Fluid Mechanics Worked Examples CARL SCHASCHKE

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F U i
MECHANICS
WORKEDEXAMPLESFOR ENGINEERS
Carl Schaschke
Fluidmechanicsis anessentialcomponentofmanyengineeringdegreecourses.
Totheprofessionalengineer,a knowledgeofthebehaviouroffluidsis ofcrucial
importancein cost-effectivedesignandefficientoperationofprocessplant.This
bookillustratestheapplicationoftheoryinfluidmechanicsandenablesstudents
newtothesciencetograspfundamentalconceptsin thesubject.
Writtenarounda seriesofelementaryproblemswhichtheauthorworksthrough
toasolution,thebookisintendedasastudyguideforundergraduatesinprocess
engineeringdisciplinesworldwide.It will alsobeof usetopractisingengineers
withonlya rudimentaryknowledgeoffluidmechanics.
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Concentratingon incompressible,Newtonianfluids and single-phaseflow
throughpipes,chaptersinclude:continuity,energyandmomentum;laminarflow
andlubrication;tankdrainageandvariableheadflow.A glossaryof termsis
includedforreferenceandallproblemsuseSIunitsofmeasurement.
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ISBN 0-85295-405-0
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Fluidmechanics
Workedexamplesforengineers
CarlSchaschke
IChemE
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--
The informationin thisbookis givenin good
faithandbeliefin itsaccuracy,butdoesnot
implytheacceptanceof anylegalliabilityor
responsibilitywhatsoever,by theInstitution,
orby theauthor,for theconsequencesof its
use'ormisusein anyparticularcircumstances.
All rightsreserved.No partof thispublication
maybereproduced,storedin aretrieval
system,or transmitted,in anyformor by any
means,electronic,mechanical,photocopying,
recordingor otherwise,withouttheprior
permissionof thepublisher.
Publishedby
InstitutionofChemicalEngineers,
DavisBuilding,
165-189RailwayTerrace,
Rugby,WarwickshireCV213HQ,UK
IChemEisaRegisteredCharity
@ 1998Carl Schaschke
Reprinted2000withamendments
ISBN 0852954050
Photographsreproducedby courtesyof British Petroleum(page112),
Conoco (page98) andEsso UK pic (pagesxviii, 60, 140,192and234)
Printedin theUnitedKingdom by RedwoodBooks, Trowbridge,Wiltshire
11
Preface
Studentscommonlyfind difficulty with problemsin fluid mechanics.They
maymisunderstandwhatis requiredor misapplythesolutions.This bookis
intendedto help.It is a collectionof problemsin elementaryfluid mechanics
withaccompanyingsolutions,andintendedprincipallyasastudyaidforunder-
graduatestudentsofchemicalengineering- althoughstudentsofallengineering
disciplineswill find it useful.It helpsin preparationfor examinations,when
tacklingcourseworkandassignments,andlaterinmoreadvancedstudiesof the
subject.In preparingthisbookI havenottriedto replaceother,fullertextson
the subje~t.InsteadI have aimedat supportingundergraduatecoursesand
academictutorsinvolvedin thesupervisionof designprojects.
In thetext,workedexamplesenablethereadertobecomefamiliarwith,and
to graspfirmly, importantconceptsandprinciplesin fluid mechanicssuchas
mass,energyandmomentum.Themathematicalapproachis simpleforanyone
with prior knowledgeof basicengineeringconcepts.I havelimitedtheprob-
lemsto thoseinvolving incompressible,Newtonianfluids and single-phase
flow throughpipes.Thereis no attemptto includetheeffectsof compressible
andnon-Newtc;mianfluids,orof heatandmasstransfer.I alsoheldbackfrom
moreadvancedmathematicaltoolssuchasvectorialandtensorialmathematics.
Many of theproblemsfeaturedhavebeenprovidedby universitylecturers
who aredirectlyinvolvedin teachingt1uidmechanics,andby professional
engineersin industry.I haveselectedeachproblemspecificallyfor thelight it
throwsonthefundamentalsappliedtochemicalengineering,andfor theconfi-
denceitssolutionengenders.
The curriculaof universitychemicalengineeringdegreecoursescoverthe
fundamentalsof t1uidmechanicswith reasonableconsistencyalthough,in
certainareas,therearesomedifferencesinbothproceduresandnomenclature.
This bookadoptsaconsistentapproachthroughoutwhichshouldberecogniz-
abletoall studentsandlecturers.
I havetailoredtheproblemskindlycontributedbyindustrialiststosafeguard
commercialsecretsandtoensurethatthenatureofeachproblemis clear.There
111
is no informationordetailwhichmightallow aparticularprocessor company
toberecognized.All theproblemsuseSI units.As traditionalsystemsofunits
arestill verymuchin usein industry,thereis a tableof usefulconversions.
Fluid mechanicshasajargonof itsown,soI haveincludedalistofdefinitions.
Thereareninechapters.They coverarangefromstationaryfluidsthrough
fluids in motion.Each chaptercontainsa selectednumberof problemswith
solutionsthatleadthereaderstepby step.Whereappropriate,thereareprob-
lemswithadditionalpointstofacilitateafullerunderstanding.Historicalrefer-
encestoprominentpioneersin fluid mechanicsarealsoincluded.At theendof
eachchapteranumberof additionalproblemsappear;theaimis to extendthe
reader'sexperiencein problem-solvingandto helpdevelopa deeperunder-
standingof thesubject.
I wouldliketoexpressmysincereappreciationtoDr RobertEdge(formerly
of StrathclydeUniversity),Mr BrendonHarty (RocheProductsLimited),Dr
Vahid Nassehi (LoughboroughUniversity), Professor ChristopherRielly
(LoughboroughUniversity),ProfessorLaurenceWeatherley(Universityof
Canterbury),Dr GraemeWhite (Heriot Watt University),Mr Martin Tims
(Esso UK plc) and Miss Audra Morgan (IChemE) for their assistancein
preparingthisbook.I amalsogratefulfor themanydiscussionswith profes-
sionalengineersfromICI, EssoandKvaernerProcessTechnology.
The texthasbeencarefullychecked.In theevent,however,thatreaders
uncoveranyerror,misprintor obscurity,I wouldbegratefulto hearaboutit.
Suggestionsfor improvementarealsowelcome.
Listofsymbols
The symbolsusedin theworkedexamplesaredefinedbelow.Wherepossible,
theyconformto consistentusageelsewhereandto internationalstandards.SI
unitsareusedalthoughderivedSI unitsor specialisttermsareusedwhere
appropriate.Specificsubscriptsaredefinedseparately.
Roman
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April2000
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areaof channelor tank
breadthof rectangularweir
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coefficient
concentration
diameter
impellerdiameter
fraction
frictionfactor
depthof bodybelowfreesurface
force
gravitationalacceleration
head
slopeof channel
constant
fundamentaldimensionfor length
length
massloading
mass
massflowrate
meanhydraulicdepth
fundamentaldimensionfor mass
channelroughness
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numberof pipediameters
rotationalspeed
specificspeed
pressure
power
wettedperimeter
volumetricflowrate
radius
frictionalresistance
radius
depth
suctionspecificspeed
thicknessof oil film
time
fundamentaldimensionfor time
torque
velocity
volume
width
work
principalco-ordinate
distance
principalco-ordinate
distance
principalco-ordinate
statichead
ratioof pipeto throatdiameter
film thickness
finitedifference
absoluteroughness
pumpefficiency
angle
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kinematicviscosity3.14159
density
surfacetension
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Fluidmechanicsand
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Fluid mechanicsformsanintegralpartof theeducationof achemicalengineer.
The sciencedealswith thebehaviourof fluids whensubjectedto changesof
pressure,frictional resistance,flow throughpipes, ducts,restrictionsand
productionof power.It alsoincludesthedevelopmentandtestingof theories
devisedtoexplainvariousphenomena.To thechemicalengineer,aknowledge
of thebehaviourof fluids is of crucialimportancein cost-effectivedesignand
efficientoperationof processplant.
Fluid mechanicsis well knownfor thelargenumberof conceptsneededto
solveeventh~apparentlysimplestofproblems.It is importantfor theengineer
tohaveafull andlucidgraspof theseconceptsinordertoattempttosolveprob-
lemsin fluid mechanics.Thereis, of course,aconsiderabledifferencebetween
studyingtheprinciplesof thesubjectforexaminationpurposes,andtheirappli-
cationby thepractisingchemicalengineer.Both thestudentandtheprofes-
sionalchemicalengineer,however,requirea soundgrounding.It is essential
thatthebasicsarethoroughlyunderstoodandcanbecorrectlyapplied.
Manystudentshavedifficultyinidentifyingrelevantinformationandfunda-
mentals,particular~yclosetoexaminationtime.Equally,studentsmaybehesi-
tantin applyingtheoriescoveredin their studies,resultingfrom eitheran
incompleteunderstandingof theprinciplesor a lack of confidencecausedby
unfamiliarity.For thosenew to thesubject,findinga clearpathto solvinga
problemmaynotalwaysbestraightforward.For theunwaryandinexperienced,
theopportunityto deviate,to applyincorrector inappropriateformulaeor to
reachamathematicalimpassein thefaceof complexequations,is all tooreal.
Thedangeris thatthestudentwill dwellonamathematicalquirkwhichmaybe
specificpurely to the mannerin which the problemhasbeen(incorrectly)
approached.A disproportionateamountof effortwill thereforebeexpendedon
somethingirrelevantto thesubjectof fluid mechanics.
Studentsdeyelopandusemethodsfor studywhicharedependenton their
own personalneeds,circumstancesand availableresources.In general,
however,aquickeranddeeperunderstandingof principlesis achievedwhena
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problemis providedwithanaccompanyingsolution.The workedexampleis a
recognizedand widely-usedapproachto self-study,providinga clear and
logicalapproachfroma distinctstartingpointthroughdefinedsteps,together
withtherelevantmathematicalformulaeandmanipulation.Thismethodbene-
fits thestudentby appreciationof boththedepthandcomplexityinvolvedin
reachingasolution.
While someproblemsin fluid mechanicsarestraightforward,unexpected
difficultiescanbeencounteredwhenseeminglysimilaror relatedsimpleprob-
lemsrequiretheevaluationof adifferentbutassociatedvariable.Althoughthe
solutionmay requirethe samestartingpoint, the routethroughto the final
answermaybequitedifferent.For example,determiningtherateof uniform
flow alonganinclinedchannelgiventhedimensionsof thechannelis straight-
forward.Butdeterminingthedepthof flow alongthechannelfor givenparame-
tersin theflow presentsaproblem.Whereastheformerisreadilysolvedanalyti-
cally,thelatteris complicatedby thefactthatthefluidvelocity,flow areaanda
flow coefficientall involvethedepthof flow.An analyticalsolutionis no longer
possible,thusrequiringtheuseof graphicalor trialanderrorapproaches.
Therearemanysimilaritiesbetweenthegoverningequationsin heat,mass
andmomentumtransportandit is oftenbeneficialto bringtogetherdifferent
branchesof thesubject.Otheranalogiesbetweendifferentdisciplinesarealso
useful,althoughtheymustbeappliedwithcare.In fluid mechanics,analogies
betweenelectricalcurrentandresistanceareoftenused,particularlyin dealing
withpipenetworkswherethesplittingandcombiningoflines canbelikenedto
resistorsin parallelandin series.
Some applicationsof fluid mechanicsrequire involved procedures.
Selectinga pump,for example,follows a fairly straightforwardsetof well-
definedstepsalthoughthelengthyprocedureneededcanbecomeconfusing.It
is importanttoestablishtherelationshipbetweentheflowrateandpressure,or
head,lossesin thepipeworkconnectingprocessvesselstogether.With fric-
tionallossesduetopipebends,elbowsandotherfittingsrepresentedby either
equivalentlengthpipeor velocityheads,pumpingproblemsthereforerequire
carefuldelineation.Any pumpcalculationis bestreducedtotheevaluationof
thesuctionpressureorheadandthenof thedischargehead;thedifferenceis the
deliveryheadrequiredfromthepump.For asizingcalculation,all thatis really
neededis to determinethedeliveryheadfor therequiredvolumetricflowrate.
As in manyprocessengineeringcalculationsdealingwith~quipmentsizing,the
physicallayoutplaysan importantpart,notonly in standardizingthemethod
for easycheckingbutalsoin simplifingthecalculations.Obviouslytherewill
becasesrequiringmoredetailbut,withabitof attention,suchdeviationsfrom
practicecaneasilybe incorporated.
Finally,theapplicationof fluidmechanicsin chemicalengineeringtoday
reliesonthefundamentalprincipleslargelyfoundedin theseventeenthand
eighteenthcenturiesby scientistsincludingBernoulli,NewtonandEuler.
Many of today'sengineeringproblemsare complex,non-linear,three-
dimensionalandtransient,requiringinterdisciplinaryapproachestosolution.
High-speedandpowerfulcomputersareincreasinglyusedtosolvecomplex
problems,particularlyin computationalfluiddynamics(CFD). It is worth
remembering,however,thatthesolutionsareonlyasvalidasthemathematical
modelsandexperimentaldatausedtodescribefluidflowphenomena.Thereis,
forexample,noanalyticalmodelthatdescribespreciselytherandombehaviour
offluidsinturbulentmotion.Thereisstillnosubstituteforanall-roundunder-
standingandappreciationoftheunderlyingconceptsandtheabilitytosolveor
checkproblemsfromfirstprinciples.
Vlll IX
I.
L
Contents
Preface III
Listof symbols
Fluidmechanicsandproblem-solving
v
VII
I Fluid statics
Introduction
1.1 Pressureata point
1.2 Pres~urewithinaclosedvessel
1.3 Forceswithinahydraulicram
1.4 Liquid-liquidinterfacepositioninasolventseparator
1.5 Liquid-liquidinterfacemeasurementbydifferentialpressure
1.6 Measurementofcrystalconcentrationbydifferentialpressure
1.7 Pressurewithinagasbubble
1.8 Pressuremeasurementbydifferentialmanometer
1.9 Pressuremeasurementbyinvertedmanometer
1.10Pressuremeasurementbysinglelegmanometer
1.11Pressuremeasurementbyinclinedlegmanometer
1.12Archimedes'principle
1.13Specificgravitymeasurementbyhydrometer
1.14Transferofprocessliquidtoaship
Furtherproblems
1
1
3
5
7
9
11
13
15
17
19
21
23
25
27
29
31
2 Continuity,momentumandenergy
Introduction
2.1 Flowinbranchedpipes
2.2 ForcesonaU-bend
2.3 Pressurerisebyvalveclosure
2.4 TheBernoulliequation
2.5 Pressuredropduetoenlargements
35
35
36
38
40
42
44
XI
2.6 Pipeentranceheadloss
2.7 Forceonapipereducer
2.8 Vortexmotion
2.9 Forcedandfreevortices
Furtherproblems
3 laminarflow andlubrication
Introduction
3.1 Reynoldsnumberequations
3.2 laminarboundarylayer
3.3 Velocityprofileinapipe
3.4 Hagen-Poiseuilleequationforlaminarflowinapipe
35 Pipediameterforlaminarflow
3.6 laminarflowthroughataperedtube
3.7 Relationshipbetweenaverageandmaximumvelocityinapipe
3.8 Relationshipbetweenlocalandmaximumvelocityina pipe
3.9 Maximumpipediameterforlaminarflow
3.10Verticalpipeflow
3.11Filmthicknessinachannel
3.12Flowdownaninclinedplate
3.13Flowdownaverticalwire
3.14Flowandlocalvelocitythroughagap
3.15Relationshipbetweenlocalandaveragevelocitythroughagap
3.16Relationshipbetweenaverageandmaximumvelocitythroughagap
3.17Shearstressforflowthroughagap
3.18Flatdiscviscometer
3.19Torqueonalubricatedshaft
3.20lubricatedcollarbearing
Furtherproblems
4 DimensionalanalysisIntroduction
4.1 Flowthroughanorifice
4.2 Flowovernotches
4.3 Scale-upofcentrifugalpumps
4.4 Frictionalpressuredropforturbulentflowinpipes
4.5 Scalemodelforpredictingpressuredropinapipeline
Furtherproblems
5 Flowmeasurementbydifferentialhead
Introduction
xii
47
49
51
53
57
5.1 Pitottube
5.2 Pitottraverse
5.3 Horizontalventurimeter
5.4 Orificeandventurimetersinparallel
55 Venturimetercalibrationbytracerdilution
5.6 Differentialpressureacrossaverticalventurimeter
5.7 Flowmeasurementbyorificemeterinaverticalpipe
5.8 Variableareaflowmeter
5.9 Rotametercalibrationbyventurimeter
Furtherproblems
115
117
119
122
124
126
128
130
134
136
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61
63
65
66
68
70
71
73
75
77
78
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81
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84
86
87
88
89
91
93
95
6 Tankdrainageandvariableheadflow
Introduction
6.1 Orificeflowunderconstanthead
6.2 Coefficientofvelocity
6.3 Drainagefromtankwithuniformcross-section
6.4 Tankprainagewithhemisphericalcross-section
65 Tankdrainagewithcylindricalcross-section
6.6 Drainagebetweentworeservoirs
6.7 Tankinflowwithsimultaneousoutflow
6.8 Instantaneoustankdischarge
6.9 Instantaneoustankinflowwithoutflow
6.10Tankdrainagethroughahorizontalpipewithlaminarflow
Furtherproblems
141
141
143
145
147
149
151
153
155
157
159
161
163
II
7 Openchannels,notchesandweirs
Introduction
7.1 Chezyformulaforopenchannelflow
7.2 Flowinarectangularopenchannel
7.3 Depthofflowinarectangularchannel
7.4 Economicaldepthofflowinrectangularchannels
75 Circularchannelflow
7.6 Maximumflowincircularchannels
7.7 Weirsandrectangularnotches
7.8 Depthofarectangularweir
7.9 Instantaneousflowthrougharectangularweir
7.10Flowthroughatriangularnotch
7.11TankdrainagethroughaV-notch
7.12Flowthroughatrapezoidalnotch
Furtherproblems
167
167
169
171
173
175
177
179
180
182
184
186
188
189
190
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99
100
102
104
106
108
110
113
113
XUl
8 Pipefrictionandturbulentflow
Introduction
8.1 Economicpipediameter
8.2 Headlossduetofriction
8.3 Generalfrictionalpressuredropequationappliedtolaminarflow
8.4 Blasius'equationforsmooth-walledpipes
8.5 Prandtl'suniversalresistanceequationforsmooth-walledpipes
8.6 Pressuredropthrougharough-walledhorizontalpipe
8.7 Dischargethroughasiphon
8.8 Flowthroughparallelpipes
8.9 Pipesinseries:flowbyvelocityheadmethod
8.10Pipesinseries:pressuredropbyequivalentlengthmethod
8.11Relationshipbetweenequivalentlengthandvelocityheadmethods
8.12Flowandpressuredroparoundaringmain
8.13Tankdrainagethroughapipewithturbulentflow
8.14Turbulentflowinnon-circularducts
8.15Headlossthroughataperedsection
8.16Accelerationofaliquidinapipe
Furtherproblems
9 Pumps
Introduction
9.1 Centrifugalpumps
9.2 Centrifugalpumpmatching
9.3 Centrifugalpumpsinseriesandparallel
9.4 Cavitationincentrifugalpumps
9.5 Netpositivesuctionhead:definition
9.6 Netpositivesuctionhead:calculation1
9.7 Specificspeed
9.8 Netpositivesuctionhead:calculation2
9.9 Effectofreducedspeedonpumpcharacteristic
9.10Dutypointandreducedspeedofacentrifugalpump
9.11Power,impellerdiameter,speedanddeliveredhead
9.12Suctionspecificspeed
9.13Reciprocatingpumps
9.14Single-actingreciprocatingpiston
9.15Dischargefromreciprocatingpumps
9.16Rotarypumps
Furtherproblems
xiv
193
193
196
197
199
200
202
204
206
209
211
214
217
219
221
223
226
228
230
Glossaryof terms
Selectedrecommendedtexts
275
282
284Nomenclatureandpreferredunits
Usefulconversionfactors 285
288
289
Physicalpropertiesofwater(atatmosphericpressure)
Lossesforturbulentflowthroughfittingsandvalves
Equivalentsandroughnessofpipes
291
292Manningcoefficientforvariousopenchannelsurfaces
Moodyplot
Index
293
294
235
235
237
239
243
244
245
247
249
251
252
254
256
260
262
264
265
268
269
xv
'Thescientistdescribeswhatis:
theengineercreateswhatneverwas.'
TheodorevonKarman(1881-1963)
'f hear, and f forget
f see,and f remember
f do, and f understand.'
Anonymous
Fluidstatics
Introduction
Fluids,whethermovingorstationary,exertforcesoveragivenareaorsurface.
Fluidswhicharestationary,andthereforehaveno velocitygradient,exert
normalorpressureforceswhereasmovingfluidsexertshearingforcesonthe
surfaceswithwhichtheyareincontact.ItwastheGreekthinkerArchimedes
(c287BC-c212BC)whofirstpublisheda treatiseon floatingbodiesand
providedasignificantunderstandingof fluidstaticsandbuoyancy.It wasnot
foranother18centuriesthattheFlemishengineerSimonStevin(1548-1620)
correctly.providedanexplanationofthebasicprinciplesoffluidstatics.Blaise
Pascal(1623-1662),theFrenchmathematician,physicistandtheologian,
performedmanyexperimentsonfluidsandwasabletoillustratethefunda-
mentalrelationshipsinvolved.
In theinternationallyacceptedSI system(SystemeInternationald'Unites),
thepreferredderivedunitsof pressureareNewtonspersquaremetre(Nm-2)
withbaseunitsof kgm-ls-l. Theseunits,alsoknownasthePascal(Pa),are
relativelysmall.Thetermbaris thereforefrequentlyusedto representone
hundredthousandNewtonspersquaremetre(105Nm-2or0.1MPa).Many
pressuregaugesencounteredintheprocessindustriesarestilltobefoundcali-
bratedintraditionalsystemsofunitsincludingtheMetricSystem,theAbsolute
EnglishSystemandtheEngineers'EnglishSystem.Thiscanleadtoconfusion
in conversionalthoughmanygaugesaremanufacturedwithseveralscales.
FurthercomplicationarisessincethePascalis arelativelysmalltermandSI
recommendsthatanynumericalprefixshouldappearin thenumeratorof an
expression.Althoughnumericallythesame,Nmm-2is oftenwronglyused
insteadofMNm-2.
It is importanttonotethatthepressureofafluidisexpressedinoneof two
ways.Absolutepressurerefersto thepressureabovetotalvacuumwhereas
gaugepressurereferstothepressureaboveatmospheric,whichitselfisavari-
ablequantityanddependsonthelocalmeteorologicalconditions.Theatmo-
sphericpressureusedasstandardcorrespondsto101.3kNm-2andisequivalent
'
FLUID MECHANICS FLUID STATICS
toapproximately14.7poundsforcepersquareinch,orabarometricreadingof
760mmHg.Thepressurein avacuum,knownasabsolutezero,thereforecorre-
spondsto a gaugepressureof -101.3 kNm-2 assumingstandardatmospheric
pressure.A negativegaugepressurethusreferstoapressurebelowatmospheric.
The barometeris a simpleinstrumentfor accuratelymeasuringtheatmo-
sphericpressure.In itssimplestformit consistsofasealedglasstubefilledwith
a liquid (usuallymercury)andinvertedin areservoirof thesameliquid.The
atmosphericpressureis thereforeexerteddownwardsonthereservoirof liquid
suchthattheliquid in thetubereachesan equilibriumelevation.Above the
liquid meniscusexistsa vacuum,althoughin actualfactit correspondsto the
vapourpressureof theliquid. In thecaseof mercurythis is a pressureof 10
kNm-2 at20°c.
In additiontogaugesthatmeasureabsolutepressure,therearemanydevices
andinstrumentsthatmeasurethedifferencein pressurebetweentwopartsin a
system.Differentialpressureis of particularusefor determiningindirectlythe
rateof flow of a processfluid in a pipeor duct,or to assessthestatusof a
particularpieceof processequipmentduringoperation- for example,identi-
fying theaccumulationof depositsrestrictingflow, whichis importantin the
caseof heatexchangersandprocessventilationfilters.
Althoughtherearemanysophisticatedpressure-measuringdevicesavail-
able,manometersarestillcommonlyusedfor measuringthepressureinvessels
or inpipelines.Variousformsof manometerhavebeendesignedandgenerally
areeitheropen(piezometer)orclosed(differential).For manometertubeswith
aboreof lessthan12mm,capillaryactionis significantandmayappreciably
raiseordepressthemeniscus,dependingonthemanometricfluid.
Finally, while fluids maybedescribedassubstanceswhichofferno resis-
tanceto shearandincludebothgasesandliquids,gasesdifferfromliquidsin
thattheyarecompressibleandmaybedescribedby simplegaslaws.Liquids
areeffectivelyincompressibleandfor mostpracticalpurposestheir density
remainsconstantanddoesnotvarywith depth(hydrostaticpressure).At ultra
high pressuresthis is not strictly true. Water, for example,has a 3.3%
compressibilityatpressuresof 69MNm-2 whichis equivalentto a depthof 7
kill. It was Archimedeswho first performedexperimentson the densityof
solidsby immersingobjectsin fluids.Thefamousstoryis toldof Archimedes
being askedby King Hiero to determinewhethera crown was puregold
throughoutor containeda cheapalloy,withoutdamagingthecrown.Suppos-
edly,whileinapublicbath,Archimedesis saidtohavehadasuddenthoughtof
immersingthecrownin waterandcheckingitsdensity.He wassoexcitedthat
heranhomethroughthestreetsnakedshouting'Eureka!Eureka!- I have
foundit! I havefoundit!'.
1.1 Pressureata point
Determinethetotalforce ona wall of anopentank2 mwidecontainingfuel
oil of density924kgm-3at a depthof2 m.
Patm
--- --------- - -- - - - - - - - - - --- -- -----
PI-L'.z
H
tP2
Solution
To determinethepressureatapointin thestaticliquidbelowthefreesurface,
considerthe equilibriumforceson a wedge-shapedelementof the liquid.
Resolving.inthex-direction
pb.yLsin8-Plb.yb.Z=0
where
sin8=b.z
L
Then
P =Pl
Resolvingin thez-direction
F +pb.yLcos8 - P2;}.xb.y=0
where
2 3
FLUID MECHANICS FLUID STATICS
cos 8 =Lix
L
Lixfiz
=p-fiyg
2
1.2Pressurewithinaclosedvessel
A cylindricalvesselwithhemisphericalendsis verticallymountedon itsaxis.
Thevesselcontainswaterofdensity1000kgm-3andtheheadspaceispressur-
izedtoagaugepressureof50kNm-2.Theverticalwall sectionof thevesselhas
a heightof 3 mandthehemisphericalendshaveradii of 1 m.If thevesselis
filled tohalfcapacity,determinethetotalforce tendingtol(it thetopdomeand
theabsolutepressureat thebottomof thevessel.
andtheweight(forcedueto gravity)of theelementis
F =mg
If theelementisreducedtozerosize,in thelimitthistermdisappearsbecauseit
representsan infinitesimalhigherorder than the other termsand may be
ignored.Thus
P =P2
/~~::~~~};/ aim
// /Im ~Notethattheangleof thewedge-shapedelementis arbitrary.Thepressurepis
thereforeindependentof8.Thus,thepressureatapointin theliquidis thesame
in alldirections(Pascal'slaw).To determinethepressureatadepthH, theequi-
librium(upwardanddownward)forcesare
p armLixfiy +pLixfiyHg - pLixfiy =0
l'v =50 kNm-2
3m -- - - ---
whichreducesto
------- --
p =Parm+pgH
The pressure(aboveatmospheric)atthebaseof thetankis therefore
H =2.5m
=18.129X 103 Nm-2
~ \:5l
p =pgH
=924x g x 2
The totalforceexertedoverthewall is therefore
F =pa
2
18.129x 103x2x2
2
=36258X 103N
Solution
Thetotalverticalforce,F, tendingtolift thedomeis thepressureappliedover
thehorizontalprojectedarea
2
F =Pv 'ITr
The totalforceis foundto be36.26kNm-2.
wherePv is thegaugepressurewithinthevessel.Thatis
F = 50 x 103 X 'ITX 12
=156X 106N
\
4 5
FLUID MECHANICS FLUID STATICS
=101.3X 103+50X ]03 +1000X g X 2j
=]75.3X 103Nm-2
1.3Forceswithina hydraulicram
A hydraulicramconsistsof a weightlessplungerof cross-sectionalarea
0.003m2andapistonofmass1000kgandcross-sectionalarea0.3m2.The
systemisfilled withoil of density750kgm-3.Determinetheforceon the
plungerrequiredforequilibriumif theplungerisatanelevationof2mabove
thepiston.
Note thatabovetheliquid surfacethepressurein theheadspaceis exerted
uniformlyon theinnersurfaceof thevessel.Be]owtheliquid,however,the
pressureonthevesselsurfacevarieswithdepth.The absolutepressure(pres-
sureabovea vacuum)atthebottomof thevesselis therefore
p =Palm +Pv +pgH
V2
Theforcetendingtolift thedomeis 1.56MN andthepressureatthebottomof
thevesselis 175.3kNm-2.
Note that,unlikethegaspressurewhichis exerteduniformlyin thehead
space,theanalysistodeterminethehydrostaticforcesactingonthesubmerged
curvedsurface(lower domedsection)requiresresolvingforcesin both the
verticalandhorizontaldirections.The magnitudeof thehorizontalreactionon
thecurvedsurfaceis equalto thehydrostaticforcewhich actson a vertical
projectionof thecurvedsurface,whilethemagnitudeof theverticalreactionis
equaltothesumof theverticalforcesabovethecurvedsurfaceandincludesthe
weightof theliquid.In thiscase,however,thevesselis symmetricalsuchthat
the hydrostaticforce is in the downwarddirection.The downwardforce
imposedby thegasandliquid is thus
2
2 27tr
F =(Pv + pgh)7tr + pg-
3
Plunger
2m
VI
x '--'5
Oil
2 2X7tx]3
=(50,000+ 1000X g X 1.5)X 7txl +1000X g X -
3
Solution
Forthepisto~,thepressureatthedatumelevationxxis
FI
Pxx =-
al=223,854N
=223.8kN whereF I is theforceof thepistonanda1is theareaof thepiston.Thispressure
isequaltothepressureappliedbytheplungeratthesamedatumelevation.That
is
F2
Pxx =- + PougH
a2
whereF2 is theforceontheplunger,a2is theareaof theplungerandH is the
elevationof theplungerabovethedatum.Therefore
Fl =F2 +PoilgH
al a2
6
7
FLUID MECHANICS
, Rearranging
F2=a2(::-PougH J
=O.o03X
(
lOOOXg -750xg X2
)0.3
=54N
Theforcerequiredforequilibriumis foundtobe54N. Notethatif nodown-
wardforceis appliedtotheweightlessplunger,theplungerwouldrisetoan
elevationof4.44m.
Thehydraulicramillustratedisanexampleofaclosedsysteminwhichthe
pressureappliedby thepistonis transmittedthroughoutthehydraulicfluid
(oil).Theprincipleof pressuretransmissionis knownasPascal'slawafter
Pascalwhofirststatedit in 1653.Hydraulicsystemssuchasrams,liftsand
jacksarebasedonthisprincipleandareusefulforliftingandmovingpurposes.
It isusualinsuchhydraulicsystemstoreplacethepistonwithcompressedair.
Theforceappliedisthencontrolledbytheappliedairpressure.Highpressures
canthereforebeachieved,asinthecaseofhydraulicpresses,inwhichtheforce
exertedagainsta pistonin turnexertstheforceovera smallerarea.For
example,theplungershowncorrespondstoadiameterof62mmoverwhichan
equilibriumpressureof 18kNm-2is applied.If it weretobeconnectedtoa
shaft18mmindiameter,thentheforceexertedovertheareaoftheshaftwould
correspondto222kNm-2- afactorof 12timesgreater.
8
~
FLUID STATICS
1.4Liquid-liquidinterfacepositionin a
solventseparator
Mixtureswhichcontaintwomutuallyinsolubleorganicandaqueousliquids
are to beseparatedin a separatorwhichconsistsof a verticalchamberwith
overflowandunderflow.Themixtureisfedslowlytotheseparatorinwhichthe
aqueousphase,of constantdensityJ 100kgm-3,is dischargedfrom theunder-
flow atthebaseof thechambertoa dischargepoint50cmbelowtheoverflow
level in thechamber.The organicphasecan vary in densityfrom 600-800
kgm-3.Determinetheminimumheightof thechamber,H, whichcanbeusedif
theorganicphaseis nottoleavewiththeaqueousphase.If theheightH ismade
equalto 3 m, determinethelowestpossiblepositionof theinteifacein the
chamberbelowtheoverflow.
Ventto
atmosphere
- Organicphase
~ I. 1U"d:~~O,m11H,.I Feed
1'1 I H \ H2
.Aql!equs
. .phas~.
.' '.
Solution
Theseparatoris assumedtooperateatatmosphericpressure.Equatingthepres-
surein thechamberanddischargepointfor themaximumpossibledepth(in
metres)for theorganicphasein thechambergives
9
FLUID MECHANICS
PogH =Paqg(H -OS)
wherePo andPaq arethedensitiesof organicandaqueoussolutions,respec-
tively.Rearranging
H = OSPaq
Paq -Po
To ensurenolossof organicphasewiththeaqueousphase,theheightof the
chamberis greatestfor thehighestpossibleorganicdensity(800kgm-3).
Therefore
H =OSx 1100
1100- 800
=2.2m
Forafixedlengthofchamberof3m,theinterfacebetweenthetwophasesis
determinedfromthepressurein thechamberanddischargepoint.Thatis
PogH] +PaqgHz=Paqg(H-OS)
where
H=H]+H2
Therefore
PogH] +Paqg(H -H])=Paqg(H -OS)
Rearranging,theinterfacepositionis at its lowestpositionwhentheorganic
phasehasa densityof 600kgm-3.Thatis
OSpaq
HI =
Paq - Po
OSx 1100
1100- 600
=l.lm
The maximumdepthis foundtobe2.2m andtheinterfacebelowtheoverflowis foundtobe1.1m.Ideally,thefeedpointtothechamber-shouldbelocatedat
theliquid-liquidinterfaceto ensurequickandundisturbedseparation.Where
thedensityof theorganicphaseis expectedtovary,eitheranaverageposition
or thepositioncorrespondingto themostfrequentlyencountereddensitymay
beused.
10
~
j
FLUID STATICS
1.5Liquid-liquidinterfacemeasurementby
differentialpressure
Aqueousnitric acid is separatedfrom an insolubleoil in a vessel.Dip legs
extendinto bothphasesthroughwhichair is gentlydischargedsufficientto
overcomethehydrostaticpressure.Determinetheposition of the inteiface
betweenthelegsif thelegsareseparateda distanceof 1mfor whichthediffer-
entialpressurebetweenthelegsis 10kNm-2.Thedensitiesofoil andnitricacid
are 900kgm-3and1070kgm-3,respectively.
+Air +Air
Solution
The useof dip legsis aneffectiveway of measuringliquid densities,liquid-
liquidinterfacepositionsanddetectingthepresenceof solidmaterialin liquids.
As it hasnomovingormechanicalpartsit isessentiallymaintenancefreeandit
hasthereforefoundapplicationin thenuclearindustryamongstothers.In this
application,thedip legsareusedto determinethepositionof theliquid-liquid
interfacein whichthedensitiesof thetwophasesareassumedto beconstant.
The differentialpressurebetweenthelegsis
/';.p=PogH] +pl1gHz
whereP0 andP11arethedensitiesof theoil andnitricacidandwherethefixed
distancebetweentheendsof thedip legsis
H=H]+Hz
=I m
11
- 1----=-1
Freesurface
1-
0-
0 I I OH0 v
.T
Hj
.:'"""'"
.1H'. 2. v'O
Nitricacid
FLUID MECHANICS
EliminatingH2 andrearranging
H _!!.p -PngH1 -
(p0 - Pn)g
- 10X 103- 1070x g x 1
(900-1070) x g
=0.30m
Thedepthisfoundtobe30cmbelowtheupperdipleg.
Notethatasinglediplegcanbeusedtodeterminethedepthof liquidsof
constantdensityin vesselsin whichthegaspressureappliedis usedtoover-
comethehydrostaticpressure.Forcasesin whichthedensityof theliquidis
likelytovary,duetochangesinconcentrationorthepresenceof suspended
solids,thedensitycanbedeterminedusingtwodiplegsofdifferentlength,the
endsofwhichareafixeddistanceapart.In themorecomplicatedcaseof two
immiscibleliquidsinwhichthedensitiesofbothphasesmayvaryappreciably,
it is possibletodeterminethedensityof bothphasesandthelocationof the
interfaceusingfourdiplegswithtwoineachphase.
Inpractice,itisnecessarytoadjustcarefullythegaspressureuntilthehydro-
staticpressureisjustovercomeandgasflowsfreelyfromtheendof thedip
leges).Sensitivepressuresensingdevicesarethereforerequiredfor thelow
gaugepressuresinvolved.Fluctuatingpressurereadingsareusuallyexperi-
enced,however,asthegasbubblesformandbreakoff theendof theleg.
Conversionchartsmaythenbeusedtoconverta meanpressurereadingto
concentration,interfacepositionorliquidvolume,asappropriate.
12
...-
FLUID STATICS
1.6Measurementofcrystalconcentrationby
differentialpressure
Theconcentrationof sodiumsulphatecrystalsin a liquidfeedto a heat
exchangeris determinedbya differentialpressuremeasurementof thesatu-
ratedliquidin theverticallegfeedingtheheatexchanger.If thepressure
measurementsareseparatedbya verticaldistanceof 1.5m,determinethe
densityofthesolutionwithcrystalsandthefractionofcrystalsfor adifferential
pressuremeasurementof22kNm-2.Thedensityofsaturatedsodiumsulphate
is1270kgm-3anddensityofanhydroussolutionsulphateis2698kgm-3.
-
+ Flow
Steam
--- Heat
exchanger
!'J.p
a
'"
II
::t:
Condensate
Solution
Assumingnodifferentialpressurelossduetofrictionintheleg,thedifferential
pressureisduetothestaticpressurebetweenthepressuremeasurementpoints.
Thatis
!!.p=pgH
whereP is thedensityof thesolution.
13
FLUID MECHANICS
Rearranging
P=b.p
gH
22x 103
g x 15
=1495kgm-3
The densityofthesolutionwithcrystalsis 1495kgm-3.This densityis greater
thanthatof thesaturatedsodiumsulphatesolutionaloneandthereforeindicates
thepresenceof crystalsfor whichthefractionalcontentis foundfrom
P =IIp s +12pc
whereps isthedensityof saturatedsolution,pc isthedensityofcrystals,andfl
and12aretherespectivefractionswhere
fI+12=l
EliminatingfI
12=P-Ps
Pc -Ps
1495-1270
2698-1270
=0.157
Thatis, thecrystalcontentis foundtobe15.7%.This is, however,anoveresti-
matesincefrictionaleffectsof theflowingliquidin thelegareignored.Where
theycannotbeignoredthedifferentialpressureis modifiedto
b.p=pg(H -HL)
whereH L is theheadlossduetofriction.
14
~-
FLUID STATICS
1.7Pressurewithinagasbubble
A smallgasbubblerisingin anopenbatchfermenterhasa radiusof0.05cm
whenit is3 mbelowthesurface.Determinetheradiusofthebubblewhenit is i
mbelowthesurface.it maybeassumedthatthepressureinsidethebubbleis
2 air abovethepressureoutsidethebubble,wherer is theradiusof thebubble
and0"is thesurfacetensionof thegas-fermentationbrothandhasa valueof
0.073Nm-l. Thepressureandvolumeofthegasinthebubblearerelatedbythe
expressionpV =c wherec is a constant.
Palm
+ + + Freesurface+
---------
HI
Gas
bubble ~
\iY
Solution
At adepthof3m,thepressurewithinthebubble,PI, isdependentonthepres-
sureatthefreesurface,thehydrostaticpressureandsurfacetensioneffect.Thus
20"
PI =Palm +pgHI +-
rl
=101.3X 103+1000X g X 3 + 2 X 0.Q73
5 X 10-4
=13l.Q22X 103Nm-2
15
FLUID MECHANICS
At adepthof 1m,thepressureinsidethebubble,P2' is
2cr
P2 =Palm +pgH2 +-
r2
=101.3X 103+1000X g xl + 2 X 0.073
rz
=111.11X 103+0.146
rz
SincepV is aconstant,then
PI VI =P2V2
wherefor a sphericalbubble
4 3 4 3
PI-1trl =pZ-1tr23 3
Thatis
3 3
Plrl =P2rz
Therefore
131.022X 103X (5X 10-4)3 =(111.11X 103+ 0.::6)x ri
Thecubicequationcanbesolvedanalytically,bytrialanderrororbyassuming
thatthesecondtermin thebracketsis substantiallysmall,reducingtheeffort
requiredforsolutiontoyieldabubbleradiusofapproximately0.053mm.
16
--
FLUID STATICS
1.8Pressuremeasurementbydifferentialmanometer
Determinethepressuredifferencebetweentwotappingpointson a pipe
carryingwaterfor adifferentialmanometerreadingof20cmofmercury.The
specificgravityofmercuryis13.6.
- .- -'i-~'
P2
Mercury
Solution
ThedifferentialorV-tubemanometerisadeviceusedtomeasurethedifference
inpressuresbetweentwopointsandconsistsofatransparentV-tube,usually
madeofglass,andcontainsamanometricfluidsuchasmercury.It istypically
usedtomeasurethepressuredropofmovingfluidsduetofrictionalongpipes
orduetoobstaclesin pipelinessuchasflowmeasuringdevices,fittingsand
changesingeometry.Thepressuredifferenceof theprocessfluidis indicated
bythedifferenceinlevelsofthemanometricfluidbetweenthetwoverticallegs
oftheV-tubewhich,atthedatumelevationxx,are
PI +pg(HI +H)=pZ +pgH[ +PHggH
wherePHg is thedensityof mercuryandP thedensityof water.
17
FIOW-I----,;, - -
HI
L-
W"", I " IH x
FLUID MECHANICS
Rearranging,thedifferentialpressuret!.pbetweenthelegsis
t!.p=PI - P2
=pgHI +PHggH -pg(HI +H)
=(pHg - p)gH
=(13,600-1000)x g x 02
=24.721X 103Nm-2
The differentialpressureis 24.7kNm-2.
Notethatthelocationof themanometerbelowthepipe,HI, is notrequiredin
thecalculation.In practiceit is importanttoallowsufficientlengthin thelegsto
preventthemanometricfluid reachingthetappingpointon thepipefor high
differentialpressures.Filled with mercury,differentialmanometerscantypi-
callybeusedto measuredifferentialsupto about200kNm-2 orwithwaterto
about 20 kNm-2. Where a temperaturevariation in the processfluid is
expected,it is importanttoallowfordensity-temperaturevariationof themano-
metricfluid, whichcanaffectreadings.
In general,the U-tube differentialmanometeras a pressure-measuring
deviceis largelyobsolete.Therearemanysophisticatedmethodsandpressure-
measuringdevicesnow usedby industry.But the differentialmanometer
continuesto beausefultool in thelaboratoryandfor testingpurposes.
18
--
FLUID STATICS
1.9Pressuremeasurementbyinvertedmanometer
A laboratoryrig is usedtoexaminethefrictional lossesin smallpipes.Deter-minethepressuredropinapipecarryingwaterif adifferentialheadof40cmis
recordedusingan invertedmanometer.
x
Flow- ,-
Solution
Theinvertedmanometeravoidstheuseof amanometricfluid andinsteaduses
theprocessfluid (waterin thiscase)tomeasureitsownpressure.It consistsof
aninvertedU-tubewitha valveintowhichairor aninertgascanbeaddedor
vented.Here,thepressureatthedatumelevationxx,in leftandrighthandlegs
IS
PI -pg(H + HI)=P2 -pgHI -PairgH
whereP is thedensityof waterandPair is thedensityof air.Rearranging,the
differentialpressuret!.pis therefore
t!.p=PI -P2
=pg(H +HI)-pgHI -PairgH
=(p - Pair )gH
19
Water
r.-
Hj
'-j- ---
FLUID MECHANICS
Sincethedensityof air is in theorderof 1000timeslessthanthatof water,it
maythereforebereasonablyassumedthatthedifferentialpressureis approxi-
matedto
!!J.p'" pgH
'" 1000x g x 0.4
=3924Nm-2
Thedifferentialpressureis foundtobe3.9kNm-2.As withthedifferential
manometer,theelevationofthemanometer,HI, isnotrequiredinthecalcula-
tion.In practice,however,it is importanttoensurea reasonablepositionof
liquidlevelsin thelegs.Thisisbestachievedbypressurizingthemanometer
withairorinertgasusingthevalve,whereforhighpressuresthedensitymay
becomeappreciableandshouldbetakenintoconsideration.Inthecaseofair,
theerrorinthecalculationisunlikelytobegreaterthan0.5%.Inthecaseillus-
trated,thedensityofwatercorrespondstoatemperatureof 10°Cforwhichthe
densityofairatatmosphericpressureis 1.2kgm-3.If thishadbeentakeninto
account,itwouldhaveyieldedadifferentialpressureof3919Nm-2oranerror
of0.12%.A moresignificanterrorislikelytobeduetotheeffectsoftempera-
tureondensityandmayaffecttheresultbyasmuchas1%.Othererrorsare
likelytobecausedbydefiningthetoplevelof themanometricfluidin the
verticallegdueto itsmeniscus.A column-heightaccuracyof 0.025mmis,
however,generallyachievablewiththekeenesteyereading.
20
~
FLUID STATICS
1.10Pressuremeasurementbysinglelegmanometer
A mercury-filledsingleleg manometeris usedto measurethepressuredrop
acrossa sectionofplantcontainingaprocessfluid ofdensity700kgm-3.The
pressuredrop is maintainedby anelectricaldevicewhichworkson an on/off
principleusinga contactarrangementin a narrowverticaltubeofdiameter2
mmwhilethesumphasadiameterof2 em.If thepressuredropacrosstheplant
is to be increasedby 20 kNm~2,determinethe quantityof mercuryto be
removedfrom thesumpif thepositionof the electricalcontactcannotbe
altered.
Plant
-r ~
- - --- -'T
Signalto
pressure
control
mechanism
Process
fluid
Sensing
device
Sump Tube
Electrical
contact
Solution
Thesinglelegmanometerusesasumporreservoirof largecross-sectionin
placeofoneleg.Whenadifferentialpressureisapplied,thelevelinthelegor
tuberisesduetoadisplacementfromthesump.Theratiooflegtosumpareais
generallyneededfor particularlyaccuratework butis ignoredfor most
purposessincetheareaofthesumpiscomparativelylargerthanthatoftheleg.
Thedevicein thiscaseoperateswhenthelevelof mercuryin thetubefalls,
breakingtheelectricalcircuit.The pressurecontrolmechanismtherefore
21
FL UID MECHANICS
receivesa signalto increasethepressuredifference.Whenthemercurylevel
rises,theoppositeoccurs.An increasein pressuredropof 20kNm-2therefore
correspondstoanincreasein differencein levelof mercuryof
H=~
(pHg - p)g
20X 103
- (13,600- 700)x g
=0.158m
The volumeofmercurytoberemovedtoensurethecontactis stilljustmadeis
therefore
2
V=1td H
4
2
- 1tx 0.02 x 0.158
- 4
-5 3=4.96x 10 m
°1
Iii
I
i
i
II
That is, thevolumeto be removedis approximately50 ml. Notethatif the
displacementofmercuryfromthesumpintothetubeis takenintoaccountthen
thiswouldcorrespondtoadropinlevelinthesump,H s' of
H =~H
s A
~(~J H
=
(
°.0002
J
2 x 0.158
0.Q2
-5=1.58x 10 m
Thisisverysmallandignoringit isjustified.
22
--
FLUID STATICS
1.11Pressuremeasurementbyinclinedlegmanometer
An oil-filledinclinedlegmanometeris usedtomeasuresmallpressurechanges
acrossanairfilter inaprocessventpipe.If theoil travelsa distanceof 12cm
alongthelegwhichis inclinedat anangleof20°to thehorizontal,determine
thegaugepressureacrossthefilter. Thedensityof oil is 800kgm-3.
P2
Solution
This instrumentis useful for measuringsmall differentialpressuresand
consistsof.a sumpof manometricfluid (oil) with a legextendeddownintoit
andinclinedatsomesmallangle.Applying a differentialpressureacrossthe
sumpandthelegresultsin adisplacementof themanometricfluid intotheleg,
thedistancethemanometricliquidtravelsupalongthelegbeingameasureof
differentialpressureandis
!1P=P]-P2
=pg(H] +H2)
If theoil is displacedfromthesumpupalongthelegby adistanceL, thecorre-
spondingdropin levelin thesump,H] , is therefore
H _aL
]-A
Also, theverticalriseof theoil is relatedtolengthbythesineof theangleof the
inclinedleg.Thatis
H2 =Lsin 8
23
FLUID MECHANICS
The differentialpressureis therefore
b.p=pgl~+Lsin e)
=pgLl~+sine)
As no detailsareprovidedregardingthedimensionsof themanometer,the
cross-sectionalareaof theoil sump,A, is thereforeassumedto beverymuch
largerthantheareaof theleg,a.Theequationthereforereducesto
b.p=pgLsine
=800x g x 0.12x sin200
= 322Nm-2
The differentialpressureis foundtobe322Nm-2.
The deviceis particularlyusefulfor measuringsmalldifferentialpressures
sinceif thetermsinsidethebracketsarekeptsmallitallowsthelengthalongthe
inclinedleg,L, tobeappreciable.If, for agivendifferentialpressure,theequiv-
alentmovementofmanometricliquidupaverticallegwouldbeh, say,thenthe
ratioof movementsL toh
L-
h ~ +sine
A
canthereforebeconsideredasamagnificationratio.
24
--
FLUID STATICS
1.12Archimedes'principle
A vesselcontainingaprocessmaterialwitha combinedtotalmassof 100kgis
immersedin waterfor coolingpurposes.Determinethetensionin thecableof
an overheadcraneusedto manoeuvrethefully immersedcontainerinto its
storagepositionif thebulkdensityof thevesselis 7930kgm-3.
- - -- - - - - --- -- -----
Container
"""""""""",,'
Solution
Considera body of massme immersed in the liquid such thatthenetdownward
force is thedifference between thedownward andupward forces. That is
F=mcg-mg
where m is the mass of water displaced. This is known as Archimedes' prin-
ciple and statesthat when a body is partially or totally immersed, there is an
upthrustequal to theweight of fluid displaced. For the immersedobject, thenet
downward force is taken by the tension in the cable and can be determined
where the massof the container andwater displaced is related to volume by
V=~
Pc
m
p
25
FLUID MECHANICS FLUID STATICS
m=m Pc-
Pc
1.13Specificgravitymeasurementbyhydrometer
A hydrometerfloats in waterwith6emofitsgraduatedstemunimmersed,and
inoilofSG0.8with4emofthestemunimmersed.Determinethelengthofstem
unimmersedwhenthehydrometerisplacedin a liquidof SG 0.9.
wherePc is thebulk densityof thecontainerandP is thedensityof water.
Rearranging,themassof waterdisplacedby thecontaineris therefore
The tensionin thecableis therefore
(
1- ~
)F=mcg Pc
(
1000
)=100x g x 1- 7930
=857N
x
Thatis,thetensionin thecableis 857N. Notethatthetensionin thecablewhen
thevesselis liftedoutof thewateris - -- 'L- --- -- --
f = mg
=1O0xg
=981N
The buoyancyeffectthereforereducesthetensionin thecableby 124N.
Weight
Solution
Hydrometersaresimpledevicesformeasuringthedensityorspecificgravityof
liquidsroutinelyusedinthebrewingindustrytodeterminequicklytheconver-
sionofsugartoalcoholinfermentation.Theyconsistofaglasstubewhichhave
aweightedglassbulbandgraduatedstemofuniformdiameterandfloatinthe
liquidbeingtested.Thedensityorspecificgravity(SG)isusuallyreaddirectly
fromthegraduatedstematthedepthtowhichit sinks.Fornonetdownward
force,theverticaldownwardforcesactingonthebodyareequaltotheupthrust.
Thus
mg =mhg
26 27
FLUID MECHANICS
wheremandmh arethemassofliquid andhydrometer,respectively.Thus,
Archimedes'principlefor a floatingbody statesthatwhena body floats,it
displacesa weightof fluid equalto its own weight.The displacementby the
hydrometeris therefore
mhg =pg(L-x)a
=pog(L-xo)a
wherea is thecross-sectionalareaof thestem,p andp() arethedensitiesof
waterandoil,andxandx0 arethelengthsofstemunimmersedintherespective
liquids.Therefore
pg(L -x)a =Pog(L -xo)a
Rearranging,thelengthof hydrometeris therefore
L =px -Poxo
Po -p
1000x 0.06- 800x 0.04
800- 1000
=0.14m
For thehydrometerimmersedin a liquidof SG 0.9(900kgm-3),letthelength
of stemremainingunimmersedbexL' Therefore
1000x g x (0.14-0.06)x a =900x g x (0.14- xL) x a
Solving,xL isfoundtobe0.051I m.Thatis,thelengthofstemabovetheliquid
ofSG0.9is5.11cm.
28
--
FLUID STATICS
1.14Transferof processliquidtoa ship
A liquidhydrocarbonmixtureofdensity950kgm-3istransferredbypipelineto
a ship at a loading terminal.Prior to transfer,the ship has an unloaded
displacementof5000tonnesanddraftof3m.Transferofthehydrocarbonisat
a steadyrateof 125m3h-l.If theseabedis ata depthof 5.5m,determinethe
quantitydeliveredandtimetakenif theshiprequiresat leastI mof clearance
betweenthe sea bed and hull to manoeuvreaway safelyfrom the loading
terminal.
-- -I !0
- - ~
1-1
TI T2 5.5ill
""""""""" """"""""" """"" " Seabed
Solution
Applying Archimedes'principle,theshipprior to transferdisplacesits own
weightof seawater.Thatis
msg =mg
=pAT]g
wheremsandm arethemassof shipandseawaterdisplaced,pisthedensityof
seawater,A is thewaterplaneareaandTj is thedepthof theshipbelowthe
waterline.After transfer
msg +mhcg =pAT2g
wheremhcis themassof hydrocarbonmixture.Aftertransfer,theshipis clear
fromtheseabedby 1m.Combiningthesetwoequations,themassof hydro-
carbontransferredis
29
FLUID MECHANICS
mhc=mSl~ -1)
=5x 106X(4: - 1)
=2.5X106kg
The transfertimeis therefore
t = mhc
PhcQ
2.5X 106
950x 125
=21.05h
Thatis, atransferof2500tonnesofhydrocarbonmixtureis completedin 21.05
hours.
It shouldbe notedthatthe approachillustratedis rathersimplistic.No
accountis madefor thedimensionsof theshipin termsof its lengthandbeam
northevariationof thewaterplaneareawithdepth.The beamis animportant
dimensionin termsof stabilitywherethestabilityis dependentontherelative
positionof theship'scentreof gravityandcentroidof thedisplacedvolume
calledthecentreof buoyancy.A shipis unstableandwill capsizewhen,for a
heelof up to 10°,a line drawnverticallyup from thecentreof buoyancyis
belowthecentreof gravity- apointknownasthemetacentre.
The safetransferof liquidstoandfromtankswithinshipsrequiresacareful
sequenceof operation.Tidal effectson mooredshipsandtheeffectsof the
liquid freesurfacein thetanksmustalsobetakenintoconsideration.It wasthe
British politician SamuelPlimsoll (1824-1898)who was responsiblefor
gettinglegislationpassedto prohibit'coffinships'- unseaworthyandover-
loadedships- beingsentto sea.The MerchantSeaAct of 1874included,
amongstotherthings,enforcementof thepaintingof lines,originallycalled
Plimsoll marksandnow knownasloadline marks,to indicatethemaximum
loadlinewhichallowsfor thedifferentdensitiesof theworld'sseasin summer
andwinter.
30
~
FLUID STATICS
Furtherproblems
(1) Explainwhatis meantby gaugepressureandabsolutepressure.
(2) A hydraulicpresshasa ramof 10cm diameteranda plungerof 1 cm
diameter.Determinetheforcerequiredontheplungertoraiseamassof 500kg
ontheram.
Answer:49.05N
(3) Thereadingof abarometeris 75.5cmof mercury.If thespecificgravity
of mercuryis 13.6,convertthispressuretoNewtonspersquaremetre.
Answer:100,792Nm-2
(4) A rectangulartank5mlongby2mwidecontainswatertoadepthof2m.
Determinetheintensityofpressureonthebaseofthetankandthetotalpressure
ontheend.
Answer:19.6kNm-2, 39kNm-2
(5) Determinethetotalpressureon a verticalsquaresluice,of 1 m square,
positionedwithits topedge3 mbelowthelevelof water.
Answer:34.3kNm-2
(6) A tubeis filledwithwatertoadepthof600mmandthen450mmofoil of
SO 0.75is addedandallowedtocometorest.Determinethegaugepressureat
thecommonliquidsurfaceandatthebaseof thetube.
Answer:3.3kNm-2,9.2kNm-2
(7) Showthatwhenabodyis partiallyortotallyimmersedinaliquid,thereis
anupthrustonthebodyequaltotheweightof theliquiddisplaced.
(8) Showthatafloatingbodydisplacesaweightof theliquidequaltoitsown
weight.
(9) A V-tubehasa left-handlegwithadiameterof 5cmandaright-handleg
withadiameterof 1cmandinclinedatanangleof240.If themanometerfluidis
oil with a densityof 920kgm-3anda pressureof 400Nm-2 is appliedto the
left-handleg,determinethelengthby whichtheoil will havemovedalongthe
right-handleg.
Answer:9.9cm
31
FLUID MECHANICS
(10) Determinetheabsolutepressurein anopentankcontainingcrudeoil of
density900kgm-3atadepthof 5 m.
Answer:145.4kNm-2
(11) An openstoragetank 3 m high containsaceticacid, of density1060
kgm-3,andis filled to half capacity.Determinetheabsolutepressureat the
bottomof thetankif thevapourspaceabovetheacidis maintainedatatmo-
sphericpressure.
Answer:117kNm-2
(12) A differentialmanometercontainingmercuryof SO 13.6andwaterindi-
catesaheaddifferenceof 30cm.Determinethepressuredifferenceacrossthe
legs.
Answer:37.1kNm-2
(13) A V-tubecontainswaterandoil. The oil, of density800kgm-3,restson
thesurfaceof thewaterin theright-handlegto adepthof 5 cm.If thelevelof
waterin theleft-handlegis 10cmabovethelevelof waterin theright-handleg,
determinethepressuredifferencebetweenthetwolegs.Thedensityofwateris
1000kgm-3.
Answer:589Nm-2
(14) A separatorreceivescontinuouslyanimmisciblemixtureof solventand
aqueousliquidswhichis allowedto settleinto separatelayers.The separator
operateswithaconstantdepthof 2.15mby wayof anoverflowandunderflow
arrangementfrom bothlayers.The positionof theliquid-liquid interfaceis
monitoredusingadip legthroughwhichair is gentlybubbled.Determinethe
positionof theinterfacebelowthesurfacefor agaugepressurein thediplegof
20kNm-2.The densitiesof thesolventandaqueousphasesare865kgm-3and
1050kgm-3,respectively,andthe dip leg protrudesto within 5 cm of the
bottomof theseparator.
Answer:90cm
(15) A hydrometerwithamassof 27ghasabulbof diameter2cmandlength
8 cm,anda stemof diameter0.5cmandlength15cm.Determinethespecific
gravityof a liquidif thehydrometerfloatswith 5 cmof thestemimmersed.
Answer:1.034
32
.-.
FLUID STATICS
(16) Two pressuretappingpoints,separatedby averticaldistanceof 12.7m,
areusedto measurethecrystalcontentof a solutionof sodiumsulphatein an
evaporator.Determinethedensityof thesolutioncontaining25%crystalsby
volumeandthedifferentialpressureif thedensityof theanhydroussodium
sulphateis 2698kgm-3andthedensityof saturatedsodiumsulphatesolutionis
1270kgm-3.
Answer:1627kgm-3,203kNm-2
(17) A vacuumgaugeconsistsof aV-tubecontainingmercuryopentoatmos-
phere.Determinetheabsolutepressurein theapparatusto whichit is attached
whenthedifferencein levelsof mercuryis 60cm.
Answer:21.3kNm-2
(18) Determinetheheightthroughwhichwateris elevatedby capillarityin a
glasstubeof internaldiameter3mmif thehydrostaticpressureis equalto4cr/d
wherecris thesurfacetension(0.073Nm-l) andd is thediameterof thetube.
Answer:9.9mm
(19) Explaintheeffectof surfacetensionon thereadingsof gaugesof small
boresuchaspiezometertubes.
(20) A shiphasa displacementof 3000tonnesin seawater.Determinethe
volumeof theship belowthewaterline if thedensityof seawateris 1021
kgm-3.
Answer:2938m3
(21) A closedcylindricalsteeldrumof sidelength2 m,outerdiameter1.5m
andwall thickness8 mm is immersedin a jacketcontainingwaterat 20°C
(density998kgm-3).Determinethenetdownwardandupwardforceswhenthe
drumis bothfull of waterat 20°C andempty.The densityof steelis 7980
kgm-3.
Answer:5.17kN, -29.4 kN
(22) An oil/waterseparatorcontainswaterof density998kgm-3toadepthof
75cmabovewhichis oil of density875kgm-3toadepthof75cm.Determine
thetotalforceontheverticalsideof theseparatorif it hasasquaresection1.5m
broad.If theseparatoris pressurizedby airabovetheoil, explainhowthiswill
affecttheanswer.
Answer:16kN
33
~
Continuity,
momentum
andenergy
Introduction
With regardtofluids in motion,it is convenientto considerinitially anideal-
izedformof fluidflow. In assumingthefluid hasnoviscosity,it is alsodeemed
tohavenofrictionalresistanceeitherwithinthefluid orbetweenthefluid and
pipe walls. Inviscid fluids in motionthereforedo not supportshearstresses
althoughnormalpressureforcesstill apply.
Therearethreebasicconservationconceptsevokedin solvingproblems
involvingfluids in motion.The conservationof masswasfirst consideredby
Leonardoda"vinci(1452-1519)in 1502withrespecttotheflow withinariver.
Appliedtotheflow throughapipethebasicpremiseis thatmassis conserved.
Assumingnolossfromor accumulationwithinthepipe,theflow intothepipe
is equaltotheflow outandcanbeprovedmathematicallyby applyingamass
balanceoverthepipesection.Theflow of incompressiblefluidsatasteadyrate
is thereforethesimplestform of thecontinuityequationandmaybe readily
appliedto liquids.
Theconservati?nof momentumis Newton'ssecondlawappliedtofluidsin
motion,andwasfirstconsideredby theSwissmathematicianLeonhardEuler
(1707-1783)in 1750.Again,by consideringinviscidfluid flow understeady
flow conditions,calculationsare greatlysimplified.This approachis often
adequatefor mostengineeringpurposes.
The conservationof energywas first consideredby the Swiss scientist
Daniel Bernoulli (1700-1782) in 1738 to describethe conservationof
mechanicalenergyof amovingfluid in asystem.Thebasicpremiseis thatthe
totalenergyof thefluid flowinginapipemustbeconserved.An energybalance
onthemovingfluid acrossthepipetakesintoaccountthereversiblepressure-
volume,kineticandpotentialenergyforms,andis greatlysimplifiedby consid-
eringsteady,inviscidandincompressiblefluid flow.
35
2.1 Flowin branchedpipes
Waterflows throughapipesectionwithaninsidediameterof 150mmata rate
of0.02m3s-l.Thepipebranchesintotwosmallerdiameterpipes,onewithan
insidediameterof 50mmandtheotherwithan insidediameterof 100mm.If
theaveragevelocityin the50mmpipe is 3 ms-l, determinethevelocitiesand
flows in all threepipesections.
Qt-
Pipe2 d2=50mm
PipeI dl=150mm/ -Q3
~ (~Pd d,- 100 mm
~ -Q2
Solution
Thecontinuityequationiseffectivelyamathematicalstatementdescribingthe
conservationofmassofaflowingfluidwherethemassflowintoapipesection
isequaltothemassflowout.Thatis
p]a]v]=P2a2v2
For anincompressiblefluidin whichthedensitydoesnotchange,thevolu-
metricflowis therefore
a]vI =a2v2
Forthebranchedpipesysteminwhichthereisnolossoraccumulationof the
incompressibleprocessfluid(water),theflowthroughtheISOmmdiameterpipe
(PipeI) isequaltothesumofflowsinthe50mm(Pipe2)and100mmdiameter
pipes(Pipe3).Thatis
Q] =Q2+Q3
rcd2 rcd2- 2 3
--v2 +-v3
4 4
Rearranging,thevelocityinthe100mmdiameterpipeistherefore
36
..-...
CONTINUITY, MOMENTUM AND ENERGY
4QI - d~V2rc
v3 = ;;z3
4 x 0.Q2- 0.052X 3
rc
0.12
=1.8ms-]
This correspondsto aflow of
rcd2
Q - 33 - -v3
4
2
= rcx 0.1 x 1.8
4
=O.oJ4m3s-1
Similarly,~hevelocityandflowcanbefoundfortheothertwopipesandare
givenbelow.
Diameter,mm
Velocity,ms-l
Flowrate,m3s-]
PipeI
ISO
1.13
0.020
Pipe2
50
3.00
0.006
Pipe3
100
1.80
0.014
37
FLUID MECHANICS
2.2 ForcesonaU-bend
A horizontalpipehasa 180°V-bendwitha uniforminsidediameterof200mm
andcarriesa liquidpetroleumfraction ofdensity900kgm-3at a rateof 150
m3h-l.Determinetheforceexertedbytheliquidonthebendif thegaugepres-
sure upstreamand downstreamof thebendare 100 kNm-2and 80 kNm-2,
respectively.
-2
~L: ) yF~ I: x,
Solution
The thrustexertedby theflowing liquid onthehorizontalbendis resolvedin
boththex- andy-directions.Assumingthatthegaugepressuresof theliquidare
distributeduniformlyin theV-bend,thenresolvingtheforcein thex-direction
gIves
Fx =Pial cos81 - P2a2cos82 +pQ(v2cos82 -vI cos81)
andin they-direction
Fy =Plalsin81 +p2a2sin 82 -pQ(v2sin82 +vlsin81)
The respectiveupstreamanddownstreampressureforcesare
PIal =1x 105 1tX022x-
4
=3141N
and
P2a2 =8 x 104 1tX 022x-
4
=2513N
38
~
CONTINUITY, MOMENTUM AND ENERGY
For theuniformcross-section,theaveragevelocityremainsconstant.Thatis
VI =v2
=4Q
1td2
4x 150
3600
1tx 022
=1.33ms-l
The momentumfluxesaretherefore
pQVl =pQv2
=900x 150 x 1.33
3600
=49.9N
For theliquidenteringthe180°bendtheangle81is0°andfortheliquidleaving
82 is 180°.Theresolvedforcein thex-directionis therefore
Fx =3141x cosoo-2513x cos1800+49.9x (cosI800-cosOO)
=5554N
Sincesin0°andsin180°areequalto zero,theforcein they-directionis
Fy =0
Althoughnottakenintoconsiderationhere,thereactionin theverticaldirection
Fz canalsobeincludedwherethedownwardforcesareduetotheweightof the
bendandthefluid containedwithinit.
39
FLUID MECHANICS
2.3 Pressurerisebyvalveclosure
A valveattheendofa waterpipelineof50mminsidediameterandlength500m
is closedin 1secondgivingrisetoa un~formreductioninflow. Determinethe
averagepressureriseat thevalveif theaveragevelocityof thewaterin the
pipelinebeforevalveclosurehadbeen1.7ms-l.
Solution
When a liquid flowing along a pipelineis suddenlybroughtto restby the
closureof avalveoranyotherobstruction,therewill bea largerisein pressure
duetothelossof momentumcausingapressurewaveto betransmittedalong
thepipe.Thecorrespondingforceonthevalveis therefore
F=m~
t
wherev/t is thedecelerationof theliquidandthemassof waterin thepipeline
IS
m =paL
Thus
v
F =paL-
t
2
=1000x 1tx 0.05 x 500x 1.7
4 1
=1669N
correspondingtoapressureonthevalveof
F
p=-
a
4F
1td2
4x 1669
1tX 0.052
=850.015X 103kNm-2
Theaveragepressureonthevalveonclosureisfoundtobe850kNm-2.Serious
anddamagingeffectsdueto suddenvalveclosurecanoccur,however,when
theflow is retardedat sucharatethatapressurewaveis transmittedback
40
~
CONTINUITY. MOMENTUM AND ENERGY
alongthepipeline.The maximum(or critical)timein whichthewatercanbe
broughtto restproducingamaximumorpeakpressureis
t =~
c
wherec is thevelocityof soundtransmissionthroughthewater.With noresis-
tanceattheentranceto thepipeline,theexcesspressureis relieved.Thepres-
surewavethentravelsbackalongthepipelinereachingtheclosedvalveata
time 2L/ c later.(Theperiodof 2L / c is knownasthepipeperiod.)In practice,
closuresbelowvaluesof 2L / c areclassedasinstantaneous.In thisproblem,
thecriticaltimecorrespondsto0.67secondsforatransmissionvelocityof 1480
ms-l andis belowthe1.0secondgiven.Thepeakpressurecanbesignificantly
greaterthantheaveragepressureonvalveclosurewiththepressurewavebeing
transmittedupanddownthepipelineuntilitsenergyis eventuallydissipated.It
is thereforeimportanttodesignpipingsystemswithinacceptabledesignlimits.
Accumulators(air chambersor surgetanks)or pressurerelief valveslocated
nearthevalvescanpreventpotentialproblems.
The peakpressureresultingfromvalveclosuresfasterthanthepipeperiod
canbecalculated(inheadform)from
H =vc
g
This basicequation,developedby theRussianscientistN. Joukowskyin 1898,
impliesthata changein flow directlycausesa changein pressure,andvice
versa.The velocity of sound transmission,c, is howevervariableand is
dependentupon the physical propertiesof the pipe and the liquid being
conveyed.Thepresenceof entrainedgasbubblesmarkedlydecreasestheeffec-
tivevelocityof soundin theliquid.In thiscase,thepeakheadis
H =1.7x 1480
g
=256.5m
which correspondsto a peakpressureof 2516kNm-2.However,the
Joukowskyequationneglectstoconsiderthepossibleriseduetothereduction
infrictionalpressurelossesthatoccurasthefluidisbroughttorest.Italsodoes
notconsiderthepressurein theliquidthatmayexistpriortovalveclosure- all
ofwhichmaywellbein excessof thatwhichcanbephysicallywithstoodbythepIpe.
41
FLUID MECHANICS
2.4TheBernoulliequation
An opentankof waterhasa pipelineof uniformdiameterleadingfrom it as
shownbelow.Neglectingall frictional effects,determinethevelocityof water
in thepipeandthepressureatpointsA, Band C.
105m
]
Freesurface
I I
B
c
'_'_'_0_'_'_'
2-T
2.0m
Solution
TheBernoulliequation(namedafterDanielBernoulli)is
2 2
PI vI P2 v2
-+-+ZI =-+-+Z2
pg 2g pg 2g
Thefirst,secondandthirdtermsof theequationareknownasthepressurehead,
velocityheadandstaticheadtermsrespectively,eachof whichhasthefunda-
mentaldimensionsof length.This is animportantequationfor theanalysisof
fluid flow in whichthermodynamicoccurrencesarenotimportant.It is derived
foranincompressiblefluidwithoutviscosity.Theseassumptionsgiveresultsof
acceptableaccuracyfor liquids of low viscosityand for gasesflowing at
subsonicspeedswhenchangesin pressurearesmall.
To determinethe velocityin thepipe, theBernoulli equationis applied
betweenthefreesurface(point1)andtheendof thepipe(point2) whichare
bothexposedtoatmosphericpressure.Thatis
PI =P2 =Palm
Thetankispresumedtobeofsufficientcapacitythatthevelocityofthewaterat
thefreesurfaceisnegligible.Thatis
VI ",0
Therefore
r-
Ii.-.
CONTINUITY, MOMENTUM AND ENERGY
v2 =J2g(ZI - Z2)
=.j2gxO.2
=1.98ms-I
Theaveragevelocityisthesameatallpointsalongthepipeline.Thatis
v2 =vA =vB =vC
The pressureatA is therefore
pA 0 +1 - ZA - :; J
[
1.982
)=1O00gx 2- 2g
=17,658Nm-2
The pressureatB is
PB ~p+ -zB - ~n
=1O00g +_1~:2J
=-1962 Nm-2
Finally, thepressureatC is
Pc =+1 - 'c - ~~]
(
1982
J=1O00gx -1.5- ~
=-16,677Nm-2
The averagevelocityin thepipelineis 1.98ms-I andthepressuresatpointsA,
Band Care 17.658kNm-2,-1.962 kNm-2and-16.677kNm-2,respectively.
43
FLUID MECHANICS
2.5 Pressuredropduetoenlargements
WaterflowsthroughapipewithaninsidediameterofSemata rateof10m3h-l
andexpandsintoapipeof insidediameter10em.Determinethepressuredrop
acrossthepipeenlargement.
, ,, ,,, ,, ,, ,
: a2:, ,
a ' ,
PI 1: VI V2 : P2~,- ~,-, ,, '
, I~: ,,
Flow-
Solution
If apipesuddenlyenlarges,eddiesformatthecornersandthereis apermanent
and irreversibleenergyloss. A momentumbalanceacrossthe enlargement
gIves
PIa2 +pQvI =P2a2 + pQV2
Rearranging,thepressuredropis therefore
P2 -PI =pQ(VI -V2)
a2
wheretheaveragevelocityin thesmallerpipeis
VI = 4Q
1td2
4x~
3600
1tX 0.052
=1.41ms-I
44
~
CONTINUITY, MOMENTUM AND ENERGY
andin thelargerpipeis
'2 =(:J,
=
(
°.05
)
2 x 1.41
0.1
=0.352ms-I
The pressuredropis therefore
10
1000x - x (1.41-0.352)
3600
P2-PI= 21tX0.1
4
=374Nm-2
ApplyingtheBernoulliequationoverthesection,theheadlossis
whichreducesto
HL =(VI -V2)2
2g
2
J
2
=~
(
1_2
2g vI
Fromcontinuityfor anincompressiblefluid
aIvI =a2v2
45
2' 2
VI -V2 PI -P2
HL = +
2g pg
2 2
pQ(vI -V2)- vI -V2 -
2g a2pg
2 2
2V2(vI -V2)_VI-V2-
2g 2g
FLUID MECHANICS
Then
2
[ ]
2
VI al
H L =2g 1- a2
or in termsof diameterfor thecircularpipe
V2
[ [
2
]
2
H L =2~ 1- ~~]
Therefore
H L =1.412X
[
1-
[
0.05
]
2
]
2
2g 0.1
=0.057m
The pressuredropis therefore
APi =pgHL
=1000x g x 0.057
=559Nm-2
The pressuredrop is 559Nm-2. Note thatfor a considerableenlargementwhere
a2»al the head loss tendsto
2
VI
HL =-
2g
Thatis, theheadlossduetoanenlargementis equaltoonevelocityheadbased
onthevelocityin thesmallerpipe.This is oftenreferredtoasthepipeexithead
loss.Notethatalthoughthereis alossof energy(orhead)theremaynotneces-
sarilybeadropin fluid pressurebecausetheincreasein cross-sectioncausesa
reductionin velocityandanincreasein pressure.
46
--
CONTINUITY, MOMENTUM AND ENERGY
2.6 Pipeentranceheadloss
Derivean expressionfor theentranceloss in headformfor afluid flowing
throughapipeabruptlyenteringapipeof smallerdiameter.
!~;,I I I: ~ I
FlowI al v I a v2 :a
~ -' I vc --=---I 2
I I I
:~~:
I I I: I I
I
Solution
The permanentandirreversiblelossof headdueto a suddencontractionis not
due to the suddencontractionitself, but due to the suddenenlargement
followingthe<:;ontraction.Consider,therefore,apipeof areaal whichreduces
toareaa2'Thefluidflowingintothenarrowpipeis furthercontractedforminga
venacontracta.At thispointtheareaavcis relatedtothesmallerpipeareabya
coefficientof contractionas
avc =Cca2
Beyondthevenacontracta,thefluid expandsandfills thepipe.The headloss
dueto thisexpansionis
2
HL=(vvc-V2)
2g
andis knownas theCarnot-BordaequationaftertheFrenchmathematicians
Lazare Nicolas MargueriteCarnot (1753-1823)and Jean Charles Borda
(1733-1799).Fromcontinuity
a2v 2 =avcv vc
=Cca2vvc
Therefore
v2
vvc - Cc
47
FLUID MECHANICS
Then
]
22
V2 ~-1
H L ~ 2,[c;
2
V2
=k-
2g
The constantk is foundby experiment.For asuddencontraction,theheadloss
is closeto
V2
HL =05~
2g
and is usuallyreferredto as theentranceheadloss to a pipe.Experimental
valuesare
a2fal
k
0.6
0.21
1.0
0
0.8
0.07
0
0.5
0.2
0.45
0.4
0.36
0.5
k
0
0 1.0
aia!
48
~
CONTINUITY, MOMENTUM AND ENERGY
2.7 Forceona pipereducer
Waterflowsthroughapipeofinsidediameter200mmata rateof100m3h-l.1f
theflow abruptlyentersa sectionreducingthepipe diameterto 150mm,for
whichtheheadloss is 0.2velocityheadsbasedon thesmallerpipe,determine
theforce requiredtoholdthesectioninposition.Upstreamof thereducer,the
gaugepressureis 80kNm-2.
: I_Fx
:aj
Flow Pj: VI~--,-,,,,,,
:a2
V2 : P2
--=---:~,,,,
i,J-Fx
Solution
Thevelocitiesinthelargerandsmallerpipeare
VI =4Q
1td2
I
4x 100
3600
1tx 0.22
=0.884 ms-j
and
V2 =4Q
1td21
4x 100
3600
1tX0.152
=157ms-l
The headlossatthereduceris basedonthevelocityin thesmallerpipeas
49
FLUID MECHANICS
2
V2
HL =02-
2g
2
=02 x 157
2g
=0.Q25m
Thepressureinthe200mmdiameterpipeis80kNm-2.Thepressureinthe150
mmdiameterpipeisfoundbyapplyingtheBernoulliequation
2 2
El +~ =P2 +~ + H L
pg 2g pg 2g
Rearranging
P 2 2
P2 =Pl +-(vl -v2)-pgHL2
=80X 103+ 1000x (0.8842-1572) -1000 x g x 0.Q25
2
=78,913Nm-2
The upstreamanddownstreampressureforcesaretherefore
3 rex 022
PIa 1 = 80x 10 x-
4
=25l3N
and
P2a2 =78,913x rex 0.152
4
=1394N
The forcein thex-directionis therefore
Fx =pQ(v2 -v I)-PIal +p2a2
=1000x 100x (157-0.884)-2513+1394
3600
=-1100N
A forceof 1.1kNm-2 in theoppositedirectionto flow is requiredto holdthe
reducingsectionin position.
50
~
CONTINUITY, MOMENTUM AND ENERGY
2.8 Vortexmotion
Derivean expressionfor thevariation of total head across thestreamlines of a
rotating liquid.
\P+f'1P /,v +dv
L ~ :::: "'Streamlinedr
~P~,? S_li"e
Solution
ConsideranelementofliquidoflengthL andwidthdrbetweentwohorizontal
streamlinesofradiirandr +drandwhichhavecorrespondingvelocitiesv and
v+dv.Thedifferencein radialforceisequaltothecentrifugalforce.Thatis
2
I1.pL=pLv dr
r
fromwhichthepressureheadis thereforededucedtobe
I1.p=v2dr
pg gr
Theradialrateof changeof pressureheadis therefore
~pd- 2pg =~
dr gr
while theradialchangeof velocityheadis
Q
2
d - 2 2
2g =(v +dv) - v
dr 2gdr
v dv
g dr
51
FL UID MECHANICS
The rateof changeof totalheadwithradiusis therefore
dE V2 v dv
-=-+--
dr gr g dr
=~(~+ ~~)
This is animportantresultbasedonahorizontalmovingfluid andcanbeused
todeterminethevariationof head(orpressure)withradiusfor bothforcedand
freevortexmotion.In afreevortex,thefluid is allowedtorotatefreelysuchas
in thecaseofawhirlwind,flowroundasharpbendordrainagefromaplughole.
Thereis aconstanttotalheadacrossthestreamline.Thus
dE =0
suchthatvr is aconstant.For freevortexflow it canbeshownthatthevelocity
increasesandpressuredecreasestowardsthecentre.In aforcedvortexin whichafluid is rotatedorstirredbymechanicalmeans,
thetangentialvelocityis directlyproportionaltothestreamlineradiusas
v =COr
wherecois theangularvelocity.For forcedvortexflow it canbeshownthatthe
freesurfaceis aparaboloid.
52
.liiio.....-
CONTINUITY, MOMENTUM AND ENERGY
2.9 Forcedandfreevortices
An impellerofdiameter50cmrotatingat60rpmabouttheverticalaxisinside
a large vesselproducesa circular vortexmotionin the liquid. Inside the
impellerregionthemotionproducesaforcedvortexandafree vortexoutside
theimpellerwith thevelocityof theforced andfree vorticesbeingassumed
equalat theimpelleredge.Determinethelevelof thefree surfaceat a radius
equaltotheimpelleranda considerabledistancefrom theimpellershaftabove
theliquidsuifacedepression.
Freevortex
-I'" Forcedvortex "1.
Freevortex
r3
1
z] Z2 Z3
Impeller
Datum
Solution
Fortwo-dimensionalflow,therateofchangeof totalheadH withradiusr is
givenby
dH =~
(
~+dV
)dr g r dr
(seeProblem2.8,page51)
53
FLUID MECHANICS
For theforcedvortex,thetangentialvelocityis relatedto angularvelocityby
v =mr
andthus
dv =mdr
Therefore
2m2rdH---
dr g
Integrating,thedifferencein totalheadbetweentwostreamlinesof radiir2and
rl IS
H2 2m2r2
f dB =- f rdr
HI g rj
to give
H2 -HI =af(r; - r;)
g
FromtheBernoulliequationappliedatthefreesurface(p1=P2)'thetotalhead
IS
2 2
v2 -VI
H2-Hl= +Z2-Z1
2g
2 2 2
cu(r2 - rl )
= + Z2 - Z1
2g
Combiningtheequationsfor totalheadgives
2 2 2
m (r2 - rl )
Z2 -ZI =-
2g
54
.......
CONTINUITY, MOMENTUM AND ENERGY
Sinceatthecentreof theparaboloidrl =0,theelevationof thefreesurfaceZ2 at
theradiusof theimpellerr2 is therefore
(mr2)2
Z2 -ZI =-
2g
= (2rrNr2)2
2g
( 60 r2x nX 60x 025
2g
=0.126m
Forthefreevortex
dB =0
Therefore
dv + dr :::00
V r
Integrating
loge 2 +loge r2 =0
vI rl
thenfor thefreesurface
c =vr
wherec is aconstant.At theedgeof theimpeller(r2) thisis
c =v2r2
2
=mr2
2
=2nNr2
60 2=2 x n X - x 025
60
=0.393
55
----
mr FLUID MECHANICS
II
IIII
II~II
The tangentialvelocityattheedgeof theimpelleris
c
v2 =-
r2
andatadistantpointfromtheimpelleris
c
v3 =-
r3
ApplyingtheBernoulliequationatthefreesurface(P2=P3)then
2 2
v2 -v3
23 -Z2 =-
2g
I
II!
I!
J~J -[~J
2g
Sincetheradiusr3 is largethen
c2
23 - z2 =-
2gri
0.3932
2x g X0252
=0.126m
The totaldepressionin therotatingliquidis therefore
z3 -z) =(Z3 -Z2)+(Z2 -z)
=0.126m+0.126m
=0252m
Thetotaldepressionisfoundtobe25.2cm.
56
.......
CONTINUITY. MOMENTUM AND ENERGY
Furtherproblems
(I) Waterflowsupwardsthroughapipewhichtapersfromadiameterof200
mmto150mmoveradistanceof1m.Neglectingfriction,determinetherateof
flowif thegaugepressureatthe200mmsectionis200kNm-2andatthe150
mmsectionis 150kNm-2.Waterhasadensityof 1000kgm-3.
Answer:0.192 m3s-1
(2) Showthatfor a liquidfreelydischargingverticallydownwardsfromthe
endof apipe,thecross-sectionalareaof thevenacontracta,a2' topipearea,aI,
separatedby a distance2 is
2gz =~-~
2 2 2
Q a2 al
(3) A pipeofinsidediameter100mmissuddenlyenlargedtoadiameterof
200mm.Determinethelossofheadduetothisenlargementforarateofflowof
0.05m3s-1.
Answer:1.16m
(4) Wateris dischargedfromatankthroughanexternalcylindricalmouth-
piecewithanareaof100cm2underapressureof30kNm-2.Determinetherate
ofdischargeif thecoefficientofcontractionis0.64.
Answer:0.0645m3s-1
(5) Waterisaddedtoaprocessvesselintheformofajetanddirectedperpen-
dicularlyagainsta flatplate.If thediameterof thejet is 25mmandthejet
velocityis 10ms-l,determinethepowerof thejetandthemagnitudeof the
forceactingontheplate.
Answer:491W,49.1N
(6) A processliquidofdensity1039kgm-3isfedcontinuouslyintoavessel
asajetatarateof12m3h-l.If thejet,whichhasadiameterof25mm,impinges
onaflatsurfaceatanangleof60°tothejet,determinetheforceontheplate.
Answer:2004N
(7) A jetofwater50mmindiameterwithavelocityof 10ms-I strikesaseries
of flatplatesnormally.If theplatesaremovingin thesamedirectionasthejetwith
avelocityof7 ms-l, determinethepressureontheplatesandtheworkdone.
Answer:30kNm-2,412W
57
FLUID MECHANICS
(8) Show thatthe efficiency11of a simplewaterwheelconsistingof flat
platesattachedradiallyaroundthecircumferencein which a fluid impinges
tangentiallyis
2(vI -v2)v2
11== 2
VI
wherevI is thevelocityof thejet andv2 is thevelocityof theplates.
(9) Determinetheefficiencyof theplatesin FurtherProblem(7).
Answer:42%
(10) Showthatthemaximumefficiencyof thewaterwheeldescribedin
FurtherProblem(8)inwhichajetimpingesnormallyonitsflatvanesis50%.
(11) Wateris dischargedthroughahorizontalnozzleatarateof25litresper
second.If thenozzleconvergesfromadiameterof 50mmto25mmandthe
wateris dischargedtoatmosphere,determinethepressureattheinlettothe
nozzleandtheforcerequiredtoholdthenozzleinposition.
Answer:150kNm-2,133N
(12) Twohorizontalpipeswithinsidediametersof4 cmareconnectedbya
smoothhorizontal90°elbow.Determinethemagnitudeanddirectionof the
horizontalcomponentof theforcewhichis requiredtoholdtheelbowata
flowrateof25litrespersecondandatagaugepressureof3atmospheres.Air at
atmosphericpressuresurroundsthepipes.
Answer:1063N,45°
(13) A pipelineof insidediameter30cmcarriescrudeoil of density920
kgm-3atarateof500m3h-l.Determinetheforceona45°horizontalelbowif
thepressureintheelbowisconstantat80kNm-2.
Answer:4122N, 22.5°
(14) A 200mminsidediameterpipecarriesaprocessItquidofdensity1017
kgm-3ata rateof 200m3h-l.Determinethemagnitudeanddirectionof the
forceactingona90°elbowduetomomentumchangeonly.
Answer:141N,45°
58
...........
CONTINUITY, MOMENTUM AND ENERGY
(15) A pipelinewith a diameterof 90 cm carrieswaterwith an average
velocityof 3 ms-I. Determinethemagnitudeanddirectionof theforceacting
ona 90°bendduetomomentumchange.
Answer:8097N, 45°
(16) Aceticacidwithadensityof 1070kgm-3flowsalongapipelineatarate
of 54m3h-l.Thepipelinehasaninsidediameterof 100mmandrisestoan
elevationof5m.Determinethekineticenergyperunitvolumeoftheacidand
thepressureheadattheelevatedpointif thegaugepressureatthelowereleva-
tionis 125kNm-2.
Answer:1951Jm-3,16.9m
(17) Distinguishbetweenafreeandforcedvortex.
(18) Showthatthesurfaceof a liquidstirredwithina cylindricalvessel
formingaforcedvortexis aparaboloid.
(19) Dete(minethedifferenceinpressurebetweenradiiof12cmand6cmofa
forcedvortexrotatedat1000rpm.
Answer:59.2kNm-2
(20) In a freecylindricalvortexof water,thepressureis foundtobe200
kNm-2ata radiusof 6 cmandtangentialvelocityof 6 ms-I.Determinethe
pressureataradiusof 12cm.
Answer:213.5kNm-2
(21) A cylindricaltankofradius1mcontainsaliquidofdensity1100kgm-3
toadepthof 1m.Theliquidisstirredbyalongpaddleofdiameter60cm,the
axisofwhichliesalongtheaxisofthetank.Determinethekineticenergyofthe
liquidperunitdepthwhenthespeedofrotationof thepaddleis45rpm.
Answer:904Jm-I
(22) Commentontheconsequenceofassuminginviscidfluidflowintermsof
flowthroughpipesandvortexmotionintanks.
59
Laminarflow
andlubrication
Introduction
Flowing fluids may exhibitone of two typesof flow behaviourthatcanbe
readilydistinguished.In streamlineorlaminarflow, fluid particlesmovealong
smoothparallelpathsor layers(laminae)in thedirectionof flow with only
minormovementacrossthestreamlinescausedby diffusion.
Over thecenturies,theexistenceof laminarandturbulentflow hasbeen
studiedextensivelyby manyprominentscientists.In 1839,it wasfirstnotedby
theGermanhydraulicsengineerGotthilfHeinrichLudwig Hagenthatlaminar
flow ceased,whenthevelocityof a flowing fluid increasedbeyonda certain
limit.With muchworkonthesubjectoverthefollowingthreedecades,Hagen
finallyconcludedin 1869thatthetransitionfromlaminartoturbulentflow was
dependenton velocity,viscosityandpipe diameter.Around the sametime,
Frenchphysicianandphysicist,JeanLouisMariePoiseuille,whilstresearching