaledby R 3 0 Theeffectofspinningaliquidbridgeisdestabilizing,spinningstrengthens theeffectoftransversecurvature.Butinthecaseofadropwehavenoevidence ofaneffectoftransversecurvatureandweimaginespinningmightbestabilizing. Thusifwehaveadropslightlyaboveitscriticalvolumewemightbeableto restoreitsloststabilitybyspinningitaboutitsaxisofsymmetry. First,weseethat 2 doesnotappearexplicitlyintheeigenvalueproblems, hencetheeffectofspinningonthestabilityofadropactsonlybychangingthe shapeofthedrop.TheeigenvalueproblemremainsthesameasEqs.31, 3and3where Z isnowthesolutionofEqs.5,5and 5. Secondthereisnosolution Z =0 at V =0 unless 2 =0. Andthirdwenolongerhave Z r =0 at r =1 atacriticalpointthatis abifurcationpoint.Hencewecannolongerdetectabifurcationpointbythe methodofstraightlineconstructionasisdescribedinFigure.3-8. Ourrsttaskistoset V =0 andndthecriticalrangeof B asfunction of 2 At V =0, 2 =0 wehave Z =0 forall B andthesolutionstothe eigenvalueproblemdonotdependon B .Hencewebeginwiththecriticalvalue of B at V =0, 2 =0 andaskwhatdoesthecriticalvalueof B doat V =0 as 2 increasesfromzero? At V =0, 2 =0 wehaveabasesolution Z 0 =0, P 0 =0 andeigenvalue 2 0 correspondingtoeigenfunctions 0 ,nomatterthevalueof B 76

PAGE 77

Writing 2 =0+ 2 1 and Z =0+ Z 1 wehave,atconstant B and V P 1 + 1 2 2 1 r 2 )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ 1 = 1 r d d r r d Z 1 d r where Z 1 =0 at r =1 and 1 0 Z 1 r d r =0 Henceweobtain Z 1 = A J 0 B 1 = 2 r + P 1 B + 1 2 2 1 B r 2 )]TJ/F24 11.9552 Tf 11.955 0 Td [(2 2 1 B 2 where A and P 1 canbefoundbyusingEqs.5and5andwehave Z 2 / 2 1 atallvaluesof B at V =0 Nowhavingthesolutions 2 0 0 totheeigenvalueat V =0, 2 =0 and writing 2 = 2 0 + 2 1 and = 0 + 1 where 2 =0+ 2 1 77

PAGE 78

and Z =0+ Z 1 wehave C 1 )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 1 0 )]TJ/F27 11.9552 Tf 11.956 0 Td [( 2 0 1 = 1 r @ @ r r @ 1 @ r + 1 r 2 @ 2 1 @ 2 where 1 =0 at r =1 and 2 0 d 1 0 1 r d r =0 Andbecause Z 1 doesnotappear,weseethat 2 1 =0 and 1 isamultipleof 0 Thustoseetheeffectof 2 1 on 2 at V =0 wemustgotosecondorder. Wewrite 2 =0+ 2 1 Z =0+ Z 1 + 1 2 2 Z 2 2 = 2 0 + 1 2 2 2 2 and = 0 + 1 + 1 2 2 2 wherewehave Z 1 and 1 .Wedonotneed 2 and Z 2 doesnotappearbecause Z 0 =0. Hence,atsecondorder,the 2 2 2 problemis C 2 )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 0 2 )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 2 0 = 1 r @ @ r r @ 2 @ r + 1 r 2 @ 2 2 @ 2 )]TJ/F24 11.9552 Tf 11.956 0 Td [(3 1 r @ @ r rZ 2 1 r @ 0 @ r )]TJ/F24 11.9552 Tf 15.233 8.088 Td [(1 r 2 Z 2 1 r @ 2 0 @ 2 where 2 =0 at r =1 and 2 0 d 1 0 2 r d r =0 78

PAGE 79

Thenweobtain )]TJ/F27 11.9552 Tf 11.956 0 Td [( 2 2 0 r d r d = )]TJ/F24 11.9552 Tf 9.299 0 Td [(3 0 @ @ r rZ 2 1 r @ 0 @ r d r d )]TJ/F45 11.9552 Tf 11.955 16.272 Td [( 0 1 r Z 2 1 r @ 2 0 @ 2 d r d wheretheRHSofEq.5is 3 2 0 d @ 0 @ r 2 r =1 Z 2 1 r r =1+ 1 0 d r 1 r Z 2 1 r 2 0 @ 0 @ 2 d anditisgreaterthanzeronomatterthevalueof B at V =0 ,thoughtitdepends on B .Wealsohave 2 = 2 0 + 1 2 2 2 2 decreasingat V =0 ,forany B ,as 2 increasesfromzero.Thusat V =0, 2 = 0 wecanincrease B to 2 0 wheretolosestabilitybutat V =0, 2 =0 2 1 welosestabilitybeforewereach 2 0 .Anditappearsatrstthatspinningis destabilizing. Hencejustasspinningabridgecutsdownthestablerangeof L = R 0 at V = R 2 0 2 L,spinningadropcutsdownthestablerangeof B at V =0 .Nowifwe drawagraphof Z vs r at V =0 2 smalland B < B crit 2 =0 ,wehavethe Figure5-7wheretheshapeisrepresentedusingtheverticalcoordinate Z asa functionoftheradius r 79

PAGE 80

Figure5-7.Stableshapeofthespinningdroprotatingatsmall 2 withzero volumeand B < B crit 2 =0 Uponincreasing 2 wecanndtheshapegivenbyFigure5-8andsooner orlatertheshapebecomesunstableviz., 2 V =0, 2 B = B .InFigure5-8 thebaseshapeisalreadyunstable. 80

PAGE 81

Figure5-8.Unstableshapeofthespinningdroprotatingatlarge 2 withzero volumeand B < B crit 2 =0 Thussetting B and 2 suchthatthedropisstableat V =0 ,weincrease V andproducea V vs P curveat B andattwovaluesof 2 for B nottoosmallin Figure5-9.Thedropat 2 > 0 breaksupatgreatervolumethan 2 =0 does. 81

PAGE 82

Figure5-9. V vs P curvefor 2 =0 and 2 > 0 Thus 2 isstabilizing:thebifurcationpointfor 2 > 0 hasagreatervolume thanthecasein 2 =0 NowmultiplyingEq.5by r andintegratingover 0 < r < 1 wehave 1 2 P + 1 8 2 + V 2 = Z r + Z 2 r 1 = 2 j r =1 andwewishtoknowif Z r r =1=0 atcritical.ToseethiswedifferentiateEq. 5withrespectto r toobtain 2 r )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ r = 1 r d d r [ r + Z 2 r 3 = 2 d Z d r ] )]TJ/F24 11.9552 Tf 15.232 8.088 Td [(1 r 2 1 + Z 2 r 1 = 2 d Z d r 82

PAGE 83

andassumingacriticalpointisabifurcationpoint,viz., m =1 ,wehave,inwriting = ^ r cos )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 ^ = 1 r d d r [ r + Z 2 r 3 = 2 d ^ d r ] )]TJ/F24 11.9552 Tf 15.233 8.088 Td [(1 r 2 1 + Z 2 r 1 = 2 ^ where ^ =0 at r =1 Toobtain Z r atcriticalpointwhere 2 = B asolvabilityconditionmustbe satised,viz., 1 0 2 r ^ r d r = Z r 1 1+ Z 2 r ^ r at r =1 andseethat Z r r =1 isnotzerounless 2 =0 ,eventhough 2 > 0 doesnot causealossofsymmetry. 83

PAGE 84

CHAPTER6 CONCLUSIONSANDFUTUREWORK 6.1Conclusions Apendentdropandaliquidbridgearetwoexamplesofliquid-bodycongurationswhoseequilibriumshapesarisefrombalancesofdifferentpotential energies.Thependentdropissimplyaliquidbodythatissuspendedfroman openingwhiletheliquidbridgeistheliquidbodythatissuspendedbetweentwo parallelsolidsurfaces.Theseequilibriumbodies,bearmanysimilaritiesintheir stabilitycharacteristicsandthisthesisexploitsthesesimilaritiestomakepredictionsonthenatureandconditionsfortheinstability.Thestabilityofpendent dropsandclosedliquidbridgesareoftechnologicalaswellasscienticinterest. Forexample,pendentdropstabilityisanimportantfactorinthemeasurements ofinterfacialtensionbetweenliquidsusingdroptensiometers.Pendentdrop shapeandstabilityisalsoofimportanceinthestudyofmigrationofdropsalong hydrophobicsurfaces.Likewiseliquidbridgestabilityisveryimportantduringthe melt-zonereningofcompoundsemiconductorssuchasGalliumArsenideand GalliumSelenide.Inaddition,liquidbridgeinstabilityplaysacrucialroleinthe disruptionofcloggedalveoliinpre-matureinfants. Notwithstandingtheirimportanceintechnology,thestabilityofequilibrium liquidbodiesisalsoofscienticinterest.Forexample,ithasbeenobserved thatsymmetricbodiessuchashangingdropsandsessiledropsaswellas liquidbridgesmaybreakcatastrophicallyeithersymmetricallyaboutaplane orunsymmetricallyaboutaplane.Determiningthemannerofbreakupand thecausesforbreakuparesomeoftheobjectivesofthethesis.Whetherthe problemofinterestisapendentdroporaliquidbridgeitisshownthatvelocity perturbationsareofnoconsiderationunderneutralstability. 84

PAGE 85

Thestabilityofthependentdropdependsononlytwoscaledgroups.These arethescaledvolumeandtheBondnumberwhenthevolumeisacontrolled input.Whenpressureisthecontrolledinputtheonlyscaledgroupsarethe scaledpressureandtheBondnumber.Whenvolumeiscontrolledwendthat thedrophasamaximumvolumebeyondwhichitmustbreakcatastrophically. Andthepatternofbreakupwhereitpinchesoffatmaximumvolumeisalways symmetricaboutaplane.HoweverdependingontheBondnumberthedrop canalsobreakbeforethevolumereachesitsmaximum,whichmeansthat theupperboundonvolumeisnottheinstabilitylimitforthedropforallBond numbers,i.e.,thereexistsacriticalBondnumber,abovewhichthedropbreaks beforetheupperboundonvolumecanbereached.Inthissituationthepattern ofbreakupisantisymmetric.Whenpressureisthecontrolledinputwend analogousresultsandindeedapredictiononthepressurecontrolledexperiment maybemadefromthecalculationsthatpredictvolume-controlledinstability.The analysisinthisthesisalsoshowsthatthedropbreakupoccurseveninthecase ofactitiousone-dimensionaldropinaqualitativemannerasthedrophanging fromacircularopening.Thisobservationimmediatelyleadstotheconclusion thattransversecurvaturehas nothing todowiththebreakup.Thisiscontrasted withthebreakupofaliquidbridgewheretransversecurvatureisallimportant andtheonlycauseforthebreakup. Inthestudyofclosedliquidbridgeitisassumedthattheliquidinthebridge hasthesamedensityasitsambient.Theonlydimensionlessgroupsweneed arethescaledvolumeandthelength-to-end-plateradiusratio.Indeedthe squareofthelengthtoradiusinthebridgeplaysthesameroleasBondnumber inthedrop.Asinthedropitisfoundthatabridgecanalsobreakupeither symmetricallyorasymmetrically,dependingonwhetheritreachesitsminimum 85

PAGE 86

scaledvolumerstorifitpinchesoffbeforetheminimumscaledvolumeis reached. Inthecaseofthedropaswellasthebridge,itisshowninthethesis,that theuidcontactstheedgetangentiallyatthepointofunsymmetricbreakup. Thisresultleadstotheremarkableconclusionthattheunsymmetricbreakup point,whichisabifurcationpoint,maybepredictedwithoutthecalculationof alineareigenvalueproblem.Thisisperhapsoneofthemoststartlingofthe conclusionsofthestudy. Whileeigenvaluesneednotbeobtainedinordertocomputetheinstability pointsortoknowthenatureoftheinstability,muchmaystillbelearntfrom eigenvaluecomputations.Todothisacompanioneigenvalueproblemtermed thediffusioneigenvalueproblemisconsidered.Theeigenvaluesassociatedwith theneutralstabilityoftheequilibriumliquidproblemsareshowntobeordered withrespecttotheeigenvaluesofthecompaniondiffusioneigenvalues.This orderingalsosayshowthevolume-controlledandpressure-controlledproblems relatetooneanother.Itislearntthattheformerismorestablethanthelatter. Manyifnotalloftheaboveconclusioncanbeobtaineddirectlyfromsimple ideasinlinearalgebra. Thethesisconcludeswithastudyontheeffectofrotationonthestabilityof pendentdropsandliquidbridges.Asurprisingresultisthatrotationstabilizesthe formerwhileitdestabilizesthelatter.Thisconclusionisnotunqualied.Inthe caseoftherotatingdropthestabilizationoccursonlyforlargeenoughvolumes whileinthecaseoftheliquidbridgethedestabilizationistrueforallliquidbridge congurations. Thegeneralresultsobtainedinthisthesismaybespecictoequilibrium liquidcongurationsexempliedbypendentdropsandliquidbridges.However 86

PAGE 87

theanalysismaybeextendedtoalargerclassofproblems.Theresultsoffer insightintobreakupmechanisms,theroleofgeometricsymmetry,andstability. Inadditiontothevariousqualitativefeaturesonstabilitythedissertationalso displaysseveralcomputationalresults.Theeffectofrotationonliquidbridgesis showntobedifferentthanthatondrops.Aninterestingcomputationalmethod thatrequirespiecewiseanalysisinthecaseofequilibriumbodieswithderivative singularitiesisputforthinanappendix. 6.2FutureWork:ModelofaRotatingPendentDropSpinningatConstant AngularVelocity Thestabilityofpendentdropsandclosedliquidbridgesarealsoaffected bytherotationabouttheirverticalcenteraxis.Howeverrotationplaysavery differentroleonthestabilityofadropversusabridge.Itdestabilizesthedropat smallvolumesandstabilizesthedropwithlargevolumes.Inbridges,italways hasadestabilizingeffect.Thisobservationneedsfurtherstudy. 87

PAGE 88

APPENDIXA THEPATCHEDMETHOD Atsomepointwemayndsomebaseshapeswhichcannotbewritten as z = Z r andwewillthenneedtointroduce r = R z butwecannot domuchwithashapewrittenlikethisbecause R z isinniteat z = z CL and becauseour Z =0 solutioncannotbesowritten.Thuswemayneeddifferent independentvariablestowritethecorrespondingpartofthebaseshapeand patchthemtogethertoformacompleteshapeshowninFigureA-1.Forthistype ofbaseshapeweusedifferentindependentvariablestorepresenteachpartof theshapeandpatchtheseparttogether,whichistheso-calledpatchedmethod. Ifthecross-sectionis1-D,wedividethebaseshapeintoveparts,i.e.,part, ,,andinFigureA-1.Ifthecross-sectioniscircular,theprocesscan bedoneusingthreedifferentparts,i.e.,part,,,andinFigure A-1. FigureA-1.Theproleofdropbaseshapewhichcannotbeexpressedbyonly oneindependentvariable. Thenwemakethevariouswaysofexpressingthesurfacettogether.Each regionhasitsownbasenonlinearequation,stabilityproblem,eigenvalue problemandcorrespondingboundaryconditions. 88

PAGE 89

A.1PatchMethodforTheFirstExperimentwith1-DCross-section Thewholeproblemcanberepresentedby5domainsshowninFigureA-1.The basenonlinearequationsforall5partsare P )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ = d d x 1 + Z 2 x 1 = 2 d Z d x )]TJ/F24 11.9552 Tf 9.298 0 Td [(1 < x < )]TJ/F48 11.9552 Tf 9.298 0 Td [(x Part P )]TJ/F48 11.9552 Tf 11.955 0 Td [(Bz = )]TJ/F22 11.9552 Tf 13.779 8.088 Td [(d d z 1 + X 2 z 1 = 2 d X d z z < z < z Part P )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ = d d x 1 + Z 2 x 1 = 2 d Z d x )]TJ/F48 11.9552 Tf 9.298 0 Td [(x < x < x Part P )]TJ/F48 11.9552 Tf 11.955 0 Td [(Bz = )]TJ/F22 11.9552 Tf 13.779 8.088 Td [(d d z 1 + X 2 z 1 = 2 d X d z z < z < z Part P )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ = d d x 1 + Z 2 x 1 = 2 d Z d x x < x < 1 Part wherewealternatelyuse x and z asindependentvariablesinourcalculation. Thebaseboundaryandvolumeconditionsare x = 1: Z =0 positions: Z x = X z d X d z = 1 d Z d x and V = x )]TJ/F49 7.9701 Tf 6.587 0 Td [(x z )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z d x +2 z z X d z )]TJ/F24 11.9552 Tf 11.956 0 Td [(2 1 x Z d x )]TJ/F24 11.9552 Tf 11.956 0 Td [(2 z x 89

PAGE 90

wheretheinput V and B andtheoutputisthebasesolution.Theequationsfor stabilityproblemforall5partsare P 1 )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ 1 = d d x 1 + Z 2 0 x 3 = 2 d Z 1 d x )]TJ/F24 11.9552 Tf 9.299 0 Td [(1 < x < )]TJ/F48 11.9552 Tf 9.299 0 Td [(x Part P 1 = )]TJ/F22 11.9552 Tf 13.779 8.088 Td [(d d z 1 + X 2 0 z 3 = 2 d X 1 d z z < z < z Part P 1 )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ 1 = d d x 1 + Z 2 0 x 3 = 2 d Z 1 d x )]TJ/F48 11.9552 Tf 9.299 0 Td [(x < x < x Part P 1 = )]TJ/F22 11.9552 Tf 13.779 8.088 Td [(d d z 1 + X 2 0 z 3 = 2 d X 1 d z z < z < z Part P 1 )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ 1 = d d x 1 + Z 2 0 x 3 = 2 d Z 1 d x x < x < 1 Part where Z 0 and X 0 denotebasesolutionsand P 1 Z 1 and X 1 aretherst-order perturbationsof P Z and X ,respectively.Theboundaryandvolumeconditions forstabilityproblemare x = 1: Z 1 =0 positions: X 1 z 0 + d X 0 d z j z = z 0 Z 1 x 0 =0 d X 1 d z j z = z 0 + Z 1 d 2 X 0 d z 2 j z = z 0 = )]TJ/F49 7.9701 Tf 10.494 4.884 Td [(Z 1 x + X 1 Z 0 xx d Z 0 d x j x = x 0 2 and 0= )]TJ/F45 11.9552 Tf 11.291 16.273 Td [( x )]TJ/F49 7.9701 Tf 6.586 0 Td [(x Z 1 d x + z z X right 1 d z )]TJ/F45 11.9552 Tf 11.956 16.273 Td [( z z X left 1 d z )]TJ/F45 11.9552 Tf 11.955 16.273 Td [( )]TJ/F49 7.9701 Tf 6.586 0 Td [(x )]TJ/F25 7.9701 Tf 6.586 0 Td [(1 Z 1 d x )]TJ/F45 11.9552 Tf 11.955 16.273 Td [( 1 x Z 1 d x where X left 1 and X right 1 aretherst-orderperturbationsof X atregionand inFigureA-1,respectively. 90

PAGE 91

A.2PatchMethodforTheFirstExperimentwithCircularCross-section Thewholeproblemcanberepresentedby3domainsshowninFigureA-1.The basenonlinearequationsforall3partsincirculargeometryare P )]TJ/F48 11.9552 Tf 11.956 0 Td [(BZ = 1 + Z 2 r 3 = 2 [ )]TJ/F48 11.9552 Tf 10.494 8.088 Td [(Z r r )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z rr )]TJ/F48 11.9552 Tf 13.15 8.088 Td [(Z 3 r r ]0 < r < r Part P )]TJ/F48 11.9552 Tf 11.955 0 Td [(Bz = 1 + 1 R 2 z 3 = 2 [ )]TJ/F24 11.9552 Tf 18.409 8.087 Td [(1 R z R + R zz R 3 z )]TJ/F24 11.9552 Tf 21.473 8.087 Td [(1 R 3 z R ] z < z < z Part P )]TJ/F48 11.9552 Tf 11.956 0 Td [(BZ = 1 + Z 2 r 3 = 2 [ )]TJ/F48 11.9552 Tf 10.494 8.087 Td [(Z r r )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z rr )]TJ/F48 11.9552 Tf 13.15 8.087 Td [(Z 3 r r ] r < r < 1 Part where r and z arechosentobetheindependentvariablesalternately.Thebase boundaryandvolumeconditionsare r =1: Z =0 positions: Z r = R z d R d z = 1 d Z d r and V =2 r 0 z )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z r d r + z z R 2 d z )]TJ/F24 11.9552 Tf 11.956 0 Td [(2 1 r Zr d r )]TJ/F27 11.9552 Tf 11.955 0 Td [( r 2 z wheretheinput V and B andtheoutputisthebasesolution.Theequationsfor stabilityproblemforall3partsincirculargeometryare m =0: P 1 )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ 1 = Z 1 rr + Z 1 r r + Z 2 0 r 3 = 2 )]TJ/F25 7.9701 Tf 13.15 4.884 Td [(3 Z 0 rr Z 0 r Z 1 r + Z 2 0 r 5 = 2 0 < r < r Part P 1 = 1 + R 2 0 z 3 = 2 [ )]TJ/F48 11.9552 Tf 9.298 0 Td [(R 1 zz )]TJ/F49 7.9701 Tf 13.151 4.884 Td [(R 0 z R 1 z R 0 )]TJ/F49 7.9701 Tf 13.469 4.884 Td [(R 1 R 2 0 )]TJ/F49 7.9701 Tf 13.15 6.251 Td [(R 2 0 z R 1 R 2 0 + 3 R 0 z R 1 z R 0 zz 1+ R 2 0 z ] z < z < z Part P 1 )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ 1 = Z 1 rr + Z 1 r r + Z 2 0 r 3 = 2 )]TJ/F25 7.9701 Tf 13.15 4.884 Td [(3 Z 0 rr Z 0 r Z 1 r + Z 2 0 r 5 = 2 r < r < 1 Part 91

PAGE 92

m 6 =0: )]TJ/F48 11.9552 Tf 9.299 0 Td [(BZ 1 = Z 1 rr + Z 1 r r )]TJ/F50 5.9776 Tf 7.782 4.394 Td [(m 2 Z 1 r 2 )]TJ/F49 7.9701 Tf 6.586 0 Td [(Z 2 0 r m 2 Z 1 r 2 + Z 2 0 r 3 = 2 )]TJ/F25 7.9701 Tf 13.15 4.883 Td [(3 Z 0 rr Z 0 r Z 1 r + Z 2 0 r 5 = 2 0 < r < r Part 0= 1 + R 2 0 z 3 = 2 [ )]TJ/F48 11.9552 Tf 9.298 0 Td [(R 1 zz )]TJ/F49 7.9701 Tf 13.15 4.883 Td [(R 0 z R 1 z R 0 )]TJ/F49 7.9701 Tf 13.469 4.883 Td [(R 1 R 2 0 )]TJ/F49 7.9701 Tf 13.151 6.25 Td [(R 2 0 z R 1 R 2 0 + m 2 R 1 + R 2 0 z R 2 0 + 3 R 0 z R 1 z R 0 zz 1+ R 2 0 z ] z < z < z Part )]TJ/F48 11.9552 Tf 9.299 0 Td [(BZ 1 = Z 1 rr + Z 1 r r )]TJ/F50 5.9776 Tf 7.782 4.395 Td [(m 2 Z 1 r 2 )]TJ/F49 7.9701 Tf 6.586 0 Td [(Z 2 0 r m 2 Z 1 r 2 + Z 2 0 r 3 = 2 )]TJ/F25 7.9701 Tf 13.15 4.884 Td [(3 Z 0 rr Z 0 r Z 1 r + Z 2 0 r 5 = 2 r < r < 1 Part where Z 0 and R 0 representbasesolutionandtherst-orderperturbationsare Z 1 cos m and R 1 cos m sothat m =0 and m 6 =0 denoteaxisymmetricand nonaxisymmetricperturbations,respectively.Theboundaryandvolume conditionsforstabilityproblemare m =0: r =1: Z 1 =0 positions: Z 1 r 0 + d Z 0 d r j z 0 R 1 z 0 =0 Z 1 r + R 1 Z 0 rr Z 2 0 r + R 1 z =0 and 0= r 0 Z 1 r 0 d r 0 )]TJ/F45 11.9552 Tf 11.955 16.273 Td [( z z R 0 R 1 d z + 1 r Z 1 r 0 d r 0 m 6 =0: r =1: Z 1 =0 positions: Z 1 r 0 + d Z 0 d r j r = r 0 R 1 z 0 =0 Z 1 r + R 1 Z 0 rr Z 2 0 r + R 1 z =0 92

PAGE 93

A.3PatchMethodforTheSecondExperimentwith1-DCross-section Thewholeproblemcanberepresentedby5domainsshowninFigureA-1.The basenonlinearequationsforall5partsinthesecondexperimentare P m + C 1 2 [ )]TJ/F45 11.9552 Tf 11.291 9.631 Td [( x )]TJ/F49 7.9701 Tf 6.586 0 Td [(x z )]TJ/F48 11.9552 Tf 11.956 0 Td [(Z d x )]TJ/F24 11.9552 Tf 11.956 0 Td [(2 z z X d z +2 1 x Z d x +2 z x ] )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ = d d x 1 + Z 2 x 1 = 2 d Z d x )]TJ/F24 11.9552 Tf 11.955 0 Td [(1 < x < )]TJ/F48 11.9552 Tf 9.298 0 Td [(x Part P m + C 1 2 [ )]TJ/F45 11.9552 Tf 11.291 9.63 Td [( x )]TJ/F49 7.9701 Tf 6.586 0 Td [(x z )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z d x )]TJ/F24 11.9552 Tf 11.955 0 Td [(2 z z X d z +2 1 x Z d x +2 z x ] )]TJ/F48 11.9552 Tf 11.955 0 Td [(Bz = )]TJ/F22 7.9701 Tf 12.79 4.707 Td [(d d z 1 + X 2 z 1 = 2 d X d z z < z < z Part P m + C 1 2 [ )]TJ/F45 11.9552 Tf 11.291 9.631 Td [( x )]TJ/F49 7.9701 Tf 6.586 0 Td [(x z )]TJ/F48 11.9552 Tf 11.956 0 Td [(Z d x )]TJ/F24 11.9552 Tf 11.956 0 Td [(2 z z X d z +2 1 x Z d x +2 z x ] )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ = d d x 1 + Z 2 x 1 = 2 d Z d x )]TJ/F48 11.9552 Tf 11.956 0 Td [(x < x < x Part P m + C 1 2 [ )]TJ/F45 11.9552 Tf 11.291 9.631 Td [( x )]TJ/F49 7.9701 Tf 6.586 0 Td [(x z )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z d x )]TJ/F24 11.9552 Tf 11.955 0 Td [(2 z z X d z +2 1 x Z d x +2 z x ] )]TJ/F48 11.9552 Tf 11.955 0 Td [(Bz = )]TJ/F22 7.9701 Tf 12.79 4.707 Td [(d d z 1 + X 2 z 1 = 2 d X d z z < z < z Part P m + C 1 2 [ )]TJ/F45 11.9552 Tf 11.291 9.631 Td [( x )]TJ/F49 7.9701 Tf 6.586 0 Td [(x z )]TJ/F48 11.9552 Tf 11.956 0 Td [(Z d x )]TJ/F24 11.9552 Tf 11.956 0 Td [(2 z z X d z +2 1 x Z d x +2 z x ] )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ = d d x 1 + Z 2 x 1 = 2 d Z d x x < x < 1 Part wherethevolumeexpression )]TJ/F45 11.9552 Tf 11.291 16.273 Td [( x )]TJ/F49 7.9701 Tf 6.586 0 Td [(x z )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z d x )]TJ/F24 11.9552 Tf 11.955 0 Td [(2 z z X d z +2 1 x Z d x +2 z x = )]TJ/F48 11.9552 Tf 9.298 0 Td [(V and V isthevolumeofdropinthisexperiment. C isaninputconstantandis chosentobegreaterthan 2 here.Thebaseboundaryconditionsare x = 1: Z =0 positions: Z x = X z d X d z = 1 d Z d x 93

PAGE 94

wheretheinputis P m andtheoutputisthebasesolution.Theequationsfor stabilityproblemforall5partsinexperimentIIare C 1 2 [ x )]TJ/F49 7.9701 Tf 6.587 0 Td [(x Z 1 d x )]TJ/F45 11.9552 Tf 11.956 9.631 Td [( z z X right 1 d z + z z X left 1 d z + )]TJ/F49 7.9701 Tf 6.586 0 Td [(x )]TJ/F25 7.9701 Tf 6.587 0 Td [(1 Z 1 d x + 1 x Z 1 d x ] )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ 1 = d d x 1 + Z 2 0 x 3 = 2 d Z 1 d x )]TJ/F24 11.9552 Tf 11.956 0 Td [(1 < x < )]TJ/F48 11.9552 Tf 9.299 0 Td [(x Part C 1 2 [ x )]TJ/F49 7.9701 Tf 6.586 0 Td [(x Z 1 d x )]TJ/F45 11.9552 Tf 11.955 9.63 Td [( z z X right 1 d z + z z X left 1 d z + )]TJ/F49 7.9701 Tf 6.586 0 Td [(x )]TJ/F25 7.9701 Tf 6.587 0 Td [(1 Z 1 d x + 1 x Z 1 d x ] = )]TJ/F22 7.9701 Tf 12.791 4.707 Td [(d d z 1 + X 2 0 z 3 = 2 d X 1 d z z < z < z Part C 1 2 [ x )]TJ/F49 7.9701 Tf 6.587 0 Td [(x Z 1 d x )]TJ/F45 11.9552 Tf 11.956 9.631 Td [( z z X right 1 d z + z z X left 1 d z + )]TJ/F49 7.9701 Tf 6.586 0 Td [(x )]TJ/F25 7.9701 Tf 6.587 0 Td [(1 Z 1 d x + 1 x Z 1 d x ] )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ 1 = d d x 1 + Z 2 0 x 3 = 2 d Z 1 d x )]TJ/F48 11.9552 Tf 11.955 0 Td [(x < x < x Part C 1 2 [ x )]TJ/F49 7.9701 Tf 6.586 0 Td [(x Z 1 d x )]TJ/F45 11.9552 Tf 11.955 9.631 Td [( z z X right 1 d z + z z X left 1 d z + )]TJ/F49 7.9701 Tf 6.586 0 Td [(x )]TJ/F25 7.9701 Tf 6.587 0 Td [(1 Z 1 d x + 1 x Z 1 d x ] = )]TJ/F22 7.9701 Tf 12.791 4.707 Td [(d d z 1 + X 2 0 z 3 = 2 d X 1 d z z < z < z Part C 1 2 [ x )]TJ/F49 7.9701 Tf 6.587 0 Td [(x Z 1 d x )]TJ/F45 11.9552 Tf 11.956 9.631 Td [( z z X right 1 d z + z z X left 1 d z + )]TJ/F49 7.9701 Tf 6.586 0 Td [(x )]TJ/F25 7.9701 Tf 6.587 0 Td [(1 Z 1 d x + 1 x Z 1 d x ] )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ 1 = d d x 1 + Z 2 0 x 3 = 2 d Z 1 d x x < x < 1 Part where Z 0 X 0 denotebasesolutionsand Z 1 X 1 aretherst-orderperturbations of Z and X ,respectively. X left 1 X right 1 aretherst-orderperturbationsof X at regionandinFigureA-1.Theboundaryandvolumeconditionsforstability problemare x = 1: Z 1 =0 positions: X 1 z 0 + d X 0 d z j z = z 0 Z 1 x 0 =0 d X 1 d z j z = z 0 + Z 1 d 2 X 0 d z 2 j z = z 0 = )]TJ/F49 7.9701 Tf 10.494 4.883 Td [(Z 1 x + X 1 Z 0 xx d Z 0 d x j x = x 0 2 A.4PatchMethodforTheSecondExperimentwithCircularCross-section Thewholeproblemcanberepresentedby3domainsshowninFigureA-1.The basenonlinearequationsforall3partsinthesecondexperimentwithcircular 94

PAGE 95

geometryare P m + C 2 [ )]TJ/F24 11.9552 Tf 9.298 0 Td [(2 r 0 z )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z r d r )]TJ/F45 11.9552 Tf 11.955 9.63 Td [( z z R 2 d z +2 1 r Zr d r + r 2 z ] )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ = 1 + Z 2 r 3 = 2 [ )]TJ/F49 7.9701 Tf 10.494 4.707 Td [(Z r r )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z rr )]TJ/F49 7.9701 Tf 13.15 5.699 Td [(Z 3 r r ]0 < r < r Part P m + C 2 [ )]TJ/F24 11.9552 Tf 9.299 0 Td [(2 r 0 z )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z r d r )]TJ/F45 11.9552 Tf 11.955 9.63 Td [( z z R 2 d z +2 1 r Zr d r + r 2 z ] )]TJ/F48 11.9552 Tf 11.955 0 Td [(Bz = 1 + 1 R 2 z 3 = 2 [ )]TJ/F25 7.9701 Tf 16.259 4.707 Td [(1 R z R + R zz R 3 z )]TJ/F25 7.9701 Tf 19.174 4.707 Td [(1 R 3 z R ] z < z < z Part P m + C 2 [ )]TJ/F24 11.9552 Tf 9.298 0 Td [(2 r 0 z )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z r d r )]TJ/F45 11.9552 Tf 11.955 9.631 Td [( z z R 2 d z +2 1 r Zr d r + r 2 z ] )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ = 1 + Z 2 r 3 = 2 [ )]TJ/F49 7.9701 Tf 10.494 4.707 Td [(Z r r )]TJ/F48 11.9552 Tf 11.955 0 Td [(Z rr )]TJ/F49 7.9701 Tf 13.151 5.698 Td [(Z 3 r r ] r < r < 1 Part Thevolumeofdropisdenedas V =2 r 0 z )]TJ/F48 11.9552 Tf 11.956 0 Td [(Z rdr + z z R 2 dz )]TJ/F24 11.9552 Tf 11.956 0 Td [(2 1 r Zrdr )]TJ/F27 11.9552 Tf 11.956 0 Td [( r 2 z Thebaseboundaryandvolumeconditionsare r =1: Z =0 positions: Z r = R z d R d z = 1 d Z d r Theequationsforstabilityproblemare C [ r 0 Z 1 r 0 d r 0 )]TJ/F45 11.9552 Tf 11.955 9.63 Td [( z z R 0 R 1 d z + 1 r Z 1 r 0 d r 0 ] )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ 1 = Z 1 rr + Z 1 r r )]TJ/F50 5.9776 Tf 7.782 4.394 Td [(m 2 Z 1 r 2 )]TJ/F49 7.9701 Tf 6.586 0 Td [(Z 2 0 r m 2 Z 1 r 2 + Z 2 0 r 3 = 2 )]TJ/F25 7.9701 Tf 13.151 4.884 Td [(3 Z 0 rr Z 0 r Z 1 r + Z 2 0 r 5 = 2 0 < r < r Part C [ r 0 Z 1 r 0 d r 0 )]TJ/F45 11.9552 Tf 11.955 9.631 Td [( z z R 0 R 1 d z + 1 r Z 1 r 0 d r 0 ] = 1 + R 2 0 z 3 = 2 [ )]TJ/F48 11.9552 Tf 9.298 0 Td [(R 1 zz )]TJ/F49 7.9701 Tf 13.151 4.883 Td [(R 0 z R 1 z R 0 )]TJ/F49 7.9701 Tf 13.468 4.883 Td [(R 1 R 2 0 )]TJ/F49 7.9701 Tf 13.151 6.25 Td [(R 2 0 z R 1 R 2 0 + m 2 R 1 + R 2 0 z R 2 0 + 3 R 0 z R 1 z R 0 zz 1+ R 2 0 z ] z < z < z Part C [ r 0 Z 1 r 0 d r 0 )]TJ/F45 11.9552 Tf 11.955 9.631 Td [( z z R 0 R 1 d z + 1 r Z 1 r 0 d r 0 ] )]TJ/F48 11.9552 Tf 11.955 0 Td [(BZ 1 = Z 1 rr + Z 1 r r )]TJ/F50 5.9776 Tf 7.782 4.394 Td [(m 2 Z 1 r 2 )]TJ/F49 7.9701 Tf 6.586 0 Td [(Z 2 0 r m 2 Z 1 r 2 + Z 2 0 r 3 = 2 )]TJ/F25 7.9701 Tf 13.15 4.883 Td [(3 Z 0 rr Z 0 r Z 1 r + Z 2 0 r 5 = 2 r < r < 1 Part 95

PAGE 96

where Z 0 and R 0 representbasesolutionandtherst-orderperturbationsare Z 1 cos m and R 1 cos m sothat m =0 and m 6 =0 denoteaxisymmetricand nonaxisymmetricperturbations,respectively.Theboundaryandvolume conditionsforstabilityproblemare r =1: Z 1 =0 positions: Z 1 r 0 + d Z 0 d r j r = r 0 R 1 z 0 =0 Z 1 r + R 1 Z 0 rr Z 2 0 r + R 1 z =0 96

PAGE 97

APPENDIXB THEDIFFUSIONEIGENVALUEPROBLEM B.1TheDiffusionEigenvalueProblemforaPendentDrop Toseewhatwecanlearnaboutdropssuspendedfromarbitrarycross sectionswiththeboundarydenotedby C ,weintroduceaneigenvalueproblem dependingonlyon Z x y nomatterhow Z isobtained.Hencewehavethe followingeigenvalueproblemEqs.BandB L + 2 =0 B where =0 on C B Now L isself-adjointandwehave A L d A = )]TJ/F45 11.9552 Tf 9.298 16.273 Td [( A d A r I + r Z r Z )-222(r Z r Z + r Z r Z 3 = 2 r = )]TJ/F45 11.9552 Tf 9.298 16.272 Td [( A jr j 2 + jr r Z j 2 + r Z r Z 3 = 2 d A Hencewedenoteoursolutions 0 < 2 1 < 2 2 < 2 3 1 > 0, 2 3, andobservethattheydependon Z andwedenotetheintegralsofthe s ,viz., A d A ,by I where I 1 > 0 Webeginbyassumingthattherstthree I s arezeroattheleastsomeof themandwesaythatourcross-sectionhaslessandlesssymmetryasmoreand moreofthe I s arenotzero.Assumingmoreofthe I s arezero,neitherthevalueof C inEq3northevalueof A d A inEq4canbezero. 97

PAGE 98

NowforexperimentIwesolveEq3byexpanding inthe s ,assuming A 2 i d x d y =1 ,andthuswehave = X c i i c i = A i d A whereuponwend c i = CI i 2 )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 i andweobtain = X c i i = C X I i 2 )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 i i Because C isnotzeroandbecause d x d y iszero,wecanobtainthe s by solving X I 2 i 2 )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 i =0 B Ifwecalculatethevalueoftheleft-hand-sideofEq.Bandplotitsprolein thegraphshowninFigureB-1.TheonlyparametersneededtodrawFigureB-1 canbeobtainedsimplybycarryingoutthecalculationtothediffusioneigenvalue problemEqs.BandB.Whichisveryhelpfultopredictthepointof stabilityforthependentdrop.Thisishowitworks:ateachzero,Eq.Bis satisedandthusweobtainalltheeigenvaluesfortheeigenvalueproblemEqs. 3,3and3,whichcanbeusedtohelpusunderstandthestability ofthedropfortherstexperiment. 98

PAGE 99

FigureB-1.Theproleoftheleft-hand-sideofEq.B,asafunctionof 2 wethenderivefromFigure.B-1 0 < 2 1 < 2 1 < 2 2 < 2 2 < 2 3 anorderingthatholdsforall B and V wherewecanobtain Z Havingsolvedthersteigenvalueproblemfor Z and P given B and V ,we turntothesecondexperimentandset P sothat P )]TJ/F48 11.9552 Tf 12.016 0 Td [(B V A hasthesamevalueas P intherstexperiment.Thenatthesamevalueof B wehavethesame Z and V ,andthereforethe 2 i s and 2 i s andwesolvethesecondeigenvalueproblem,Eq 3,againexpanding inthe s .Doingthisweobtain c i = I i 2 )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 i B 1 A A d A 99

PAGE 100

whereuponourequationfor 2 isnow X I 2 i 2 )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 i = 1 B B Thevalueofleft-hand-sideiscalculatedandrecordedinFigureB-2asafunction of 2 .Ateachvaluewheretheleft-hand-sideisequalto 1 B ,Eq.Bissatised andthusweobtainalltheeigenvaluesfortheeigenvalueproblemofthesecond experimentEq.4,whichcanbeusedtohelpusunderstandthestabilityof thedropforthesecondexperiment. FigureB-2.Theproleoftheleft-hand-sideofEq.B,asafunctionof 2 Againwend 0 < 2 1 < 2 1 < 2 2 < 2 2 < andweconclude: 2 i II < 2 i I 100

PAGE 101

Atlargevaluesof B : 2 i II 2 i I ,atsmallvaluesof B : 2 i II 2 i Because B mustbegreaterthan B wemaywishtoknowwhether 2 1 is greaterorlessthan B .If B isgreaterthan 2 2 2 1 islessthan B Tondoutifadenitedropshape, Z x y B V ,isstabletoasmall perturbationwemustdecidewhetherornotthehomogeneousproblems,Eqs 3,3,3,hasasolution Z 1 P 1 otherthanzero.Ifitdoes, V isatits criticalvaluegiventhevalueof B .Fromtheperspectiveofeigenvalueproblems, ifwesetavalueof B ,theshapes Z x y B V or Z x y B P arestableto smallperturbationssolongas B isnotoneoftheeigenvalues 2 i I B V or 2 i II B P Inbothexperimentsweonlyneed I 1 6 =0, I 2 6 =0 tokeep 2 1 and 2 2 apartand thedropisalwaysunstabletoa 1 mode.Butifthecross-sectionisarectangle oranellipsethesymmetryof Z andthusthesymmetryof 2 wouldmake I 2 =0 Asaresult,wecanalwayshaveonesolutiontotheeigenvalueproblemEqs. 3,3and3denotedas 2 atany B and V suchthat 2 = 2 2 = 2 A 1 d x d y =0, C =0 B Hencecalculating P i =1,3,... I 2 i 2 )]TJ/F28 7.9701 Tf 6.587 0 Td [( 2 i andplottingitsvalueasafunctionof 2 ,neglectingtheterncorrespondingto i =2, wendasolution 2 lyingbetween 2 1 2 3 showninFigureB-3butwedonotknowwhere 2 2 is.InFigureB-3 2 1 and 2 3 are twodistinctasymptotesbetweenwhichliesaeigenvalueproblemsolution 2 .It isalsofoundthatone 2 ispinnedon 2 2 whiletheothermaylietoitsleftorright, andwhetheritliesleftorrightmaydependon B 101

PAGE 102

FigureB-3.Valueof P i =1,3,... I 2 i 2 )]TJ/F28 7.9701 Tf 6.586 0 Td [( 2 i asafunctionof 2 Thusthesymmetrysolutiontotheeigenvalueproblemofadroppinned toacurveboundingasymmetriccross-sectioncanbeobtainedbysolvingthe correspondingdiffusioneigenvalueproblemandweexpecttondcross-sections forwhich I 2 = A 1 d x d y =0 andhenceoneofthesolutionstotheeigenvalue problem,Eqs.3,3and3andEqs.4and4is 2 = 2 2 = 2 A d x d y = C =0 Theeigenvaluesweareinterestedinarethesmallestandthenextsmallest,one ofwhichispinnedon 2 2 andtheotherliesbetweentheinterval 2 1 2 3 .The correspondingeigenfunctionshaveeithersymmetricorantisymmetricshapes 102

PAGE 103

andtheoneassociatedwithsmallesteigenvaluecanbeusedtorepresentthe patternatdropbreakup. B.2TheDiffusionEigenvalueProblemforaClosedLiquidBridge ForliquidbridgewithgivenLand V andthecorrespondingsolutiontothe bridgeshapeproblem R 0 z and P 0 weintroduceaneigenvalueproblemas 1 R 0 d d z [ R 0 + R 2 0 z 3 = 2 z ]+ 1 R 2 0 1 + R 2 0 z + 2 =0 B where =0 at z = LB anddenoteitssolutions 2 1 2 2 ,... B 1 2 ,... B Whereuponwend L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L [ R 0 + R 2 0 z 3 = 2 2 z d z ] )]TJ/F45 11.9552 Tf 11.955 16.272 Td [( L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L 1 R 2 0 R 0 + R 2 0 z 1 = 2 2 d z = 2 L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L R 0 2 d z B Thersttermontheright-hand-sideofEq.Bderivesfromlongitudinalcurvature,whilethesecondfromtransversecurvatureandiftransversecurvatureis strongenoughatleastone 2 canbenegative. Nowifforany wehave I = L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L R 0 d z =0 B Thenthereisa and 2 suchthat = 2 = 2 and C =0 .Otherwisefor those 'ssuchthat I 6 =0 andthose 'ssuchthat C 6 =0 ,weassume L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L R 0 2 d z =1 B 103

PAGE 104

andweexpandthesolutionstoeigenvalueproblemfortheliquidbridgeEqs. 5,5and5insetoffunctions i .Thuswehave = X c i i c i = L )]TJ/F22 7.9701 Tf 6.586 0 Td [(L R 0 i d z andmultiplyingEq.5by R 0 i ,Eq.Bby R 0 ,substitutingandintegratingover )]TJ/F22 11.9552 Tf 9.299 0 Td [(L z Lwend CI i = 2 )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 i c i B andfor C 6 =0 ,using L )]TJ/F22 7.9701 Tf 6.587 0 Td [(L R 0 d z =0 thenweobtain X I 2 i 2 )]TJ/F27 11.9552 Tf 11.955 0 Td [( 2 i =0 B andequationforthe 2 'swhosesolutionsareillustratedasfollowsinFigure B-4.TheonlyparametersneededtodrawFigureB-4canbeobtainedsimplyby carryingoutthecalculationtothediffusioneigenvalueproblemEqs.Band Bwhichisveryhelpfultopredictthepointofstabilityfortheclosedliquid bridge.Thisishowitworks:ateachzero,Eq.Bissatisedandthuswe obtainalltheeigenvaluesfortheeigenvalueproblemfortheliquidbridgeEqs. 5,5and5,whichcanbeusedtohelpusunderstandthestability oftheclosedliquidbridge. 104

PAGE 105

FigureB-4.Theproleoftheleft-hand-sideofEq.B,asafunctionof 2 Ifweintroducegravitytotheproblem,viz., )778(! g = )]TJ/F22 11.9552 Tf 9.299 0 Td [(g )778(! k ,wewilladdaterm whichis )]TJ/F25 7.9701 Tf 10.494 5.479 Td [( g )]TJ/F28 7.9701 Tf 6.586 0 Td [( z totheleft-hand-sideofEq.5andarriveatthispointwith noobviouschange.However R z wouldnotbesymmetricabout z =0 and mostlikelynoneofthe 'swillintegratetozero, 2 1 willremainalwaysbetween 2 1 2 2 2 2 willremainalwaysbetween 2 2 2 3 ,etc.Thebridgewillbreakinthe patternof 1 andtheproblemwillhavelittleinterest. 105

PAGE 106

APPENDIXC RESULTSFOR B =0 At B =0, i.e.,g =0 ,for1-Dcross-sectionwehavethebaseas P = d d x Z x + Z 2 x 1 = 2 C IntegrateEq.Cwithrespectto x ,weobtain Px = Z x + Z 2 x 1 = 2 C Hence x 2 Z 2 x + x 2 = 1 P 2 C Sinceweknowforacircle Z 2 + x 2 = a where a isaconstant,ifwecalculate derivativeonbothsideswithrespectto x ,wehave ZZ x + x =0 Thus x 2 = Z 2 Z 2 x andhenceweknowthesolutiontoEq.Cispartofacirclewithradius 1 P .For circularcross-sectionif B =0 wehavesimilarresults.Thesecirclesolutions arealwaysstabletosmallperturbationsnomatterthevalueof V .Thus P and V canbeobtainedintermsoftheradiusofthecircleandour V vs P graphat B =0 isthenshowninFigureC-1wherethereisamaximum P beyondwhich nosolutiontothebaseproblemEqs.3,3and3. P =1 isthe maximumvalueforabasesolutiontoexistandwecanproveanalyticallyitisthe maximum. 106

PAGE 107

FigureC-1. V vs P curvefor B =0 107

PAGE 108

APPENDIXD DOESTHEVELOCITYPERTURBATIONVANISHATCRITICAL? TheFigure.D-1belowisasimpliedpictureoftherstexperiment. FigureD-1.TheFirstExperiment:volumecontrolexperiment. Thereisnolightuid.Noting z = Z x y t ournonlinearequationsare @ v @ t + v r v = r T )]TJ/F27 11.9552 Tf 11.955 0 Td [( r and n = k )-222(r Z + r Z r Z 1 = 2 k r Z =0, r v =0 onthedomain.Atallwallswehave n v =0 .Along z = Z x y t wehave n v = Z t + r Z r Z 1 = 2 nn :T+ 2 H =0 and tn :T=0 anytangent 108

PAGE 109

Thebasesolutionis Z = Z 0 x y v = 0 d p 0 d z = )]TJ/F27 11.9552 Tf 9.298 0 Td [( g p 0 = )]TJ/F27 11.9552 Tf 9.298 0 Td [( gz + const and T 0 = )]TJ/F48 11.9552 Tf 9.299 0 Td [(p 0 I Nowasmallperturbationisintroducedandwewrite Z = Z 0 + Z 1 andwehave n = n 0 + n 1 where n 0 = k )-222(r Z 0 + r Z 0 r Z 0 1 = 2 and n 1 = I + r Z 0 r Z 0 + k )-222(r Z 0 + r Z 0 r Z 0 3 = 2 r Z 1 Along z = Z 0 x y wehave n 0 v z :T 1 + gZ 1 + 2 H 1 =0 t 0 n 0 :T 1 =0 and n 0 v 1 = Z 1 t + r Z 0 r Z 0 1 = 2 Andonthereferencedomainwehave @ v 1 @ t = r T 1 r v 1 =0 109

PAGE 110

Duetoatconstantvolumewehave A Z 1 d x d y =0 D Weassumeasolution v x y z t = e t v x y z etcandthuswehave Z 1 t = Z 1 and v 1 = r T 1 r v 1 =0 whereuponwendupondenotingacomplexconjugateby v 1 v 1 = r T 1 v 1 )]TJ/F24 11.9552 Tf 11.955 0 Td [(2 D:D andintegratingoverthereferencedomainweobtain V 0 v 1 v 1 d V 0 = )]TJ/F45 11.9552 Tf 9.299 16.272 Td [( S 0 d A 0 n 0 T 1 v 1 )]TJ/F24 11.9552 Tf 11.956 0 Td [(2 V 0 D:D d V 0 where n 0 isinwardandwhere S 0 denotesthesurfaceofthereferencedrop. Atthesurface, v 1 denesitsowntangentandwewrite v 1 = n 0 n 0 v 1 + t 0 t 0 v 1 whereuponweobtain V 0 v 1 v 1 d V 0 = S 0 d A 0 2 H 1 + g Z 1 Z 1 + r Z 0 r Z 0 1 = 2 )]TJ/F24 11.9552 Tf 11.955 0 Td [(2 V 0 D 1 :D 1 d V 0 Likewisewehave V 0 v 1 v 1 d V 0 = S 0 d A 0 2 H 1 + g Z 1 Z 1 + r Z 0 r Z 0 1 = 2 )]TJ/F24 11.9552 Tf 11.956 0 Td [(2 V 0 D 1 :D 1 d V 0 110

PAGE 111

where S 0 d A 0 1 + r Z 0 r Z 0 1 = 2 [ 2 H 1 + g Z 1 Z 1 ]= A d x d y [ 2 H 1 + gZ 1 Z 1 ] Whatwewishtoknowis:if Re =0= Re is j v 1 j =0 ?Nowwehave 2 H 1 = L Z 1 whereuponwecanwrite Im V 0 j v 1 j 2 d V 0 = A d x d y L Z 1 + g Z 1 )]TJ/F55 11.9552 Tf 9.299 0 Td [(Im Z 1 )]TJ/F24 11.9552 Tf 11.956 0 Td [(2 V 0 d V 0 D 1 :D 1 Im [ V 0 j v 1 j 2 d V 0 + A d x d y Z 1 L Z 1 + g j Z 1 j 2 = )]TJ/F24 11.9552 Tf 9.299 0 Td [(2 V 0 d V 0 D 1 :D 1 where Z 1 L Z 1 ispositive.Likewisewehave )]TJ/F55 11.9552 Tf 9.298 0 Td [(Im [ V 0 v 1 2 d V 0 + A d x d y Z 1 L Z 1 + g j Z 1 j 2 = )]TJ/F24 11.9552 Tf 9.298 0 Td [(2 V 0 d V 0 D 1 :D 1 where Z 1 L Z 1 = Z 1 L Z 1 Thusweconclude Im =0 when Re =0 andhence v 1 mustbezeroat critical,i.e.,at =0 .Andthisisthereasonwhyweonlylookathydrostatic perturbations. 111

PAGE 112

APPENDIXE THEMETHODOFFROBENIUSTODERIVEESTIMATESOFTHEFIRST ROOTOFJ 0 R TheexpansionofJ 0 is J 0 r =1+ a 2 r 2 + a 4 r 4 + a 6 r 6 + E where a 0 =1, a 2 = )]TJ/F24 11.9552 Tf 10.494 8.088 Td [(1 4 2 a 4 = )]TJ/F24 11.9552 Tf 13.632 8.087 Td [(3 16 2 Z r =0+ 1 64 4 E a 6 = )]TJ/F24 11.9552 Tf 10.494 8.088 Td [(1 8 2 [ 1 4 Z r =0+ 1 3 Z r =0 Z r =0]+ 1 48 4 Z r =0 )]TJ/F24 11.9552 Tf 21.777 8.088 Td [(1 2304 6 E andanestimateoftherstpositiverootis 1 z 6 1 = )]TJ/F24 11.9552 Tf 9.299 0 Td [(3 a 6 +3 a 2 a 4 )]TJ/F48 11.9552 Tf 11.955 0 Td [(a 3 2 E 112

PAGE 113

REFERENCES [1]Whichresolutionissuitablefordropshapeanalyticalsystem. http://www.uskino.com/news/50.html [2]ShengtaiLiandHuiLi.Parallelamrcodeforcompressiblemhdorhd equations. LosAlamosNationalLaboratory.Retrieved ,pages09,2006. [3]HuePLe.Progressandtrendsinink-jetprintingtechnology. Journalof ImagingScienceandTechnology ,42:49,1998. [4]WJHeidegerandMWWright.Liquidextractionduringdropformation:effect offormationtime. AIChEjournal ,32:1372,1986. [5]JamesWJenningsJrandNRPallas.Anefcientmethodforthedeterminationofinterfacialtensionsfromdropproles. Langmuir ,4:959, 1988. [6]YRotenberg,LrBoruvka,andAWNeumann.Determinationofsurface tensionandcontactanglefromtheshapesofaxisymmetricuidinterfaces. Journalofcolloidandinterfacescience ,93:169,1983. [7]ALopezdeRamos,RARedner,andRLCerro.Surfacetensionfrom pendantdropcurvature. Langmuir ,9:3691,1993. [8]AlvinUChen,PatrickKNotz,andOsmanABasaran.Computationaland experimentalanalysisofpinch-offandscaling. Physicalreviewletters 88:174501,2002. [9]RMSMSchulkes.Theevolutionandbifurcationofapendantdrop. Journal ofFluidMechanics ,278:83,1994. [10]EPitts.Thestabilityofpendentliquiddrops.part1.dropsformedina narrowgap. JournalofFluidMechanics ,59:753,1973. [11]EPitts.Thestabilityofpendentliquiddrops.part2.axialsymmetry. Journal ofFluidMechanics ,63:487,1974. [12]JFPadday,GeorgesPtr,CGRusu,JGamero,andGWozniak.The shape,stabilityandbreakageofpendantliquidbridges. Journalofuid mechanics ,352:177,1997. [13]SRMajumdarandDHMichael.Theinstabilityofplanependentdrops. JournalofColloidandInterfaceScience ,73:186,1980. [14]DHMichaelandPGWilliams.Theequilibriumandstabilityofaxisymmetric pendentdrops.In ProceedingsoftheRoyalSocietyofLondonA:Mathematical,PhysicalandEngineeringSciences ,volume351,pages117. TheRoyalSociety,1976. 113

PAGE 114

[15]DHMichaelandPGWilliams.Theequilibriumandstabilityofsessiledrops. In ProceedingsoftheRoyalSocietyofLondonA:Mathematical,Physical andEngineeringSciences ,volume354,pages127.TheRoyalSociety, 1977. [16]JBBostwickandPHSteen.Stabilityofconstrainedcapillarysurfaces. AnnualReviewofFluidMechanics ,47,2014. [17]IMartinez.Stabilityofaxisymmetricliquidbridges. ESASP ,191:267, 1983. [18]JLEspino,JMeseguer,andAnaLavern-Simavilla.Anexperimental studyofthebreakageofliquidbridgesatstabilitylimitofminimumvolume. PhysicsofFluids ,14:3710,2002. [19]WilliHHager.Wilfridnoelbondandthebondnumber. JournalofHydraulic Research ,50:3,2012. [20]LewisEJohnsandRangaNarayanan. Interfacialinstability .Springer,2002. [21]WeidongGuo,GrardLabrosse,andRangaNarayanan. Theapplication oftheChebyshev-spectralmethodintransportphenomena ,volume68. Springer,2013. 114

PAGE 115

BIOGRAPHICALSKETCH XinLin receivedhisPh.D.fromUniversityofFloridainthespringof2016. HegothisMasterofSciencedegreeinchemicalengineeringfromUniversity ofFloridainthespringof2013andbachelor'sdegreeinautomationatthe DepartmentofControlScienceandEngineeringfromZhejiangUniversity Hangzhou,Chinain2011.InJanuary2011hejoinedtheprocesscontrol researchgroupundertheguidanceofDr.OscarD.Crisalle.Asamemberof thegroup,hisresearchconcentrationlayonthedesignandimplementationof theincrementalandintegralcontrollersintheMATLAB/Simulinkenvironment.In July2013hejoinedtheowinstabilitygroupundertheguidanceofDr.Ranga Narayananandconductedresearchontheinstabilityofapendentdropanda closedliquidbridge. 115