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,whilstresearchingtheeffectsof blood flow in veins,reportedsimilarindependentwork on the
viscosityandpressuredropof waterincapillariesandreachedthesamemathe-
maticalconclusionsasHagen.
It wasnotuntil1883thatBritishscientistOsborneReynoldsshowedbyboth
dimensionalanalysisandexperimentthatthetransitiondependson a single
dimensionlessparameterwhichbearshisname.This is givenby
Re =pvd
~
whereReis theReynoldsnumber,p is thedensity,v theaveragevelocityand
~theviscosityof thefluid, andd is thediameterof pipe.
Reynolds'experimentsinvolvedinjectingatraceof colouredliquidintothe
flow of waterin a horizontalglasstube.At low flowratesthecolouredliquid
was observedto remainasdiscretefilamentsalongthetubeaxis, indicating
flow in parallelstreams.At increasedflow, oscillationswereobservedin the
filamentswhicheventuallybrokeupanddispersedacrossthetube.Thecritical
61
FLUID MECHANICS
valueof Reynoldsnumberfor thebreak-upof laminarflow wasfoundto be
about2000whileturbulentflow wasnotfoundtooccuruntilabove4000.The
so-calledtransitionzone,whichlies betweenReynoldsnumbersof 2000and
4000, was found to be a region of fluid streamlineinstability and
unpredictability.
The majorityof mathematicalproblemsinvolvinglaminarflow tendto be
straightforward.The basisof mostcalculationsinvolvesapplyinga simple
equilibriumforcebalanceto theflowing fluid with fully developedlaminar
flow. Withoutconsideringinertialforces,theaccelerationof particlesin the
fluidcausedbythepressuregradientis retardedbytheviscousshearstressesset
up by the velocity gradientperpendicularto the directionof flow. This
approachformsthebasisof theindependentworkofbothHagenandPoiseuille,
fromwhenceit is possibletodetermineimportantflow detailssuchasvelocity
profiles andratesof flow throughpipes,gapsandchannelsas well as the
viscousandlubricatingeffectsof fluidsin bearings.
62
...1lIIIo..-
LAMINAR FLOW AND LUBRICATION
3.1 Reynoldsnumberequations
Establishequationsfor Reynoldsnumberfor theflow of a fluid withaverage
velocityv in apipeof insidediameterd, in termsof kinematicviscosity,volu-
metricflowrate,massflowrateandmassloading.
solution
TheReynoldsnumberis animportantdimensionlessparameterandis valuable
for identifyingwhethertheflow of a fluid is eitherlaminaror turbulent.The
Reynoldsnumberis traditionallygivenas
Re =pvd
/1
It canbealternativelyexpressedin termsof kinematicviscositywhichis
v=f.:!-
p
to giveaReynoldsnumberexpressionof
Re =vd .
v
Sincetheaveragevelocityv is relatedto volumetricflowrateby
v =4Q
1td2
thentheReynoldsnumbercanbegivenby
Re =4pQ
/11td
Likewise,sincevolumetricflowrateis relatedto massflowrateby density
Q=m
p
thenReynoldsnumberis
Re =4m
/11td
63
FLUID MECHANICS
Finally, sincemassloadingis relatedto massflowrateby flow area
L =4m-
nd2
TheReynoldsnumberistherefore
Re =Ld
~
It shouldbenotedthattheReynoldsnumberisofprimaryimportanceinconsid-
eringtheflow of fluidsthroughpipesandopenchannels,andaroundobjects.
As a dimensionlessparametertheReynoldsnumber(Re =pvd/~)represents
theratioof inertialtoviscousforcesinaflowingfluidandis directlyassociated
with theboundarysurfaceoverwhich thefluid passes.For laminarflow in
circularpipes(Re<2000),theviscousforcesdominatewhiletheinertialforces
areof little significance.In turbulentflow thereverseis truewiththelaminar
sublayerbeingdestroyedandtheviscousforcesbeingof littlesignificance.
The Reynoldsnumberis alsoof importancein consideringtheviscousdrag
onsubmergedobjects.For aReynoldsnumberbelowavalueof 0.5theprecise
shapeof theobjectis of lessimportancethantheviscosityof thefluid andthe
velocityof flow. For suchconditionsit is possibleto derivearelationshipfor
thedragonobjectsin whichtheinertialeffectsarenotincluded.For Reynolds
numbersabove0.5, however,the reverseis againtrue- the influenceof
viscositydiminisheswhile the influenceof inertiaprogressivelydominates
withincreasingvaluesofRe.
64
~
LAMINAR FLOW AND LUBRICATION
3.2 Laminarboundarylayer
Determinethefurthestdistanceintoapipewithan insidediameterof8mmand
a well-roundedentranceatwhichfully developedlaminarflow begins.
Developingboundarylayer Fully developedlaminarflow
}
I
L. I. ~
Solution
Whenafluid withuniformvelocityflow entersa well-roundedpipe,entrance
particlesclosetothepipewall arebroughttorestduetotheviscousproperties
of thefluid. This is knownas theno-slipcondition.A steepvelocitygradient
thenexistsin thefluidvaryingfromzeroatthepipewalltotheuniformvelocity
in thebulk of thefluid. As thefluid movesfurtherinto thepipe,theviscous
retardationof theparticlesin adjacentlayersgraduallyincreasesin thickness.
For thesteadyflow throughthepipe,thefluid nearthecentreof thepipeaccel-
eratesuntil an equilibriumconditionis reachedfor fully developedlaminar
flow withaparabolicvariationof velocitywithpiperadius.Theregionwhere
thevelocityis changingandonewhereit is uniformis knownasthelaminar
boundarylayer.Sincetheuniformvelocityis approachedasymptotically,the
edgeoftheboundarylayerisdefinedasapointwherethefluidvelocityreaches
99%of itstheoreticalmaximumvalue.Theregionpriortothesectionatwhich
laminarflow isfully developedisknownastheentrancetransitionlengthL', for
which a theoreticalformulaproposedby Henry L. Langhaarin 1942and
supportedby goodexperimentalagreementis givenby
L' =0.058Red
Sincethemaximumentrancetransitionlengthoccursfor aReynoldsnumberof
2000then
L' =0.058x 2000x 0.008
=0.928m
Thefurthestdistancethefluidcanflowintothe8 mminsidediameterpipe
beforefullydevelopedlaminarflowcanexistis0.928m.
65
FLUID MECHANICS
3.3 Velocityprofilein a pipe
Derivean expressionfor thelocal velocityof afluid flowing withfully devel-
opedlaminarflow througha horizontalpipeof radiusR andsketchthevelocity
profile.
,I
,
,2
,
R
PI- r P21---.-.-. .-.-.-.
<1--
, ,, ,, L ,,- ~,, ,
Solution
Consideracylindricalelementofthefluidtobewellawayfromthewallonthe
centrelineof thepipewherethedifferenceofpressureforceacrosstheends
providesthedrivingforcetoovercomethefrictiononitsoutersurface.For
uniformflow,anequilibriumforcebalanceontheelementistherefore
2
(PI -P2)rtr ='t2rtrdL
Thatis
2
(
bP
)
2
PIrtr - PI + bL dL rtr ='t2rtrdL
The viscousshearstressis givenby
dvx
'T=/-l-
dr
Let
bp =f1p
bL L
66
~
LAMINAR FLOW AND LUBRICATION
andapplyingtheno-slipconditionatthewall (vx =0atr =R) then
Vx -1 f1p r
fdvx =-- f rdr
0 2/-l L R
wheref1pis thepressuredropoverthelengthof element(p1-P2)'
Integrationgives
[
2
]
r
-1 f1p r
vx=2/-lL2R
-1-f1p(R2 -r2)
- 4/-l L
Thevelocityof thefluidthereforehasaparabolicvariationwithpiperadius.
R
r
--(f:-
CI)
;:J
Q 0 1- -'-
< -.--.
i:>::
R
0
VELOCITY
Vmax
Note thatthe maximumvelocityof the fluid, vmax'occursat the furthest
distancefromthewall (r =0).Thatis
v =1- f1pR2
max 4/-l L
67
FLUID MECHANICS
3.4 Hagen-Poiseuilleequationfor laminarflow
in a pipe
A processvesselis to besuppliedwithglycerolof SG 1.26andviscosity1.2
Nsm-2from a storagetank60maway.A pipewithan insidediameterof 12.6
cmisavailabletogetherwithapumpcapableofdevelopingadeliverypressure
of 90 kNm-2overa widerangeofflows. Determinetheglyceroldeliveryrate
usingthisequipment.Confirmthattheflow is laminar.
vx
--_.
R
--fL.
Solution
To determinetherateof flow of thefluid throughthepipewith laminarflow
considera thin annularelementbetweenradiusrand r +dr for which the
velocity can be takenas constant.The elementalvolumetricflowrate is
therefore
dQ =vx21trdr
Thetotalrateof flowcanthereforebefoundbyintegratingacrosstheentire
piperadius
Q R
fdQ =fvx 21trdr
0 0
Sincethelocalvelocityis relatedtoradiusby
-~ b.p(R2 - r2)
vx-4flL
(seeProblem3.3,page66)
theintegrationis therefore
Q R
fdQ =~ b.pf(R2 - r2)rdr
0 2flLo
68
..........
LAMINAR FLOWAND LUBRICATION
whichgives
Q=~b.p
[
R2r2 r4
]
R
211 L --r 2 4 0
=~ b.pR4
8fl L
ThisequationwasderivedindependentlybybothHagenin1839andPoiseuille
in 1840andis knownastheHagen-Poiseuilleequation.Therateof flowis
therefore
Q =~ x 90x 103x (0.126
)
4
8x1.2 60 l 2
=0.00773m3s-]
As acheckfor laminarflow, theReynoldsnumberis
Re =4pQ
fl11;d
4x 1260x 0.00773
12x1txO.126 (seeProblem3.1,page63)
=82
andisbelow2000,thereforeconfirminglaminarflow.
69
FLUID MECHANICS
3.5Pipediameterforlaminarflow
In thedesignof a small-scalebioprocessplant,a shearsensitiveNewtonian
liquid of density1100kgm-3andviscosity0.015Nsm-2is to betransported
alonga lengthofpipeata rateof4 litresperminute.Determinethediameterof
thepiperequiredif thepressuredropalongthepipe is nottoexceed100Nm-2
per metrelength.Commenton thevalueof theReynoldsnumber.
Solution
Assuminglaminarflow,theHagen-Poiseuilleequationis
Q =~ !1pR4
811L
(seeProblem3.4,page68)
Rearranging,thepiperadiusis
R~l:~ r
-
[
8x oms x 0.004
)
t
- 60
1tX 100
=Om26m
The diameterof thepipeis therefore2S.2mm.To confirmlaminarflow, the
Reynoldsnumberis
Re=4pQ
1tl1d
4x 1100x O.oO,!
60
1tx oms x 0.Q2S2 (seeProblem3.1,page63)
=247
TheflowisthereforelaminarwithaReynoldsnumberof-247.
70
............
LAMINAR FLOW AND LUBRICATION
3.6 Laminarflowthroughataperedtube
A lubricatingoil of viscosity0.03Nsm-2is deliveredtoa machineat a rateof
10-7 m3s-1througha convergent,taperedtubeof length50 em with an
upstreamdiameterof10 mmanddownstreamdiameterof5mm.Determinethe
pressuredifferentialwhichwill maintaintheflow. Entranceand exitlosses,
andinertiaeffectsduetothechangein velocityin thetube,maybeneglected.
L NottoscaleI"
10mm
Flow .-.-.-.----
50em .1I"
Solution
Forlaminarflow,therateofflowisgivenbytheHagen-Poiseuilleequation
Q =~ !1pR4
811L
(seeProblem3.4,page68)
Overthelengthof thetaperedtube,thedifferentialpressure!1pincreaseswith
decreasingtuberadius.Fromgeometry
R =O.oOS(l-L)
whereRandL arebothmeasuredinmetres.SubstitutingforR, thetotalpressure
differentialis thereforeobtainedbyintegratingoverthelengthof tube.Thatis
811Q 0.5 dL
!1p=1to.oos4l(1-L)4
This is madeeasierusingthesubstitution
u=l-L
71
FLUID MECHANICS
Thus
du=-dL
The integrationthenbecomes
/1p- -811Q f du
1tO.oOS4U4
Integratingwithrespectto u gives
A 811Q u-3
tip =
1tO.oOS43
Thatis
[ ]
0.5
/1p- 811Q 1
-31tO.oOS4 (1-L)3 0
_8xO.Q3xlO-7 X
[
I -1
]3x 1tx O.oOS4 (1- 0.5)3
=28.5Nm-2
The pressuredifferentialis foundtobe28,SNm-2,
.
72
~
LAMINAR FLOW AND LUBRICATION
3.7 Relationshipbetweenaverageandmaximum
velocityina pipe
Deducetherelationshipbetweentheaverageandmaximumvelocityfor afluid
flowing withfully developedlaminarflow througha horizontalpipe,
Solution
Theaveragevelocityof thefluidQ canbedeterminedfromthetotalrateof
flowthroughthetotalcross-sectionalareaaavailableforflow
v=~
a
Sincetheflow Q is givenby theHagen-Poiseuilleequation(seeProblem3.4,
page68),then
~~£R4
811Lv=-
1tR2
- ~ /1pR2
- 811 L'
Also, sincethemaximumvelocityoccursatthefurthestpointfromthewall
(r =0)(seeProblem3,3,page66)
v =~ /1pR2
max 411 L
thentheratioof averagevelocitytomaximumvelocityis
1 /1p 2--R
~=8I1L
vmax ~ /1pR2
411L
I
2
That is, theaveragevelocityof a flowing fluid with laminarflow is half the
maximumvelocity,
73
FL UID MECHANICS
I vmax ~ I
I ~ Iv
0:::
(I)
~
0-'--- --
<t:
~
0:::
0 VELOCITY
-<t-
Notethattheaveragevelocitycanbealternativelyexpressedin termsof pipe
diameteras
v =~ !1pd2
3211L
andmaximumvelocityas
I !1pd2
v max =1611L
II'
74
~
LAMINAR FLOW AND LUBRICATION
3.8 Relationshipbetweenlocalandmaximum
velocityina pipe
Fuel oil of density900kgm-3andviscosity0.1Nsm-2flows throughapipe of
insidediameter100mm.Thevelocityat thecentreline is 2 ms-l. Derivean
expressionrelatingthelocal tomaximumvelocityof afluid flowing withfully
developedlaminarflow througha horizontalpipe, and determinetheshear
stressin thefuel oil at thepipe wall andthepressuredropper unit lengthof
pipe.
Solution
Thelocalvelocityisgivenby
- J !1p(R2 - r2)
vx-4l1L
(seeProblem3.3,page66)
andthemaximumvelocityoccursatthefurthestpointfromthepipewall (r =0).
v =J !1pR2
max 411 L
Therefore
J !1p(R2 - r2)
~=4I1L
vmax J !1pR2
411 L
=l-(ir
Sincetheviscousshearstressis
dvx
1:=11~
thevelocitygradientis thereforeobtainedby differentiationto give
dvx =-2v ~
dr maxR2
Thus
r
1:=-211vmaxR2
75
FLUID MECHANICS
At thepipewall (r =R),theshearstressis therefore
-2/1vmax
't =
W R
-2 x 0.1x 2
0.05
=(-)8 Nm-2
The negativesignindicatesthattheshearstressis in theoppositedirectionto
flow. To verifylaminarflow, theReynoldsnumberis
Re =pvd
/1
Vmaxd
PT
/1
2
900x - x 0.1
2
0.1
=900
whichis lessthan2000.Finally,thepressuredropalongthepipeis obtained
from
I1p- 4/1vmax
L- R2
4x 0.1x 2
0.052
=320Nm-2m-l
The pipewall shearstressis foundto be8 Nm-2 andpressuredropto be320
Nm-2 permetrelengthof pipe.
76
~
LAM IN AR FLOW AND L UB RICA TION
3.9 Maximumpipediameterfor laminarflow
A lightoil of viscosity0.032Nsm-2andSG 0.854is tobetransferredfromone
vesseltoanotherthrougha horizontalpipe.If thepressuredropalongthepipe
is notto exceed150Nm-2per metrelengthto ensurea satisfactoryflow and
flow is alwaystobe laminar,determinethemaximumpossibleinternaldiam-
eterof thepipe.
Solution
The maximumvalueof Reynoldsnumberatwhich laminarflow canexist is
2000.Thatis
Re =pvd
/1
=2000
wheretheaveragevelocityis givenby
v =~ I1pR2
8/1 L
(seeProblem3.7,page73)
Substituting,theReynoldsnumberis therefore
Pl ~ I1p(~)
2
Re = 8/1L l2
/1
=~l1p ~
32/12Ld'
or rearrangingin termsof pipediameter
d=
[
32/12Re
j
t
P I1pL
I
=
[
32X 0.0322x 2000
]
3
854 x 150
=0.08m
The maximuminsidediameteris foundtobe8 em.
77
FL UID MECHANICS
3.10Verticalpipeflow
A viscousliquid,ofdensity1000kgm-3andviscosity0.1Nsm-2,flowsbetween
twostoragetanksthrough10mofverticalpipewhichhasan insidediameterof
5 cm.Therateofflow is controlledbya valveata rateof14m3h-l.Determine
thepressuredropacrossthevalvewhenthedifferencein levelsbetweenthetwo
storagetanksis 12m.Bothtanksareopentoatmosphere.Neglectlossesin the
pipefittings.
------
r
-----
L
-----------
Solution
Gravityhasno effectfor laminarflow in a horizontalpipe.For verticalflow,
however,gravitationaleffectsarein theformof hydrostaticpressurewhere,in
thiscase,thepressuredropacrossthevalveis
f:..Pvalve=pg(L + H) - 128Q/-lL
nd4
14
128x - x 0.1x 10
=1000x g x (10+2)- 3600
nx 0.054
=92,368Nm-2
Thepressuredropacrossthevalveis foundtobe92.4kNm-2.
78
L
LAMINAR FLOW AND LUBRICATION
3.11Filmthicknessin a channel
Deriveexpressionsfor thevelocitydistribution,totalflow andfilm thicknessof
aNewtonianliquidflowingundergravitydownasurfaceinclinedatananglea
to thehorizontal.Assumethattheflow is laminarandfully developed.
Solution
Assumingsteadyflow, anegligibleshearstressatthegas-liquidinterfaceanda
uniformpressurein thedirectionof flow (x-direction), anequilibriumforce
balanceontheelementgives
WL(8 - y)pgsina ='TWL
wheretheshearstressis givenby
dvx
'T=/-ld:Y
The velocityprofilethroughthefilm canbefoundusingtheno-slipboundary
condition(y =0atvx =0).Therefore
VIdvx =pgsinas(8- y)dy
0 /-l 0
Integrating
,,~pg~ne[8Y -y:]
79
FLUID MECHANICS
The volumeflowrateof theliquid is givenby
Q 0
fdQ =fWvxdy
0 0
Thereforesubstitutingforvx
Q =Wpgsin8J
(
6Y- L
}
y
fl 0 2
Integrating
Q =Wpgsin8
[
Oy2 - L
]
0
fl 2 6 0
- Wpgsin803
3fl
Rearranging,thefilm thicknessis thereforerelatedtoflowrateby(
3flQ
J
+
0=
Wpgsin8
The film thicknesscanalsoberelatedtotheaveragevelocityby
Q
v =oW
= Wpgsin803
3flOW
=pgsin802
3fl
Rearranging,thethicknessof thefilm of liquidis therefore
(
I
0= 3flv -
pgsin8J
This equationhasbeenfoundby experimentto hold for a Reynoldsnumber
(Re =pvO/fl)lessthan500.
80
l
LAMINAR FLOW AND LUBRICATION
3.12Flowdownaninclinedplate
In thefirst oftwoexperimentswater,ofdensity1000kgm-3andviscosity0.001
Nsm-2,waspumpedtothetopofaflat plateandevenlydistributedalongone
sideof it. Theplatewasangledatapproximately60°tothehorizontalandthe
flowratewas3x10-6m3s-1permetrelengthofplate.It wasfoundthatthefilm
thicknesswas 1.016xlO-4m. In thesecondexperiment,a differentliquid of
density1500kgm-3waspumpedto thetopof theplate at 3x1o-6m3s-1per
metrelengthof plate and thefilm thicknesswasfound to be 1.25x1o-4m.
Determinetheviscosityof thesecondliquidandtheexactangleof theplate.
Solution
The rateof flow of liquiddowntheplateis givenby
Q =Wpgsin803
3fl
(seeProblem3.11,page79)
For thefirstexperimentinvolvingwater,theexactangleof inclinationis found
by rearrangingtheequationto
sin 8'= 3Qfl
Wpg03
3x 3x 10-6x 1x 10-3
1x 1000x g x (1.016x 10-4)3
=0.875
The exactangleof inclinationwastherefore
8=61°
For thesecondexperiment,theviscosityis foundby rearrangingtheequation
for flow to
Wpgsin 8 03fl=
3Q
- 1x 1000x g x 0.875x (1.25x 10-4)3- 63x 3x 10-
3 -2=2.795x 10- Nsm
The exactanglewas61°andtheviscosityof thesecondliquidwasfoundtobe
2.795x 10-3Nsm-2.
81
FLUID MECHANICS
3.13Flowdownaverticalwire
Afilm ofa Newtonianfluidflowsdowntheoutsideofa stationaryverticalwire
whichhasa radiusR. Deduceanexpressionfor thelocal velocityof thefluid,
vz' downthewire.
6
II~
III I
~ II
II
I "1
L
Film Wire
Solution
Consideraforcebalanceonanelementof thefluidbetweenradiirandR+8
assumingnosheareffectsatthefreesurface.Then
pgL1t((R+ 8)2 - r2) =t21trL
wheretheviscousshearstressis givenby
dv
,. =!l~
dr
Substitutingandnotingthatthefluid is stationary(no-slipcondition)atthewire
surface(vz =0atr =R),thelocalvelocityatradiusr measuredfromtheaxisof
thewire canbefoundfrom
82
l
LAMINAR FLOW AND LUBRICATION
V
I pgr
[
CR+8)2
}
dvz =- f - r r
0 2!lR r
=~g
(
(R + 8)2Jdr - J rdr
]!l Rr R
Integratingwithrespectto r gives
Vz =pg
[
CR+8)210ge[r]~_
[
C
]
r
2!l 2 R
=~~(2CR+8)210gel~)+(R2-r2»)
While thevelocityof thefluidis zeroatthewiresurface,it is atitsgreatestatthe
freesurface.Theaveragevelocity,rateof flow andshearstresswithinthefluid
cansimilarlybedeterminedandareusefulresultswhenconsideringtheprocess
ofwirecoatil}g.This involvesthedrawingofawireverticallyupthroughabath
of viscousliquid takingwith it a film of theliquid. In practice,however,the
coatingliquid- suchasapolymermelt- isnotusuallylikelytoexhibit
Newtonianbehaviourand,ondrying,densitychangeandshrinkagemustalso
betakenintoconsideration.
83
III
FLUID MECHANICS
3.14Flowandlocalvelocitythrougha gap
Derivean expressionfor thelocalvelocityandtotalflow ofa Newtonianfluid
withfully developedlaminarflow betweentwoflat horizontalplatesseparated
byadistance2H.
I I
""""'~""""";""""'t:~I"
Flow ----
1:
~~~~~~~~~~~1~~~~~~~~~~i~~~~~~~~~:~~~~~~, ",,"
Solution
Assumingthattheplatesaresufficientlywidethatedgeeffectscanbeneglected
(thatis,W»2H), anequilibriumforcebalanceontheelementis
2yW(pl - P2) =21WdL
Thatis
2YWlPl -(PI + ~~dL))=21WdL
wheretheshearstressis givenby
dvx
1=11-
dy
Let
bp =b.p
bL L
The local velocityvx at a distancey can thereforebe found applyingthe
no-slipboundarycondition(vx =0aty =H). Thatis
Vx -1b.pY
fdvx =- - fydy
0 I1LH
Integrationwithrespecttoy gives
84
I
I
L
LAMINAR FLOW AND LUBRICATION
v =-I b.p
[
C
]
Y
x ilL 2H
- ~ b.p(H2 - y2)
- 211 L
The velocityof the fluid thereforehas a parabolicvariationwith distance
betweentheplates,withthemaximumvelocityaty =0.
- - --4.-
The total,flow of fluid can be determinedby consideringthe sum of the
elementalflows througha narrowslit of thicknessdy. From symmetryabout
themid-pointbetweentheplatestherearetwosuchnarrowslits.Thus
Q H
fdQ =2Wfvxdy
0 0
Substitutingforvx
Q H
fdQ =W b.pf(H2 -y2)dy
0 11 L 0
andintegrationwithrespecttoy gives
Q =W b.P
[
YH2 _L
]
H
11 L 3 0
- 2W b.pH3
- 311L
In additionto assumingthatedgeeffectscanbeneglected,for incompressible
viscousflow of Newtonianfluids throughgapsor betweenplatesit is also
assumedthatthereis nopressuregradientacrossthefluidlayerandthatgravita-
tionalforcesarenegligible.
85
H
T
v
>LI I................:........
u
L - _td -Z
'"
is
H
0 v
VELOCITY max
FLUID MECHANICS
3.15Relationshipbetweenlocalandaveragevelocity
throughagap
Derivean expressionrelatingthelocal velocityof a Newtonianfluid flowing
withfullydevelopedlaminarflow betweentwohorizontalparallelplatessepa-
ratedbya distance2H in termsof theaveragevelocity.
Solution
Thetotalflowisrelatedtoaveragevelocityby
Q =2WHv
The averagevelocityis thereforeobtainedby substitutingan expressionfor
laminarflow betweentheplates(seeProblem3.14,page84)
2W b.pH3
Q =3; L
togive
2W b.pH3
3/1 L
v = 2WH
- ~ b.p H2
- 3/1 L
The localvelocityvx betweentheplatesis givenby
- ~ b.p(H2 - y2)
Vx -2/1 L
(seeProblem3.14,page84)
Therefore
Vx
~ b.p (H2 - y2)
2/1L
= ~ b.pH2
3/1L
v
The localvelocityis thereforerelatedtoaveragevelocityby
'<=~[l-[~n
wherey isthedistancemeasuredawayfromthemiddleofthegap.
86
L
LAM IN AR FLOW AND L UB RICA TION
3.16Relationshipbetweenaverageandmaximum
velocitythroughagap
Deducetheratioofmaximumvelocitytoaveragevelocityfor aNewtonianfluid
flowing withlaminarflow througha gapbetweentwoflat horizontalplates.
Solution
The maximumvelocityoccursatthefurthestpointfromeithersurface(y =0)
(seeProblem3.14,page84).Thatis
v =~b.p H2
max 2/1 L
Sincetheaveragevelocityis givenby
1 b.pH2
v =3/1 L
(seeProblem3.15,page86)
Then
~b.pH2
v max=°2/1L
v ~b.p H2
3/1 L
3
2
Themaximumvelocityis oneandahalftimestheaveragevelocitythroughthe
gap.
~ ::t:u
z
1=5
U)
is ::t:
0 v Vmax
VELOCITY
87
FLUID MECHANICS
3.17Shearstressforflowthroughagap
Oil, with a viscosity of 1.5 Nsm-2, flows with fully developed laminar flow
through a gapformed by two horizontal parallel plates set20 mmapart. Deter-
mine the magnitude and direction of the shear stresses that act on theplates
when theaverage velocity is 0.5 ms-l.
Solution
The velocityof thefluid is givenby
" =3;(1- (~n (seeProblem3.15,page86)
The velocitygradientis therefore
d;;=dt;(l-(~ r]J
=-3yv
H2
Theviscousshearstressateitherplatesurface(wall)wherey =H istherefore
't ='tw
=~(d:;1
3v
=-~-H
=-1.5x 3x 0.5
0.Q2
2
=-225Nm-2
Theshearstressatthewallis 225Nm-2andactsintheoppositedirectionto
flow.
88
.........
LAMINAR FLOW AND LUBRICATION
3.18Flatdiscviscometer
A device used to measure theviscosity of a viscous liquidconsistsofaflatdisc
whichrotatesonaflat surfacebetweenwhichis sandwichedtheliquidunder
investigation.Thedischasadiameterof 5 cmandproducesashearstressof
400Nm-2for a rotationalspeedof 600 rpm.Determinetheviscosityof the
liquidandthetorqueontherotatingdiscif theclearancebetweenthediscand
surfaceis 2 mm.
<L
(0
Solution
The viscosityof liquids can be determinedexperimentallyusing a device
knownasaviscometer.Therearevarioustypesof viscometeravailableandall
arebasedontheprinciplesof laminarflow. Theflatdiscviscometeris perhaps
oneof thesimplestforms,andcanbeusedfor quickestimatesofviscosity.For
theliquidsandwichedbetweenthesurfaceandtherotatingdisc,theshearstress
is givenby
dvx
't =~-
dy
orintermsofangularvelocityandclearancebetweendiscandplate
't =11<or
twhere
89
FLUID MECHANICS
co=2nN
=2 x n X 600
60
=62.8rads-l
Rearrangingin termsofviscosity
/1=~
COr
400x 2x 10-3
62.8x 0.D25
=0509Nsm-2
The torqueis givenby
R
T =2nf Tr2dr
0
Substitutingfor shearstress
R
T =2n/1cofr3dr
t 0
Integrating
T ~ 2.~t:I
=n/1coR 4
2t
n x 0509x 62.8x 0.D254
- 3- 2 x 2 x 10-
=9.8x 10-3Nm
Theviscosityandtorquearefoundtobe0.509Nsm-2and9.8xlO-3Nm.
90
~
LAMINAR FLOW AND LUBRICATION
3.19Torqueona lubricatedshaft
A shaft100mmindiameterrotatesat30rpsina bearingoflength200mm,the
surfacesbeingseparatedbyafilm ofoil 0.02mmthick.Determinetheviscous
torqueonthebearingandpowerrequiredtoovercomethefrictional resistance
if theviscosityof theoil is 0.153Nsm-2.
~~~~~~~~~~~~~~~~~~~l~~~~~~~~~~~~~~~~~~~
S'of<-8- - - --- _1- - 1t-
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I.. L .1
Solution
It is assumedthattherearenegligibleendeffectsandthattheshaftis lightly
loadedsothattheshaftrunsconcentrically.Theresistanceperunitareaorshear
stressis
dv
T=/1-
dy
Theoil film is verythinsuchthattheviscousshearstressmaybegivenin terms
of angularvelocityatradiusr
cor
T =/1-
t
2nNr
=/1-
t
The tangentialresistance(viscousdrag)onthebearingis
F =unrL
91
FLUID MECHANICS
The resistingtorqueis therefore
T =Fr
=unrLr
2nNr
=~-2nrLr
t
4n2r3NL
=~
=0.153x 4x n2x 0.053x 30x 02
2x 10-5
=226Nm
Thepowerof theshaftis
P =Tm
=T2nN
=226x 2x nx 30
=42,600W
The torqueon thebearingis foundto be 226Nm andthepowerrequiredto
overcomethefrictionalresistanceis 42.6kW.
Notethatif theradialclearancebetweentheshaftandbearingis notsmall,
theadjustmenttotorqueis made
2 2
4n rl r2NL
T=~
r2 - rl
whererl andr2 aretheradiiof theshaftandbearing,respectively.
92
~
LAMINAR FLOW AND LUBRICATION
3.20Lubricatedcollarbearing
Theaxial thrustona shaftis takenbya collar bearing,theoutsideandinside
diametersof which are 200 mmand 125mm,respectively.The bearingis
providedwithaforcedlubricationsystemwhichmaintainsafilm ofoil ofthick-
ness0.5mmbetweenthebearingsurfaces.Ata shaftspeedof300rpmthefric-
tionlossofpoweramountsto22 W.Determinetheviscosityof theoil.
Not to scale
~
--r
I R]
I R2
r
~ dr
Solution
Bearingsaremechanicaldevicesusedtoreducethefrictionbetweentwoparts
in contactwithoneanotherandwhichmovewithrespecttooneanother.Collar
bearings are tobe found on vertical rotating shafts and take a load on a
stationarysupport.To reducefriction,a film of lubricatingfluid separatesthe
rotatingcollar and stationarysupport.The viscousresistanceof the collar
bearingcanbeobtainedbyassumingthefaceof thecollartobeseparatedfrom
thebearingsurfaceby athinfilm of oil of uniformthicknesswheretheviscous
shearstressis givenby
dv
1:=~-
dy
For thethinfilm of oil, theshearstressmaythereforebeexpressedin termsof
angularvelocityas
1:=~mr
t
93
FLUID MECHANICS
The tangentialviscousforcechangeswithradius.For athinring ataradiusr
='t21trdr
COr
=fl-21trdr
t
Themomentoftangentialforceontheringistherefore
COr
=fl-21trdr r
t
For thewholebearingto consistof similarconcentricrings,thetotaltorque
requiredto overcomethe viscous resistanceof the bearingis found by
integrating
R[
T =21tflCOf r3dr
t R2
=1tflCO(R:- R;)2t
The powerlossis relatedtotorqueby
P =Tco
Thereforerearrangingin termsof viscosity
2tP
fl=
1tCO2(R4- R4)I 2
2x 0.0005x 22
(
300
)
2
1tX 2x 1tx '60 X (0.14- 0.06254)
=0.0837Nsm-2
The viscosityof theoil is foundtobe0.0837Nsm-2.
94
.-.
LAMINAR FLOW AND LUBRICATION
Furtherproblems
(1) Explainwhatis meantby aNewtonianfluid.
(2) A cubicblockof volume1000cm3is supportedby afilm of oil 0.1mm
thickonahorizontalsurface.Determinetheviscousdragif theblockis moved
acrossthesurfacewith a velocityof 1 ms-I. The viscosityof theoil is 0.05
Nsm-2.
Answer:5N
(3) A blockis supportedbyafilmof oil 0.2mmthickonahorizontalsurface.
Theareaof theoil film is 100cm2,theviscosityof theoil is0.05Nsm-2.Deter-
minetheviscousdragatavelocityof 1IDS-I.
Answer:2.5N
(4) A fluid flows at a rateQ in a pipeof diameterdl andhasa Reynolds
numberReI' If thefluidpassestoapipeofdiameterd2'deducetherelationship
betweenReI andRe2.
(5) Showthatthelocal velocityof a liquid with laminarflow alongahori-
zontalchannelof widthW in termsof flowrateQ anddepthbcanbegivenby
3Q
[
2
Vx =- by-L
Wb3 2
wherey is thedepthmeasuredfromthebaseof thechannel.
(6) Showthatthemaximumvelocityof a liquid with laminarflow alonga
horizontalchannelof widthWand depthbcanbegivenby
3Q
vmax- 2V1b
(7) Sketchthevelocityprofile andwritedowntheboundaryconditionsfor
the laminarmovementof a Newtonianliquid betweentwo largeflat plates
whicharehorizontal,paralleland1mmapart,if thetopplatemovesat+0.01
ms-I andthelowerplatemovesat-0.01 ms-I andthereis nonetflow of fluid
betweentheplates.
(8) Outlinetheassumptionsonwhichtheflow of fluid withfully developed
laminarflow throughapipeis based.
95
FLUID MECHANICS
(9) Glycerol,of SG 1.26andviscosity1.2Nsm-2,flows betweentwo flat
plates2 emapart.If thepressuredropalongtheplatesis 500Nm-2 permetre
length,determinethemaximumvelocityof theglycerol.
Answer:0.021ms-l
(10) Show thatthe local velocityof a Newtonianfluid with laminarflow
flowingthroughanannularspacebetweentwotubescanbegivenby
v =~ !1.p
[
2 log [
L
]
'4~L(R-y2)_(R2_,2)
.
eR
log e [~]
whereR is theinnerradiusof theoutertube,r is theouterradiusof theinner
tube,y is aradiusbetweenRand r, ~is thedynamicviscosityof thefluid and
!1.p/L is thepressuredropperunitlengththroughtheannulus.
(11) Sketchthevelocityprofileof a Newtonianfluid throughanannulargap
withlaminarflow.
(12) Sketchthevelocityprofilesfor theflow oftwo immiscibleliquidsA and
B alonga horizontalchannel.The viscositiesandthicknessof layersmaybe
assumedtobeidenticalandwherethedensityof liquidA is lessthanthedensity
of liquidB.
(13) Showthattheflowrateof theNewtonianliquidflowingdowntheoutside
of averticalstationarywireof radiusR maybegivenby
Q =~~"[R'(R +0)'<-(R +8)4[~-IOg,( R; O)J]
where0isthethicknessoftheliquidfilm.
(14) Showthatthelocalvelocityvx ofNewtonianliquidflowingwithfully
developedlaminarflow downa plateinclinedatanangle8 to thehorizontal
maybegivenby theexpression
vx =pgsin8
[
( 3~~
)
* y - ~
]~ lpgsm8 2
wherey is thedepthmeasuredfromthesurfaceof theplate.
96
~
LAMINAR FLOW AND LUBRICATION
(15) Deriveanexpressionfor thevelocitydistributionfor a Newtonianfluid
flowingwithfullydevelopedlaminarflow throughanarrowgapformedbytwo
verticalparallelplatesandsketchqualitativelythevelocityandshearstress
profiles.
(16) A liquid flows in laminarflow alonga horizontalrectangularchannel
whichhasawidthW andadepth0,whereW»o. Showthatwellawayfromthe
entrance,exit and side walls of the channel,the liquid velocityalong the
channelis givenby
6Q 2
vx =- (oy- y )
W03
whereQ is thevolumetricrateof flow andy is thedistancefromthebottomof
thechannel.
(17) A liquid flows in laminarflow atarateQ alongahorizontaltubewhich
hasaninsidediameterd. Showthatundersteadyflow conditionsandfaraway
fromtheentrancetoorexitfromthetube,theliquidvelocityvx ataradiusr is
givenby theexpression
v =32Q
(
~- r2
]x nd4 4
whereQ is thevolumetricrateof flow andd is thediameterof thetube.
(18) A cylindricalbobwithadiameterof ]00mmandlength150mmrevolves
withinacupwith.aclearanceof ] mmwiththeannularspacefilled withglyc-
erine.If thebobrotatesat 120rpmfor whichatorqueof 2.3Nm is recorded,
determinetheviscosityof theglycerine.
Answer:0.079Nsm-2
(19) Waterflowsin laminarflow alongahorizontalchannelwhichhasawidth
of 4 m. It is observedthatthedepthof waterin thechannelis 3 emandthat
smallgasbubbles,whichformatthefloorof thechannel,take220secondsto
reachthesurfaceafterdetachment.Itis alsoobservedthatthebubblesreachthe
watersurface7.32m, measuredin the directionof flow, from thepoint of
formation.Determinetherateof flow of wateralongthechannel.
Answer:0.004m3s-1
97
1
Dimensional
analysis
Introduction
Dimensionalanalysisis a particularlyusefulway of obtainingrelationships
betweenvariablesthatpredicttheoccurrenceof naturalphenomenausefulfor
scalingupmodels.Thetechniquesinvolved,alsotobefoundunderthetitlesof
similitude,theoryof dimensions,theoryof similarityandtheoryofmodels,are
basedon thephysicalrelationshipbetweenthevariablesbeingrequiredto be
dimensionallyperfect.By organizationof thevariables,it maybepossibleto
determine.a fundamentalrelationshipbetweenthem.
Dimensionalanalysisis encounteredinmanyotherbranchesofengineering.
In termsof fluidmechanics,problemsinvolvingthefundamentaldimensionsof
mass,lengthandtime canbe appliedto all forms of fluid resistance,flow
throughpipes,throughweirsandorifices.Thetechniquesrequireidentification
of all therequisitevariablesby which a phenomenonis affected;missingor
wrong variableswill lead to incorrectconclusions.The requirementfor
forming dimensionlessgroupsis that they shouldbe independentof one
anotherwithin~heset,butallpossibledimensionlessgroupsoutsidethesetcan
beformedasproductsof powersof thegroupswithinthecompleteset.
Therehaslongbeena preoccupationwith dimensionalanalysisemploying
Lord Rayleigh'smethodof indicesof 1899andthen theoremof Buckingham
of 1915(althoughfirststatedin 1892byA. Vaschy).Therearelimitationswith
thesetheories,however,largelyassociatedwith theneedto identifythevari-
ablesresponsiblefor predictinga particularphenomenontogetherwith the
interpretationof thedimensionlessgroups.A numberoftechniqueshavesubse-
quently been developedthat betterrecognizethe physical featuresof a
particularsituation.Nonetheless,dimensionalanalysisreasoningremainsa
useful tool in formulatingmodels,particularlywherethelaws governinga
particularphenomenonarenotknown.
99
I
FLUID MECHANICS
4.1 Flowthroughanorifice
Deducebythemethodofdimensions,anexpressionindimensionlesstermsfor
therateofflow Q ofa liquidofdensityp andviscosity11inapipeofdiameterd
withanorificeofdiameterdo andpressuredropf:,.p.
I'
Flow ~td"- - - - f
Solution
TheRayleighmethodofdimensionalanalysiswasproposedin 1899byBritish
physicistLord Rayleigh(1842-1919)todeterminetheeffectof temperatureon
a gas.The method,whichhasfoundwideapplicationin engineering,involves
forminganequationin whichA is somefunctionof independentvariables,A I '
A2' ...etc,whicharedimensionallyconsistentin theform
A =k(A~,A~, ...)
wherek is adimensionlessconstant.Thedimensionlessgroupsareobtainedby
evaluatingtheexponentsa,b, ...etc,andgroupingthosevariableswiththesame
power.For theorificein apipe,therelationshipis thereforeassumed
Q =k f:,.pad~ d CP d 11e
wherethecorrespondingfundamentaldimensionsare
L3T-I =k(MCIT-2)a LbLc(MC3)d(MCIT-I)e
EquatingindicesforM, LandT, respectively
O=a+d+e
3 =~ +b +c - 3d - e
-l=-2a-e
The governingvariablesarepipediameterandviscosity.A rearrangementof
theindicesin termsof c ande gives
100
l
DIMENSIONAL ANALYSIS
1 e
a=---
2 2
b=2-c-e
d =_l - ~
2 2
Thepowerrelationshipis therefore
QOCf:,.ptd2p-1
[
III l
]
e
[
~
)
c
df:,.p2p2 do
TheconstantkthereforecorrespondstoaformofReynoldsnumberandratioof
pipetoorificediameter.Thus
Q =kd2f¥-
This resultis expectedastherateof flow dependsonthevelocitythroughthe
orifice.Sincethepressuredropis relatedto headby
f:,.p= pgH
thentherateof flow canalternativelybeexpressedas
Q =kd2fiR
Note thatcomparingthisequationwith thatfor flow throughan orifice (see
Problem5.4,page122),it canbededucedthat
Cd'n
4
k~m:
whereCd is thedischargecoefficientanda0 is theareaof theorifice.Much
experimentalwork hasbeendonecorrelatingCd with Reynoldsnumberfor
differentratios of pipe to orifice diameter.For Reynoldsnumbersabove
10,000,however,Cd approachesvaluesbetween0.6 and0.65irrespectiveof
dido'
101
FLUID MECHANICS
4.2 Flowovernotches
Showbythemethodofdimensionsthattheflow of liquidQ overa notchcanbe
givenbythedimensionlessgroups
Q =kg~H~
[
~ 1 r~
]pg2H2 ~pgH2
whereH is thehead,p is thedensity,11theviscosityandcrthesurfacetensionof
theliquid,g is thegravitationalaccelerationandk is a constant.
Solution
Forproblemsinvolvingflowovernotchesit is knownthattheexperimental
coefficientsarenotconstantforallheadsbutvarywithfluiddensity,viscosity,
andsurfacetension.Assumingnotchestobegeometricallysimilar- thatis,
theyhavethesamecharacteristicdimensionswheretheangle8is thesamefor
all notchestested- thepowerrelationis assumedtobe
Q =kHUgbllcpdcre
then
Q = HUg b11CPdcrC
The fundamentaldimensionsare
L3T-l =(L)u (LT-2)b (MC1T-l)c (MC3)d (MT-2)e
EquatingtheindicesforM, Land T, respectively
O=c+d+e
3=a +b - c - 3d
-1=-2b-c-2e
The governingvariablesare viscosity and surfacetension.Obtainingthe
indicesin termsof c andegives
5 3a =- - - c - 2e
2 2
1 1
b=---c-e
2 2
d=-c-e
102
L
DIMENSIONAL ANALYSIS
Therefore
[ ]
c
1 5 e
Q=g2H2 11 cr
pglHf [pgH2]
The coefficientof dischargefor notchesthereforevarieswith fluid density,
viscosity,headandsurfacetension.Thatis, theconstantk is a functionof
[pJlIf lpg:' ]
This is expectedsincethenow ofnuidswhichhaveafreesurfaceis innuenced
in somewaysby surfacetensionandviscosity.This is truefor openchannel
now, now over notchesandweirs, andmultiphasefluid flow. The precise
natureof theinnuenceis notknownandgenerallyis onlyappreciableforcondi-
tionsof shallownow orlow headoverobstructions.Surfacetensioneffectsare
thereforeignoredfor mostpracticalpurposesandareotherwiseincorporated
intothecoef~icientof discharge.It shouldbenotedthatsurfacetensioneffects
donotappearinproblemsconcerningthesingle-phasenow of liquidsthrough
pIpes.
103
FLUID MECHANICS
4.3 Scale-upof centrifugalpumps
Showbythemethodofdimensionsthatthescale-upofcentrifugalpumpscanbe
basedonthedimensionlessgroups
[
2
~=f J.L ~ pND
pN3D5 ND3' N2D2' /.l
whereP is thepowerrequiredforpumping,N is therotationalspeedandD the
diameterof theimpeller,H is theheaddeveloped,Q is theflow delivered,p is
thedensityand/.l theviscosityof thefluid, andg is theaccelerationdueto
gravity.
Solution
The scale-upof centrifugalpumpsis a commonchallengein the process
industries.Experimentsmaybecarriedoutona laboratoryscaleorpilotplant
scaleasabasisfor designingafull-sizeplant.Alternatively,datamaybeavail-
ablefromanexistingfull-sizeplantandmayneedconversionto anotherfull-
size plant of differentcapacityand operatingconditions.The variablesof
importancecanbecombinedto givedimensionlessgroupsof power,capacity
andheadcoefficient.This is basedon the fact thatthe powerrequiredfor
pumpingis afunctionof theheaddevelopedandvolumetricflowrateaswell as
pumpsizein termsof impellerdiameteranditsrotationalspeed,andtheproper-
tiesof thefluidin termsof densityandviscosity.For thepurposeof theanalysis
H is combinedwithg wherethepowerrelationshipis assumedtobe
P =kQa (gH)bNc Ddpe/.lf
then
P =Qa (gH)b NcDdpe/.lf
In termsof thefundamentaldimensionsM, Land T
ML2T~3 =(L3T-I)a U3T-2)b (T-I)c (L)d (MC3)e (MCIT-I!
EquatingtheindicesforM, Land T, respectively
l=e+f
2 =3a+2b+d - 3e - f
-3=-a-2b-c-f
Thegoverningvariablesareflowrate,gravitationalacceleration,head(gH) and
104
1
DIMENSIONAL ANALYSIS
viscosity.Thereforerearrangingin termsof theindicesa,bandf
e=l-f
c=3-a-2b-f
d =2 - 3a - 2b + 3e +f
=2 - 3a - 2b + 3(1- f) +f
=5 - 3a- 2b - 2f
Hence
P =Qa(gH)b N3-a-2b-fD5-3a-2b-2f pi-f/.lf
or rearrangedintodimensionlessgroups
~=f
[
J.L ~ PND2
]pN3D5 ND3' N2D2' /.l
wherethepowercoefficientis a functionof thecapacitycoefficientCQ' head
coefficientCH ' andaformof Reynoldsnumber,respectively.( )
a
( )
b
[ J
P pN3D5 Q gH /.l
= ND3 N2D2 pND2
where
Q
CQ =ND3
(seeProblem9.7,page249)
and
CH=~
N2D2
105
FLUID MECHANICS
4.4 Frictionalpressuredropfor turbulentflowin pipes
Show,bytheBuckinghamITmethodofdimensionalanalysis,thatthefrictional
pressuredropI'1.pf for afluid ofdensitypandviscosity~flowingwithavelocity
vthroughacircularpipeof insidediameterd, lengthL andsurfaceroughness£
canbegivenbythedimensionlessgroups
I'1.Pf=f
(
~~~
JpvZ d'd'pvd
Solution
The BuckinghamITmethodof dimensionalanalysisproposedby EdgarBuck-
ingham(1867-1940)in 1915statesthatif, in adimensionallyconsistentequa-
tion, there are n variablesin which there are containedm fundamental
dimensions,therewill ben -m dimensionlessgroups.Buckinghamreferredto
thesegroupsasITI, ITz, etc,suchthat
f(ITI' ITz,... ITn-m)=0
In thiscase,thefunctionaldependencyof frictionalpressuredropfor thefluid
is
I'1.Pf=f(L, £,~,d,p,v)
Thus,thefunctionalrelationis
f(I'1.Pf'L, £,~, d, p, v)=O
Any dimensionlessgroupfromthesevariablesis thereforeof theform
IT- A a Tb c d de f g-UPf,L,£,~, ,p',v
for whichthefundamentaldimensionsof IT aretherefore
IT =(MCIT-Z)a (L)b(L)c (MCIT-I)d (L/ (MC3 f (LT-I)g
For thesevenvariableswiththreefundamentaldimensions(M, Land T) it is
expectedthattherewill be four dimensionlessgroups.The procedureis to
selectrepeatingvariablesby choosingvariablesequaltothenumberof funda-
mentaldimensions(three)to providea successionof IT dimensionlessgroups
which have dimensionswhich containamongstthemall the fundamental
dimensions.It is usuallysuitableto selectp,v andd. Thus
106 I
1
DIMENSIONAL ANALYSIS
IT _ pal bl d c1AI - v uPf
ITz =paZvbZdcZL
IT3 =pa3vb3dc3£
IT4 =pa4vb4 dc4~
Thatis
0 =(MC3)al(LT-I)bl(L)c1(MCIT-Z)
0 =(MC3)aZ(LT-I)bZ(L)CZ(L)
0 =(MC3/3 (LT-I)b3 (L)c3(L)
0 =(MC3)a4(LT-I)b4(Lt4(MCIT-I)
TreatingeachseparatelyusingtheRayleighmethodgives
ITI =I'1.Pf
pvZ.
ITz =~
d
IT3=~d
IT4 =~
pvd
The fourdimensionlessgroupsmaybethereforewrittenas
I'1.Pf-
(
L £ ~
)pvZ -f d' d' pvd
107
FLUID MECHANICS
4.5 Scalemodelforpredictingpressuredrop
in a pipeline
A half-scalemodelis usedto simulatetheflow of a liquidhydrocarbonin a
pipeline.Determinetheexpectedpressuredropalongthefull-scalepipe if the
averagevelocityof thehydrocarbonin thefull-scalepipe is expectedtobe1.8
ms-l. Themodeluseswaterwherethepressuredropper metrelengthis noted
to be4 kNnc2. The respectivedensitiesof waterandhydrocarbonare 1000
kgm-3and800kgm-3,andviscositiesare 1x10-3Nsm-2and9x10-4Nsm-2.
Solution
Bydimensionalanalysisitcanbeshownthatthepressurecoefficientisafunc-
tionofReynoldsnumberofthefluidinthepipe.Thatis
~-
(
~
]
-f -
pv2 pvd
Boththemodelandfull-scalepipearedynamicallysimilarsincetheresistance
toflow is duetoviscosity.TheReynoldsnumberforbothmodelandfull-scale
pipemustthereforebethesame.Thatis
(pvd
] (
PVdI
l~ model = ~ )1illlscale
Rearrangingin termsof velocitythroughthemodelis therefore
vmodel~ (~]1>11""'"
[
dfullscale
](~
]
dmodel
l~model
800X 1.8
=9 X 10-4 X 2
1000
1X 10-3
=32 ms-l
SincetheReynoldsnumberis thesamein bothmodelandfull-scalepipe,then
so toois thepressurecoefficient.Thatis
108
I
L
~
DIMENSIONAL ANALYSIS
[
~
]
-
(
~
]pv 2 model pv2 /fullscale
Rearranging,thepressuredropperunitlengthinthefull-scalepipeis therefore
2
~ - ~ (pv )fullscale
pf - Pf model 2
(pv )model
=4000X 800 X 1.82
1000X 322
=1012Nm-2
The pressuredropperunitlengthin thefull-scalepipeis expectedtobe 1.012
kNm-2.
109
DIMENSIONAL ANALYSIS
Furtherproblems
(1) StateBuckingham'sI1theorem.
(7) Showby dimensionalanalysisthataformof frictionalpressuredropfor
theflow of afluid throughgeometricallysimilarpipescanbeexpressedby
IJ.Pf =f
(
PVd,~
}pv2 ).l d
(2) Outline the procedureused in the Rayleigh methodof dimensional
analysis.
(3) Showthatby themethodof dimensionalanalysistheflow overarectan-
gularweircanbegivenby thedimensionlessgroups
wherep isthedensity,).ltheviscosityandv thevelocityof thefluid,andd is the
diameterandL thelengthof pipe.
Q -
[
).l <J
]
1 1 -f 1 3'~
Bg2H2 pg2H2 pgH
(8) Showby themethodof dynamicalanalysisthatthevolumetricflow of a
fluid througha centralcircularorificelocatedin apipecanbeexpressedas
Q =kA)¥whereB is thebreadthof theweir,H is thehead,pis thedensity,).ltheviscosity
and<Jthesurfacetensionof theliquid,g is thegravitationalaccelerationandk
is aconstant.
(4) ReworktheproblemofdimensionlessgroupsforflowoveraV-notchin
whichtheangleeis notnecessarilythesameforallnotchestoyield
wherep is thedensityof thefluid,A is theareaof theorifice,k is acoefficient
whichdependson thepipeandorificedimensionsandtheReynoldsnumber,
andIJ.pis thepressuredropacrosstheorifice.
Q -
[
).l <J
]gtH~-f pgtHf pgH2'e
(9) Showthattherateof flowofaliquidofkinematicviscosityvovera90°
V-notchcanbegivenby
whereH is thehead,pisthedensity,).ltheviscosityand<Jthesurfacetensionof
theliquid,g is thegravitationalaccelerationandk is aconstant.
--h =f
[
Htgt
]H2 g 2 V
whereH is theheadandg theaccelerationduetogravity.
(5) Show,by themethodof dimensions,thattheviscousresistanceof anoil
bearingis dependentonthelineardimensionsof thebearingd, theviscosityof
theoil 1-1,speedof rotationN andpressureonthebearingp, andis givenby
~-f
(
P
}I-1Nd3- I-lN
(10) For liquidsflowing alongpipelinesabovea criticalvelocity,showthat
pressuredropduetofrictionperunitlengthis givenby
2
IJ.pf =~ f(Re)
L d
IJ.Pf =f
(
PVd
Jpv2 1-1
wherep is thedensityandv thevelocityof thefluid, d is thediameterof the
pipeandRe is theReynoldsnumber.Henceshowthatthefrictionfactorf in the
formula4fLv2/2gd for frictional headloss is a functionof the Reynolds
number.
(6) Showby dimensionalanalysisthataformof frictionalpressuredropfor
theflow of afluid throughapipeis
wherep is thedensity,1-1theviscosityandv thevelocityof thefluid, andd is
thediameterof thepipe.
110
III
1
Flowmeasurement
bydifferentialhead
Introduction
Themeasurementof flow ofprocessfluidsis anessentialaspectof anyprocess
operation,notonlyfor plantcontrolbutalsofor fiscalmonitoringpurposes.A
widevarietyofflowmetersis availableandit is importanttoselectcorrectlythe
flowmeterfor aparticularapplication.This requiresaknowledgeandcompre-
hensionof thenatureof thefluid to bemeasuredandanunderstandingof the
operatingprinciplesof flowmeters.
Beforethe.adventof digitalcontrolsystemswhich collectandstoreflow
information,therateof flow of fluids wasusuallymeasuredby instruments
(flowmeters)usingtheprincipleof differentialpressure.Today,thereis awide
varietyofflowmetersavailable.For single-phase,closedpipeflow, flowmeters
are broadly classifiedinto thosewhich are intrusiveand thosewhich are
non-intrusivetotheflow of thefluid. Collectively,theclassificationsinclude
differential pressuremeters, positive displacementmeters, mechanical,
acousticandelectricallyheatedmeters.
Differentialflowmetersindirectlymeasurevelocity,andthereforeflow,of a
fluid by measuringa differentialhead.Consequently,theyarealsoknownas
headorratemeterswiththemaingroupof metersbeingventuri,orifice,nozzle
andPitottubes- althoughthereareothers.Suchmetersarebasedontheprin-
ciplethatwhenliquid flows througharestriction,itsvelocityincreasesdueto
continuity.The increasein kineticenergyevolvesfromthereductionin pres-
surethroughtherestrictionandit is thisrelationshipwhichallowsa measure-
mentof pressuredroptoberelatedto thevelocityandthereforeflowrate.The
relationshipis derivedfromtheBernoulliequationandis essentiallyanappli-
cationof thefirstlaw of thermodynamicsto flow processes.
Venturimetersaresimplefluid flow measuringdeviceswhichoperateby
restrictingthe flow of fluid, therebyincreasingvelocity and consequentlyreducingpressureatthepointof restriction.By measuringthedifferentialpres-
suredropatthepointofrestrictiontheflowratecanbereadilydetermined.The
113
FLUID MECHANICS FLOW MEAS UREMENT BY DIFFERENTIAL HEAD
device,whichhasno movingparts,consistsof a rapidlytaperedsectionto a
throatand gentledownstreamexpansionsection.This designpreventsthe
phenomenonof separationandthus a permanentenergyloss. For a well-
designedventuri,thedischargecoefficientshouldlie betweena valueof 0.95
and 0.98. If the dischargecoefficientis not known or is unavailablefor a
particularventuri,a valueof 0.97maybereasonablyassumed.Althoughthe
venturimetercanbeinstalledin anyorientation,caremustbetakento usethe
appropriateequationfor flow derivedfromtheBernoulliequation.
The orificemeteris cheaperto manufacturethantheventurimeter,buthas
thedisadvantageof a higherpermanentenergyloss.The deviceconsistsof a
plateordiaphragmideallymanufacturedfromacorrosionanderosion-resistant
material,positionedbetweentwo flangesin a pipeline.The plateis often
centrallydrilled,althougheccentricandsegmentaldiaphragmsareused.The
meteroperatesby increasingthevelocityof thefluid asit flows throughthe
restrictionandmeasuringthe correspondingdifferentialpressureacrossthe
device.Thereis,however,aregionofhighturbulencebehindtheorificegiving
risetoahighpermanentenergylossandconsequentlythedischargecoefficient
is considerablylessthanthatobtainedusingtheventurimeter.A valueof0.6is
frequentlyusedfor highflowratesalthoughthecoefficientvarieswiththesize
of orificerelativeto thepipediameterandrateof flow.
Of themechanicalflowmeterswhichfunctionon simplefluid flow princi-
ples,therotameteroperatesusingafixeddifferentialheadbuta variableflow
area.It consistsof averticaltaperedtubethroughwhichafluid flowsupwards
andwheretheelevationof thefloatcontainedwithinthetaperedtubeprovides
anindicationof therateof flow.
5.1 Pitottube
A Pitot tubeis usedtodeterminethevelocityofair atapointinaprocessventi-
lationduct.A manometercontainsafluid ofSG 0.84andindicatesa differen-
tial headreadingof 30 mm.Determinethelocal velocityin theduct if the
densityofair is 1.2kgm-3andmaybeassumedconstant.
Flow- 2
1"
Solution
The Pitot tube,pamedaftertheeighteenthcenturyFrenchengineerHenri de
Pitot (1695-1771)who inventedit, is a deviceusedto measurethelocal or
point velocityof a fluid in a pipeor duct.It can alsobe usedto determine
flowrateby measurementsof localvelocitiesin thecross-sectionof thepipeor
ductandis particularlyusefulthereforewherethevelocityprofileis irregular.
Thedeviceoperatesbymeasuringthedifferencebetweentheimpactandstatic
pressuresin thefluid andusuallyconsistsof twoconcentrictubesarrangedin
parallel;onewithanopeningin thedirectionof flow, theotherperpendicularto
theflow. ApplyingtheBernoulliequation,wherethereis virtuallynoheadloss
dueto theproximityof thetwopoints
2
El +~ =P2
pg 2g pg
114 115
FLUID MECHANICS
The tubeis also assumedto be small with respectto pipe size; otherwise,
correctionsfor disturbancesandflow areareductionarerequired.Sincethe
fluid atpoint2 is stationary,P2 thereforecorrespondsto theimpactpressure.
Rearranging
VI = J2(P2 p- PI)
For thedifferentialmanometer,thedensityof themanometricfluid, Pm' is
considerablygreaterthanthe densityof air, p. The approximatedifferential
pressureis thus
P2 - P I =PmgH
=840x g x 0.03
=247Nm-2
The velocityof theair in theductis therefore
vI=J2X24712
=2029ms-I
The velocityis foundtobe20.29ms-I. Notethatthedensityof theaircanbe
readilydeterminedapplyingtheidealgaslawfor aknownbarometricpressure
andtemperature.It mayalsobenotedthattheactualvelocitymeasuredby the
Pitot is givenby
V =C-j2gH
whereC is a coefficientwhich is approximatelyunity for largepipesand
smoothPitotsbutis appreciablylessfor low Reynoldsnumberflow.
116
L
FLOW MEASUREMENT BY DIFFERENTIAL HEAD
5.2 Pitottraverse
Determinetherateofflow andaveragevelocityofair inaprocessventpipeof
50 cm diameterfor which local readingsof velocityfrom a Pitot tubeare
recordedbelow.
Radiusr (m) 0 0.05
Velocityv (ms-l) 19.0 18.6
0.15
16.3
0.10
17.7
0.20
14.2
0.225
12.9
0.25
0
20
IS
-
or;
§ 10
"",'"
~N
5
0
0 0.100.05 0.15 0.20 0.25
RADIUS r (m)
Solution
The distributionof velocityacrosstheprocessventpipeusinga Pitot tube,
knownasaPitottraverse,canbeusedtodetermineflowrateeithernumerically
orby graphicalintegration.The elementalflowratefor thedeviceataradiusr,
recordingthelocalvelocityvx ' is therefore
dQ =vx21trdr
The totalflow maythereforebedeterminedgraphicallyfromaplotof 21tvx r
versusr. Thus
Radiusr (m) 0
Velocityv (ms-I)19.0
21tvxr(m2s-I) 0
0.20
14.2
17.8
0.225
12.9
18.2
0.25
0
0
0.05
18.6
5.8
0.10
17.7
11.1
0.15
16.3
15.4
117
FLUID MECHANICS
Fromtheplot,theareaunderthecurveis foundtobe2.74m3s-l.The average
velocityis foundfromthetotalflow acrosstheflow area.Thatis
4Qv=-
1td2
4 x 2.74
1tx 052
=13.94ms-I
The averagevelocityis foundtobe 13.94ms-I. Notethattheaveragevelocity
is notthesumof thevelocitiesreporteddividedby thenumberof readings.
This procedureis applicableonlyto symmetricalvelocitydistributions.For
unsymmetricalflow or flow in non-circularducts,theprocedurecanbemodi-
fied toevaluatethetotalflow overtheflow section.For rectangularductsthis
involvesdividingthecross-sectioninto regularsizedsquaresandmeasuring
thelocalvelocitiesatthesepoints.Alternatively,log-linearfor circularcross-
sectionsor log-Tchebychevpositionsfor bothcircularandrectangularcross-
sectionscanreducethecomputationsinvolvedbutrequireaccuratepositioning.
Numericalvelocity-areaintegrationtechniquesare,however,nowpreferredto
graphicaltechniqueswherethe accuracyof the final resultdependson the
numberofmeasurementsmade.Ideally,forcircularcross-sectionsthereshould
beabout36pointswith6 oneachequi-spacedradiiandnotlessthan12with3
oneachof 4 equi-spacedradii.
118
~
FLOW MEAS UREMENT BY DIFFERENTIAL HEAD
5.3 Horizontalventurimeter
A horizontalventurimeterwitha dischargecoefficientof0.96is tobeusedto
measuretheflowrateofwaterupto0.025m3s-1inapipelineof internaldiam-
eter100mm.Themeteris connectedto a d(fferentialmanometercontaining
mercuryof SG 13.6.1fthemaximumallowabled(fferencein mercurylevelsis
80 em,determinethediameterof thethroatandtheshortestpossibleoverall
lengthof themeter.
Flow--- I 2
~-
Solution
This fluid flow measuringdevice, first namedafter the Italian physicist
GiovanniBattistaVenturi(1746-1822)byClemensHershelsin 1886,consists
of a taperedtube which constrictsflow so that the differentialpressure
producedby theflowingfluid throughthethroatindirectlygivesameasureof
flowrate.Therateofflow canbedeterminedbyapplyingtheBernoulliequation
atsomepointupstreamof theventuri(pointI) andatthethroat(point2).For a
horizontalventuri
2 2
PI vI P2 v2-+-=-+-
pg 2g pg 2g
119
FLUID MECHANICS
Fromcontinuityfor anincompressiblefluid
a]vI =a2v2
Substitutingforv2 andrearranging
PI~ =vi
[[
~
]
2 -1
]
pg 2g a2
Rearranging,thevelocityin thepipev I is therefore
VI = 2(PI - P2)
{[:;J - 1)
The actualflow throughtheventuriincorporatesacoefficientof dischargeCd
to allowfor frictionaleffectsandis definedastheratioof actualtotheoretical
flow. Therefore
Q=CdavI
=Cda.l2(PI - P2)
p(~4-1)
where~is theratioof pipediametertothroatdiameteranda is theflow areaof
thepipe.For apipediameterof 0.1m thepipeflow areais
1tx 0.12
a=
4
=7.85x 10-3m2
Rearranging,theflow equationin termsof~is
l
4
2(PI - P2) +1~=1
[
SL
]
2
P Cda
120
I
~
FLOW MEASUREMENT BY DIFFERENTIAL HEAD
For themercury-filleddifferentialmanometerthedifferentialpressureis
PI -P2 =(PHg -p)gH
=(13,600-1000) x g x 0.8
=98,885Nm-2
~istherefore
l
4
~=I 2x 98,885
[
2 +1
1000x 0.Q25
0.96X 7.85X ]0-3 ]=2.087
d]
d2
The diameterof thethroatis therefore
d]
d2 =13
0.1--
2.087
=0.0479m
Thatis,thethroatdiameteris foundtobe48mm.Thedimensionsof theventuri
areimportanttominimizepermanentenergylosses.Therecommendeddimen-
sionsof a venturisuggestan inletentranceconeto haveanangleof 15°-20°
with an exit cone angle of 5°_7.5°. The throat length is 0.25 to 0.5
pipe-diameterswith tapping points located between 0.25 and 0.75
pipe-diametersupstream.In thiscase,theshortestpossibleoveralllengthof
venturiis thereforeanentranceconeof 7.1cmlength(20°),athroatof 2.5cm
(0.25pipe-diameters)andanexitconeof 19.7cm(7.5°)givinganoveralllength
of 29.3cm.
121
FLUID MECHANICS
5.4 Orificeandventurimetersin parallel
An orificeplatemeteranda venturimeterareconnectedinparallel in a hori-
zontalpipe of insidediameter50 mm.Theorificehasa throatdiameterof25
mmanddischargecoefficientof0.65whiletheventurihasa throatdiameterof
38 mmand dischargecoefficientof 0.95.Determinetheproportionof flow
througheithermeter.
Qo
Orifice
Q Q
Venturi
Solution
It is assumedthattheliquid flows alongafrictionlesshorizontalpipeandthat
lossesdueto fittingscanbeneglected.For liquid to flow acrossbothinstru-
ments,thepressuredropacrossthemmustbe thesame.Due to thedifferent
characteristicsof thesetwometers,theflow throughtheorificeshouldbeless
thanthatthroughtheventurisincetheoverallpressuredropfor agivenflow is
higherthroughanorificethanaventuri.Thusfor bothmeters,theflowrateis
Qo =Cd(o]G
2!J.p
p[(J-]] (seeProblem5.3,page119)
Qv =Cd(v]G
2!J.p
p[(a:)'-I:
wheretheratioofpipetoorificeareaintermsofdiameteris
~ =
(
~
)
2
ao do
122
.-..
FLOW MEAS UREMENT BY DIFFERENTIAL HEAD
andratioof pipeto venturithroatareain termsof diameteris
~ =
(
~
)
2
av dv
Sincethedifferentialpressureacrossbothmetersis thesame,then
( ]
2
[
4
] [ ;
2
[
4
]
~ ~ -1 - ~ ~ -1
Cd(o) (do) - Cd(v) ( dv )
Thatis
( )
2
[( )
4
] ( )
2
[( )
4
J
Qo 0.05 1 - Qv 0.05
0.65 0.Q25 - - 0.95 0.038 -1
whichreducesto
Qv =4Qo
Fromcontinuity,thetotalflow Q isthesumoftheflow throughthemeters.That
IS
Q=Qo+Qv
=Qo +4Qv
=5Qo
Thus 20%of theflow passesthroughtheorificemeterwhile80%of theflow
passesthroughtheventuri.Notethatfor thesamethroatsize,thedifferencein
volumetricflowrateis entirelydueto thepermanentfrictionallossesimposed
by bothmeters.
123
FLUID MECHANICS
5.5Venturimetercalibrationbytracerdilution
A venturimeterwitha throatdiameterof 6 cmis usedtomeasuretheflowrate
ofwaterofdensity1000kgm-3alonga horizontalpipelinewithaninsidediam-
eterof100mm.Theflowmeteriscalibratedusingasolutionofsaltwaterwitha
. saltconcentrationof20gl-l andaddedcontinuouslyupstreamofthemeterata
rateof3 litresperminute.Determinethedischargecoefficientof themeterif a
pressuredrop acrossthethroatof 3.7 kNm-2is recordedanda dilutedsalt
concentrationof0.126gl-l isfoundbyanalysisdownstream.
Salt
I Flow Q
t ConcentrationCa
Water
flow -.. -.. Flow Qi+Qa
Qi
Solution
The calibrationof flowmetersby themethodof tracerdilutioninvolvesthe
additionof extraneousmaterialwhosepresencecanbe quantitativelydeter-
minedby ananalyticaltechnique.The extraneousmaterialmayor maynotbe
alreadypresentasanimpurity.In thiscase,theflowrateof waterthroughthe
meteris relatedtodilutedsaltconcentration.Fromamaterialbalanceonthesalt
QaCa =(Qi +Qa)Co
whereQi andQa aretheratesof flow ofwaterandsaltsolution,andCa andC0
are the addedupstreamand diluteddownstreamsalt concentrations.Rear-
ranging,theflowrateof wateris therefore
Q =Q
[
Ca - Co
)1 a C0
=0.003x
[
20- 0.126
)60 0.126
=7.89X 10-3 m3s-1
124
.........
FLOW MEASUREMENT BY DIFFERENTIAL HEAD
The totalflowQ throughthemeterduringthecalibrationis therefore
Q =Qi +Qa
=7.89X 10-3 +5 x 10-5
=7.895X 10-3m3s-1
The rateof flow throughahorizontalventurimeteris givenby
Q =Cda.1 2!1p
p(p4-1)
wherePistheratioofpipetothroatdiameter
p=10
6
=1.67
Rearrangingin termsof dischargecoefficient
Q
Cd =. I 2!1p
a p(p4 -1)
Jrx0.12
4
7.895X10-3
2x 3700
1000x (1.674-1)
x
=0.962
Thecoefficientofdischargeisfoundtobe0.962.
125
FLUID MECHANICS
FLOW MEAS UREMENT BY DIFFERENTIAL HEAD
5.6 Differentialpressureacrossaverticalventurimeter
A processliquidof density850kgm-3flows upwardat a rateof 0.056m3s-l
througha verticalventurimeterwhichhasan inletdiameterof 200mmand
throatdiameterof 100mm,withdischargecoefficientof 0.98.Determinethe
differencein readingof twopressuregaugeslocatedat therespectivetapping
pointsa verticaldistanceof 30cmapart.
I"
Rearrangingin termsof differentialpressure
PI -P2 =pg[;~((:;r -}" -ZI]
The velocityis relatedtoflowrateby
Q =CdavI
andfor thecircularcross-section
(:;J ~ (:;J
therefore
2 PI -P2
~ ~[[c:aJ[( :J -1 +2g(Z2 - Z,)]
=850x
I[
0.056
]
' x
[
(0.2
)
4 -1 +2x g x 0.3
2 7Ix 0.22 l0.1
0.98x 4
_22
I
t
Flow 21
Datum =23,594Nm-2
Solution
ApplyingtheBernoulliequationbetweentheupstreamposition(1)andthe
throat(2)
The differentialpressureis foundto be 23.6kNm-2. Note thata differential
manometerwouldnotrecorda differencein levelof manometricfluid when
thereis no flow throughtheventuriwhereastwoindependentpressuregauges
wouldrecorda staticpressuredifferenceequalto
PI -pz =pg(zz -Zl)2 2
PI vI P2 v2
-+-+Zl=-+-+Zz
pg 2g pg 2g
alv I =aZv2
Fromcontinuityfor anincompressiblefluid
126 127
FLUID MECHANICS
5.7 Flowmeasurementbyorificemeterinaverticalpipe
Oil ofdensity860kgm-3flows upa verticalpipesectionofdiameter225mm.A
manometerfilled withfluid ofdensity1075kgm-3is usedtomeasurethepres-
suredropacrossan orificeplatewitha throatdiameterof 75mm.Determine
theflowrateof oil if thedeflectionof themanometerfluid is 0.5m.Assumea
dischargecoefficientofO.659for theorifice.
2
Orifice
t
Flow
Datum
128
22
21
I,
20
I
1
FLOW MEASUREMENT BY DIFFERENTIAL HEAD
Solution
ApplyingtheBernoulliequationbetweenpoints1and2
2 2
PI v I P2 v2
-+-+ZI =-+-+Z2
pg 2g pg 2g
Fromcontinuityfor anincompressiblefluid
aivi =a2v2
Thereforesubstitutingforv2 andrearranging
p, -P2 +pg(Zl-Z2)~p~i[(:;J -1]
Forthemanometer
PI -P2 =pg(Z2 -ZI)+(Pm -p)gH
The theoreticalvelocitythroughthepipeis therefore
2gH (pm - p)
vl=.) p(~4-1)
2 x g x 0.5x (1075- 860)
860x
[
(0.225
)
4 - 1
]lom5
=0.523I1)S-1
Notethatthisexpressiondoesnotcontaintermsinz.Thevelocityandtherefore
flowrateis independentof theorientationof thepipe.The actualflowrateis
then
Q=Cdalvl
2
=0.659x 1tx 0.225 x 0.523
4
=0.014m3s-1
The rateof flow is foundtobe0.014m3s-1.
129
FLUID MECHANICS
5.8Variableareaflowmeter
A rotameterusedtomeasuretheflow ofwaterconsistsofafloat withamassof
30g setina taperedglasstube20cmin length.Thetubehasan internaldiam-
eterof 22mmat itsbaseand30mmat its topend.Determinetherateofflow
whenthefloat is at mid-heightin thetube.Therotameterhasa coefficientof
dischargeof0.6andthedensityof thefloat is 5100kgm-3.
d{
- d~- -
17
e
2
H IHnfl"
db I L
t
Flow
Solution
Oftenreferredtoasarotameter,thevariableareaflowmeterisusedtomeasure
therateof flowof a fluidbyvirtueof theelevationof asolidfloatwithina
verticaltaperedtube.Thetubeistransparentandisusuallymadeofglasswhile
thefloatismadeofmetal,ceramicorplasticandis usually'bomb'shaped,in
thatithasacylindricalbodywithacone-shapedbottomandshortflattoppiece.
Somefloatsaregroovedwhichencouragesthefloattospintherebyimproving
stabilityduetoagyroscopiceffect.Thisdevicehaseffectivelyafixedperma-
nentpressuredropandavariableflowarea.Asthefluidflowrateis increased,
thefloatmovesupthetubeuntilanequilibriumpositionof forcesisreached.
130
........
FLOW MEASUREMENT BY DIFFERENTIAL HEAD
FromArchimedes'principle,thisisapositionwheretheupthrustisequaltothe
weightof fluid displaced.Thatis
!J.pAj =PjVjg -pVjg
whereAj is thecross-sectionalareaof float,pj is thedensityofthefloat,pis the
densityof fluidandVj is thevolumeof float.Rearrangingin termsof !J.pthen
!J.p =Vrg(pr - p)
Aj
where!J.pis relatedtoflowratebytheBernoulliequation.At theelevationof the
float
2 2
p+!J.p Vt P va-+-=-+-
pg 2g pg 2g
wherevt andva arethevelocitiesof thefluid in thetubeandannulus,respec-
tively.Therefore
p 2 2
!J.p=-(Va -Vt)
2
Sincethevelocityof thefluid throughtheannulusis significantlygreaterthan
thatin thetube,thisapproximatesto
2
A PVa
tip =-
2
Combiningbothequationsfor !J.p,thevelocityof thefluid throughtheannulus
is therefore
- 2Vrg(pr - p)
va -,I pAl
IntroducingadischargecoefficientCd toallowforlossesduetofriction,the
actualmassflowrateisthereforegivenby
m=Cdpaava
=Cdpa /2Vrg(Pr - p)a,
pAl
=Cdaa~2VrgP(Pr-p)Aj
131
FL UID MECHANICS
If thedensityof thefluid is notinfluencedbytemperatureorcomposition,then
thesquareroottermremainsconstant.Further,if thedischargecoefficientdoes
notvarygreatly,thenthereis analmostlinearvariationbetweenmassflowrate
andtheareaof theannulus.To relatetheareaof theannulustothefloatposition
abovethebottomof thetube,thefloatis assumedto sitperfectlyin thebottom
of thetubeof internaldiameterdb whenthereis no flow. Fromgeometry,the
internaldiameterof thetubed for anyfloatpositionabovethebottomH, for an
angleof taper8,is therefore
8
d =db +2H tan-2
for whichtheareaof theannulus,aa' aroundthefloatis
a =~(d2 - db2)
a 4
=~[(db + 2H [onH - dt]
"~[dE+4dbHt'"~+4H2(tanmr -dE]
Thisapproximatesto
8
aa =ndbH tan-2
The massflowrateis therefore
m=CdndbHtan!!.!2Vtgp(Pt -p)2nd;
4
=CdHtan!!.J8VIgp(PI -p)n2
Thatis,themassflowrateis approximatelylinearwithfloatpositioninthetube.
In practice,a scaleis markedonthetubeandtherotameteris suppliedwith a
calibrationcurvefor a particularfluid, temperatureandfloat.In thiscase,the
t1oathasavolume
132
1
FLOW MEASUREMENT BY DIFFERENTIAL HEAD
V -'!!L1-
PI
0.03
5100
=5.88x 10-6m3
The half-angleof taperis relatedto topandbottomdiametersseparatedby a
heightH Tube
8 dt - dbtan-=-
2 2HTube
0.03- 0.22
2 x 0.2
=0.Q2
Themasst1owratethroughtherotameterfor at1oatatthemid-pointin thetube
(10cm)is therefore
m =0.6x 0.1x 0.Q2x ~8x 5.88x 10-6 x g x 1000x (5100-1000) x n
=0.0925kgs-l
correspondingtoa volumetricflow of 5.55litresperminute.
133
FLUID MECHANICS
5.9 Rotametercalibrationbyventurimeter
A horizontalventurimeterwitha throatdiameterof2.5 cmis usedtocalibrate
a verticalrotameterin a pipelinecarryingwater.Thepipelinehasan inside
diameterof5 cm.Therotameterhasa coneangleof2°andusesafloat witha
volumeof 100cm3anddensity8000kgm-3.Determinetheflowrateof water
andthedischargecoefficientfor therotameterifa differentialpressurereading
of 1.2kNm-2is recordedfor theventuriwhenthefloat is atanelevationof 15
cm.Thedischargecoefficientfor theventuriis 0.97.
Ro"m~T13}
Venturi
Flow Q
Solution
The flow of waterthroughthehorizontalventuriis
Q =Cda.1 211p
p(~4-1)
where~istheratioofpipetothroatdiameter
~=~
d2
0.05
0.Q25
=2
134
........
FLOW MEASUREMENT BY DIFFERENTIAL HEAD
Therefore
Q =0.97x 1tx 0.052
4 x
2x 1200
1000x (24-1)
=7.62X 10-4 m3s-1
The flow throughtherotameteris givenby
C H e I
Q =~tan-\l8Vfgp(Pf -p)1tP 2
Rearranging,thedischargecoefficientistherefore
Q
Cd =H e
f811v gp(Pf -p)1t-tan--yOYf .P 2
7.62X 10-4
0.15 2° ~ -4- x tan- x 8x 1x lOx g x 1000x (8000-1000)x 1t1000 2
=0.70
The coefficientfor therotameteris foundtobe0.70.
135
FLUID MECHANICS
Furtherproblems
(1) Waterflowsdownaverticaltaperingpipe2 mlong.Thetopof thepipe
hasa diameterof 10cm andthediameterof thebottomof thepipe is 5 cm.
Determinethedifferenceofpressurebetweenthetopandthebottomendsof the
pipeif theflow rateis 1m3min-l.
Answer:1766Nm-2
(2) Thevelocityof waterin apipewithaboreof 250mmis measuredwitha
Pitottube.The differencein headatthecentreof thepipeis foundtobe10cm
of water.Determinetheflowrateof waterperminuteif theaveragevelocityof
wateris two-thirdsthevelocityatthecentre.The coefficientof thePitottube
maybetakenasunity.
Answer:0.0458m3s-1
(3) A liquid of density800kgm-3flows througha horizontalpipewith an
inside diameterof 150mm undera pressureof 400 kNm-2. Assumingno
losses,determinetheflow whenthepressureata75mmdiameterreductionis
200kNm-2.
Answer:0.102m3s-1
(4) StartingwiththeBernoulliequation,showthatthevolumetricflowrateof
anincompressiblefluidthroughahorizontalventurimeterwherethethroatand
upstreampositionsareconnectedbyaV-tubecontainingafluidofdensityPm
canbegivenby
Q =Cda
2gH(P; - 1)
(~4-1)
whereCd is thedischargecoefficient,a is thepipeflow area,p is thefluid
density, ~is the ratio of the pipe diameter to throat diameter, g is the gravita-
tionalaccelerationandH is thedifferencein levelsof thefluid in theV-tube.
(5) A venturimeterof inlet diameter10cm andthroatdiameter5 cm is
installedin a verticalpipethroughwhichwaterflows upwards.The meteris
calibratedwhereit is notedthatfor a flow rateof 11.5litresper second,the
differencein readingsof thepressuregaugesconnectedtotheinletandthroatis
136
l
FLOW MEASUREMENT BY DIFFERENTIAL HEAD
20 kNm-2. If thedifferencein theheightof thetwo tappingpointsis 30cm,
determinethedischargecoefficientfor theventurimeter.
Answer:0.971
(6) A venturimeteris installedinahorizontalpipewithaninsidediameterof
100mmcarryinganorganicsolventof density1200kgm-3.The only equip-
mentavailablefor measuringpressuredifferencesis a manometercontaining
mercuryof SO 13.6,withamaximumallowabledifferenceof levelsof 80cm.
Assumingadischargecoefficientis 0.97,determinethemaximumflowrateof
solventwhichcanbemeasured.
Answer:0.0251m3s-1
(7) Theflow in apipelineof diameterd is measuredby anorificemeterwith
an orifice diameterof 0.5d.The dischargecoefficientis 0.62.Calculatethe
throatdiameterof ahorizontalventurimeterwhichwouldgivethesamepres-
suredifferenceatthesamerateof flowin thepipe.Thedischargecoefficientfor
theventurimetermaybetakenas0.96.
Answer:0:405d
(8) Discussbrieflythemeritsof boththeorificeplatemeterandtheventuri
meterasdifferentialpressuremeters.
(9) Describe the types'of orifice plate which are available and their
application.
(10) A horizontalventurimeterwitha5cmdiameterthroatisusedtomeasure
theflow of slightlysaltwaterin apipeof insidediameter10cm.The meteris
calibratedby adding2 litresperminuteof 0.5molarsodiumchloridesolution
upstreamof themeterandanalysinga sampleof waterdownstreamfromthe
meter.Beforetheadditionof thesaltsolution,a 1litresamplerequired10cm3
of 0.1molarsilvernitratesolutionin atitration.Afteraddition,a I litresample
required24.4cm3of 0.1molarsilvernitrate.Determinethedischargecoeffi-
cientof themeterfor adifferentialpressureacrosstheventuriof 17.4kNm-2.
Answer:0.96
(11) Identifysuitabletypesof tracersubstanceswiththeirappropriatedetec-
tion systemswhichcanbeusedtocalibrateflowmeters.
137
FLUID MECHANICS
(12) An organicliquidof density980kgm-3flowswithanaveragevelocityof
3 ms-l alonga75 mmdiameterhorizontalpipe.A restrictionis placedin the
pipeinwhichthereis a50mmdiameteropeningforthewatertoflow.Thepres-
sureofthewaterin the75mmpipeis 150kNm-2.Determinethepressureatthe
restriction.
Answer:144.2kNm-2
(13) Thedifferenceinheadregisteredbetweentwolimbsof amercurygauge,
withwaterabovethemercuryandconnectedtoahorizontalventurimeter,is 20
cm.If theventurihasapipeandthroatdiameterof 15cmand7.5cm,respec-
tively, determinethe dischargethroughthemeterassuminga coefficientof
dischargeof 0.97.
Answer:0.031m3s-1
(14) Thepositionof afloat,ofdensity3000kgm-3,in arotameterfor apartic-
ularflow is 15cmabovethebottomof thetaperedtube.If thefloatisreplaced
with anidenticalsizeof floatbutwithadensityof 2500kgm-3,thentheposi-
tionof thefloatin thetubeis increasedby 2.5cm.Determinethedensityof the
fluid.
Answer:1115kgm-3
(15) Highlighttheadvantagesanddisadvantagesof usingrotametersfor the
measurementof fluid flow.
(16) A uniformpipelineof circularcross-sectionis to carrywateratarateof
0.5m3s-1.If anorificeplateis to beusedto monitortheflowratein theline,
determinethe differentialpressurereadingacrossthe plate at the design
flowrate.The pipe hasan internaldiameterof 45 cm andtheorifice has a
concentricholeof 30cm.If thedifferentialpressureis tobemeasuredusinga
mercury-filledmanometer,determinethedifferencein levelbetweenthetwo
legs.The coefficientof dischargefor theplatemaybetakenas0.6.
Answer:55.8kNm-2,45.1cm
(17) Anundergraduatelaboratoryexperimentinvolvesave!1turimeterwhich
is usedtomeasuretherateof flowofwaterthroughahorizontalpipewithan
internaldiameterof5cm.A calibrationis providedforthemeterandisgiven
by
Q =O.o96m
138
a
l
FLOW MEASUREMENT BY DIFFERENTIAL HEAD
whereQ is theflowratein litrespersecondandH is themanometricheadin a
water/mercurymanometermeasuredin millimetres.Determinethe throat
diameterof theventuriif thecoefficientof dischargemaybe assumedto be
0.97.
Answer:16.9mm
(18) A Pitottubeis usedtomeasurethevelocityofwateratapointin anopen
channel.If thePitot is connectedto an openair-filled invertedmanometer,
determinethe velocityof the waterfor a differencein liquid levels in the
manometerlegsof 10mm.
Answer:0.44ms-l
(19) A Pitot traverseis usedto recordthelocal velocitiesof air acrossthe
entranceofacircularductof diameter91cm.ThePitotis connectedtoawater-
filled differentialmanometerand measurementsof the differencein levels
betweenthelegsaregivenbelow:
Distancefrom'pipewall, cm
Differencein levels,mm
0
0
5
7
10
10
20
12
30
14
40
15
45.5
15
If theairhasastaticpressureof 101.3kNm-2 andtemperatureof 20°C,deter-
minetherateof flow andtheaveragevelocity.Themeanmolecularmassof air
is 29kgkmol-l andtheUniversalgasconstantis 8.314kJkmol-lK-l.
Answer:7.5m3s-1,11.5ms-l
(20) OutlinetheadvantagesanddisadvantagesofPitottubesforthemeasure-
mentof fluid velocityandflow.
(21) A venturimetermeasurestheflow of waterin a 100mminsidediameter
horizontalpipe.The differencein headbetweentheentranceandthethroatof
themeteris measuredby a U-tube,containingmercury(SG 13.6)with the
spaceabovethemercuryin eachlimb beingfilled with water.Determinethe
diameterof thethroatof themetersuchthatthe differencein thelevelsof
mercuryshallnotexceed300mmwhenthequantityof waterflowing in the
pipeis 10kgs-l. Assumethedischargecoefficientis 0.97.
Answer:39mm
139
i
.........
Tankdrainage
andvariable
headflow
Introduction
The rateat whichfluidsfreelydischargefromtanksandvesselsthrough
orificesandconnectingpipesis dependentonthepressureorheadwithinthe
tankandonfrictionalresistance.Theshape,sizeandformoftheorificethrough
whichthefluiddischarges,andthelengthanddiameterofpipeandfluidprop-
erties,alsoinfluencetherateofdischargeduetotheeffectsoffrictionalresis-
tance.Asthefluidpassesthroughtheorificeissuingasafreejet,itcontractsin
area,thusfurtherreducingtherateofdischarge.Thecontractioniscausedby
theliquidin thevesselin thevicinityof theorificehavingamotionperpen-
diculartothatof thejetandexertinga lateralforce.Thesectionof thejetat
whichthestreamlinesbecomeparallelis knownasthevenacontracta,and
thereisnofurthercontractionbeyondthispoint.Owingtothereductioninboth
velocityandflowareaof thejet,theactualrateof flowis muchlessthanthe
theoreticalprediction:therelationbetweenthemisknownasthecoefficientof
discharge.
The simplestmethodof determiningthecoefficientof dischargeis to
measurethequantityoffluid(liquid)dischargedforaconstantheadanddivide
bythetheoreticaldischarge.Intermsofenergyconversionforafreelydraining
tank,thepotentialenergyisconvertedtokineticenergyor,intheheadform,the
staticheadis thusconvertedtovelocityhead.It is thereforededucedthatthe
theoreticalvelocityof thejet is proportionalto thesquarerootof thehead,
describedinwhatiscalledtheTorricelliequationaftertheItalianmathemati-
cianEvangelistaTorricelli(1608-1647).Thedifferencebetweentheactualand
theoreticalvelocities,knownasthecoefficientofvelocity,isduetofrictionat
theorificeandis smallfor sharp-edgedorifices.Thecoefficientof velocity
variesdependingonthesizeandshapeoftheorificeaswellastheappliedhead.
Nevertheless,typicalvaluesarein theorderof0.97andmaybefoundexperi-
mentallyforavertically-mountedorificeinthesideofatankbymeasuringthe
horizontalandverticalco-ordinatesoftheissuingjet'strajectory.
141
FLUID MECHANICS
Fortankdrainageproblemswhichdonotinvolvesimultaneousinflowto
maintainaconstantheadabovetheorifice,thereisacontinuouslossofcapacity
andconsequentlyof head.Therateof dischargeis thereforenotconstantbut
variable,withtheheadbeingdependentonthegeometryofthetank.Thetime
takenfortankdrainagecanbeconsideredbyapplyinganunsteadystatemass
balanceoverthetankandequatinftherateof dischargewiththelossof
capacityfromthetank.Analyticalsolutionscanbededucedfor mosttank
configurationsin termsof tankgeometry,forsimultaneousinflow,flowinto
adjacentvesselsthroughopenings,submergedorifices(sluices),andforflow
throughconnectingpipeswithlaminarflow.Problemsin tankdrainagewhich
involvetheflow throughconnectingpipeswith turbulentflow arenot,
however,mathematicallystraightforward.
In simpletreatmentsof tankdrainageproblems,it is necessarytoemploy
pertinentassumptions.Theseincludeawell-ventilatedtankwheretheapplied
pressure(usuallyatmospheric)isthesameatboththefreesurfaceoftheliquid
inthetankandatthejet,wherethereis nofreevortexformationandwherethe
dischargecoefficientis constant.In practice,however,eachof thesehasan
influenceandshouldbe allowedfor whereappropriateby moredetailed
calculations.
142
~
TANK DRAINAGE AND VARIABLE HEAD FLOW
6.1 Orificeflowunderconstanthead
A watercoolingtowerreceives5000m3ofwaterperday.If awaterdistribution
systemis to bedesignedto covertheareaof thetowerusing1 cmdiameter
orificesin thebaseof theductinginwhichthewaterwill beata depthof20 cm,
determinethenumberof orifices required.A dischargecoefficientof 0.6 is
assumedfor theorifices.
Solution
To determinetherateatwhichtheliquiddischargesthroughtheorifices,it is
necessarytodeterminethevelocityoftheliquidinthefreejets.Assumingthat
thereis no pressurechangeacrossthefreejet, thestaticheadis directly
convertedtovelocityheadas
2
H=~
2g
Rearranging,thetheoreticalvelocitythroughtheorifice,v,istherefore
v =~2gH
andisknownastheTorricelliequation.Theactualvelocityvact'however,is
lessthanthistheoreticalpredictionduetopermanentandirreversibleenergy
lossesandcanbefoundbyintroducingthecoefficientof velocityCv' which
representstheratioof theactualtotheoreticaljetvelocity.Thatis
Cv =V act
V
As liquid eme~gesfromtheorifice, thestreamlinesconvergeto forma vena
contractajustbeyondtheorifice,atadistanceof halftheorificediameter.The
cross-sectionalareaof thevenacontractacanbefoundbyintroducingacoeffi-
cientof contractionCc' andis theratioof actualto theoreticalflow area.The
coefficientof dischargeCd is thereforedefinedastheproductofCv andCc
where
Qact =CvCcao~2gH
=Cdao~2gH
Thetotalflowthroughalltheorificesistherefore
Q=nCdao~2gH
wheren isthenumberoforifices.Rearranging
143
FLUID MECHANICS
Qn=-
C daD.J2gH
5000
- 3600x 24
nx 0.012 I
0.6x x '\j2x g x 0.2 .4
=620
A totalof 620orificesis required.Notethatatthevenacontractathestream-
linesareno longerconvergingastheflow passesthroughtheorifices,butare
parallel.The internalpressurein thejet will havedecreasedfromitsupstream
pressureto atmosphericwhich is imposedon thefreejet. At thispoint, the
velocityacrossthejet is essentiallyuniformandequalto thetotalhead.
144
I
~
TANK DRAINAGE AND VARIABLE HEAD FLOW6.2 Coefficientofvelocity
To determinethecoefficientof velocityofa smallcircularsharp-edgedorifice
in thesideofa vessel,thehorizontalandverticalco-ordinatesof thetrajectory
of thejet weremeasuredfor a headof20em.Thehorizontalco-ordinatefrom
thevenacontractawasfoundtobe86emwhilsttheverticalco-ordinatewas96
em.Determinethecoefficientof velocity.
---
1
~-
H
I'" x -I
Solution
Thecoefficientof velocityfor sharp-edgedorificeslocatedin thesideof atank
canbereadilydeteJ;minedfromthetrajectoryof theissuingfreejet, in which
thehorizontaldistancetravelledbythejetmeasuredfromthevenacontractais
x =V actt
throughaverticaldistance
1 2
Y =:- gt
2
Eliminatingtimet, theactualvelocityof thejet is
Vact =X~
145
FLUID MECHANICS
The theoreticalvelocityis givenby Torricelli's equationas
v =-J2gH
The coefficientof velocityis therefore
Cv =Vact
V
x/g
- f2y--
-J2gH
x
- 2Jiii
0.86
2X -J0.96X 0.2
=0.981
The coefficientof velocityfor theorificeis foundto be0.981.Notethatthe
relationshipbetweenvelocityandheadwhichwasfirstdevelopedbyTorricelli
in around1645is a simpletreatmentof theBernoulliequationin whichstatic
headis convertedtovelocityhead.Sincethestaticheadvariesuniformlywith
depthin thevessel,it is possibleto predictthetrajectoryof thefreejet from
sharp-edgedorificesfor otherelevationsonthesideof thevessel.
The velocityof the dischargingjet is assumedto be uniformat thevena
contracta(seeProblem6.1,page143).This is truefor sharp-edgedorificesbut
notfor roundedorifices.Roundedorificeshavetheeffectof slightlyreducing
thecoefficientofdischarge.Significantly,thecoefficientofcontractionis more
markedwithsharp-edgedorifices.Precisemeasurementsof theareaofthevena
contracta,however,arenot alwayspossible.Betterestimatescanusuallybe
obtainedfrom estimatesof the coefficientof dischargeand coefficientof
velocity such thatC c =Cd ICv .
146
~
TANK DRAINAGE AND VARIABLE HEAD FLOW
6.3 Drainagefromtankwithuniformcross-section
A cylindricaltankofdiameter1mmountedon itsaxiscontainsa liquidwhich
drainsthrougha 2 cmdiameterhole in thebaseof thetank.1fthetankorigi-
nally contains1000litres,determinethetimetakenfor totaldrainage.The
dischargecoefficientfor theholemaybetakenas0.6.
Solution
Considera tankof uniformcross-sectionAt, throughan orifice of areaao'
locatedin thebaseof thetank.An unsteadystatemassbalanceonthetankfor
no inflow of liquid relatestherateof flow fromthetankthroughtheorificeto
thechangeof capacityof thetank
C dao -J2gH =-A dH
dt
Rearranging,thetimetakenfor tankdrainagefrom a headH I to H2 canbe
foundby integration
t
fdt= -A H2 _I
0 CdaoJii f H 2dHHI
Completingtheintegration
t =~ fIrH~ -Hf
)Cda(}fil
147
A,
- - - - I ldH
01an
J}'IH'
+Flow out
FLUID MECHANICS
For totaldrainage,H 2=0.Therefore
A ~2HIt =Cdao g
Theinitialliquidlevelabovetheorifice,HI' is obtainedfromthevolumeof the
tankVI ' wheredl is thediameterof thetank
HI =4VI
1td2I
4xl
1tX12
=1.27m
Therefore
1tX12
t = ~ x ~2x 1.27
0.6x 1tx 0.Q22 g
4
=2120s
Total drainageis foundto take35minutesand20 seconds.
148
~
TANK DRAINAGE AND VARIABLE HEAD FLOW
6.4 Tankdrainagewithhemisphericalcross-section
A hemisphericaltankof4 mdiametercontainsa liquidandis emptiedthrough
a holeof diameter5 em.If thedischargecoefficientfor thehole is 0.6,deter-
minethetimerequiredtodrain thetankfrom aninitial depthof 1.5m.
- - -- -- --
ldH
} H, IH'
Solution
Applyinganunsteadymassstatebalanceonthetank,therateofdischargefrom
theholeis equalto thechangein capacityin thetank
Cdao.J2gH=-A dHdt (seeProblem6.3,page147)
Thedepthof liquidinthetankdoesnotchangeuniformlywithcross-sectional
areaof thetank.ThedepthH is relatedtothesurfaceareaof theliquidinthe
tankwhere
A =1tX2
The distancex is relatedtodepthH by Pythagoras
R2 2 + 2=x y
where
R =y +H
149
TANK DRAINAGE AND VARIABLE HEAD FLOW
Therefore
1II11
X2 =R2 - (R - H)2
=2RH - H 2
6.5 Tankdrainagewithcylindricalcross-section
DeriveanexpressionforthetimetodischargeliquidfromadepthH ina cylin-
drical tankof radiusR andlengthL positionedon itssidethroughanorificeof
areaao locatedontheundersideof thetank.
The balancethereforebecomes
t H
fdt = -1t fz(2RH - H2)
0 CdaoJii ! dHHI Hz
- -1t Hz
- CdaoJii 1 (2RHt -Ht)dHI
t - 1t
(
4 3 3
- -R"2"2 2 i 5
CdaoJii 3 (HI -H2 )-5(Hl -Hi))
-::1f: iHJ,
f
Thereforeon integration
To emptythetank,H2=0.Therefore
t = 1t
(
~RHt - ~H ~
)CdaoJii 3 I 5 I
(
4 3 2 5
)X 3"X2X1SZ -5x!Sz
xiii
= 1t
0.6x 1tx0.052
4
Solution
Thetimetakenforthetanktodraincanbedeterminedfrom
Cdao.J2gH=-A dH
dt (seeProblem6.3,page147)
=2286s Theareaoftheliquidsurfacevarieswithdepthwhere
A =2xL
Thetimetodrainthetankisfoundtobeapproximately38minutes.Notethatif
thevesselwereinitiallyfullandiscompletelyemptied,then
HI=R
The distancex is berelatedto liquiddepthby Pythagoras
R2 =x2 + y 2
H2 =0 where
togive R =y +H
5
141tR2t =
15CdaJii
Therefore
x =J2RH -H2
150
151
FLUID MECHANICS
The balanceequationis therefore
Cdao-J2gH =-2-J2RH - HZL dHdt
Rearranging,thetotaltimefordischargeisfoundbyintegrating
t
Jdt = -2L H2 I
0 Cda f2-- J (2R- H)2dH0 -v",g H I
to give
t = 4L 1
3C r;c((2R-H
)2 (
3
dao-v2g Z - 2R -Hl)2)
Notethatif thevesselwashalf full andwascompletelyemptiedthroughthe
orificethen
Hl=R
Hz =0
thenthetimefordrainagewouldbe
1
4LR2
t =
3CdaDf2i
152
~
TANK DRAINAGE AND VARIABLE HEAD FLOW
6.6 Drainagebetweentworeservoirs
A reservoirbeneatha smallforcedconvectionwatercoolingtoweris4 mlong
and 2 m wide.Beforerecirculation,thewaterflows into a smalleradjacent
reservoirofdimensions2 mby2 mbywayofa submergedopeningwithaflow
areaof100cm2anddischargecoefficientof0.62.Duringnormaloperationthe
levelin thelargerreservoiris30cmabovethatin thesmallerreservoir.Deter-
minetherateofflow betweenthereservoirsduringnormaloperationandif the
recirculationis halted,determinethetimetakento reducethedifferencein
levelsto 10cm.
ldh2
T
Solution
Forsteadyoperationwithacontinuousflowbetweenreservoirstherateofflow
throughthesubmergedopeningisgivenby
Q =Cda-J2gH
=0.62x 0.01x ~2gx 0.3
=0.015m\-l
Thus,theflow understeadyconditionsis foundtobe0.015m3s-1.Whenthe
recirculationis haltedthedrainagefrom thelargertankto thesmallertank
proceedsandthedifferencein levelsH changesby incrementsofdH whilein
thelargerandsmallertanksbydh1anddhz, respectively
H -dH =H -dhl -dhz
153
Al
- - -- - - - -- - -
-
- - -
- -
- -
lhl
T
-
H
FLUID MECHANICS
The incrementallossin capacityof thelargertankis equalto theincremental
gainin capacityin thesmallertank.Thatis
A]dh] =A2dh2
The incrementaldifferencein levelsis therefore
dH =dh{1+~~J
Thechangeincapacityof thelargertankis givenbytheunsteadystateequation
(seeProblem6.3,page147)
Cda-J2gH =-A] dh]dt
Substitutingfordh]
Cda-J2gH = -A] dH
[,<:t
Rearranging
t A H2 1
fdt= -] fH-2dH
0 Cda-J2i[1 + ~~JHl
Integratinggives
1 1
2A](Hf - H~)t =
Cda-J2i[1 +~~J
1 1
- 2x4x2x (0.32- 0.12)
- 0.62x omx -J2i x
(
I +4x 2
)2x2
=45s
The timetakenis foundtobe45seconds.
154
~
TANK DRAIN AGE AND V ARIAB LE HEAD FLOW
6.7 Tankinflowwithsimultaneousoutflow
Deriveanexpressionfor thetimetakenfor theliquid in a tankto reacha new
levelif theliquiddrainsfromanorificewhilethereis a constantflow of liquid
intothetank.
Solution
Theunsteadystatemassbalanceonthetankis
Q =Cdao-J2gH+A dHdt (seeProblem6.3,page147)
whereQis therateof flow intothetankanddH/dt is therateof changein liquid
level.For convenience,k is usedtogrouptheconstants
k =Cdao-J2i
Therefore
Q -km =AdH
dt
To determinetheheadof liquid in thetankat anygiventime,it is therefore
necessaryto integratebetweenthelimitsH]andH 2
t H2
fdt=Af ~ I
0 H]Q-kH2
155
Flowin
A, + Q--- ---
I ldH
--- -- --
01ao
11 }'IH'
+ Flow out
FLUID MECHANICS
To simplifythisintegrationthesubstitutionis used
I
u=Q-kHz
Rearranging
H=(Q~UJ
anddifferentiatingwithrespecttou
dH =-2(Q - u)du
k2
The integrationbecomes
t
fdt =-2A f
Q -u-du
0 k2 u
=-2A
(QfdU - fdU )k2 u
Completingtheintegrationgives
l r
1
1 ;
-2A Q - kH2 I 1
t =- Q log 2 + k(H Z - HZ )
k2 e 1 2 IQ - kH2I
orinfull
t = -2A
[
Q 10
[
Q -Cdao~
]
1 I
](Cdaoj2i)2 ge Q-Cdao~2gHI +Cdaoj2i(H~ -Hf)
156
~
TANK DRAINAGE AND VARIABLE HEAD FLOW
6.8 Instantaneoustankdischarge
A concretetankis ]5 m longby 10m wideand its sidesare vertical.Water
entersthetankata rateof200litrespersecondandisdischargedfromasluice,
thecentrelineofwhichis50cmabovethebottomof thetank.Whenthedepthof
waterin thetankis 2 m,theinstantaneousrateofdischargeis observedtobe
400litrespersecond.Determinethetimefor thelevelin thetanktofall] m.
Solution
Foratankofuniformcross-sectionwhichreceivesasteadyflowofliquidata
rateQandisallowedtodrainfreelythroughanopeningorsluice,thetimetaken
toalterthelevelcanbeshown(seeProblem6.7,page155)tobe
l r
1
1 ]
-2A Q - kH2 I I
t =- Qlog 2 + k(HZ -HZ )
k2 e 1 2 IQ - kH2I
Thedischargecoefficientandtheareaof thesluicearenotprovided.However,
therateof flow throughthesluiceQs is relatedtoheadby
I
Qs =kHz
or
-1
k =QsH 2
Whenthelevelwas2.0mtheflowwas0.4m3s-1.Thatis adepthof liquid
abovethecentrelineof thesluice,HI' of 1.5m.Thus
-1
k =0.4x 1.52
5
=0.326mzs-l
Thefinaldepth,H 2,is 0.5mabovethecentrelineof thesluice.Thetimetaken
for thelevelto fall is therefore
157
FLUID MECHANICS
[ [
1
] )
2A Q - kH2 1 1
t =~ Q log e ~ + k(H I - H !)
k Q-kH2 1
=-2 x 15~10x
[
0.2x loge
[
0.2- 0.326x 0.5~
]
+0.326x (O.5t-1.5t)
]0.326 0.2- 0.326x 1.52
=1536s
The timeis foundto beapproximately25){minutes.
Notethatalthoughtheareaof thesluiceanddischargecoefficientarenot
provided,theconstantk is usedto substitute
k =Cdaf2i
Assumingareasonabledischargecoefficientfor thesluiceof 0.62,theareaof
thesluicea canbefoundby rearranging
k
a--
- Cd f2i
0.326
=0.62x f2i
=O.ll9 m2
This wouldcorrespondtoasluiceof squarecross-sectionwithsidesmeasuring
34 cm.Careshouldbetaken,however,whenconsideringdischargesthrough
largeorificeswheretheheadproducingflow maybesubstantiallylessatthetop
thanatthebottom.
158
~
TANK DRAINAGE AND VARIABLE HEAD FLOW
6.9 Instantaneoustankinflowwithoutflow
A tankof uniformcross-sectionis providedwith a circular orifice50 mmin
diameterin thebottom.Waterflows into thetankat a uniformrateand is
dischargedthroughtheorifice.It is notedthatit takes90secondsfor thehead
in thetanktorisefrom 60cmto 70cmand120secondsfor it torisefrom 120
cmto125cm.Determinetherateof inflowandthecross-sectionalareaof the
tankassuminga dischargecoefficientof0.62for theorifice.
::t:
Q
~
""
:r:
dH
:dt
I
I
I
I
I
I
I
I
TIMEt
Solution
Anunsteadystatemassbalanceonthetankwhichreceivesaflowofwaterata
constantrateQ is
Q =Cdao)2gH +A dH
dt
(seeProblem6.7,page155)
Rearranging
dH =~- Cdao)2gH
dt A A
1
=~_kH2
A A
where
159
FLUID MECHANICS
k =CdaoJii
1tX 0.052
=0.62X X Jii4
5
=5.39x 10-3m2s-1
Foranaverageheadof0.65m,thechangein levelis0.1mover110seconds.
Thatis
0.1 dH
90 dt
I
=~- 5.39X 10-3 X 0.652
A A
For anaverageheadof 1.225m,thechangein levelis 0.05m taking120
seconds.Thatis
0.05 dH-=-
120 dt
I
=~- 5.39X 10-3 X 1.2252
A A
Solvingthesimultaneousequations
A =2.34m2
Q =6.94X 10-3 m3s-1
The areaof thetankandflowratearefoundto be2.34m2and6.94litresper
second,respectively.
160
~
TANK DRAINAGE AND VARIABLE HEAD FLOW
6.10Tankdischargethrougha horizontalpipewith
laminarflow
A viscousNewtonianliquidofdensity1100kgm-3andviscosity0.08Nsm-2is
fed toaprocessfrom a vesselofdiameter1.2mthrougha 3 mlengthofhori-
zontalpipe withan insidediameterof 25mmattachednearto thebaseof the
vessel.If theinitial levelof liquid in thevesselis 1.5mabovethepipeandthe
flow throughthepipeis laminar,determinethetimetofeedO.75m3oftheliquid
totheprocess.
A,
HI
ldH
Hil -------------
- - ---
td~ L 1 Flow,f
Solution
Neglectingentranceandexitlosses,anunsteadymassstatebalancefor the
vesselis
av =-A dH
dt
If theheadin thevesselis usedtoovercometheviscousresistanceof thepipe,
theaveragevelocityof theflowingliquid is thereforegivenby
t:J.pd2v=-
32/-LL
(seeProblem3.7,page73)
=pgHd2
32/-LL
161
~
IIII
III
I
FLUID MECHANICS
Therefore
2
apgHd =-A dH
32!-IL dt
Thetotaltimetakenfortheleveltofall fromaheadH 1toH2 is thereforefound
from
t H
fdt =-321lLAl dH
0 apgd2 H H1
On integrationthisbecomes
t =-321lLA loge
[
H 2
]apgd2 HI
where the final level in the vessel,H 2' is found from the total volume
dischargedto theprocess
4V
H2=Hl--
1td2t
=1.5- 4x 0.75
1tx 122
=0.84m
Thetimetakenistherefore
-32 x 0.08x 3x 1tX122
t= 4-
1tx 0.0252 x log [
0.84
]
4 x 1100x g x 0.0252 e l.s
=1521s
Thetimeisfoundtobeapproximately25~minutes.
162
........
TANK DRAINAGE AND VARIABLE HEAD FLOW
Furtherproblems
(1) Statethe assumptionsuponwhich simpletankdrainageproblemsare
based.
(2) A cylindricalvesselmountedverticallyonitsaxishasacross-sectional
areaof 1.2m2andcontainswaterwhichisallowedtodischargefreelythrough
anorifice,withacross-sectionalareaof 1oOcm2,positionedin thesideofthe
tank4mbelowthesurface.If thetankisopentoatmosphere,determinethetime
takenfortheleveltofallby2m.Thedischargecoefficientmaybetakenas0.6.
Answer:53s
(3) A rectangularorificeinthesideofatankis 1mbroadand50cmdeep.
Thelevelofwaterinthetankis50cmabovethetopedgeoftheorifice.Deter-
minetheflowthroughtheorificeif thecoefficientofdischargeis 0.62.Note
thatthevelocitythroughtheorificemaynotberegardedasconstantasthe
variationinheadatdifferentdepthsoftheorificewillbeconsiderable.
Answer:1.18m3s-1
(4) A rectangularorificeinthesideofalargewatertankhasabreadthof1m
anddepthof0.5m.Thewaterononesideof theorificeis atadepthof 1m
abovethetopedge;thewaterlevelontheothersideof theorificeis 25cm
belowthetopedge.Determinetherateofdischargeif thedischargecoefficient
maybetakenas0.62.
Answer:1.49m3s-1
(5) A sphericalstoragevessel3 min diametercontainsaprocessliquidat
halfcapacity.Determinethetimetakentodrainthevesseltoadepthof 1m
throughanorificewithadiameterof 25mmatthebottomof thevessel.The
dischargecoefficientoftheorificeis0.6.
Answer:2341s
(6) Deriveanexpressionforthetimetodrainanopenhemisphericalvessel
of radiusR throughanorificeof areaa0' if thevesselhasaninflow of liquidof
constantrateQ.
(7) Showthatthe timetakento draina sphericalvesselof radiusR
completelyfromfull throughanorificeof areaa locatedatthebottomof the
vesselis
163
FLUID MECHANICS
5
141tR2
t -
- 15Cdafii
whereCd isthecoefficientofdischargeandg theaccelerationduetogravity.
(8) A cylindricalvesselof diameter3 m verticallymountedon its axis
containsa liquidataninitialdepthof 4 mandis allowedtodischargetoa
similarnearbyvesselofdiameter4mbywayofapipewithaninsidediameter
of 100mm.Determinethefinaldepthinbothvesselsandthetimetoreachthis
condition.Ignoreanylossesduetofriction.
Answer:1.44m,502s
(9) A rectangulartank6mlongby2mwideisdividedintwopartsbyaparti-
tionsotheareaofonepartistwicetheareaoftheother.Thepartitioncontainsa
sluiceofarea100cm2.If thelevelofliquidinthesmallerdivisionis2mabove
thatofthelarger,determinethetimetoreducethedifferenceinlevelto50cm.
Assumeadischargecoefficientof0.6.
Answer:142s
(10) Twoidenticalopentanksofcross-sectionalarea4m3areconnectedbya
straightlengthofpipeoflength10mandinternaldiameter5cm.A viscousoil
withaviscosityof0.1Nsm-2anddensity900kgm-3isinitiallyatadepthof2min oneof thetankswhiletheothertankis empty.If avalvein theconnecting
pipeisfullyopened,determinethetimeforthedifferenceinleveltofallto5mm.
Neglectpipeentranceandexitlossesandassumetheconnectingpipeisinitially
fullofoil.
Answer:5100s
(11) In anexperimenttodeterminethecoefficientsof contraction,velocity
anddischargefor acircularorificeof 9 mmdiameter,waterwasdischarged
throughtheorificemountedverticallyinthesideofatank.A constantheadof
waterof 1.2mwasmaintainedabovethecentrelineoftheorifice.Theissuing
jetwasfoundtostrikeatargetplateahorizontaldistanceof850mmfromthe
venacontractaa verticaldistanceof 155mmbelowthecentrelineof the
orifice.Determinethevaluesofthecoefficientsfortheorificeif themeasured
dischargefromtheorificewas91litresin470seconds.
Answer:0.636,0.985,0.627
164
~
I
I
L
TANK DRAINAGE AND VAlUABLE HEAD FLOW
(12) Showthatthetimetakenfor liquidof densityp andviscosity/-lto drain
withlaminarflow fromtankofdiameterdt anddepthH throughaverticalpipe
of insidediameterd andlengthL, attachedtotheundersideof thetankis given
by
32/-lLd;1 (
H +L
)t = 4 og e -pgd L
(13) Show,wheretheareaofthetankisnotappreciablygreaterthanthearea
oftheorifice,thatthetheoreticalvelocityofajetofliquidv,foraconstanthead
H, flowingfromanorificeinthesideofatank,canbegivenby
~
'=v1-(~f
wherea is theareaof theorificeandA is thecross-sectionalareaof thetank.
(14) The.bottomof aprocessvesselhasaconicalsectionandcontainsaliquid
whichisrequiredtobedrainedthroughanopeningwithadiameterof25 mm.If
theliquidin thesectionis initiallyatadepthof 1.5mabovetheopeningcorre-
spondingtoadiameterof thesectionof 1.73m,determinethetimetodrainthe
sectioncompletely.Stateanyassumptionsused.
Answer:883s
(15) A processvesselwithauniformareaof 1.2m2receivesa liquidata
steadyrateof0~04m3s-1andissimultaneouslydischargedthroughanopening
atthebottomof thevessel.Whenthedepthof liquidwas0.5mtheinstanta-
neousrateof dischargewasnotedasbeing0.03m3s-l.Determinethetime
takenfortheleveltoriseby20m.
Answer:70s
(16) Waterflowsatasteadyrateof 360m3h-1intoavertical-sidedtankof
area10m2.Thewaterdischargescontinuouslyfromasluiceofarea0.0564m2.
If theinitiallevelinthetankis2.5mabovethesluice,determinethefinaldepth
after5minutesassumingadischargecoefficientof0.6forthesluice.
Answer:0.6m
165
~
Openchannels,
notchesandweirs
Introduction
Unlikethesingle-phaseflow of fluidsthroughpipes,theflow of liquids
throughopenchannelsorflumesischaracterizedbyafreesurfacenormallyat,
ornear,atmosphericpressure.Riversandartificialcanalsareexamplesofopen
channels.Openchannelflow alsoapplies,however,to theflowof liquids
throughpipeswhicharenotrunfull,asinthecaseofseweragepipes.Weirsare
verticalobstructionswhichlieacrossopenchannels,whilstnotchesareopen-
ings,normallyrectangularortriangular,cutinsuchweirs.Inpractice,theliquid
inthechannelbuildsupbehindtheweiruntilflowovertheweiroccurswithan
equilibriumheadof liquidbeingameasureofflowrate.
Therearenumerousapplicationsofopenchannelflow,notchesandweirs.
Althoughprimarilyencounteredintheprocesseffluentindustry,therearemany
otherapplicationsinthechemical,powerandoil industries.Theperforated
traysof distillationcolumns,for example,involvetheopenflowof liquid
throughwhichvapourisbubbledupfromthetraybeneath,withthedepthof
liquid maintainedby a weir overwhichthe liquiddischargesinto the
downcomer.
Manysophisticatedmathematicalmethodsandprocedureshavebeendevel-
opedandappliedtovarioustypesof openchannelflow.Traditionalmethods
arestillwidelyused,however,basedonthepioneeringworkof theFrench
engineerAntoineChezyin 1775andtheexperimentalworkoftheIrishengi-
neerRobertManningin 1891.In itssimplestform,thechannelis assumedto
haveagentleslopeinwhichtherateof flowundertheinfluenceofgravityis
balancedby frictionallossesandsuchthatthestreamlinesin theliquidrun
paralleltothefloor(orbed)ofthechannel.Underconditionsofuniformflow,
thereisthereforenochangeinvelocityintheflowingliquid(noacceleration)
andthedepthisconstant.Unlikesingle-phasepipeflow,thepressurecondi-
tionswithintheliquidaredeterminedbyhydrostaticprinciplesandconstant
atmosphericconditionsatthefreesurface.
167
FLUID MECHANICS
Thegeometryof channelsvariesfromwideandshallowtonarrowanddeep
with sidewalls thatmaybevertical,inclinedatan angleor curved.As open
channelshavea freesurface,openchannelflow thereforediffers frompipe
flow in thatthecross-sectionof flow is notconstrained.For increasedrateof
flow thereis an increaseddepth.This affectsnot only theflow areabutalso
wettedsurface(bothbedandwall)of thechannel.An importantparameterused
tocharacterizethegeometryof channelsis themeanhydraulicdepthdefinedas
theratioof flow areato wettedperimeter.
For problemsinvolvingopenchannels,determiningtherateof flow for a
given depthof liquid is straightforward.Arriving at the depthfor a given
flowrate,however,usuallyrequirestrial anderrorapproachessincethedepth
influencesboththeflow areaandthewettedsurface.Thecommonmethodis to
guessavaluefor thedepthandcalculatethecorrespondingflowrate,repeating
theprocedureasufficientnumberof timesto arriveattheanswer.
It shouldbenotedthatmanyof theequationsassociatedwithopenchannel
flow areempiricallybased.Many of the availabledatahavebeenobtained
usingwaterastheliquid medium.Empiricalrelationshipsusedto predictthe
flow for otherliquidsshouldthereforebeappliedwithcaution.
168
~
OPEN CHANNELS, NOTCHES AND WEIRS
7.1 Chezyformulaforopenchannelflow
Derive the Chezyformulafor the uniformflow of a liquid along an open
channelinclinedwitha slopei.
I L
-=--~ I
-- --=-L--=-~
~p8
F=mgsin8
Solution
Considera liquid of densityp flowing in anopenchannelof uniformcross-
sectionalareaA, inclinedata smallangle8.Understeadyuniformflow, the
gravitationalforceperpendicularto thebaseof thechannelis balancedby the
viscousresistanceforcesactingin theoppositedirection.Themassof theliquid
in thechannelcanbeexpressedin termsof liquiddensityandvolume
F =mgsine
=pALg sin 8
The resistanceforceswhichretardflow areobtainedby consideringthemean
wall shearstress'w of theliquid in contactwiththewettedchannelwall.
F ='wPL
Forequilibrium
pALgsin8='wPL
If themeanwallshearstresscanberelatedtothekineticenergyperunitvolume
'w =Lpv22
169
FLUID MECHANICS
wheref isaformoffrictioncoefficient.Notingalsothatforsmallangles,sine
is approximatelyequaltotaneortheslopeof thechanneli, anddefiningthe
ratioofflowareaA towettedperimeterP asthemeanhydraulicdepth
A
m=-
P
theaveragevelocityof theliquidflowingin thechannelcanthereforebe
expressedas
v =~2~mi
Thatis, for a givenfrictionfactorandmeanhydraulicdepththevelocityis
proportionaltothesquarerootoftheslopei.Definingalso
c=ff
The flow throughthechannelmaythereforebegivenby
Q =CAJ;;i
ThisisknownastheChezyformulawhereC istheChezycoefficient.Experi-
mentalvaluesandseveralcorrelationsareavailable,themostcommonof
whichistheManningformulagivenby
1
C=~
n
wheren isadimensionalroughnessfactor,themagnitudeofwhichdependson
thetypeof surface.Examplesof typicalroughnessfactorsaregivenonpage
292.Notethatastheroughnessof thechannelincreases,thevalueof n also
increases,reducingthevalueoftheChezycoefficientandthereforeflow.
170
-
L
OPEN CHANNELS, NOTCHES AND WEIRS
7.2 Flowina rectangularopenchannel
Waterfor coolingis deliveredtoa coolingtoweralonga rectangularconcrete
channell mwide.Determinetherateofdeliveryfor uniformflow ata depthof
30 em.The slopeof thechannelis 1..1000andtheroughnessfactorfor the
concreteis 0.014m-l/3s.
--------
I" 1m "1
Solution
ThedeliveryofwaterthroughtheopenchannelisgivenbytheChezyformula
Q =CAJ;;i (seeProblem7.1,page169)
wherei istheslope,A istheflowareaandC istheChezycoefficientgivenby
theManningformula(seeProblem7.1,page169).Thedischargeisthus
1
m6
Q =-AJ;;i
n
2
=:!m3.fi
n
Forthedimensionsofthechannel,theflowareaisA =1x 0.3
=0.3m2
andwettedperimeteris
P =2x 0.3+1
=1.6m
171
FLUID MECHANICS
The meanhydraulicdepthis therefore
A
m=-
p
0.3
1.6
=0.1875m
Thus,therateof deliveryis
03 ~
Q =~ X 0.18753X ~0.001
0.014
=0222 m3s-1
The deliveryof waterthroughthechannelis foundtobe0.222m3s-1.
" I
172
.110....-
OPEN CHANNELS. NOTCHES AND WEIRS
7.3 Depthofflowina rectangularchannel
A concrete-linedrectangularchannel12 m widehasa slopeof 1 in 10,000.
Determinethedepthofwaterflowing in it if thevolumetricflowrateis60m3s-l.
ObtaintheChezycoefficientfrom theManningformula takingtheroughness
factor as0.015m-1I3s.
12III -II'"
Solution
TheflowthroughthechannelisgivenbytheChezyformula
Q =CA& (seeProblem7.1,page169)
wherei is the slopeandC is the Chezy coefficientgivenby theManning
formula
L
C=~
n
wheren isthero!lghnessfactor.Theflowareaofthechannelin termsofdepthis
A =12H
andthewettedperimeter
P=12+2H
The meanhydraulicdepth(MHD) is therefore
A
m=-
P
12H
12+2H
6H-~
6 +H
173
FLUID MECHANICS OPEN CHANNELS, NOTCHES AND WEIRS
7.4 Economicaldepthofflowin rectangularchannels
A rectangularchannelis to be designedfor conveying300 m3of waterper
minute.Determinetheminimumcross-sectionalareaof thechannelif theslope
is 1 in 1600andit canbeassumedthat
An analyticalsolutionfor depthH is notpossible.It is thereforenecessaryto
useagraphicalor trialanderrorapproachillustratedbelow.
62
61
r':' 60
is
c)i
""'
~ 59
~
~
~ 58
57
174
v =70&
wherev is thevelocityof waterin thechannel,mis themeanhydraulicdepth
andi is theinclinationof thechannel.
I~
B
-I
Solution
Foraconstantflowthroughthechannelthemaximumflowoccurswhenthe
wettedperimeteris aminimum.Thewettedperimeteris
P=B+2H
andflow area
A =HB
4.0 4.1 4.13
HEAD H, ill
4.2 In termsof depth,thewettedperimeteris therefore
A
P=-+2H
H
Differentiatingthewettedperimeterwithrespecttodepthandequatingtozero
dP - -~ +2
dH - H2
=0
Therefore
A =2H2
175
Head Area Perimeter MHD Flowrate
H,m A,m2 P,m m,m Q, m3s-1
4.0 48 20 2.4 57.36
4.1 49.2 20.2 2.44 59.38
4.2 50.4 20.4 2.47 61.39
4.13 49.56 20.26 2.45 59.98
The depthis foundtobeapproximately4.13m.
FLUID MECHANICS OPEN CHANNELS, NOTCHES AND WEIRS
B =2H
7.5 Circularchannelflow
Determinethedepthofflow ofwaterina sewerpipeofdiameter0.9144mand
inclination1:200whenthedischargeis 940m3h-I.TheChezycoefficientmay
betakenas 100mII2s-I.
Thus
That is, for theminimumcross-sectionalarea,thebreadthof thechannelis
twicethechanneldepth.The meanhydraulicdepthis therefore
A
m=-
p
2H2
4H
H
2
Fromtheequationprovidedfor velocity,theflow throughthechannelis
Q =70A j;;;i /~ !
.:::::-~-- '>-.
~
Solution
Letr betheradius,8thehalfanglesubtendedatthecentrebythewaterlevel
andd thedepth.Fromgeometry
R-H
cos8=-
R
In termsof depthH, theflow canthereforebeexpressedas
Q =140Ht H
Rearrangingin termsof channeldepth
2
H=[I;o~J
[
(lQQl
!
t
= ~x ..)2x 1600
140
p
from which the angle8 may be obtainedin radians.The areaof wetted
cross-sectionis givenby
2
28R 2.
A =- -R sm8cos8
R
=R2(8- sin228)
=1.6m
Theflowareaistherefore
A =2H2 P =2R8
andwettedperimeteris
=2x 1.62 for whichthemeanhydraulicdepthis
=5.12m2
Am=-
P
The minimumflow areais foundtobe5.12m2.
176 177
FLUID MECHANICS
Q =CA~
The flow throughthechannelis givenby theChezyfonnula
(seeProblem7.1,page169)
whereCis theChezycoefficientandi is thechannelslope.A graphicalor trial
anderrorapproachis usedfor guessedvaluesof depth:
:: 1000
6 940
d
p:.i
~
p::
~
ti 500
178
0
0 0.1 0.2
DEPTH H, m
0.201
0.3
I
L
OPEN CHANNELS, NOTCHES AND WEIRS
7.6 Maximumflowin circularchannels
Determinethetheoreticaldepthof liquidina closedcircularchannelof radius
Rfor maximumvelocity.
Solution
The velocityof liquid in a circular channeldependson the depth.As the
velocityis proportionaltothemeanhydraulicdepth,itsmaximumvaluemaybe
obtainedby
pdA -A dP =0
d8 d8
wherefromgeometry,thewettedareaof flow is
A =R28- R2 sin28
2
=R2l8 - sin228)
(seeProblem7.5,page177)
andthewe~tedperimeteris
P =2R8
DifferentiatingAlP andequatingtozero
3 3( sin28
)2R 8(l-cos28) =2R l8 - ~
Therefore
28=tan28
The solutionto whichis
28 =257SO
The depthfor maximumvelocityis therefore
H =R R 257.5
0
- cos-
2
=R(l +0.62)
=1.62R
The depthfor maximumvelocityis 0.81timesthechanneldiameter.
179
H,m cos8 8,rad A,m2 P,m m,m Q, m3h-1
0.1 0.781 0.674 0.039 0.616 0.063 248.7
0.15 0.671 0.834 0.070 0.763 0.092 543.2
0.2 0.562 0.973 0.106 0.890 0.119 932.2
0.25 0.453 1.101 0.146 1.006 0.145 1412.9
0.201 0.560 0.975 0.107 0.891 0.120 940.0
The depthin thechannelis foundtobe0.201m.
1500
FLUID MECHANICS
7.7 Weirsandrectangularnotches
Theweir abovea downcomerin therectifyingsectionofa distillationcolumn
hasabreadthof1.2moverwhichliquidflowswitha headof18mm.Determine
therateofflow if thecoefficientofdischargefor theweir is 0.56.
FI"",-- -- ~== ~~- - B .I- - - -==c pz-[\ """""""'~ TdZ lH
\
Solution
To determinetheflowrateofliquid overweirs(andnotches)considertheflow
of liquidalongahorizontalplaneatdepthz belowthefreesurface.It is assumed
thattheupstreamvelocity,sometimesknownas thevelocityof approach,is
negligiblein comparisonto theflow overtheweir.At a pointjust beyondthe
weir,thevelocityof theliquidwill haveincreasedtosomevalue,v, inwhichthe
staticheadis convertedtovelocityhead
2
vz=-
2g
Therefore
v =~2gz
Thus,thevelocityisproportionaltothesquarerootofthehead.To determine
therateof flow,consideranelementaldepthdz of theflowovertheweir
betweendepthszandz+dz.SincetheelementalflowareaisBdz,thetheoretical
elementalflowratedQistherefore
dQ =~2gzBdz
ISO
..0100..-
OPEN CHANNELS, NOTCHES AND WEIRS
Thetotalflowrateof liquidQ is foundby integratingoverthedepthof theweir
or rectangularnotchH. Thatis
Q H 1
fdQ =Bj2i fz2dz
0 0
to give
2 3
Q =-Bj2iH23
In practice,therearepermanentenergylossesduetoedgeeffectsgivingriseto
a flowratewhichis lowerthanthetheoretical.Introducinga dischargecoeffi-
cientCd, theactualflowrateis
2 3
Q =-CdBj2iH23
In thiscase
2 3
Q =- x0.56x1.2x j2i xO.olS23 .
=4.79x 10-3m3s-1
This correspondsto arateof flow of liquidovertheweir of 17.25m3h-l.
Notethatthedischargecoefficienttakesintoaccountthecontractionof the
overflowingjet as well as the effectsof viscosityand surfacetension(see
Problem4.2,page102).Little is knownof theseparateinfluenceof thesetwo
fluid parametersexceptthattheybecomeappreciablewhentheheadon the
weir andsizeof theweirdecrease(seeProblem7.9,pagelS4).
IS 1
FLUID MECHANICS
7.8 Depthofa rectangularweir
A hydroelectricpowerstationis locatedona dammedreservoirandis largely
usedtogenerateelectricityat timesofpeakdemand.In additionto thewater
whichpassesthroughtheturbines,aflow of watercontinuouslybypassesthe
turbinesandis dischargedfromthereservoirovera rectangularweir intothe
riverbelow.Whentheturbinesareoperatingatpeakdemand,thebypassflow
is ata minimumof34m3h-landthefall fromthereservoirtotheriveris5.5m.
Whenthepowerstationoperatesat low demand,theflow of waterbypassing
theturbinesis ata maximumof217 m3h-landthefall is 8.5m.Determinethe
heightof theweircrestabovethesurfaceof theriver.
- ---
Flow-
f' I",I",Riverbed
Solution
The flow overtheweirof heightHI intotheriverbelowfor bothcasesis
2 1
Ql =-CdBj2i(HI -HoF
3
(seeProblem7.7,page180)
and
2 1
Q2 =-CdBj2i (H2 - H of3
whereB is thebreadthof theweir.Assumingaconstantdischargecoefficient
Cd thenQI is relatedtoQ2 by
QI =
[
HI -Ho
]
t
Q2 H2 -Ho
182
~
Rearrangingin termsof H 0
[Qt J
Ho =H2 ~2 -HI
[~~J -1
2
=55x (¥iY - 85
2(~1:Y-1
=4.27m
OPEN CHANNELS, NOTCHES AND WEIRS
The weir is foundtobeaheightof 4.27m abovetheriverbed.
183
FLUID MECHANICS
7.9 Instantaneousflowthrougha rectangularweir
Waterusedas a coolingmediumin a processplant is storedin a largeopen
tankwhichhasa rectangularoverflowweir. Whenthetankis overfilledthe
watercascadesovertheweir.Determinethetimetakento lower thelevelof
waterto a headof 1 cmabovetheweir crestif it is notedthatit takes600
secondsfor theleveltofall froman initial headof 8 cmto 6 cmovertheweir
crestwhentheinflowtothetankhasceased.
Solution
Theelementalrateof flowthroughtheweiris
2 3
dQ =-CdBj2i H2dt3
(seeProblem7.7,page180)
andthedischargefromthetankis
dQ =-AdH
Therefore
2 3
-CdBj2iH2dt =-AdH3
Iii
III
I
I
The dischargeovertheweir is thereforegivenin generaltermsby
kHt =_dH
dt
Foranaverageheadof0.07mthen
k x omt ={
-om
)600
Thus
k=1.8x 10-3m-l/2s-1
184
.-..
OPEN CHANNELS, NOTCHES AND WEIRS
Thetotaltimet todrainovertheweirfromaninitialheadHI toafinalheadH 2
is foundby integratingwithrespecttoH
t Hz 3
f
1
f
--
dt =-- H zdH
0 k HI
2 1 1
=-(H z - HZ)
k 2 1
2 1 1
- X (0.01 Z - 0.08 Z)
1.8x 10-3
=7182s
Thetimeis foundtobeapproximately2 hours.Notethatit wouldtakeaninfi-
nitelengthof timeto emptythetank,H 2 =0.This equationthereforedoesnot
hold for verysmallvaluesof H 2 becausea layerof wateradherestotheweir
crestdueto surfacetension.
::t::
CI
<r:
gj
dHI
dt
TIMEt 00
185
FLUID MECHANICS
7.10Flowthroughatriangularnotch
A weaksolutionofcausticsodaisfed toanabsorbertoremovesulphurdioxide
from a processgas.Thedistributorto theabsorberconsistsof channelswith
90°V-notchesthroughwhichthecausticsodasolutiondischarges.Determine
therateofflow througheachnotchif theheadabovetheroot(bottom)is 3 em.
A dischargecoefficientof O.62 is assumedforthenotches.
2Htan{:!
I- -I
- ---- --
Solution
Triangularnotchesaregenerallyusedto measuresmallratesof flow in rivers
andchannelsandcanbeusedtochannelflow for distributionpurposesasillus-
tratedin thequestion.For thegeneralcasefor flow overatriangularnotch,let
the notchsemi-anglebe 8 for which the elementalarealies at somedepth
betweenz andz+dzbelowthefreesurface.The theoreticalelementalrateof
flow,dQ, throughtheelementalarea,is thereforegivenas
dQ =~2gz2(H- z)tan 8dz
IntroducingadischargecoefficientCd, thetotalactualflowrateof liquid over
thenotchis thereforeobtainedby integratingfrom z=0at thefreesurfaceto
z=H atthebottom(orroot)of thenotch.Thatis
Q H 1 3
fdQ =Cd2.figtan8f(Hz.2 -z2)dz
0 0
186
........
OPEN CHANNELS. NOTCHES AND WEIRS
Integratingwithrespecttoz gives
~
[
2.l 2 ~
]
H
Q =Cd -V2g 2 tan 8 - Hz z - - Z2
3 5 0
8 ~
=-Cd.fig tan8Hz15
If thesemi-angleis45°(90°notch),theequationfor liquidflowratereducesto
8 5
Q =-Cd.fig H.215
8 5
=- x 0.62x .figx 0.03.215
=228X 10-4 m\-I
This correspondsto arateof flow througheachnotchof 0.822m3h-l.
Notethat,aswithweirsandrectangularnotches(seeProblem7.7,page180)
anddischargingorifices(seeProblem6.2,page145),thedischargecoefficient
Cd allowsforenergylossesandthecontractionof thestreamcross-section.The
actualdischargethroughthetriangularorV-notchis thereforefoundmymulti-
plyingthetheoreticalflow (ordischarge)byCd'
In thisanalysisthevelocityof theliquidapproachingthenotchis assumedto
beconsiderablylessthantherateof flow throughthenotch.Thekineticenergy
of theapproachingliquid is thereforeneglected.Thatis, thevelocitythrough
thehorizontalelementalareaacrossthenotchis assumedtobedependentonly
on the depthof flow below the free surface.If, however,the velocityof
approachcannotbeconsideredtobenegligible,asmaybethecaseforanarrow
channelwitha singlenotch,thetotalheadproducingflow will beincreasedby
thekineticenergyof theapproachingliquidsuchthatthetotalflow throughthe
notchis foundfrom
Q=I~Cd~l'ner[H+~!J-[~!n
wherev1is theaveragevelocityof theapproachingliquidin thechannel.
187
FLUID MECHANICS
7.11TankdrainagethroughaV-notch
A sharp-edgedV-notchin thesideofa rectangulartankmeasuring4 mlongby
2 mbroadgivesa calibration:
5
Q =1.5HZ
whereQ is measuredin cubicmetresper secondandH is measuredinmetres.
Determinethetimeto reducetheheadin thetankfrom J 5 cmto 5 cmif the
waterdischargesfreely throughthenotchandthereis no inflowtothetank.
Solution
ConsiderthelevelofwatertobesomedepthH abovethebottomofthenotch.
A smallquantityof flowdQwouldreducethedepthbyaleveldH in timedt.
Then
2-
dQ =1.5H 2dt
withachangeincapacityof thetankof
dQ =-AdH
ThetimetolowerthelevelinthetankfromaninitialdepthHI toafinaldepth
H2is thereforefoundfrom
t H2 5
Jdt =-A J H-zdH
0 1.5HI
Integrating
2 A -1 -1
t=--(H 2 -H 2)
31.5 2 1
3 -1
=2x 4x 2 x (0.05- 2 - 0.152)3x 15
=257s
Thetimeisfoundtobeapproximately4~minutes.
Notethattheconstant1.5isequalto
15=~Cdjii15 (seeProblem7.10,page186)
correspondingtoadischargecoefficientCd of0.635.
188
~
OPEN CHANNELS. NOTCHES AND WEIRS
----
7.12 Flowthroughatrapezoidalnotch
Deduceanexpressionfor thedischargethrougha trapezoidalnotchwhichhas
abaseB andaheadH, andthesidesofwhichmakeanangle8tothevertical.
H
I.. B --I
Solution
Theflowcanbeconsideredastheflowthroughatriangularnotch(seeProblem
7.10,page186)andarectangularweir(seeProblem7.7,page180)wherethe
respectivedischargesare
8 2-
Qnotch = -Cdjii tan8H215
2 1
Qweir =-CdBjii H 23
Combiningthedischarges
8 5 2 3
=-Cd jii tan8Hz +-CdB.,j2gH215 3
8 3
( 5B)=-CdjiiH2 Htan8+-15 4
Qtrap.notch
Thisequationmayalsobeobtainedfromfirstprinciplesforanelementalflow
throughthetrapezoidalsectionandintegratingoverthetotaldepthof flowH.
189
FLUID MECHANICS
Furtherproblems
(1) Determinethetheoreticaldepthof liquid in aclosedcircularchannelfor
maximumdischarge.
Answer:0.95d
(2) A concrete-linedrectangularchannelis tobebuilttotransportwaterfrom
anearbyriverto apowerstationasa supplyof make-upwaterfor thecooling
towers.Determinetheminimumdepthof thechannelif therateof flow is not
expectedtoexceed650m3min-1in achannel5 mwidewithafall of 1in5000.
AssumetheManningformulacanbeusedto determinetheChezycoefficient,
wheretheroughnessfactorfor theconcretemaybetakenas0.014m-l/3s.
Answer:2 m
(3) Explainwhy thereshouldbea depth,lessthanfull capacity,atwhicha
pipewill carrywateratamaximumrateofnow andsuggestawayin whichthis
canbedeterminedanalytically.
(4) Determinetheangleof thewalls andtherelationshipbetweenbreadth
and depthof a channelof trapezoidalcross-sectionnecessaryto deliver
maximumnow.
Answer:30°,B =2/-J3H
(5) Waterfromareservoirdischargesoverarectangularweirof breadth10
m intoa sluicebelow.Whenthereservoiris atmaximumcapacity,therateof
dischargeovertheweir is 5000m3h-1 andthedistancefromthesurfaceof the
reservoirdowntothesluiceis 2 m.Determinetheheightof theweircrestand
thedischargecoefficientif thedroptothesluiceis 1.9mfor arateofdischarge
overtheweirof 1000m3h-1.
Answer:1.85m,0.81
(6) Experimentaldataarerecordedforthedischargethrougha90°triangular
notchfor differentheads.Plot thesedataon log-log paperanddeterminethe
coefficientof dischargefor thenotch.
Flow, m3h-1
Head,m
2.99
0.05
6.95
0.07
9.70
0.08
13.02
0.09
16.94
0.10
0.30
0.02
Answer:0.63
190
..-..
OPEN CHANNELS, NOTCHES AND WEIRS
(7) Deriveanexpressionfor thedischargethrougha weirwheretheareaof
the channelthroughwhich the waterapproachesthe weir is such thatthe
velocityof approachis notinsignificant.
(8) Deriveanexpressionfortherateof flow of liquidoverarectangularweir
of breadthB in termsof headH.
(9) Determinethedischargecoefficientfor a rectangularweir placedin a
channell mwideof inclination1:2500withaChezycoefficientof 66ml/2s-1
if thedepthin thechannelis50cmandtheheightabovetheweircrestis 32cm.
Answer:0.62(10) Determinetherateof now througha 60°V-notchif theheadof liquid
abovetherootis 5 cm.A dischargecoefficientof 0.6maybeassumed.
Answer:1.65m3h-1
(11) Deduceanexpressionfor thenow of liquidthroughatriangularnotchof
angle8.
(12) Showthattheheadof liquidovera weiron a sievetrayin a distillation
columnis givenby
H =O.715[~)\
whereQis therateof now (m3s-1)andB is thebreadthof theweir(m).
(13) DeterminethecoefficientofdischargefortheweirinFurtherProblem(12).
Answer:0.56
(14) An aqueoussolutionof sodiumhydroxideisfedcontinuouslytoapacked
columntoremovehydrogenchloridefromaprocessgas.Theliquiddistributor
abovethepackingconsistsof channelsandthe aqueoussolutionovernows
through60identical60°V -notches.Determinetheheadabovetherootof the
notchesin the distributorif the liquid now to the columnis 58 m3h-1.A
dischargecoefficientof 0.62for thenotchesmaybeassumed.
Answer:4 cm
191
..-..
Pipefrictionand
turbulentflow
Introduction
Most applicationsconcernedwith fluids in pipelinesinvolveturbulentflow.
Greateffortshavebeenmadeby scientistsandengineersover theyearsto
developempiricalrelationshipsthatpredictthenatureof turbulentfluidflow in
bothroughandsmooth-walledpipes.But in spiteof this,noexactsolutionsare
availablewhichwill predictpreciselythenatureof theturbulentflow.
FrenchengineerHenryPhilibertGaspardDarcyin 1845andGermanengi-
neerandscientistJulius Weisbachin 1854,aftermuchexperimentalwork,first
proposed.africtionfactorequationthatexpressedthepressurelossina piping
systemin termsof velocityhead.In 1911,Paul RichardHeinich Blasius,a
studentof Ludwig Prandtl(1875-1953),demonstratedthatfor smooth-walled
pipesthefrictionfactorwasdependentonlyontheReynoldsnumber.Blasius
producedthefirstplotof itskindwith frictionfactorversusReynoldsnumber
for theempiricalrelationship
-1
f =0.079Re 4
valid for Reynoldsnumbersbetween4x1O3and Ix 105.Three yearslater,
British engineersSir ThomasErnestStanton(a formerassistantof Osborne
Reynolds)andJ.R. Pannellestablishedthe friction factor/Reynoldsnumber
relationshipto be independentof thefluid. Over thefollowing yearsseveral
otherrelationshipswerealsoestablishedof theform
f =a+bRec
193
a b c Validity
Lees(1924) 1.8xlO-3 0.152 -0.35 4x1O3<Re<4x105
Hermann(1930) l.35xlO-3 0.099 -0.3 4x1O3<Re<2x106
Nikuradse(1932)8xlO-4 0.055 -0.237 4x1O3<Re<3.2x1O6
FLUID MECHANICS
To predictthefrictionof fluidsin smoothpipeswithturbulentflow, Prandtl
developeda numberof empiricalmodelsbasedon boundarylayer theory,
mixinglengthandwall effects.Thesemodelsledto Prandtl'suniversalresis-
tanceequationfor turbulentflow in smoothpipesgivenby
1
jj =410glO(Rejj)-0.4
andvalidfor Reynoldsnumbersbetween5xl03 and3.4xl06.
In 1858,Darcyconductednumerousdetailedexperimentsinvolvingpipes
madefromvariousmaterials.In 1932,GermanengineerJohannesNikuradse
analysedDarcy'sdataandnotedthatfor turbulentflow in rough-walledpipes,
the friction factor varied only slightly with Reynolds number,showinga
decreasein frictionfactorwith increasingReynoldsnumber.Above a certain
limit, thefriction factorwas foundto be independentof Reynoldsnumber.
Nikuradsealso conductedhis own work using artificiallyroughenedpipes.
Gluingcarefullygradedsandontotheinsideof smoothpipes,heestablisheda
scaleof relativeroughnessinwhichthediameterof sandparticlesin thepipe,E,
is takenastheratioof theuncoatedpipediameter,d. Usingthedataof Niku-
radsein 1921,the Hungarian-bornaerodynamicistTheodorevon Karman
(1881-1963),aco-workerof Prandtl,developedanempiricalrelationshipfor
turbulentflow throughrough-walledpipesin themodifiedform
Jy. =210glOl~) +2.28
In 1939c.F. Colebrook,workingin collaborationwithc.M. White,devel-
opeda mathematicalfunctionwhich agreedcloselywith valuesobtainedfor
naturallyroughcommercialpipes.This involvedusingbothPrandtl'ssmooth
pipelawof frictionandcombiningit withvonKarman'sfully roughpipelawof
frictionintoasingleempiricalexpressionin theform
1 (E 4.675
)jj =2.28-410glOld +Rejj
It wasLouis F. Moodywho,in 1944,firstpresentedacompositeplotof fric-
tionfactorwithReynoldsnumberwhichincludedthestraightlinerelationship
for laminarfrictionfactor,smoothpipeturbulentfrictionfactorandroughpipe
turbulentfriction factor, togetherwith the conceptof relativeroughness.
Known astheMoodyplot(seepage293),it continuestobeavaluabletool for
evaluatingfriction factorsfor pipe flow in which it is evidentthat in the
194
L
PIPE FRICTION AND TURB ULENT FLOW
turbulentregion,the slopeof the curverelatingfriction factorto Reynolds
numberdecreaseswithincreasingReynoldsnumber,becomingindependentof
Reynoldsnumberathighvalues.
Nomenclatureandscientificreasoningarebothmajordifficultiesencoun-
teredin thefieldofpipefriction.Thereareseveralwidely-acceptedapproaches
usedby industry,eachhavingavaliditywhich- irrespectiveof procedure-
providesidenticalfinal answers.To the unwary,the approacheshave the
propensityto confuse.As acomparison,thefrictionfactordevelopedin 1893
by theAmericanengineerJohn ThomasFanning(1837-1911),andknownas
theFanningfrictionfactorf, is relatedto othercommonly-encounteredforms
andsymbolsfor thefrictionfactorwhere
- ~=~=~ =2~
f =Cf =iF - 4 4 pv2 pv
A commonproblemin calculatingpressurelossthroughpipesin whichthe
flow of fluidisknowntobeturbulentbutofunknownrelativeroughnessand/or
flowrate,and thusunknownReynoldsnumber,is thatthe Fanningfriction
factorcannotbereadilyobtained.In suchcasesit is usualto assumeeithera
valueof0.005tosimplifytheproblem,ortousethisasabasistobeginanitera-
tiveprocedureconvergingona solution.
195
FLUID MECHANICS
8.1 Economicpipediameter
Explainwhatis meantbyeconomicpipediameter.
Minimum
cost
f-<
CI'J
0
U
Operating
costs
EPD
DIAMETER
Solution
Theeconomicpipediameteristhediameterofpipewhichgivestheminimum
overallcostforanyspecificflowrate.Ingeneral,however,pipesaredetermined
byselectingareasonablefluidvelocitywhichprovidesareasonablepressure
dropandisvirtuallyindependentofdiameter.It ispossible,however,todeter-
minetheeconomicpipediameterbasedon economicprinciplesby which
capitalchargesforpipesandvalvesarebalancedagainstpumpinvestmentand
operatingcosts.An economicoptimumcanbereadilydeterminedwherethe
costofpipematerialappropriatetohandleaparticularfluidis correlatedwith
pipediameter.Onceacalculationhasbeenmade,it isnecessarytodetermine
theReynoldsnumbertoensurethatflowis turbulentandthatotheressential
requirementsaremet,suchasnetpositivesuctionheadin thesuctionlinesto
centrifugalpumps.
Theminimuminvestmentis calculatedforexpensiveorexoticpipemate-
rialssuchasalloys,pipelineslargerthan300mmindiameterandcarbonsteel
lineswithalargenumberofvalvesandfittings.Thepipeschedulingisselected
bydeterminingeithertheinnerorouterdiameterandthepipewallthickness.
Theminimumwallthicknessisafunctionofallowablestressofthepipemate-
rial,diameter,designpressure,andcorrosionanderosionrates.
In thecaseofhighlyviscousliquids,pipelinesarerarelysizedoneconomic
considerations.
196
.a ,.
PIPE FRICTION AND TURBULENT FLOW
8.2 Headlossduetofriction
Deriveanequationfor thepressureandheadlossduetofrictionfor afluid
flowingthroughapipeoflengthL andinsidediameterd.
,... L -,
1:w
Flow PI-- -.-- -'-'-.
Pz - J~<L-'-'-'
Solution
Letthepressuredifferenceordropbesolelyduetofrictionmanifestasawall
shearstre.ss,'tw.Forsteadystateconditions,aforcebalanceonthefluidinthe
cross-sectionof thepipeis
nd2 nd2 nd2
PI--P2-=!!,Pj-4 4 4
='twnLd
Thepressuredropduetofrictionistherefore
!!'Pj =4'twfd
If thewallshearstressisrelatedtothekineticenergypervolume,then
't =Lpv2w 2
wheref is theFanningfrictionfactor.Thepressuredropduetofrictionmay
thereforebeexpressedas
!!,pj=2fPv2Ld
197
~
\1
FLUID MECHANICS
or in headform
2
4jL~
Hf =d 2g
This is knownastheFanningorDarcyequation.It wasmuchearlier,however,
thatDarcyin 1845andWeisbachin1854firstproposedthefrictionfactorequa-
tionaftermuchexperimentalwork,givingrisetotheheadlossduetofrictionin
theform
AL~
Hf =d 2g
This is knownastheDarcy-Weisbachequationwherethefrictionfactoris
relatedtotheFanningfrictionfactorby
A=4f
198
~
PIPE FRICTION AND TURBULENT FLOW
8.3 Generalfrictionalpressuredropequation
appliedto laminarflow
Deducea relationshipbetweenthefrictionfactor usedin theDarcy equation
andReynoldsnumberfor afluid withfully developedlaminarflow.
Solution
Theaveragevelocityofafluidwithlaminarflowthroughthepipeisgivenby
v =~ I1p R2
811 L
(seeProblem3.7,page73)
Rearranging,thepressuredropalongthepipeis therefore
A 811LvLJ.p=-
R2
orintermsofpipediameter
I1p=3211Lv
d2
TheDarcyequationforfrictionalpressuredropalongapipeis
I1Pf=2fPv2Ld
Combiningthetwoequationsin termsofpressuredrop
2fpv 2L =3211Lv
d d2
which,in termsof theFanningfrictionfactor,reducesto
f =1611
pvd
16--
Re
That is, for laminarflow thefrictionfactoris inverselyproportionalto
ReynoldsnumberandwhenthelogofReynoldsnumberisplottedagainstthe
logof thefrictionfactor(Moodyplot)givesastraightlinewithagradientof
-1. Notethat,usingtheDarcy-Weisbachequation,thisisequivalentto
A= 64
Re
199
FLUID MECHANICS
8.4 Blasius'equationforsmooth-walledpipes
Ethylbenzene,witha densityof 867kgm-3andviscosityof7.5xI0-4 Nsm-2,is
tobetransferredata rateof12m3h-1througha smooth-walledpipeline200m
longundertheactionofgravity.Determinetheinternaldiameterofthepipeline
if thereis afall in elevationalongthepipelineof 10 m.AssumetheBlasius
equationcanbeappliedtodeterminethefrictionfactor.
Solution
Theheadlossduetofrictionisequaltothestatichead.Thatis
H =Hf
=4fL~
d 2g
wherethefrictionfactoris givenby theBlasiusequation
-1
f =0.D79Re 4
andwheretheReynoldsnumberexpressedin termsof volumetricflowrateis
Re =4pQ
1td~
Therefore
4x 0.D79
H=
4x 0.D79x
12
1
-t
r
12 ,24x 867x --- 4x -
4x 0.D79x I 3600 x 200x - 3600I x [l'f
1tx75 x 10-4 1t
10=
2g
200
l
PIPE FRICTION AND TURBULENT FLOW
This reduces to
19
1.44x 106=d-4
6 -19
loge 1.44x 10 =- loge d
4
loge d =-2.985
Solving, thediameterof thepipelineis foundto be 0.0505m (50.5mm).A
checkfor Reynoldsnumbergives
Re =4pQ
1td~
124x 867x -
- 3600
1tx 0.0505x 75 X 10-4
=97,153
This valueli~sbetween4x103and1x105,confirmingthevalidityof usingthe
Blasiusequationfor smooth-walledpipes.
201
FLUID MECHANICS
8.5 Prandtl'suniversalresistanceequationfor
smooth-walledpipes
A benzenemixture,withdensityof873kgm-3andviscosityof8.8x10-4Nsm-2,
isfedfromanopenoverheadtanktoaprocessthrougha smoothpipeof25 mm
boreandlength18m.If theprocessoperatesatapressureof55kNm-2above
atmospheric,determinetheminimumallowableheadof theliquidsurfacein
thetankabovetheprocessfeedpoint if theflow isnottofall below40kgmin-l.
Allow for entranceand exitlossesand applyPrandtl's universalresistance
equationfor turbulentflow in smoothpipes.
--- ---- - -- -- --
Opentank
H
Solution
Assumingthatthevelocityin theoverheadtankis smallin comparisonto the
velocity in the pipe (vI =0), thenapplyingthe Bernoulli equationbetween
points1and2
2
PJ P2 v2
-+zl =-+-+Z2 +HL
pg pg 2g
The velocityv2 canbedeterminedfrom
4m
v2=-
pnd2
4x 40
60
873x nx 0.Q252
=155ms-J
202
.) 1
PIPE FRICTION AND TURB ULENT FLOW
For Reynoldsnumber
pvdRe=-
~
873x 155x 0.Q25
8.8x 10-4
=38,441
which satisfiesthe criteriafor Prandtl's equation(5xIO3<Re<3.6xIO6).The
Prandtlequationis
J-.r =410g10(ReFn - 0.4
=410glO(38,441-H)-0.4
By trialanderror
f LHS
0.004 15.81
0.005 14.14
0.006 . 12.91
0.0055 13.48
RHS
13.14
13.33
13.49
13.42
Thatis, a frictionfactorof approximately0.0055.Thetotalheadlossis dueto
pipefriction,entranceandexitlosses
2 2 2
4fL v2 v2 v2
H L =- - +05- +1.0-
d 2g 2g 2g
=v;
(
4:fL+15
)2g d
RearrangingtheBernoulliequation,theminimumallowableheadis therefore
2 2
H =P2 - PI +2 +2
(
4fL +15
)pg 2g 2g d ~
55X 103 1552 1552
(
4x 0.0055x 18
15}
+-+-x +
873x g 2g 2g 0.Q25
=8.66m
The minimumallowableheightof liquid in thetankabovetheprocessfeed
pointis 8.66m.
203
FLUID MECHANICS
8.6 Pressuredropthrougha rough-walled
horizontalpipe
Twoethanolstoragetanksareconnectedby100mo.fstraightpipe.Bothtanks
are opentoatmosphereandtheconnectingpipehasan insidediameterof 50
mmand relativeroughnessof 0.002.Determinethepressuredrop overthe
lengthofpipe if theflowrateis 15m3h-l andestimatethedifferencein levelof
ethanolbetweenthetwotanks.Thereare no extrafittingsin thepipe but the
entrancetothepipeandtheexitfromthepipeshouldbetakenintoaccount.The
densityofethanolis 780kgm-3anditsviscosityis 1.7x1o-3Nsm-2.
--- ---- - -- -- --
H
- - -- -- --
Solution
The averagevelocityof ethanolthroughthepipeis
4Qv=-
rtd2
4x~
3600
rtX0.052
=2.12ms-l
whichcorrespondstoaReynoldsnumberof
Re =pvd
/.l
780x 2.12x 0.05
1.7x 10-3
=48,635
204
.......
PIPE FRICTION AND TURB ULENT FLOW
The relativeroughnessfor thepipe,£/d, is 0.002.FromtheMoody plot (see
page293)thefrictionfactoris 0.0065.The headlossdueto frictionalongthe
pipeis thereforefoundfrom
2fpv2L
!:J.Pf =- d
2x 0.0065x 780x 2.122x 100
0.05
=91,146Nm-2
The differencein levelsbetweenthetwotankscanbedeterminedby applying
theBernoulliequationatthefreesurfaceofbothtanksfromwhichtheheadloss
is thereforeequalto thestatichead.Thatis
H =HL
= Hf + H exit+H entrance
2 2 2
- 4fL ~ +1.0~ +0.5~
- d 2g 2g 2g
=~
l
4fL +1.5
)2g d
=2.122 X
[
4XO.o065X100+1.5
)2g 0.05
=1225m
Thepressuredropdueto frictionis 91.1kNm-2 andthedifferencein levelsis
12.25m.
205
FLUID MECHANICS PIPE FRICTION AND TURB ULENT FLOW
8.7 Dischargethrougha siphon
A liquid of density1150kgm-3is siphonedfrom an openvatusinga tubeof
internaldiameter25 mmand length40 m. Thetuberisesverticallyfrom its
upperendadistanceJ 0 mtothehighestpoint.Thedischargeendis 2 mbelow
theupperendand is opentoatmosphere.Determineboththeminimumheight
of liquid allowablein thevatabovetheopenend,and therateofflow. The
minimumallowablepressureof theliquid is J 8kNnr2.Assumea Fanningfric-
tionfactor of0.005.
surface.Applying theBernoulliequationbetweenthefreesurface(1)andthe
toppointin thetube(2),then
2
PI P2 v2
-+Zl =-+-+Z2 +HL
pg pg 2g
wheretheheadlossdueto frictionandentrancelossis
2 v2
4fL~ +05-
H L =d 2g 2g
=4x 0.005x 10x ~+0.5x ~
0.Q25 2g 2g
10m
2
=8.5~
2g
~IIl
!'m
L=40m
d=2Smm
The Bernoulliequationis therefore
101.3x 103
H _18XI03 v~ 10 85v2+ - +-+ +.-
1150x g' 1150x g 2g 2g
whichreducesto
2
v2H =2.62 +9.5-
2g
Solution
Siphoningis ausefultechniquefordecantingaliquidfromavesselwherethere
maybealayerof sedimentwhichmustnotbedisturbed.It involvesthetransfer
of liquidtoanothervesselatalowerelevationbymeansof apipeortubewhose
highestpointis abovethesurfaceof theliquidin theuppervessel.Thedeviceis
startedby filling thesiphontubewithliquidby applyingpressureontheupper
surfaceorsuctionattheoutlet.Liquidthencontinuestorisetothetoppointand
dischargecontinuouslyundertheinfluenceof gravityto thelowervessel.A
limitation,however,is thatthepressureatthehighestpointmustnotfall below
thevapourpressureof theliquid. Shouldthisoccur,vapourwill bereleased
fromsolution(boiling)andcausea breakin theliquid stream.The siphoning
actionwill alsoceaseif thepipeentranceis no longerbelowtheupperliquid
ApplyingtheBernoulliequationbetweentheliquidsurfaceanddischargepoint
whicharebothata,tmosphericpressure
2
v2
H+2=-+HL
2g
wherethelossesareduetotubefriction,entranceandexitlosses
4fL v2 v2 v2
H L =~ - +0.5- +1.0-
d 2g 2g 2g
=4x 0.005x 40x ~+1.5x ~
0.Q25 2g 2g
2
=33.5~
2g
206 207
FLUID MECHANICS PIPE FRICTION AND TURBULENT FLOW
22 2
2.62+95~ +2=~ +335~
2g 2g 2g
8.8 Flowthroughparallelpipes
A liquidflows througha shortpipe whichbranchesintotwoparallelpipesA
andB eachwitha lengthof50 mandwithinsidediametersof25 mmand50
mm,respectively.Theendsofthepipesareconnectedtogetherbyanothershort
pipe.Determinetheflow througheachpipe if theyhavea drop inelevationof
3 m.Assumea constantFanningfrictionfactor in bothpipesof 0.005.
Therefore
Solving, the velocitythroughthe tubeis found to be 1.9ms-I. The corre-
spondingflowrateis therefore
Q =nd2
4V2
2
nx 0.025 x 1.9
- 4
L =50m ..1I-
=9.33 X 10-4 m3s-1
d =50mm
-Q" ~
~iPd -Q, d,~25~J ~
FjOW~iPe" ~
andtheminimumdepthin thevatis
2v
H=2.62+95~
2g
1.92
=2.62+95x -
2g
Solution.
Theheadlossdueto frictionin pipeA is
2
4fL vA
HI =--
d 2g
=4.37m
Theminimumallowableheightofliquid inthevatis4.37m.Theflowratein the
siphon tubeis found to be 9.33xlO-4 m3s-]. Note that this is a problem
involvingvariableheadflow. Theflowrateis thereforenotconstantin thetube
andis atamaximumwhenthelevelin thevatis atitshighestpositionabovethe
openend.
2
4 x 0.005x 50 vAx-
0.025 2g
V2
=40~
2g
andfor pipeB is
2
4fL vB
HI =-'-----
d 2g
2
4 x 0.005x 50 vBx-
0.05 2g
V2
=20~
2g
208 209
FLUID MECHANICS
As thepressureis thesameattheendof eachpipethen
2 2v v
40i =20--.!L
2g 2g
=H
=3m
fromwhichthevelocitiesvA andvB arefoundtobe 1.21ms-I and1.71ms-I
respectively.The rateof flow throughthepipesis therefore
nd2A
QA =-vA
4
2
nx 0.D25x 121
- 4
=5.94X 10-4 m3s-1
and
nd2
Q
- B
B - -vB
4
2
nx 0.05 x 1.71
- 4
=3.36x 10-3m3s-1
The ratesof flow throughparallelpipesA andB aretherefore2.14m3h-I and
12.08m3h-l, respectively,correspondingto atotalflow of 14.22m3h-l.
210
~
"f
PIPE FRICTION AND TURB ULENT FLOW
8.9 Pipesinseries:flowbyvelocityheadmethod
Twowaterreservoirsareconnectedbya straightpipe1 kmlong.For thefirst
halfof itslengththepipeis12cmindiameterafterwhichit is suddenlyreduced
to 6 cm.Determinetheflow throughthepipe if thesurfaceof thewaterin the
upperreservoiris 30 m abovethatin thelower.Assumeafrictionfactor of
0.005for bothpipes.
--- ---
- --
- - -- -- --
H =30ill
--- ---- - -- -
Solution'
Fromcontinuity,thevelocityinbothsectionsofpipeofcircularcross-sectionis
relatedtoflowfortheincompressiblefluidas
nd2 nd2I 2
-vI =-v2
4 4
Therefore
d2
- 2
vI - -v2
d2I
- 0.062--v
0.122 2
V2
4
Basing the calculationson the smallerdiameterpipe, the headloss at the
entrancetothelargerpipeis
2 2v v
05 -.L =0.03125~
2g 2g
(seeProblem2.6,page47)
211
FLUID MECHANICS
The headlossin the12cminsidediameterpipeduetofrictionis givenby
2
4fLI VI
Hf =--
d 2g
or in termsof thesmaller6 cminsidediameterpipe
2
=4fLI ~
16d2g
2
4 x 0.005x 500 v2x-
16x 0.12 2g
V2
=52~
2g
The entranceheadlossatthesuddencontractionHe betweenpipesis found
usingthechartprovidedin Problem2.6onpage47,anduses
v2
H =k~
c 2g
inwhichk is foundbyinterpolationtobe0.44,notingthattheratioofpipeareas
a2/al is equalto0.25.Thatis
2
v2
He =0.44-
2g
The headlossdueto frictionin the6 cminsidediameterpipeis
Hf =4fL2 v~
d 2g
2
4 x 0.005x 500 v2x-
0.06 2g
V2
=166.7~
2g
212
..........
PIPE FRICTION AND TURB ULENT FLOW
Theheadlossattheexitis
2
v2
H exit =1.0-
2g
(seeProblem2.5,page44)
The totalheadlossH betweenthetworeservoirsis therefore
V2
H =~ x (0.03125+52 +0.44+166.7+1.0)
2g
v2
=173.4~
2g
This is equaltothestaticheadbetweenthereservoirs.Thatis
2v
30=173.4~
2g
Rearranging,thevelocityin thesmallerpipeistherefore
v2 = 30x 2x g
173.4
=1.84ms-1
Therateof flow betweenreservoirsis thus
2
1td2
Q=-v2
4
2
1tX 0.06 x 1.84
- 4
=52 X 10-3 m3s-1
Therateof flow is foundtobe5.2x1O-3m3s-1or 1.872x1O-2m3h-l.
213
FLUID MECHANICS
8.10Pipesinseries:pressuredropbyequivalent
lengthmethod
Kerosene,with densityof 815 kgm-3and viscosityof 7xlO-3 Nsm-2,flows
throughapipesystemat a rateof 12x103kgh-l. Thepipesystemconsistsof
50mof50mmborestraightpipewithtwo90°elbows,followedbya reducerto
a 38mmboresection50min lengthwithtwofurther90°elbows.Thepipehas
anabsolutewall roughnessof0.02mm.Determinethepressurelossduetofric-
tionin thesystemif theequivalentlengthofeachelbowis40pipediametersand
thereduceris equalto 0.2velocityheads.AssumethatthevonKarmanequa-
tionfor rough-walledpipesapplies.
Flow--
SOm SOm
Solution
Thevelocityinthe50mmboresectionis
4mv=-
p1td
4 x 12X 103
3600
815x 1tX0.052
=2.08ms-l
214
........
PIPE FRICTION AND TURB ULENT FLOW
ApplyingthevonKarmanequation
ff; =210glO(:1 )+ 228
=2x lOglO
(
50
)
+228
0.D2 .
givesafrictionfactorof 0.0121.The frictionalheadlossthroughthepipeand
two90°elbowsexpressedin equivalentlengthis givenby
4flLeq v2
HSOmm=~2g
4x 0.Dl21x (50+2x 50x 0.05) 2.082= x-
0.05 2g
=11.64m
Thevelocityin the38mmboresectionis
4x 12X 103
v = 3600
815x 1tx 0.0382
=3.61ms-l
ThefrictionfactorisobtainedfromthevonKarmanequation
ff; =210glO(d: )+ 2.28
=2x lOglO
[
~
)
+2.28
0.D2
togiveavalueof0.0128.Theheadlossduetofrictionin termsofequivalent
lengthistherefore
H 4x 0.Dl28x (50+2x 50x 0.038) 3.612
38mm= 0.038 x ~
=48.15m
215
FLUID MECHANICS
The headlossatthereducer,H r ' basedonthesmallerborepipeis
v2
H r =0.2-
2g
=0.2X 3.612
2g
=0.13m
The totalheadlossis therefore
H L =11.64+48.15+0.13
=59.92m
The pressurelossduetofrictionis therefore
I1Pf =pgHL
=815x g x 59.92
=479.07x 103Nm-2
The totalpressuredropdueto frictionthroughthepipesystemis 479kNm-2.
216
---
PIPE FRICTION AND TURBULENT FLOW
8.11Relationshipbetweenequivalentlengthand
velocityheadmethods
Determinetheequivalentfrictionfactor relatingtheheadlossforflow ofafluid
arounda 90°elbowwhichis expressedasboth1.2velocityheadsand60pipe
diameters.
Solution
Theheadlossfortheelbowexpressedintermsofvelocityheadis
V2
H L =12-
2g
andexpressedintermsofequivalentlength,theequationforfrictionalheadloss
for theelbowis
2
4fLeq ~
HL =~2g
wheretheequivalentlengthof pipeis
Leq =60d
Notethattheequivalentlengthof fittingis thatlengthof pipewhichgivesthe
samepressuredrop as thefitting.Sinceeachsize of pipe (or fitting) would
requireadifferentequivalentlengthfor aparticularfitting,it is usualtoexpress
equivalentlengthassomanypipediameterswhichis thereforeindependentof
pipe.Therefore
2v
H L =240f-
2g
Combiningthetwoequationsfor headloss
2 2
]2~ =240f~
2g 2g
Therefore
1.2
f =240
=0.005
2]7
FLUID MECHANICS
That is, theequivalentFanningfrictionfactoris foundto be0.005.This rela-
tionshipbetweenvelocityheadandequivalentlengthof pipeholdswell for
highReynoldsnumbers.At low Reynoldsnumbersin theturbulentregionthe
two methodsdeviatedueto theinfluenceof pipesurfaceroughness.In 1943,
AmericanhydraulicsengineerHunterRouse gavea limiting equationthat
distinguishedbetweenthetransitionregionandthefully roughregimewhere
Re=~
FJ( ~)
BeyondtheReynoldsnumberpredictedby this equation,thefriction factor
becomesessentiallyindependentof theReynoldsnumber.TheRouselimitline
canbe includedontheMoodyplot(seepage293).
218
.-..
PIPE FRICTION AND TURB ULENT FLOW
8.12Flowandpressuredroparounda ringmain
Wateris suppliedtoa smalllaboratoryfour-sideringmainABCD atA andB,
withflowrates3x1o-3m3s-1and0.5x10-3m3s-I,respectively,andisdrawnat
rateC andD withtheflowrateatD being0.9x10-3m3s-I. Determinetherate
offlow fromB to C andthelowestpressuredropin themain.Theconnecting
pipeseachhavelengthsof10mandinternaldiametersof25 mm.TheFanning
frictionfactor is assumedtobe0.005.
H
)"51>'
3.0Is.l
A C
2.6Is.l
0.9Is.1
D
Solution
The pressuredroparound the main is
f..pAB + f..pBC =f..pAD + f..pDC
where
A 2fpv2L
tip AD =
d
=2fPL
(
4(3x 10-3_Q)
]
2
d 'ITd2
2 x 0.005x 1000x 10x 16x 10-6 x (3 _Q)2
'ITx 0.0255
=16,600(3_Q)2
219
FLUID MECHANICS
Likewise
t1p AB = 16,600Q2
t1pBC =16,600(Q+05)2
t1pDC =16,600(2.1_Q)2
whereQisexpressedin1itrespersecond.Therefore
Q2 +(Q +05)2=(3 _Q)2 +(2.1-Q)2
whichreducesto
l12Q -13.16=0
Solving,theflowrateQis 1.175litrespersecond.TherateofflowfromB toCis
therefore1.675litrespersecond.Thecorrespondingpressuredropsthroughthe
pipesaretherefore
t1PAD=555kNm-2
t1PAB=22.8kNm-2
t1PBC=46.4kNm-2
t1PDC=14.3kNm-2
withthelowestpressuredropin thepipeconnectingC andD. Notethatwhere
thefrictionfactors,lengthsanddiametersarenotequal,aquadraticequationin
termsof flowratearises.
220
.......
PIPE FRICTION AND TURBULENT FLOW
8.13Tankdrainagethroughapipewithturbulentflow
Waterflowsfromanopencylindricaltankofcross-sectionalarea4 m2through
a 20mlengthofhorizontalpipeof25mminsidediameter.Determinethetime
tolowerthewaterlevelinthetankfrom2mto0.5mabovetheopen endof the
pipe.Assumeafrictionfactor ofO.005andallowfor entryandexitlosses.
ldH
H'IH11 Flow,f
Solution
Atmosphericpressureis exertedbothatthefreesurfacein thetankandatthejet
issuingfromtheopenpipe.ApplyingtheBernoulliequationwhereit is reason-
ableto assumethatthevelocityof thewaterin thetankis far less thanthe
velocityin thepipe,thestaticheadis therefore
v2
H=-+HL
2g .
wheretheheadlossforpipefriction,exitandentrancelossisgivenby
4fL v2 v2 v2
H L =- - +1.0- +05-
d 2g 2g 2g
=~
(
4fL +15
)2g d
Therefore
V2 v2
(
4fL
)H =2g +2g d +15
221
A,-- - - -- - ---- -
Id L,------------- ,
Ii
FLUID MECHANICS
Rearranging
2gH
v = 14jL+25
d
Applyinganunsteadystatemassbalanceoverthetank,thechangeincapacity
of thetankisequaltotherateof flowthroughthepipe.Thatis
dHav =-A-
dt
Substitutingforpipevelocityandrearranging,thetotaltimetolowerthelevel
fromHI toH2is
~4jL +25 Hz -1
t -A d f H zdH
fdt = aJ2i HI0
Integrating
~4jL+25 1 t
2A d (Hf-H2)
t = aJ2i
=
4x 0.005x 20+25 1 1
2x4x./ 0.025 X(22_05Z)
2
1tx 0.025 x J2i4
=7912s
Thetimetakenisfoundtobe2hoursand12minutes.
222
........
PIPE FRICTION AND TURBULENT FLOW
8.14Turbulentflowin non-circularducts
Hot water,atatemperatureof80Dewithacorrespondingdensityof972kgm-3
andviscosity3.5x10-4Nsm-2,flows at a rateof 50m3h-l throughan itemof
processplantwhichconsistsofa horizontalannulusconsistingof twoconcen-
tric tubes.Theoutertubehasan innerdiameterof 150mmandtheinnertube
hasan outerdiameterof 100mm.Determinethepressurelossduetofriction
perunitlengthif thesurfaceroughnessofthetubingis0.04mm.UsetheMoody
plot toobtainthefrictionfactor.
I~ -I
FI~ ::~I::==:=-:£::=::]~:::
di Ido- - - - - -._-
Fl~::B::~::::::::::~::::::::::]~:::
Solution
Fortheflowthr.oughanon-circularpipeorduct,considerahorizontalductof
anyshapewithuniformcross-section.A forcebalanceon thefluid in the
cross-sectiongives
(PI - P2)a =!!.Pfa
=T.wPL
wherePisthewettedperimeterofthewall,L isthelengthofelementandaisthe
uniformcross-sectionalarea.Relatingthewallshearstresstofrictionfactorin
theform
T.w=Lpv2
2
223
FLUID MECHANICS
Then
/';.Pt=!pV2LP2a
Comparingthis to theequationfor pressuredropdueto frictionin a circular
pipe
/';.pt=2!PV2Ld
it is notedthat4a/P hasreplacedd. Thereforedefiningthetermdeq, knownas
theequivalenthydraulicdiameter,as
4a
deq =p
thefrictionalpressuredropequationfor non-circularpipescanbeexpressedin
theform
2+. 2
/';.Pt=~
deq
The flow areafor thetubeannulusis
11: 2 2
a=-(do -di)
4
= 7r,X (0.152-0.12)
4
=9.82X 10-3m2
andthewettedperimeteris
P =n(do +di)
=11:x (0.15 + 0.1)
=0.785m
The equivalenthydraulicdiameteris therefore
d =4 x 9.82X 10-3
eq 0.785
=0.05m
224
~
PIPE FRICTION AND TURBULENT FLOW
Therelativeroughnessbasedontheequivalenthydraulicdiameteristherefore
I:: 0.04
---
d 50
=8X 10-4
The averagevelocitythroughtheannulusis
V=~
a
50
3600
9.82x 10-3
=1.41ms-l
The Reynoldsnumberis
pvdeqRe=-
1.1
972x 1.41x 0.05
35x 10-4
=195,788
From theMoody plot (page293),theFanningfrictionfactoris 0.0052.The
pressurelossperunitlengthdueto frictionis therefore
/';.Pt =2!pv2,
L deq
2x 0.0052x 972x 1.412
0.05
=402Nm-2m-l
That is, thepressuredropdueto frictionis foundto be400Nm-2 permetre
lengthof tube.
225
FLUID MECHANICS
8.15Headlossthrougha taperedsection
Waterflows througha conicalsectionofpipewhichnarrowsfroman internal
diameterof 30 cmto an internaldiameterof iO cm. Thetotal lengthof the
sectionis 3 m.Determinethefrictionalheadlossfor aflow of0.07m5s-1if the
FanningfrictionfactorisO.OOS.ignoreentranceandexitlossesandeffectsdue
to inertia.
L(3m) II- ..
Not to scale
FlO":'-f° Oill - - - - - - - - - - - -t'~'m
Solution
The headlossduetofrictionis givenby
2
H - 4fL~
f - d 2g
or in termsof flowrate
(~ 12
Hf =4fL~d 2g
=32fLQ2
7(2gd5
Fromgeometry,thediameterof thesectionis relatedto lengthby
d =0.3- 0.0666L
wherediameterd andlengthL aremeasuredin metres.The totalfrictionalloss
is thereforeobtainedby integrationoverthelengthof thesection.
226
.......
PIPE FRICTION AND TURB ULENT FLOW
Thatis
H 32fQ2 3
f
dL
fdHf =~ 0(0.3-O.o666L)50
The integrationis simplifiedusingthesubstitution
u =0.3- O.o666L
forwhichthedifferentialis
du =-O.o666dL
The integrationrequiredis therefore
H
f -32fQ2 f dudH - 5
f - 0.06667(2g u0
Thus,integratingwithrespectto u gives
Hf = -32fQ2 u-4
0.06667(2g 4
Thatis
Hf = -32fQ2
[
(0.3-O.o666L)-4
]
3
0.06667(2g -4 0
=32x 0.005x 0.072x
[
(0.3- 0.0666x 3)-4 - 0.3-4
]0.0666x 7(2x g 4
=0.3m
The frictionalheadlossthroughthesectionis foundtobe0.3m.
227
~
U""'--
8.16Accelerationofaliquidinapipe
Two largestoragetankscontainingaprocessliquidofdensity1100kgm-3are
connectedbya 100mminsidediameterpipe50min length.Thetanksoperate
with a constantdifferencein levelof 4 m andflow is controlledby a valve
locatednearthereceivingtank.Whenfully open,thevalvehasaheadlossof13
velocityheads.If thevalveissuddenlyopened,determinethetimetakenforthe
processliquid toreach99%of thefinal steadyvalue.Assumeafrictionfactor
of 0.006andneglectentranceandexitlosses.
L
fd Valve
Solution
Theprocessliquidisassumedtobeinelasticsothatthereisnopressurewave.
Theaveragesteadyvelocityof theliquidunderfull flowmaybedetermined
fromthecasewhere,for thefreelydischargingliquid,thestaticheadis
balancedbythefrictionalresistancetoflowduetothepipewallandthevalveas
2 2
H=4jL~+13~
d 2g 2g
Rearranging
2gH2_-
vl-4jL+13
d
2xgx4
4x 0.006x 50+13
0.1
=3.14m2s-2
228
I
I
........
PIPE FRICTION AND TURBULENT FLOW
Therefore
vI =1.77 ms-I
For theacceleratingliquid,theinertiaheadis alsoincludedwhereit is assumed
thattheliquid is inelastic.Thus
2 2
H =4jL~ +13~ +~dv
d 2g 2g g dt
Rearranging
2gH 2 2L dv=v +
4jL +13 4fL +13dt
d d
2 2 2x 50 dvv =v +
I 4x 0.006x 50+13dt
0.1
=V2 +4dv
dt
Rearranging
4dv
dt=~2
vI -v
=:1 [v I 1+v +vI 1-v Jv
Integratingbetweenthelimitsofnovelocity(stationaryliquid)and99%ofthe
finalsteadyvalueforvelocity,thetimetakenis
[ [ )~
O.99V1
2 vI +v
t =- loge -
vI vI -v 0
2
[I [
1.77+0.99x 1.77
]J
=- x oge1.77 1.77-0.99x1.77
=5.98s
Thetimetakenis foundtobeapproximately6secondsfortheliquidtohave
reached99%of itsfinalvelocity.
229
FLUID MECHANICS
Furtherproblems
(1) Determinethepressuredrop dueto friction alonga 1 km pipelineof
insidediameter254mmusedfordrawingseawateratarateof500m3h-l intoa
chemicalplantforuseasaprocesscoolingmedium.Thefrictionfactoris 0.005
anddensityof seawater1021kgm-3.Answer:302kNm-2
(2) Determinethepressuredropduetofrictionalongapipeof insidediam-
eter100mm,500min length,carryingaprocessliquidofdensity900kgm-3at
arateof 46.8m3h-l.The Fanningfrictionfactoris assumedto be0.005.
Answer:123.3kNm-2
(3) Using theDarcy equation,determinethepressuredrop alonga 10 m
lengthof pipeof insidediameter25.4mmcarryingaprocessliquidof density
1010kgm-3andviscosity0.008Nsm-2atarateof 0.015m3min-l.
Answer:1.91kNm-2
(4) Determinethepressuredropduetofrictionalonga 1kmlengthofpipeof
insidediameter200mmcarryinga processliquidatarateof 125m3h-l. The
processliquidhasa densityof 950kgm-3andviscosityof 9xlO-4Nsm-2and
thepipewall hasanabsoluteroughnessof 0.04mm.
Answer:48.8kNm-2
(5) Determinetheheadlossduetofrictionin termsof thenumberofvelocity
headsfor aflowingfluidin apipelineof length2.5kmandinsidediameter100
mmwith aFanningfrictionfactorof 0.006.
Answer:600v2/2g
(6) An experimentaltestrig wasusedto determinetheparametersa andb
relatingthefrictionfactorto Reynoldsnumberin theform
b
f =aRe
The rig consistsof 10m of smooth-walledpipewith aninsidediameterof 16
mmanduseswaterastheliquidmedium.Resultsfromtwotrialsgavefrictional
pressuredropsof 13.9kNm-2and20.0kNm-2forflowsof0.94m3h-land1.16
m3h-l , respectively.Determinethevaluesfora andb.
Answer:a=0.079,b=-0.25
230 1
PIPE FRICTION AND TURBULENT FLOW
(7) A distillationcolumntoweroperatingatagaugepressureof 700kNm-2
receivesa hydrocarbonfeedof density780 kgm-3 and viscosity0.6xlO-4
Nsm-2atarateof486m3h-l fromaneighbouringcolumnoperatingatagauge
pressureof 1.2MNm-2. The connectingpipeworkconsistsof 40 m of 25cm
insidediameterpipewithanabsoluteroughnessof 0.046mmandhasatotalof
sevenelbowsandasinglecontrolvalve.If thelevelof hydrocarbonmixturein
thebaseof thefeedcolumnis steadyat4 mabovethegroundandthefeedpoint
of thereceivingcolumnis atanelevationof 15mabovetheground,determine
thepressuredropacrossthecontrolvalve.Allow for entranceandexitlosses
andalossfor eachbendequivalentto0.7velocityheads.
Answer:387kNm-2
(8) Show thattheheadloss dueto friction,Hf' for theflow of fluid in a
circularpipeis givenby
Hf =4fL~
d 2g
wheref i~thefrictionfactor,L is thelengthofpipe,v is theaveragevelocity,d
is thediameterof thepipeandg thegravitationalacceleration.
(9) A smoothpipeof internaldiameter150mmis usedfortransportingoil of
density854kgm-3andviscosity3.28xlO-3Nsm-2.Thepipehasalengthof 1.2
kmandanelevationof 15.4m.If theabsolutepressuresatthelowerandupper
endsare850kNm-2 and335kNm-2,respectively,determinetherateof flow
assumingtheBlasiusequationappliedfor smoothpipes.
Answer:0.0445m3s-1
(10) Explainwhatis meantby thetermequivalentheadin theevaluationof
energylossesacrosspipefittings.
(11) Wateratarateof 1m3s-1andwithviscositylxlO-3 Nsm-2is tobetrans-
ferredfromanopenreservoirto anotherone60m belowit througha smooth
pipe2400mlong.Determinetheinternaldiameterof thepipeusingtheBlasius
equationfor smoothpipes.
Answer:0.48m
(12) A pipeline,500 m long and with an internaldiameterof 150mm,
connectstwo largestoragetanks.If thedifferenceof liquid level in thetwo
231
FLUID MECHANICS
tanksis 30m,determinethefreeflow throughthepipe.Ignoreentranceandexit
lossesbutassumeafrictionfactorof 0.01.
Answer:0.021m3s-1
(13) Draw a diagram,with a brief description,to illustratehow thefriction
factorvarieswithReynoldsnumber.
(14) Describetheprincipleof thesiphonandincludea sketch.
(15) Explainthelimitingfactorsof a siphonusedto transfera liquid.
(16) A four-sideringmainABDCis suppliedwithwateratA fordistributionin
aprocessplantatpointsB, C andD atratesof 0.25m3s-1, 0.1m3s-l and0.05
m3s-l. Determinetheflowratein pipeAB if thepipedetailsare:
Pipe Length,m Diameter,m
AB 1000 0.5
BC 1500 0.3
CD 500 0.3
DA 1000 0.4
The Fanningfrictionfactoris assumedtobe0.005in eachof thepipes.
Answer:0.274m3s-1
(17) Determinethepressuredrop alonga 40 m lengthof pipe in termsof
velocityheadandequivalentlengthapproachesfortheflow ofwateratarateof
20 m3h-l, if thepipecontainsa gatevalve(open)andthree90°elbows.The
internaldiameterof thepipeis 100mm,thevelocityheadvaluesfor thevalve
andelbowsare0.15and0.7velocityheadsandtheequivalentlengthofpipeare
7 and35pipediameters,respectively.The Fanningfrictionfactoris 0.006.
Answer:2.96kNm-2and3.07kNm-2
(18) Explainwhycalculationsof thepressuredropduetofrictionforflow of a
liquid througha pipeusingthe velocityheadandequivalentlengthof pipe
approachesdonotnecessarilyprovidethesameresult.
(19) Wateris suppliedtoathree-sideringmainABC withaflow of0.1m3s-1
atA andis drawnfromBand C atratesof 0.06m3s-1and0.04m3s-l, respec-
tively.Theinternaldiametersof thepipesare:AB 50mm,AC 38mmandBC 38
232
l
PIPE FRICTION AND TURBULENT FLOW
mm,andeachhasa lengthof 50m.Determinetherateof flow inpipeBC if the
frictionfactoris thesameandconstantfor eachpipe.
Answer:0.007m3s-1
(20) If, in thepreviousproblem,thefriction factoris givenbytheBlasius
formula for smooth-walled pipes(j =O.o79Re-1/4), commentontheeffect this
wouldhaveanddeterminetheflow in pipeBe.
Answer:0.0061m3s-1
(21) A fermentedbrothofdensity990kgm-3is decantedfromafermentation
vesseltoareceivingvesselby siphon.Thesiphontubeis 10mlongwithabore
of 25.4mm.Determinetherateof transferif thedifferenceinheadbetweenthe
two vesselsis 2 m. Determinethepressureatthehighestpointofthesiphon
tubewhichis 1.5mabovethebrothlevelandthetubelengthtothatpointis 4m.
Assumeafrictionfactorof 0.004andstandardatmosphericpressure.
Answer:1.136xI0-3m3s-l, 74.3kNm-2
(22) A 0".155m internaldiameterpipelineis fedwith waterfroma constant
headtanksystem.Thewateris dischargedfromthepipeline12.5mbelowthe
levelof thewaterin theheadtank.The pipelinecontainsfour90°elbows,a
fully openglobevalve,a half opengatevalveanda threequartersopengate
valve.The respectiveequivalentlengthfor elbow,open,halfopenandthree
quartersopenvalvesare40, 300,200 and40 pipe diameters,respectively.
Determinetherateof flow if theFanningfrictionfactoris 0.025.
Answer:3.018xl0-2m3s-1
(23) A cylindricaltankwithadiameterof 1.5mmountedonitsaxiscontainsa
liquid ata depthof 2 m.The liquid is dischargedfromthebottomof thetank
througha 10m lengthof pipeof internaldiameter50mmatanelevation1m
belowthebottomof thetank.Flow is regulatedby avalve.Determinethetime
todischargehalftheliquidif thevalveis opensuchthattheenergylossthrough
the valveis equivalentto 40 pipe diameters.The Fanningfrictionfactoris
assumedtobe0.008.
Answer:412s
233
~
Pumps
Introduction
Pumpsaremachinesfor transportingfluidsfromoneplaceto another,usually
alongpipelines.They are classifiedas centrifugal,axial, reciprocatingand
rotaryandmaybefurthergroupedaseitherdynamicorpositivedisplacement
pumps.Dynamicpumps,whichincludecentrifugalandaxialpumps,operate
by developinga high liquid velocity(kineticenergy)andconvertingit into
pressure.To producehighratesof discharge,dynamicpumpsoperateathigh
speeds,althoughtheiroptimalefficiencytendstobelimitedtoanarrowrange
of flows. Positive displacementpumpsoperateby drawing liquid into a
chamberorcylinderbytheactionofapiston;theliquidisthendischargedinthe
requireddirectionby theuseof checkvalves.This resultsin a pulsedflow.
Positivedisplacementpumps,however,arecapableof deliveringsignificantly
higherheadsthandynamicpumps.Rotarypumpsareanotherformof positive
displacementpumpcapableof deliveringhigh heads;in this casefluids are
transportedbetweentheteethof rotatingandcloselymeshinggearsor rotors
andthepumpcasingor stator.Unlike reciprocatingpumpstheflow is contin-
uous,althoughr~tarypumpstendto operateat lower speedsthandynamic
pumpsandtheirphysicalsizeis likely tobelarger.
In thedecision-makingprocessforthespecificationandselectionof apump
toprovidesafe,reliableandefficientoperationfor aparticularapplication,the
generalprocedurefollowsmanywell-definedsteps.Flow regulationandhead
requirementsarethe two most obvious fundamentalfactorsthat mustbe
considered.Equally,however,thephysicalcharacteristicsof thefluids-
viscosityandlubricationproperties,solidscontentandabrasiveness,aswell as
corrosionanderosionproperties- significantlyinfluencethechoiceofpump
type.A completeanalysisof thesystemto discovertherequirementfor the
numberofpumpsandtheirindividualflow is alsonecessary.Thisanalysismust
includebasiccalculationsfor differentialstaticheadto provideanestimateof
thenecessarypumphead.
235
FLUID MECHANICS PUMPS
Onceanappropriatetypeofpumphasbeenidentifiedfor aparticularfluid, it
is thennecessaryto establishtherelationshipbetweenheadandflowratethat
canbe deliveredby thepumpto matchthesystemrequirementsin termsof
staticheadandfrictionallosses.Therearemanyadditionalcalculatwnsand
checksthatmustbemadebeforetheanalysisis complete.It is essential,for
example,to ensurethatpumpsarecorrectlylocatedto ensureavoidanceof
undesirableeffectssuchascavitationin centrifugalpumpsandseparationin
reciprocatingpumps.Clearly,notall pumpinginstallationsdemandthesame
level of attention.Pumpsrequiredto deliverclean,cold waterwith modest
deliveryheads,for example,arelesscomplexthanthosewhosedutiesinvolve
hightemperatureandpressure,highviscosityandabrasiveliquids.
In practice,therearenumerousadditionalandimportantconsiderationsthat
mustbetakenintoaccounttoensurethesafe,reliableandefficientinstallation
andoperationof apump.It is importanttoconsidermaintenancerequirements,
implicationsandlikelihoodof leaks,availabilityof spares,economicfactors
suchas capitalandoperatingcosts,safetyfor personnelandenvironmental
considerationsincludingnoise.For theengineernewtopumpspecificationand
selection,thedecision-makingprocessmayappearcomplicated,confusingand
even,at times,conflicting.Experienceandfamiliaritywith pumpselection
eventuallyresultsin confidence,therebyreducingthe effortrequiredin the
successfulprocurement,installationandoperationof apump.
9.1 Centrifugalpumps
Describetheessentialfeaturesandmeritsof centrifugalpumpsandpossible
reasonsfor vibrationduringoperation.
Discharge
t
Volute
Solution
Centrifugalpumpsareby far themostcommonlyusedtypeof pumpin the
processindustries,becausetheyareversatileandrelativelyinexpensive.They
are very suitablefor handlingsuspendedsolids,able to continueoperating
whenthedelivery'lineis blocked,havelow maintenancecostsandareeasily
fabricatedin a widerangeof corrosion-resistantmaterials.However,theyare
unableto develophigh headsunlessmultiplestagesareusedandtheyalso
requireprimingby ancillaryequipment.Theyofferreasonableefficiencyover
only a limitedrangeof conditionsandarenot particularlysuitablefor very
viscousfluids.
Thedesignofacentrifugalpumpconsistsofaseriesofbladesattachedatthe
centreofashaftknownasanimpellerandrotatedathighspeedinsideacasing.
Fluid is fedin axiallyatthecentre(eye)of theimpellerandis thrownoutin a
roughlyradialdirectionby thecentrifugalaction.The largeincreasein kinetic
energywhichresultsis convertedintopressureenerg~atthepumpoutleteither
by usinga volutechamberor a diffuser;thelatteris moreefficientbutmore
expensive.Thereareconsiderablevariationsin impellerdesign,butalmostall
havebladeswhich arecurved,usuallybackwardto thedirectionof rotation.
236 237
FLUID MECHANICS
This arrangementgivesthemoststableflow characteristic.Centrifugalpumps
donotoperatebypositivedisplacementandit is importanttonotethatthehead
theydevelopdependsnotonlyonthesizeandrotationalspeedof theimpeller
butalsoonthevolumetricflowrate. -
There are, in fact, many differenttypes of centrifugalpump. Each is
intendedto performa specificduty,suchashandlingvarioustypesof liquids
includingslurries,operatingathigh temperatureor deliveringhigh headsor
flow rates.A variationof the singleimpellerpumpshownis thetwo-stage
centrifugalpump.This pumphastwoimpellersmountedontheshaftsuchthat
theoutputfromthefirst impelleris fedintothesecondimpeller.By operating
impellersin series,higherheadscan be developed.Multi-stagecentrifugal
pumpshavethreeor moreimpellersmountedin serieson thesameshaftand
thesepumpsarethereforecapableof producingconsiderablyhigherheads.
In general,centrifugalpumpscanbeclassifiedaccordingto theirconstruc-
tion andlayoutin termsof impellersuction,typeof volute,nozzlelocation,
shaftorientation,bearingsupportandcouplingto thedriver.Single suction
pumps,for example,havea suctioncavityon one sideof theimpelleronly,
whereasdouble suctionpumpshave suctioncavitiesat eitherside of the
impeller.Single suctionpumps,however,are subjectto higheraxial thrust
imbalanceduetoflow cominginononesideof theimpelleronly.Pumpswitha
horizontalshaftarepopularduetoeaseof servicingandmaintenance,although
pumpswithashaftin theverticalplanetendtobeusedwherespaceis perhaps
limited.Shaftsthatareunsupportedby a bearingareknownasoverhungand
canbefedexactlyattheeyeof theimpeller.Shaftswithbearingsupportonboth
endssuchthattheimpelleris locatedinbetweenthebearings,however,provide
lessshaftdeflectionalthoughtheshaftactuallyblockstheimpellereye.
It is notuncommonfor centrifugalpumpstobetroubledbyvibrationduring
operation.This maybe dueto thedestructivephenomenonof cavitationin
whichvapouris releasedfromsolutionleadingtoalossof dischargeanddeliv-
eredhead.The sourceof thevibrationmay,however,bedueto operatingthe
pumpdry,againstablockedor closeddeliveryline. It mayalsobedueto the
shaftbeingoutof alignment,anunbalancedimpeller,bearingor sealwear,or
the pumpbeing incorrectlywired.Vibrationmay also be dueto poor pipe
support,theeffectsofsolids,includingforeignbodies,inthefluid,andairlocks.
238
~
PUMPS
9.2 Centrifugalpumpmatching
A centrifugalpumpis usedtotransfera liquidbetweentwoopenstoragetanks.
A recycleloopmixesthecontentsofthefeedtankanda restrictionorificein the
recycleline is usedtolimittheliquidvelocityto2 ms-l. Determinethenumber
ofvelocityheadsacrosstherestrictionorificewhena valvein thetransferline
tothereceivingtankisclosed.If thevalveisfully open,showthattheflowrateis
approximately1.lx10-2 m3s-1whenthelevelof theliquidin thereceivingtank
is 12mabovethecentrelineof thepumpandthelevelin thefeedtankis 5 m
abovethecentreline. Thepumpcharacteristicis
H =12- 70Q- 4300Q2
All pipelineshavean insidediameterof 100mmandthesuction,recycleand
transferlineshaveequivalentlengthsof 2 m,10m and20 m,respectively.A
Fanningfrictionfactor of 0.005maybeassumed.
Not to scale
Transfer
Recyc1e Restriction
orifice
2{
Valve
Datum
Suction
Solution
To determinetheheadloss acrosstherestrictionorificeconsiderthecasein
whichthevalveis closed.Therateofflow throughtherecyclelineis therefore
nd2
Q=-v
4 r
2
nx~x2
4
=0.0157m3s-l
239
FLUID MECHANICS PUMPS
wherevr is theaveragevelocityin therecycleline limitedto 2 ms-l by the
restrictionorifice.
Fromthepumpcharacteristic,thedeliveredheadbythecentrifugalpumpis
thus
2 2 2 2 2
H=4jLr ~+ 4jLs (Vr +Vt)+k~+1.0~
d 2g d 2g 2g 2g
H =12-70Q -4300Q2
=12- 70x 0.0157- 4300x 0.01572
Likewise,applyingtheBernoulliequationtoboththesuctionandtransferline,
thesystemheadis equaltothefrictionalheadlossin thesuctionline,transfer
lineandpipeexitlossaswellasthedifferencein staticheadbetweentheliquid
levelsin thetwotanks.Thatis
=9.84 m
2 2 2 2
4jLs(vr+Vt) 4jLtVt 10VtH =Zt -Zs +- +--+ . -
d 2g d 2g 2g
At the duty point this is equalto the systemcharacteristic.Applying the
Bernoulliequationtothesuctionandrecyclelines,theheadlossis dueto fric-
tion,lossthroughtherestrictionorificeandexitlossfromthepipe.Thatis
2 2 2
H =4f(Ls +Lr) ~ +k~ +1.0~
d 2g 2g 2g
.
) Equatingandrearranging,thevelocityin therecyclelineis therefore
whereLs andLr are the suctionandrecycleequivalentlengthsof pipes,
respectively.Rearranging
II
=2x g x 9.84- 4x 0.005x (2+10)-1
22 01
[
2x g x (12- 5)+ 4x 0.005x 20
- . 01 +1
4 x 0.005x 10
--cu---+44.9+1
X ]4' j
k =2gH - 4f(Ls +Lr) -1
2 dvr
=1.75ms-I
=44.9
That is, theheadloss acrosstherestrictionorifice is equalto 44.9velocity
heads.
For thecaseof thevalve,beingfully openallowingtransferof liquid tothe
receivingtank,thevelocityin thetransferline,Vt,is
This thereforecorrespondstoaheadof
2 2 2 2 2
H =4jLr ~ + 4jLs (Vr +Vt) + k~ +1.0~
d 2g d 2g 2g 2g
4Qt
Vt =-
rrd2
4x 1.1x 10-2
rrxO.12
- 4jLs vi
(
4fLr + Ls k
l)
V;---+ + + -
d 2g d 2g
4xO.o05x2 1.42
(
4XO.o05X(10+2)
49
1)
1.752= x-+ +4. + x-
0.1 2g 0.1 2g
=7.49m
=1.4ms-l
whereQt is the rateof transfer.Applying the Bernoulli equationto both
suctionandrecyclelines,thesystemheadis equaltothefrictionalheadlossin
therecycleline,suctionline,acrosstherestrictionorificeandpipeexit.Thatis
240 241
111!.1 ~
FLUID MECHANICS
The totalflowratedeliveredby the pump(throughthe transferandrecycle
lines)is therefore
Q =Qt +Qr
1td2
=-(Vt +vr)
4
2
_1tXO.l X (1.75+1.4)
- 4
=0.0247m3s-1
which,fromthepumpcharacteristicequation,correspondstoadeliveredpump
headof
H =12- 70Q - 4300Q2
=12- 70x 0.Q247- 4300x 0.Q2472
=7.65m
Thedeliveredheadthereforecloselymatchesthesystemheadattheflowrateof
1.1xlO-2m2s-1,correspondingto thedutypoint.
It shouldbenotedthatwhile thedutypointcorrespondsto themaximum
flow of liquiddeliveredby thepumpforthesystemillustrated,itmaynotcorre-
spondto themostefficientoperationof thepump(bestefficiencypoint,or
'bep').Also, toensurethatcavitationis avoided,detailsof thepump'srequired
netpositivesuctionhead(NPSHR) arealsorequired(seeProblem9.5,page
245).
242
........
PUMPS
9.3 Centrifugalpumpsinseriesandparallel
Highlightthebenefitsof usingcentrifugalpumpsin seriesandparallel.
Solution
Thereareoccasionswhenit is desirableto usecentrifugalpumpsto delivera
higherheadthanis normallypossiblefor afixedflowrate.Thiscanbeachieved
by usingtwopumpsin series.The totalpumpcharacteristicis thesumof the
individualcharacteristics.Pumpsin seriesareoftenusedfor high pressure
applicationssuchasboilerfeedpumps.Specialdesignsareoftenused- for
example,runningtheimpellerson acommonshaftwiththecasinglaid outso
thatthevoluteof onepumpleadsintotheeyeof thenext.
It is possibleto deliverhigh flowratesfor a givenheadby usingsimilar
pumpsin a parallelarrangementwherethepumpcharacteristicsareadded
together.
Head
Parallel' Q]
_~~Q2
~
Flow
Head
Series Pump 1+Pump2
/
--d-d-
Flow
243
FLUID MECHANICS
9.4 Cavitationin centrifugalpumps
Explainwhatis meantbycavitationin centrifugalpumps.
Solution
Vapourformsin anyliquidwhenthepressurein theliquidis lessthanthe
vapourpressureattheliquidtemperature.Thepossibilityofthishappeningis
muchgreaterwhentheliquidis in motion,particularlyonthesuctionsideof
centrifugalpumpswherevelocitiesmaybehighandthepressurecorrespond-
inglyreduced.Havingbeenformed,thevapourbubblestravelwiththeliquid
andeventuallycollapsewithexplosiveforcegivingpressurewavesof high
intensity.Thiscollapse,orcavitation,occursontheimpellerbladescausing
noiseaswellasvibrationanderosionof theblades,whichmayeventually
resultinatypicallypittedappearancesimilartothatofcorrosion.Apartfrom
theaudibleanddestructiveeffectofcavitation,anothersignisarapiddecrease
indeliveredheadandpumpefficiency.Asaremedy,athrottlevalveshouldbe
placedin thedeliverylineandwhenanysymptomsofcavitationareobserved
thevalvecanbe partiallyclosed,therebyrestrictingthethroughputand
reducingthekineticenergy.
Cavitationis morelikelytooccurwithhigh-speedpumps,hotliquidsand
liquidswitha highvolatility(highvapourpressure).Problemscanalsobe
encounteredwithslurriesandliquidswithdissolvedgas,andinparticularingas
scrubberswhichusuallyoperatewithliquidssaturatedwithgas.Notethatthe
relatedphenomenonofseparationcanoccurinreciprocatingpumpsduetothe
accelerationof theliquid.Thisis morelikelywhenthedeliverylinesarelong
andthecorrespondingkineticenergyis large.Apartfromprovidingsufficient
netpositivesuctionhead(seeProblem9.5,page245),separationcanbe
avoidedbyattachingairvesselstoboththedeliveryandsuctionlines.
244
~
PUMPS
9.5 Netpositivesuctionhead:definition
Explainwhatis meantbyavailableandrequirednetpositivesuctionhead
(NPSH).
p
--- -
H
Discharge
Suction
Solution
To avoidproblemsof cavitationin centrifugalpumpsit is necessaryforthe
lowestpressureof theliquidin thepump,whichis usuallyattheeyeof the
impeller,toexceedthevapourpressureof theliquidbeingdelivered.Should
thepressurefall belowthevapourpressureof theliquid,localvaporization
(boiling)is likelyto occuralthoughtheprecisemechanismof cavitation
inceptionis notfully understood.Sincethevapourpressureof liquidsis a
functionof temperature,it maybethecasethatcavitationis morelikelyto
occurduringthehotmonthsof summerthanin coldwinters.Atmospheric
conditionsmayalsoaffectthelikelihoodof cavitationoccurringat low
barometricpressures.The requirednetpositivesuctionhead(NPSH) or
NPSHRisthereforethepositiveheadnecessarytoovercomethepressuredrop
inthepumpandmaintaintheliquidaboveitsvapourpressure.It isafunctionof
pumpdesign,impellersizeandspeed,andtherateofdelivery.Topreventcavi-
tationfromoccurringthismustbeexceededbytheavailableNPSHorNPSHA
andis afunctionof thesysteminwhichthepumpoperates.Thisincludesthe
minim,umworkinggaugepressureofthevapouronthesurfaceoftheliquidsp,
theminimumatmosphericpressuretakenas0.94ofstandardatmosphericpres-
surepa' thevapourpressureoftheliquidatthemaximumoperatingtempera-
turePv' thestaticheadabovethepumpcentreH, andtheheadlossdueto
frictionandfittingsH L .TheavailableNPSHorNPSHAatthepumpsuctionis
then
245
FL UID MECHANICS
NPSHA =p +Pa -Pv +H -HL
pg
andis effectivelya measureof thetotalheadavailableat thepumpsuction
abovethevapourpressure.To ensurethatvapourbubbleformationattheeyeof
theimpeller- andthusproblemsof cavitation- areavoided,it is essential
thattheNPSHA is greaterthantheNPSHR. Thatis
NPSHA >NPSHR
In the caseof reciprocatingtype positivedisplacementpumps,the same
approachto calculatingtheavailableNPSH is used.This allowsfor thestatic
liquid head at the pump inlet, minimum barometricpressure,minimum
workingpressureandvapourpressureof theliquid atthemaximumoperating
temperature.It shouldbenoted,however,thatthemaximumfrictionalheadin
thepumpsuctionlineoccursatthepointofpeakinstantaneousflow andnotthe
averageflow. This is dependenton theconfigurationof thepumpandhas
greatesteffectfor singlecylinder(or simplex)pumps(seeProblem9.15,page
265).Additionally,aheadcorrectionisrequiredtoallowfortheaccelerationof
thefluid in thesuctionlineduringeachpulsationcycle.
As with reciprocatingpumps,theavailableNPSH for rotarypumpswhich
operatewithpulsedflow requirescorrectionforbothfrictionalheadlossin the
suctionlineatpeakinstantaneousflow andheadrequiredtoacceleratethefluid
in thesuctionlineduringeachpulsationcycle.
NotethatwhiletheavailableNPSH is afunctionof thesystemin whichthe
pumpoperates,thevapourpressureof theliquidbeingpumpedis notalways
specificallyincludedin theNSPHA calculation.In thiscasethevapourpres-
sureheadis includedsuchthat
NPSHA >NPSHR+~
pg
Detailsof therequiredNPSH areusuallysuppliedby thepumpmanufacturer.
246
'\ ~
PUMPS
9.6 Netpositivesuctionhead:calculation1
A centrifugalpumpis usedto delivera liquid of density970kgm-3from an
openstoragevesselata rateof5 m3h-l.Thestoragevesselhasa diameterof3
mandis initiallyata depthof2.5m.Thepumpis locatedatanelevationof3 m
abovethebottomof thevesselandthefrictional headlossin thesuctionpipe is
0.5m.Thevapourpressureof theliquidat thetemperatureof operationis 18
kNm-2andtheNPSH is 5m.Determinethequantityof liquiddeliveredandthe
time takenbeforecavitationoccurs.Allow for worst case meteorological
conditions.
--- -
Zl Zs
Solution
Applying theBe~noulli equation betweenthe free surface of the liquid (l) and
the suction point(subscript s) in thepump
2
PI P s v s H- + ZI =- + - + Zs + L
pg pg 2g
Rearranging,thesuctionpressureheadis therefore
2
££=~+ZI-Zs -~-HL
pg pg 2g
The minimumsuctionheadbeforecavitationoccursis
, 2
££ =NPSH-~ +~
pg 2g pg
247
FLUID MECHANICS
Thus,for cavitationnotto occur
2 2
Vs Pv PI VS
NPSH -- + - =- + ZI -Zs -- -HL
2g pg pg 2g
Rearranging,andnotingthatthepressureonthefreesurfaceis takenas0.94of
thestandardatmosphericpressure,thedepththatcanbepumpedbeforecavita-
tionoccursis
Pv - PI
Zj =NPSH + + Zs + H L
pg
3
5 (18- 0.94x IOU) x 10 05= + +3+
970x g
=0.375m
correspondingtoavolumeof
v =~X 32x (25-0.375)
4
=15.Q2m2
The timetakento reachthislevelis therefore
V
t =-
Q
15.Q2
5
=3h
Thequantityofliquid deliveredis thereforeIS m3,taking3 hoursbeforecavi-
tationoccurs.
248
~
PUMPS
9.7 Specificspeed
Explainwhatis meantbyspecificspeedof a centrifugalpump.
Solution
The specificspeedis usefulfor theselectionandscale-upof centrifugalpumps
andis ameasureof pumpperformancein termsof pumpdischargeQ, headH
andimpellerspeedN. It is alwaysevaluatedat thepoint of maximumeffi-
ciency,otherwiseknownasthebestefficiencypoint.Thegeneralprocedureis
thatoncetheheadandtheflow areestablishedfor aparticularduty,thepump's
specificspeedcanbe determinedto ensurethe selectionof the appropriate
pumpin termsof impellertypeandcharacteristicswithoptimaloperatingeffi-
ciency.This is determinedfor a headcoefficientCHand capacitycoefficient
CQ' which,for twogeometricallysimilarpumps,are
gHI
CH=-
N2D2I I
(seeProblem4.3,page104)
- gH2--
N2D2
-2 2
and
QI
CQ =
N D3
I j
- Q2--
3
N2D2
Fromtheheadcoefficients,theratioof pumpimpellersis
I
EL =N2
[
!iL
J
2
D2 NI H2
andfromthecapacitycoefficients,theratioof dischargesis
~ =~
[
EL
J
3
Q2 N2 D2
249
FLUID MECHANICS
The ratioof dischargesis therefore
9l =~
[
N2
[
lil
J
I
]
3
Q2 N2 Nl H2
~(~:J[z;J
or rearrangIng
1 1
N lQf =N2Q~
1.
H4
1
1.
H4
2
and is calledthe specificspeed,N s' Althoughthe pump specificspeedis
usuallydescribedasbeingdimensionless,it hasthedimensionsof L3/4y-312.
The numericalvalueof thespecificspeedcorrespondstothetypeof pumpfor
requiredconditionsof rotationalspeed,t1owrateandhead.As aguide
Ns <0.36 Multi-stagecentrifugal/positivedisplacement
0.36<Ns<1.10 Single-stagecentrifugal(radial)
1.l0<Ns<3.60 Mixed t1owpumps
3.60<Ns<5.50 Axial t1owpumps
The maximumefficiencyfor eachpumptypespansanarrowrangeof specific
speedswithlittleoverlapbetweenpumptypes.The specificspeedis therefore
valuableto identifythepumptypefor aparticularapplication.
~
>="
»u"0)
'u
tE
~
Maximumefficiency Maximum.efficiencyMaximumefficiency
rangeof centrifugal
pumps
rangeof mixed-flow
pumps
rangeof axial-flow
pumps
SPECIFIC SPEED Ns' m3/4s-3/2
250
..010..-
PUMPS
9.8 Netpositivesuctionhead:calculation2
A low levelalarm is to be installedin a vesselcontainingwarmwaterat a
gaugepressureof50kNm-2topreventcavitationatthedesigntemperatureof
40°c. Determinetheminimumlevelof thealarmabovethecentreline of the
pumpif separatetestsshowthattheminimumNPSH for thepumpis givenby
i
NPSH =2.8N] H
whereN s is thespecificspeedof thepumpat maximumefficiencyandhasa
valueof 0.14(m3/4s-3/2),andH is thedifferentialheadandis 30m.Thefric-
tionallossesin thesuctionlineamountto0.2m.
Solution
Theminimumnetpositivesuctionheadis
i
NPSH=2.8N]H
i
=2.8xO.143 x30
:J!:6.11m
The NPSH is givenby
NPSH =Pl -Pv + 21 -HL
pg
wherePv is thevapourpressureof thewaterat40°C(7380Nm-2)anddensity,
p, is 992kgm-3.Rearranging
21 =NPSH + Pv - PI + H L
pg
3
=6.11+(7.38- 50)x10 +02
992x g
=1.93m
Thepositionof thelow levelalarmis 1.93mabovethecentrelineof thepump.
251
FLUID MECHANICS
9.9 Effectof reducedspeedonpumpcharacteristic
Thecharacteristicofa centrifugalpumpis givenbya manufacturerin dimen-
sionalformas
H =40 -140Q -4200Q2
whereQ is theflowrate(m3s-1)andH is thehead(m).Obtainanexpressionfor
thepumpcharacteristicfor thesamepumpoperatingat 75%of its testspeed.
Solution
A centrifugalpumpcanoperateoverarangeof flowratescorrespondingtoa
rangeof deliveredhead,knownasthepumpcharacteristic.Theshapeof the
curvedependsonthelayoutoftheimpeller,bladesandvolute.Inthiscase,the
specificspeedof thepumpatfulloperating,N, andreducedspeed,N1,is
Nl =0.75N
The headcoefficientfor thepumpin bothcasesis therefore
cH=~
N2D2
= gHI
N2D21
(seeProblem4.3,page104)
Therefore
HN2D2
H - 11-
N2D2
- H(0.75N)2 D2
N2D2
=0.5625H
Thecapacitycoefficientis
cQ=LND3
= Ql
NID3
252
......
PUMPS
Therefore
Ql =QNID3
ND3
- Q(0.75N)D3
ND3
=0.75Q
Thepumpcharacteristicfor thetestconditionsis therefore
~=40-140x Ql -4200X
(
QI
J
2
0.5625 0.75 0.75
Thatis
2
HI =22.5 -1O5Ql -4200Ql
Thepumpcharacteristicforbothfull operatingandtestconditionsis therefore
DischargeQ,m3s-1 0 0.01 0.02 0.03 0.04 0.05 0.06 0.08
Diff. head-H,m 40.00 38.1835.5232.0227.6822.50 17.88 1.92
Diff. headHI' m 22.5021.03 18.7215.5711.58 6.75 2.13 -
Both deliveredhead and dischargeare thereforereducedat the lower
impeller speed.Centrifugal pumps usually operatewith a fixed speed.
However,mechanicallywornpumpsmayoperateatreducedspeedsandmay
thereforebeunabletoperformtheirrequiredduty.
40
30
§
Q 20<t;
gj
10
0
0 0.02 0.04 0.06
FLOW (m3s.I)
0.08
253
FLUID MECHANICS
9.10 Duty point and reduced speed of a
centrifugal pump
A centrifugalpumpwith an impellerspeedof 1200rpm has thefollowing
characteristic:
DischargeQ, m3s-10.000 0.002 0.004 0.006 0.008 0.010
HeadH, m 40.0 39.5 38.0 35.0 30.0 20.0
Thepumpis usedtodeliveraprocessliquidalongapipeof length100mand
insidediameter50mmanddischargesfrom theendof thepipeatatmospheric
pressureatan elevationof10mabovetheopenfeedtank.Determinetheduty
pointandthespeedofthepumpthatwouldresultina reductioninflow of25%.
Assumeafrictionfactor of 0.005for thepipe andneglectentranceandexit
losses.
Solution
The duty point is the maximumflowratethatthe pump can achievefor a
particularpowerdemandandformstheintersectionbetweenthepumpand
systemcharacteristiccurve(aplotofdischargeagainsthead).Thesystemchar-
acteristicis usuallyparabolic,sincethefrictionalheadloss is proportionalto
thesquareofthefluidflowrate.ApplyingtheBernoulliequation,theheadtobe
developedby thepumpis
v2
H=-+Z2-Z1+HL
2g
- v2-
2
-+Z2-Z +
4fLv2
g 1-- d 2g
or in termsof flowrate
H =z2 - Zl + 16Q2
(
4fL +1
)2grr?d4 d
=10+ 8xQ2
[
4X 0.005x 100
Jg x IT?x 0.054 0.05 +1
=10+5.42x 105Q2
254
..0lIl...-
Thepumpandsystemcharacteristicaretherefore
DischargeQ,m3s-1 0.000 0.002 0.004 0.006
Head(pump),m 40.0 39.5 38.0 35.0
Head(system),m 10.0 12.2 18.7 29.5
40 System
34
30
g 23
~20
~
10
0
0 0.002 0.004 0.006 0.008 0.010
FLOW (m3s'1)
PUMPS
0.008 0.010
30.0 20.0
44.7 64.2
Fromthegraph,thedutypointcorrespondstoaflowrateof0.0066m3s-1anda
headof 34m.A 25%reductionin flow is therefore0.00495m3s-1witha l1ead
of 23m.Sincetheheadcoefficientfor thepumpatthetwospeedsis
gHl
CH=-
N2D21
(seeProblem4.3,page104)
- gH2--
N2D22
thenthereducedspeedis
N, ~ N{::1
1
=1200x(~~J
=987rpm
The speedof thepumptoreducetheflow by 25%is 987rpm.
255
FLUID MECHANICS
9.11Power,impellerdiameter,speedand
deliveredhead
A centrifugalpumpis usedtotransferwaterfroma vesselatanabsolutepres-
sureof 50kNm-2toanothervesselat anabsolutepressureof 150kNm-2.The
elevationof thewaterin eachvesselabovethecentreline of thepumpis 5 m
and15 m,respectively.Theconnectingpipeworkon thesuctionanddelivery
sidesof thepumphasequivalentlengthsof5 mand20m,respectively.In addi-
tion,thereis a controlvalvein thelinewhichhasa lossof 25% of thepump
d!fferentialhead;theflow controllermaintainstheflow ata rateof0.05m3s-l.
Datafroma similargeometrypumpwithan impellerdiameterof0.15manda
rotationalspeedof /500 rpmare
DischargeQ, ls-l
D!fferentialheadH, m
Power inputP, kW
0
9.25
5
8.8/
0.96
10
7.85
1.03
/5
6.48
1.19
20
4.81
1.26
25
2.96
1.45
For a liquid velocityof2 mrl, determinethedifferentialhead,theimpeller
diameterandrotationalspeedrequiredforoperationat maximumefficiency,
andthepowerinputtothepump.Assumeafrictionfactor of O.005.
2
~ ~ -L IZ2=---- -
I
FC
~
-
- - I
I
I
Zj ---
-
Datum
Solution
ApplyingtheBernoulliequationatpoint1andpoint2
2
PI P2 v2
- +Z1+H =- +- +Z2+H L
pg pg 2g
256
.......
PUMPS
whereH L is theloss acrossthevalveandfrictionalloss in thepipelinefor
whichthediameteris
d=(::j
j
=
(
4x 0.05y
nx2 )
=0.178m
The suctionline lossesare
H - 4fLs~
f(s) - d 2g
4x 0.005x 5 22
x-
0.178 2g
=0.114m
andthedeliverylinelossesare
~
4fLd~
Hj(d) = d 2g
4x 0.005x 20 22
x-
0.178 2g
=0.458m
RearrangingtheBernoulliequation
2
P2 - P 1 v2
H= +-+Z2 -z] +Hf(s) + Hf(d) +0.25H
pg 2g .
The differentialheadis therefore
H=~
[
P2-Pl +v~ +Z2-Z1+Hf(S)+Hf(d)
]
1- 0.25 pg 2g
[
5 2
- ~ x (15- 05)x 10 +~ +15- 5 +0.114+0.458
- 1- 0.25 1000x g 2g
=27.96m
257
FLUID MECHANICS
The actualpowerinputfor themodelpumpis
p =pgQH
11
Rearranging,theefficiencyis therefore
11=pgQH
P
andcalculatedfromthepumpdataas
The maximumefficiencyis therefore80%for whichthespeedis determined
fromthepumpheadandcapacitycoefficients(seeProblem9.7,page249)
N~N{:, J(~1
3 I
=1500x
(
27.96
]
4
(
°.015Y
60 6.48 0.05)
=41rps
The diameterof thepumpimpelleris therefore
D=DI
(
QNI
]
t
QIN
[
0.05x 1500
)
t
=0.15x 60
0.015x 41
=0.19m
258
~
PUMPS
The actualpoweris therefore
P =pgQH
11
- 1000x g x 0.05x 27.96
0.80
=17.143x 1O3W
The powerinputis foundto be 17.1kW for apumpwith animpellerof 19cm
androtationalspeedof2460rpm.In practice,theNPSH calculationshouldalso
becompletedtoensurecavitationis avoided.
Notethatatmaximumefficiency,thespecificspeedis
I
N NIQ2s=~ 3
H41
1
2
1500x 0.015
60
1
6.484
=0.75(m3/4s-3/2)
correspondingtoaradialsingle-stagecentrifugalpump(seeProblem9.7,page
249).
259
FlowrateQ,Is-I o 5 10 15 20 25
DifferentialheadH, m 9.25 8.81 7.85 6.48 4.81 2.96
PowerinputP,kW 0.96 1.03 1.19 1.26 1.45
Efficiency11,% o 45 75 80 75 50
FLUID MECHANICS
9.12Suctionspecificspeed
A centrifugalpumpis todeliver50m3h-1ofaprocessliquid.If thetotalheadat
thebestefficiencypointfor thepumpis 5 m,determinethesuctionspecific
speedasa dimensionlessnumberfor an impellerspeedof2000rpm.
Solution
It hasbeenthepracticetodefinethespecificspeedforapumpin tenusof its
cavitationcharacteristicssothatanimpellerdiameterD, rotatingwithaspeed
N, anddeliveryvolumetricflowrateQ, atoptimalefficiencywouldproduce
thesamepumpinletconditionsforageometricallysimilarpumpundersimilar
operatingconditions.Thus,theheadcoefficientCHand capacitycoefficient
CQ forthetwogeometricallysimilarpumpsare
gH
CH=~ND
(seeProblem4.3,page104)
Q
CQ =ND3
EliminatingD andrearrangingintermsoftheimpellerspeed
1
N =c~(gH)i
2 1
C~Q2
DefiningthesuctionspecificspeedSn asthedimensionlessgroup
1
S - C~n -- 3
C4H
1
NQ2--
3
(gH)4
whereN is theimpellerspeed(rps),Q is theflow throughthepumpandH is the
netpositivesuctionhead,then
Sn - 3
(g x 5)4
=021
260
~
PUMPS
The suctionspecificspeedprovidesanindicationof impellerspecificationas
well asthelikelihoodof cavitationwhereasa guide
Sn<0.12
0.12<5n<0.4
0.4<5n<0.7
Sn>0.7
Pumpsarefreefromcavitation.This is ausefullimit
for pumpsusedtodeliververycorrosivegas-freeliquids
wherematerialsof constructionwhichareresistantto the
destructiveeffectsof cavitationcannotbeused
(seeProblem9.4,page244).
The pumpinletboremaybethesameastheboreof the
pumpinletconnection.An inletreduceris thereforenot
required.It is alsorecommendedthatbendsconnectedto
theinlethavearadiusinexcessof fivetimestheinletbore.
Materialsof constructionshouldberesistanttocorrosion
aspumpsoperatein thisrangewithincipientcavitation.
This rangeistypicalforcommercialpumpshandlingclean
liquids.
Appliesonlytopumpswhichareusedtohandlecleannon-
corrosiveliquids.As cavitationis likely,it is essentialthat
pumpsuctioncalculationsandpumplayoutareexamined
closely.Inletpipesshouldbestraightandfreefrom
obstructions,andinletisolationvalves(suchasball
valves)shouldnotrestrictflow.
261
FLUID MECHANICS
9.13Reciprocatingpumps
Describe,with theaid of a diagram,theoperationof reciprocatingpumps,
theirapplicationsandlimitations.
Solution
In general,reciprocatingpumpstransportfluid by theactionof apistononthe
fluid containedwithina cylinder.The pistonis providedwith areciprocating
motion by meansof a connectingrod and crank where the dischargeis
dependentonthesweptvolumeandthestrokefrequency;thesweptvolumeis
theamountof fluid displacedoneachstrokedependingonthecross-sectional
areaof thecylinderandthelengthof stroke.Checkvalvesareneededtoensure
thatflow is in thecorrectdirection.
Althoughtheyaccountfor a relativelysmallproportionof all thepumps
usedon processplants,reciprocatingpumpsareapplieduniversallyacrossa
widerangeof industries.They areparticularlyusefulfor deliveringfluids into
high pressureprocessstreams.Indeed,reciprocatingpumpsare capableof
deliveringheadsup to severalthousandbar and are limited only by the
mechanicalstrengthof thepumpandslightleakage.Theyarealsoself-priming.
The flow is notconstantbutpulsed,althoughtheuseof gas(usuallynitrogen)
vesselsandotherpulsationdamperssmoothestheflow. Reciprocatingpumps
provide unsatisfactoryperformancewith very viscous liquids due to slow
262
1
PUMPS
checkvalveaction,andwith suspendedsolidsdueto attritionbetweenthe
pistonandcylinder.
The volumetricefficienciesof reciprocatingpumpsaregenerallyhigh and
dependenton thestrokefrequencyandpressure- limitedlargelyby check
valve leakagecausedby wear, corrosion or mis-seatingdue to trapped
particulates.Anotherlimitationis thephenomenonof separationin whichthe
pressureatthefaceof thepistonfallsbelowthevapourpressureof theliquid.
The liquid thereforebecomesseparatedfrom thepistonby vapour.Further
problemscan ariseinvolving vapouror gaswhich entersthecylinderspace
fromaroundaleakingpistonseal.This resultsin afall in therateof discharge,
audiblenoiseanddamagetothepump.
In additionto air vesselson boththedeliveryandsuctionlines, theflow
characteristiccanbemadesmootherby theuseof multi-cylinderpumps.The
totalflow is thusthesumof theinstantaneousflowsfromthecylinderswhere
oddnumbersof cylinders(threeor triplex,or five or quintuplex)producethe
smallestpulsations(seeProblem9.15,page265).
Abrasive,toxic and corrosiveliquids, includingslurriesandradioactive
fluids,canbehandledby a specialtypeof reciprocatingpump,thediaphragm
pump,whichconsistsof a flexiblediaphragmfixedattheedgesandflexedto
andfro. Thediaphragmpump,however,requiresprimingpriorto operation.
Anothervariationis thecylinderor rampumpin whicha plungerreplaces
thepistonandhasaclearancebetweenthecylinderwall. It is moreexpensive
thanthepistonpumpalthoughleakageis easierto detect,thepackingis easier
to replaceandit is moresuitablefor deliveringhighheads.
263
FLUID MECHANICS
9.14Single-actingreciprocatingpiston
A single-actingreciprocatingpistonhasapistonareaof 100cm2anda stroke
of30cmandis usedtoliftwatera totalheightof 15m.lfthe speedofthepiston
is 60 rpmandtheactualquantityof waterliftedis 10.6m3h-l,determinethe
coefficientofdischargeandthetheoreticalpowerrequiredtodrivethepump.
Solution
Theexpectedvolumedisplacedinthepistoncylinderis
v =ALN=0.01x 0.3x 60
60
=3x 10-3m3s-1
The actualvolumedischargedis
v = 10.6
act 3600
=2.94x 10-3m3s-1
Thecoefficientofdischargeistherefore
C - Vactd-- V
2.94x 10-3
3x 103
=0.98
andthepowerrequiredis
P =pgQH
10.6
=1000x g x - x 15
3600
=433W
Thecoefficientofdischargeandtheoreticalpowerrequiredare0.98and433
W,respectively.
264 1
PUMPS
9.15Dischargefromreciprocatingpumps
Determinetheratioofpeaktoaveragedischargefrom single,two,three,four
andfive-cylinderreciprocatingpumps.
Solution
It isassumedthatthepistonineachofthesepumpsoperateswithperfectsimple
harmonicmotionandthatthevolumetricefficiencyandaccelerationeffectsof
thefluidareignored.Thus,fora single-cylinderpump,thedisplacementof
liquidis
x =r - r cos8
=r (1- cos8)
=r(l - cos(Of)
The velocityof thepistonis
v =dx
dt
=(Orsin (Ot,
=(Orsin8
The volumedisplacedonthepressurestroke(between0 andn) is therefore
1t
V =Awrfsin8d8
0
=-A(Orcos(n- 0)
=2A(Or
whereA istheareaof thepiston.Theaverageflowoverthecycle(between0
and2n)istherefore
265
FL UID MECHANICS
Q=~
21t
2Acor
21t
Acor
1t
Sincethepeakflow occursat1t/2,then
Qpeak = Acor sin 2<2
=Acor
The ratioof peaktoaverageflow is therefore
Qpeak=Acor
Q Acor
1t
=1t
=3.14
Likewise, for duplex (two-cylinder),triplex (three-cylinder),quadruplex
(four-cylinder)andquintruplex(five-cylinder)pumps,theratiosarefoundto
be 1.57,1.05,l.ll and1.02,respectively.
266
~
Triplex
Quadruplex
Quintuplex .
PUMPS
1.05Q
-" Qm m
0.91Qm
- -" 1.11Qm
Qm
0.79Qm
1.02Q
Qm m
0.96Qm
0° 90° 1800 2700 3600
267
FLUID MECHANICS PUMPS
9.16Rotarypumps
Describebriefly,witha labelleddiagram,thelayoutandoperationofrotarypumps.
Furtherproblems
(1) Describebriefly,withtheuseof clearlylabelledsketches,theessential
features,usesandlimitationsof rotarypumps.
(2) Explainthephenomenonof cavitationappliedto centrifugalpumpsand
describepracticalmeasureswhichcanleadto itsreduction.
(3) Constructaflowchartforthedecision-makingprocessfortheselectionof
centrifugalpumps.
(4) DeterminetheadditionalavailableNPSH of asystemfor apumpusedto
transferwaterthatallowsfor worstcasemeteorologicalconditions.
Answer:0.62m
(5) Describetheessentialfeaturesof thecentrifugalpump;includeasketch.
Rotarypump(geartype)
(6) A centrifugalpump is used to deliver a dilute solutionof sodium
hydroxidewith a densityof 1050kgm-3from a largeopenbuffertankto a
headertan~10maboveit.Thevelocityof thesolutionin thesuctionpipetothe
pumpis 1.5ms-I. The suctionpipehasan internaldiameterof 3 cm andthe
deliverypipehasaninternaldiameterof 4 cm.If thefrictionalheadlossis 1.5
m,determinethetheoreticalpowerrequirementfor thepump.
Solution
In rotarypumps,portionsof liquidarecarriedfromtheentryporttothedelivery
portin compartmentswithinthepumpformedbetweentherotorandthestator.
They aremanufacturedtoa veryclosetoleranceto reduceback-leakage.They
are thereforeexpensiveto fabricate.The flowratedependson the size and
numberof compartmentsaswell asthefrequencyof rotation.They havethe
capabilityof deliveringagainsthigh heads,are self-primingand provide
smoothand continuousflow. They offer higherspeedsthanreciprocating
pumpsandaresuitablefor viscousliquidssinceno checkvalvesarerequired.
But theyrequirecarefulmaintenanceandareunsuitablefor liquids without
lubricatingpropertiessuchaswater,andforslurriesduetosmallclearancesand
possibilityof friction.
Therearevariousformsof rotarypump.TheMonopump,forexample,hasa
helical worm rotatinginsidea similarlyshapedflexible stator.This typeof
pumpcanhandlecorrosiveandgrittyliquids,andis commonlyusedto feed
slurriesto filter presses,but it mustneverbe allowedto run dry.The Roots
blower is usedfor transportinglargevolumesof gasat low gaugepressure,
usingrotorsin theformof cycloidallobes.
Notethatit is essentialtoprovideall positivedisplacementpumps(recipro-
catingandrotarytypes)with a safetyreliefvalveto avoidexcessivepressure
build-upin theeventofthedeliverylinebeinginadvertentlyclosedorblocked.
Answer:126W
(7) Explainthesignificanceof netpositivesuctionhead(NPSH).
(8) Sulphuricacidwithaspecificgravityof 1.83andaviscosity0.04Nsm-2
is pumpedintoa largevesselthrougha 25cminsidediameterpipe30m long
inclinedatanangleof 30°to thehorizontal.Determinethepowerrequiredto
pump0.064m3s-1intothevessel.A frictionfactorof 0.008maybeassumed.
Answer:17.6kW
(9) Sulphuricacidwith a specificgravityof 1.84is to bepumpedfroman
opentankto a processcolumnatarateof 25kgs-I. The columnoperatesata
pressureof 125kNm-2 andtheacidis sprayedfrom a nozzlelocated20 m
abovetheacidlevelin thetankata linearvelocityof 3 ms-I. If thetotaladdi-
tional lossesare equivalentto 2.75 m of waterhead,determinethepower
requiredtopumptheacidif thepumphasanefficiencyof 65%.
Answer:9.01kW
268 269
FLUID MECHANICS
PUMPS
(10) 75% sulphuricacid, of density 1650kgm-3 and viscosity8.6xlO-3
Nsm-2,ispumpedatarateof 10000kgh-I throughan800msectionof smooth
pipewithanintern,aldiameterof 50mmwithariseof 12m. If thepumpeffi-
ciencyis 50%,determinethepowerrequirement.AssumetheBlasiusequation
applies.
powerrequirementof therefluxpumpandmotorassemblyassuminganeffi-
ciencyof 75%.Allow for entranceandexitlosseswhereeachbendandvalve
hasa lossequivalentto0.7and0.15velocityheads,respectively.Theconstant
frictionfactormaybetakenas0.006.
Answer:14.4kW
Answer:1.74kW
Answer: 11.5kW
(14) An LPG mixture,of density600kgm-3andviscosity6xlO-4 Nsm-2,is
pumpedatarateof 10kgs-l froma storagevesselheldata gaugepressureof
500 kNm-2 to a distillationcolumnfor separation.The pipelineis 50 m in
lengthandhasaboreof 10cmandsurfaceroughnessof0.046mm.If thegauge
pressureatthefeedtrayin thecolumnis 1.1MNm-2 andis located20mabove
thecentrelineof thepumpwhilethesurfaceof theLPG mixturein thestorage
vesselis 3 m abovethepump,determinethepowerrequirementof thepump
andmotorassemblyassuminganefficiencyof 60%.Allow for entranceand
exitlosses;thepipehaseight90°elbowsandthreeopengatevalvesof 0.7and
0.15velocityheadseach,respectively.
(11) Liquid, of density850kgm-3, is pumpedat a rateof 10kgs-l froman
openstoragetankatatmosphericpressureto adistillationcolumnwhichoper-
atesatapressureof 500kNm-2aboveatmosphericpressure.Theinsidediam-
eterof thepipesupplyingthecolumnis 10cm,andthesurfaceof theliquid in
thetankis 5 m abovethegroundlevelandcanbe consideredconstant.The
columnfeedpointis 24mabovetheground.Determinethepowerrequiredfor
thepumpif theoverallefficiencyofpumpandmotoris 70%.Lossesduetofric-
tionamountto 3 m.
Answer:20.2kW
(12) The overheadcondensatefrom a distillationcolumn of density600
kgm-3andviscosity0.6xlO-4Nsm-2isfedatarateof5 kgs-l fromtheconden-
satedrumheldatagaugepressureof 12barto apressurizedstoragevesselata
gaugepressureof 10.2barpriortofurtherprocessing.Thepipelinehasalength
of 2.5kmandaninsidediameterof7.5cmwithawall roughnessof 0.046mm.
If thedifferencein level of thecondensatein thecondensatedrumandthe
dischargeof thepipe into the storagevesselis 1 m, determinethe power
requirementof thepumpandmotorassemblyif theircombinedefficiencyis
65%.Allow for entranceandexitlossesandwherethepipecontainsfifty-six
90°elbowsof0.7velocityheadseachandfouropengatevalvesof0.15velocity
headseach.
Answer:7.3kW
(15) An ethanoldistillationcolumnoperatingatagaugepressureof50kNm-2
receivesa liquid feedof density950kgm-3andviscosity9xlO-4 Nsm-2at a
rateof 39 m3h-1from an open storagevessel.The connectingpipework
consistsof 50m of 10cminsidediameterpipewith anabsoluteroughnessof
0.04mm.The pipehasatotalof 12elbowsandthreegatevalves.The levelof
liquidfeedin thestoragetankis steadyat4 mabovethecentrelineof thepump
andthefeedpointonthedistillationcolumnis atanelevationof 5 mabovethe
pump.If theoverallefficiencyof thepumpandmotoris 70%,determinethe
powerrequirementof thepump.Allowfor entranceandexitlossesandwhere
thelossfor eachbendandvalveis equivalentto 0.7and0.12velocityheads,
respectively.
Answer: 1.2kW
(13) Ethaneis removedfromanLPG mixtureinadistillationcolumninwhich
theethane-richoverheadvapouris condensed,andsomeis returnedto thetop
of thecolumnasreflux.The topof thecolumnoperatesatagaugepressureof
12bar.The gaugepressurein therefluxdrumis 11.5bar.The differencein
elevationbetweenthetoptrayof thecolumnandthelevelof liquidin thereflux
drumis 10m. The connectingpipeworkconsistsof 50 m of 0.254m inside
diameterpipe with 27 bends,two gatevalvesandonecontrolvalve.If the
flowrateof refluxedliquid,of density600kgm-3,is 299m3h-l, determinethe
(16) A methanol-waterliquidmixtureis tobetransportedfromastoragetank
to a distillationcolumnfor separation.The mixture,of density895kgm-3,is
storedwithinanenclosedtankmaintainedat anabsolutepressureof 1 atmo-
sphereandis tobepumpedtothecolumnatarateof 11kgs-l throughapipeof
insidediameter12cm.The column,whichoperatesata gaugepressureof 50
kNm-2,hasafeedpoint15mabovethelevelof theliquidmixturein thestorage
tank.Ifthe overallefficiencyof thepumpandmotoris 65%andthelossesdue
270
271
FLUID MECHANICS PUMPS
to frictionamountto4 m,determinethepowerrequirementof thepump. head in the suctionpipe is 0.3 m. Allow for worst casemeteorological
conditions.
Answer:4.1kW
Answer:6.77m
(17) Showthattheworkexertedonafluidbyacentrifugalpumpisgivenby
W =NF
53.3
(18) Experimentaldatawereobtainedfrom laboratorytestsof a centrifugal
pumpoperatingat2500rpmusingwaterasthetestfluid.F is theforceexerted
onatorquearmmeasuredfromthecentrelineof theshaftwhichhasaradiusof
179mm.
(21) A liquidhydrocarbonliquidmixture,of density538kgm-3andviscosity
3.04xlO-4Nsm-2, is pumpedfromthebottomof a distillationcolumnoper-
atingat120kNm-2throughareboileratarateof 14x1O3kgh-l. Thecentreline
of thepumpis a verticaldistanceH belowthecolumntake-offpointandthe
suctionlinehasatotallengthofH +3 metresandaninsidediameterof65mm.
ThereboilerisatanelevationofH +0.5metresabovethepump.If thepressure
dropthroughthereboileris 14kNm-2,determinethedifferentialheadacross
the pump.Neglectkinetic energyheadloss and lossesdueto fittings,and
entranceandexitlosses.Assumea constantFanningfrictionfactorof 0.004.
The NPSH for thepumpis 3 m.
whereW is thework (Watts),N is thespeedof theimpeller(revolutionsper
minute)andF is theforce(Newtons)onatorquearmmeasuredfromthecentre
lineof theshaftwhichhasaradiusof 179mm.
Answer:30.11m
Flowrate,Q(xlO-3 m3s-1) 1.47 1.31 1.18 1.00 0.91 0.71 0.48
Diff. pressure,b.p(kNm-2) 18.8 36.0 44.8 61.5 63.6 71.5 79.7
Forceonarm,F (N) 4.6 4.1 4.0 3.7 3.3 3.0 2.4
(22) A centrifugalpumpdeliverswateratarateof 0.02m3s-1againstahead
of 10m.If theimpellerspeedis 1500rpm,determinethespecificspeedNs.
Answer:0.63m3/4s-3/2
Determinetheflowratedeliveredatmaximumpumpefficiency.Commenton
thegeneralperformanceof thepump.
(19) Wateris tobetransportedfromonetanktoanotherusinga 10mpipeline
of insidediameter25 mm.A centrifugalpumpis to beusedto transportthe
waterandis positionedbelowthefirsttankwhosefreesurfaceis 2 mabovethe
freelydischargingoutlet.Determinethedutypointfor thepumpcharacteristic
given below. The friction factor may be assumedto be 0.005.Allow for
entranceandexitlosses.
(23) DeterminethesuctionspecificspeedSnfor acentrifugalpumpoperating
atanimpellerspeedof 40rpsdeliveringaflow of 110m3h-lagainstaheadof
14m.
Answer:0.174
Discharge,m3min-l
Head,m
0.000
3.5
0.008
3.2
0.012
3.0
0.018
2.4
0.024
1.4
(24) Determinethenumericalvalueof k whichrelatesthespecificspeedof a
centrifugalpump,N" to suctionspecificspeed,Sn,as
S =Ns
n k
for apumpwithadischargemeasuredin litresperminute.
Answer:0.02m3min-l Answer:176
(20) A centrifugalpumpis usedtotransferaliquidofdensity790kgm-3from
anopenstoragevessel.Thecentreof thepumpis locatedatanelevationof 2 m
abovetheliquidin thevessel.DeterminetheavailableNPSH if thevapourpres-
sureof theliquidatthetemperatureofoperationis 25kNm-2 andthefrictional
272 273
Glossaryofterms
Absolutepressure
Pressuremeasuredaboveavacuum.
Absoluteroughness
Averageheightof undulationsandimperfectionsontheinnersurfaceof apipe
wall.
Angularvelocity
Rotational.speedexpressedin radianspersecond.
Archimedes'principle
Statesthattheupthruston apartiallyor totallyimmersedbodyis equalto the
weightof liquiddisplaced.
Averagevelocity
Definedasthetotalflowrateperunitflow area.
Boundarylayer
Regionbetweenawallandapointintheflowingfluidwherethevelocityisata
maXImum.
Cavitation
Destructivecollapseofvapourinliquidinregionsofhighpressure.
Centrifugalpump
Mechanicaldeviceusedto transportfluids by way of an enclosedrotating
impellerimpartingvelocitytothefluid.
Coefficientof contraction
Ratioof areaof venacontractato orificearea.
275
FLUID MECHANICS
Coefficientofdischarge
Ratio of actual flowrate to theoreticalflowrate throughan opening or
restriction.
Coefficientof velocity
Ratioof actualto theoreticalvelocityfor a dischargingjet of fluid.
Density
Measureofthequantityofasubstanceperunitvolume.
Dynamicviscosity
Phenomenonin whichliquidsshearbetweenadjacentlayers.
Dynamicalsimilarity
Similaritybetweensystemswhenappliedforceshavethesameratio.
Elbow
A curved90°sectionofpipework.
Entranceandexitlosses
Irreversibleenergylosswhenfluidsenteror leavepipes.
Entrancetransitionlength
Distancebetweenpipeentranceandpointof fully developedlaminarflow.
Equivalenthydraulicdiameter
Usedinplaceofdiameterd inturbulentflowequationswhenthepipeorduct
cross-sectionis notcircular.Equaltofourtimestheflowareadividedbythe
wettedperimeter.Doesnotapplytolaminarflow.
Equivalentlength
Thelengthof straightpipethatwouldgivethesamepressuredropasthefitting
of thesamenominaldiameterexpressibleinpipediametersof thestraightpipe.
Frictionfactor
Empirical factor and is a functionof Reynoldsnumberand relativepipe
roughness.
Fluid
A substancewhichoffersnoresistancetochangeofshapebyanappliedforce.
276
-
GLOSSARY OF TERMS
Forcedvortex
Circularstreamofliquid thewhirlofwhichis causedbyanexternalsourcesuch
asa stirreror impeller.
Freejet
Dischargingliquidfromapipeororificewhichisnolongerinfluencedbythe
appliedhead.
Freesurface
Interfaceof a liquidwitha vapourorgasabove.
Freevortex
Streamlinesthatmovefreelyinhorizontalconcentriccircleswithnovariation
of thetotalenergyacrossthem.
Frictionalhead
Headrequiredby a systemtoovercomeresistancetoflow in pipesandassoci-
atedfittings.
Gatevalve
Deviceusedtoregulateflowinapipe,consistingofaverticalmovingsection
acrosstheflowarea.
Gaugepressure
Pressuremeasuredaboveatmospheric.
Geometricalsimilarity
Systemswhichhavesimilarphysicaldimensionssuchthattheirratiois a
constant.
Head
Pressureofaliquidexpressedastheequivalentheightacolumnof theliquid
wouldexert.
Laminarflow
Flowthatoccurswhenadjacentlayersoffluidmoverelativetooneanotherin
smoothstreamlines.FlowoccursinpipesforReynoldsnumbersbelow2000.
Massflowrate
Massflow of a fluid perunittime.
277
FLUID MECHANICS
Mass loading
Rateofmassflowperunitarea.
Meanhydraulicdepth
Ratioof flow areatowettedperimeterin anopenchannel.
Netpositivesuctionhead(NPSH)
Total suctionheaddeterminedatthesuctionnozzleof a centrifugalpumpless
thevapourpressure.
NPSHA
NPSH availableis theabsolutepressureattheinletofacentrifugalpumpandis
afunctionof elevation,temperatureandpressure.
NPSHR
TheNPSHrequiredbyacentrifugalpumpisafunctionofthepumpcharacter-
isticsandmustbeexceededbyNPSHA.
Newtonianfluids
Classof fluids in whichviscosityis independentof shearstressandtime.
Non-Newtonianliquids
Liquids in whichtheviscositydependsonshearstressand/ortime.
No-slipcondition
Assumptionthatthefluid layerin contactwitha surfaceis stationary.
NotchRectangularortriangularcutinaweir.
Openchannel
Conduitcarryingliquidwith afreesurface.
Orifice
Smallholeoropening.
Orificemeter
Deviceusedtomeasureflowratebythepressuredropthroughasmallhole.
278
~
GLOSSARY OF TERMS
Piezometer
Deviceusedtomeasurepressure,consistingof anopenvertical(glass)tube
attachedtothesideofahorizontalpipe.
Pipeline
Longsectionoflargeborepipe.
Pitottube
Deviceusedtomeasurevelocityheadof aflowingfluid.
Power
Workdoneperunittime.
Pressure
Forceofafluidappliedoveragivenarea.
Pressuredifferential
Differencein pressurebetweentwopoints.
Pressurehead
Pressure-volumeenergyof afluid expressedin headform.
Relativeroughness
Ratioofabsolutepipewallroughnesstopipeinsidediameterinconsistentunits.
Reynoldsnumber
Dimensionlessnumberexpressingtheratioof inertialto viscousforcesin a
flowing fluid.
Rotameter
Instrumentusedto measureflowrateby theverticalelevationof a float in a
taperedtube.
Separation
Phenomenonof fluid streamlineschangingdirection due to changesin
boundaryshapeas a resultof fluid inertiaor velocitydistributionnearthe
boundarysurface.
Separator
Vesselusedtoseparatetwoliquids,usuallyaslayers.
279
FLUID MECHANICS GLOSSARY OF TERMS
Shear stress
Forceofaliquidappliedoverasurface.
Velocitygradient
See'Shearrate'.
Shearrate
Changeinvelocityofafluidperpendiculartoflow.
Velocityhead
Kinetic energyof afluid expressedin headform.
Sluice
Submergedopening.
Venacontracta
Minimum flow areaformeddownstreamby a fluid flowing througha sharp-
edgedopening.
Specificgravity
Ratioof massof a liquid tomassof wateroccupyingthesamevolume.
Venturimeter
Deviceusedtomeasureflowratebythepressuredropthroughataperedsection
of tubeorpipe.
Specificspeed
Dimensionlessgroupusedto classifycentrifugalpumpimpellersat optimal
efficiencywithrespectto geometricsimilarity.
Statichead
Potentialenergyof a liquidexpressedin headform.
Viscosity
See'Dynamicviscosity'.
Steadyflow
Nochangeoffluidvelocitywithrespecttotime.
Viscousflow
See'Laminarflow'.
Streamline
An imaginerylinewhichliesinthedirectionofflow.
Volutechamber
Spiralcasingofcentrifugalpumpsurroundingimpeller.
Siphon
Transferofliquidfromonevesseltoanotheratalowerelevationbymeansofa
pipewhosehighestpointisabovethesurfaceoftheliquidintheuppervessel.
Weir
Verticalobstructionacrossachannel.
Weircrest
Topofaweiroverwhichaliquidflows.
Transitionflow
Flowregimebetweenlaminarandturbulentflow(2000<Re<4000).Velocity
fluctuationsmaybepresentandimpossibletopredict.
Turbulentflow
Flow regimecharacterizedby fluctuatingmotionanderraticpathsabove
Reynoldsnumbersof 4000.Flow occurswheninertial.forcespredominate
resultinginmacroscopicmixingof thefluid.
Uniformflow
Nochangein fluidvelocityatagiventimewithrespecttodistance.
280 281
SELECTED RECOMMENDED TEXTS
Selectedrecommendedtexts
Mott, RL., 1994,AppliedFluid Mechanics,4thedition(PrenticeHall, New Jersey).
Munson,B.R., Young,D.F. andOkiishi,T.H., 1994,FundamentalsofFluid Mechanics,
2ndedition(John Wiley, New York).
Pnueli, D. andGutfinger,C, 1997,Fluid Mechanics (CambridgeUniversity Press,
Cambridge).
Roberson,J.A. andCrowe,CT., 1997,EngineeringFluid Mechanics,6thedition(John
Wiley, New York).
Acheson,DJ., 1996,ElementaryFluid Dynamics,4thedition(OxfordUniversityPress,
Oxford).
Sharpe,G.T., 1994,SolvingProblemsin Fluid Dynamics(Longman,Harlow).
Boxer,G., 1988,WorkoutFluid Mechanics(MacMillan PressLtd, London).
Street,R.L., Watters,G.Z. andVennard,J.K., 1996,ElementaryFluid Mechanics,7th
edition(JohnWiley, New York).
Currie, I.G., 1993,Fundamentalsof Fluid Mechanics,2ndedition(McGraw-Hill, New
Jersey).
Streeter,V.L. andWylie, E.B., 1985,Fluid Mechanics,8thedition(McGraw-Hill Book
Co, London).
Darby,R, 1996,ChemicalEngineeringFluid Dynamics(MarcelDekker). Tritton,D.J., 1997,PhysicalFluid Dynamics,2ndedition(Oxford SeriesPublication,
Oxford).
Douglas,IF., Gasiorek,J.M. andSwaffield,J.A., 1996,Fluid Mechanics,3rdedition
(Longman,Harlow). Turner,I.C, 1996,EngineeringApplicationsof PneumaticsandHydraulics (Edward
Arnold, London).
Douglas, J.F. and Mathews, RD., 1996, Solving Problems in Fluid Mechanics,
Volumes I and2, 3rdedition(Longman,Harlow). Vardy,A., 1990,Fluid Principles (McGraw-Hill Book Co, London).
Fox, R.W. andMcDonald, AT, 1992,Introductionto Fluid Mechanics,4th edition
(John Wiley, New York).
Widden,M., 1996,Fluid Mechanics(MacmillanPressLtd, London).
White,F.M., 1994,Fluid Mechanics,3rdedition(McGraw-Hill, New York).
French,R.H., 1994,OpenChannelHydraulics(McGraw-Hill, New Jersey).
Granet,I., 1996,Fluid Mechanics,4thedition(PrenticeHall, New Jersey).
Young, D.F., Munson, B.R and Okiishi, T.H., 1997,A Brief Introductionto Fluid
Mechanics(JohnWiley, New York).
Holland, F.A. and Bragg, R., 1995,Fluid Flow for Chemical Engineers (Edward
Arnold, London).
Hughes,W.F. andBrighton,J.A., 1991,Schuam'sOutlineof TheoryandProblemsof
Fluid Dynamics,2ndedition(McGraw-Hill, New Jersey).
Ligget,J.A., 1994,Fluid Mechanics(McGraw-Hill, New York).
Massey,B.S., 1989,Mechanicsof Fluids, 6thedition(ChapmanandHall, London).
282 283
---
Nomenclatureandpreferredunits Usefulconversionfactors
284 285
Quantity Name Symbol Typeof In termsof Quantity Multiplier SI or preferredunit
SI unit baseSI
Acceleration
Area m2 Base
cms-2 ms-2
Degree planeangle Allowable 1t/180rad I
0.01
Density kgm-3 Base
fts-2 0.3048 ms-2
Force Newton N Derived kgms-2 Area
Length metre m Base ! mm2 0.000001 m2
Length kilometre km Allowable 103m cm2 0.0001 m2
Mass kilogram kg Base
I
in2 0.0006451 m2
Massflow kgs-l Base ft2 0.0929 m2
Mass loading kgm-2s-1 Base yd2 0.836 m2
Pressure bar bar Allowable 105kgm-ls-2
I DensityPressure Pascal Pa Derived kgm-Is-2
Pressure Nm-2 Derived kgm-Is-2
SO 1000 kgm-3
gcm-3 1000 kgm-3Radian planeangle rad Supplementary Ibm.ft-3 16.018 kgm-3Revolution angulardisplacementrps Allowable 21tfad
Ibm.UKgal-l 99.77 kgm-3Shearstress Nm-2 Derived kgm-ls-2
Surfacetension Nm-l Derived kgs-2
Ibm.USgal-1 119.83 kgm-3
Time hour h Allowable 3600s Energy
Time minute mm Allowable 60s kJ 1000 J
Time second s Base Nm 1 J
Torque Nm Derived kgm2s-2 11 Ibf-ft 1.356 J
Velocity ms-I Base Force
Viscosity Nsm-2 Derived kgm-Is-l kN 1000 N
Volume m3 Base
dyne 0.00001 NVolume litre I Allowable' 10-3m3
Ibf 4.448 N
Volumeflow m3s-1 Base
UKtonf 9964 N
Work joule J Derived kgm2s-2 I UStonf 8896 N
kgm2s-3Watt power W Derived
USEFUL CONVERSION FACTORS
286 287
FLUID MECHANICS
Length
mm 0.001 m
em 0.01 m
km 1000 m
III 0.0254 m
ft 0.3048 m
yd 0.9144 m
Mass
g 0.001 kg
oz(troy) 3.1103 kg
oz(avdp) 2.8349 kg
Ibm 0.4536 kg
Planeangle
degC) 0.01745 fad
Power
kW 1000 W
Js-I 1 W
ft.lbf.s-I 1.356 W
hp 745.7 W
Pressure
Pa 1 Nm-2
kgm-ls-2 1 Nm-2
mmHg 133.3
bar 100,000 Nm-2
SId aIm 101,300 Nm-2
dyne.em-2 0.1 Nm-2
Ibf.in-2(psi) 6895 Nm-2
Pressuredrop
Pa.m-I 1 Nm-2m-1
psi.fc 1 0.022621 Nm-2m-1
Surfacetension
kgs-I 1 Nm-I
Ibf.ft-I 14.59 Nm-I
dyne.em-I 0.001 Nm-I
dyne.m-1 0.00001 Nm-I
Time
mill 60 s
h 3600 s
d 86,400 s
Velocity(linear)
ems-I 0.01 ms-I
fts-l 0.3048 ms-l
Velocity(angular)
IpS 0.10472 rads-I
rpm 6.2832 rads-1
degs-l 0.017453 rads-1
Viscosity(dynamic)
Pa.s 1 Nsm-2
kgm-ls-1 1 Nsm-2
Ibf.s.in-2 6894.7 Nsm-2
eP 0.001 Nsm-2
P 0.1 Nsm-2
Viscosity(kinematic)
ft2s-1 0.092903 m2s-1
cSt 0.000001 m2s-1
St 0.0001 m2s-1
Volume
ml 0.000001 m3
dm3 0.001 m3
litre '0.001 m3
in3 0.0000164 m3
ft3 0.02831 m3
yd3 0.7645 m3
bbl (42USgal) 0.1589 m3
UKgal 0.004546 m3
USgal 0.003785 m3
Volumeflow
m3h-l 0.0002777 m3s-1
ft3s-1 0.028317 m3s-1
bbl.h-I 0.0004416 m3s-1
UKgal.h-1 0.0045461 m3s-1
USgal.h-1 0.003754 m3s-1
Physicalpropertiesofwater(atatmosphericpressure)
Temperature,
O(
0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
288
Density,
kgm-3
1000
1000
1000
999
998
997
996
994
992
990
988
985
983
980
978
975
972
969
965
962
958
Viscosity,
Nsm-2
0.00790
0.00152
0.0013]
0.00114
0.00100
0.00089
0.00080
0.00072
0.00065
0.00060
0.00055
0.00051
0.00047
0.00044
0.00040
0.00038
0.00035
0.00033
0.00031
0.00030
0.00028
Lossesforturbulentflow
throughfittingsandvalves
Vapourpressure,Surfacetension,
Nm-2 Nm-I
611 0.076
872 0.075
1230 0.075
1700 0.074
2340 0.074
3170 0.073
4240 0.072
8600 0.071
7380 0.070
9550 0.069
]2,300 0.068
]5,740 0.067
20,000 0.067
25,010 0.066
3],200 0.065
38,550 0.064
47,400 0.063
57,800 0.062
70,]00 0.061
84,530 0.060
101,300 0.059
Fittingorvalve
E]bow45°standard
E]bow45°longradius
Elbow 90°standard
Elbow 90°longradius
Elbow 90°square
Bend180°closereturn
Tee,standard,alongrun
Tee,standard,throughbranch
Losscoefficient,k
0.35
0.2
0.75
0.45
1.3
1.8
0.4
1.5
0.5
0.05
1.0
0.5
0.15
0.9
4.5
20.0
2.3
2.6
4.3
21.0
6.0
9.0
3.5
0.5
289
Pipeentry sharp
well rounded
Borda
Pipeexit sharp
Gatevalve fully open
%open
Yz open
Xopen
Diaphragmvalve fully open
%open
Yz open
Xopen
G]obevalve fully open
Yz open
Anglevalve fully open
Butterflyvalve fully open
FLUID MECHANICS
Checkvalve,swing
Pumpfootvalve
Sphericalplugvalvefully open
Flow meters orifice ~=0.2
~=0.5
~=0.8
nozzle ~=0.2
~=0.5
~=0.8
venturi
turbine
Cyclone
290
2-3.5
1.5
0.1
2.6
1.8
0.6
1.0
0.55
0.15
0.15
6.0
Equivalentsandroughnessofpipes
Type
Glass
Drawntubing
Steel,wroughtiron
Asphaltedcastiron
Galvanizediron,new
Galvanizediron,old
Castiron
Wood stave.
Concrete
Rivetedsteel
10-20
Equivalentsandroughness,£ (mm)
0.0003
0.0015
0.0457
0.1219
0.1524
0.2743
0.2591
0.1829
0.3048-3.048
0.9144-9.144
291
Manningcoefficientforvarious
openchannelsurfaces
Type
Weldedsteel
Planedwood
Unplanedwood
Concreteculvert
Unfinishedconcrete
Castiron
Straightsewer
Rivetedsteel
Galvanizediron
Earthchannels
Naturalstreams
Rivers
292
clean
stony
unmaintained
clean
stony
deeppools
noboulders
irregular
Moodyplot
Manningcoefficient,n(m-I/3s)
0.012
0.012
0.013
0.013
0.014
0.015
0.015
0.016
0.016
0.018
0.035
0.080
0.030
0.045
0.100
0.025
0.035
0.05
0.04
0.D3
0.02
0.015 :g
0.01 uJ
0.008 CI)
0.006 ~
0.004 ::r::"
0.002 ~
0
0.001 ~
0.0008'-I.1
0.0006:>
0.0004~
0.0002j
0.0001 [:;j
0.00005
)04 105 106
REYNOLDS NUMBER, Re
PlotoffrictionfactorversusReynoldsnumberdevelopedbyLouisF. Moody.
Laminarflow (Re<2000)
Transitionregion(2000<Re<3000)
Turbulentregion(Re>3000)
For rough-walledpipes
Re = 100
~(~)
(seeProblem8.11,page217)
293
0.025
0.020
0.015
'-
i:i 0.010
0 0.009
t;0.008
0.007
0.006
60.005
"" 0.004
0.003
0.002
103
Index
A
absolutepressure
definition
absoluteroughness,definition
acceleration
accumulators
angularvelocity,definition
ARCHIMEDES
Archimedes'principle
definition
atmosphericpressure
averagevelocity,definition
axialpumps
B
1,2,5,6
275
275
244,246
41
275
1,2
25,28,29
275
1,2
275
235,250
barometers 2
bearings 93
BERNOULLI, D. 35
Bernoulliequation 42,43,45,50,
115,119,126,129,131,
202,207,221,240,247
BLASIUS, P.R.H. 193
Blasiusequation 200
BORDA, J.e. 47
boundarylayer,definition 275
bubbles
pressurewithin
BUCKINGHAM. E.
BuckinghamII method
buoyancy
294
15
99,106
106
26
C
calibrationofflowmeters
capacitycoefficient
capillaryaction
CARNOT, LN.M.
cavitation
definition
centrifugalforce
centrifugalpumps
characteristic
definition
dutypoint
efficiency
in seriesandparallel
matching
scale-up
channels
circular
economicaldepth
open
rectangular
CHE:ZY, A.
Chezycoefficient
Chezyformula
circularchannelflow
coefficient
discharge 122,128,134,
143,146,276
head 104,105,249,252,258,260
Manning 292
power 104,105
pressure 108
velocity 141,143,145,276
COLEBROOK, e.P. 194
collarbearings 93
conservationof
energy
mass
124,132,134
104,105,249,
252,258,260
2
47
236,238,244,245,
247,259,260,261
275
51
237,244,245,
249,256
252
275
254
258
243
239
104
momentum
continuity
continuityequation
contractioncoefficient
definition
conversionfactors
criticaltime
crystalconcentration
measurementof
cylinderpumps
D
DARCY, H.P.G.
Darcyequation
Darcy-Weisbachequation
daVINCI, L
decelerationof li<J.uid
density,definition
diaphragmpumps
differentialhead
flow measurementby
differentialmanometers
differentialpressure
acrossa venturimeter
dimensionalanalysis
BuckinghamII method
for centrifugalpumps
for notches
for orifices
Rayleighmethod
dip legs
177,179
175
167-191
171-176
167
170,17l, 173,177
169,171,173,178
177,179
capacity104,105,249,252,258,260
Chezy 170,171,173,177
contraction 47, 143,146,275
35
35,36
35
35-59
35,36
47,143,146
275
285
41
193
198,199
198,199
35
40
276
263
113-139
17,20,119
2, 11,13,17,23
126
99-111
99, 106
104
102
100
99, 100
densitymeasurement
interfacemeasurement
INDEX
11-12
11-12
discharge
from reciprocatingpumps 265
dischargecoefficient 128,134,143,146
definition 276
drainage
betweentworeservoirs
drainagefrom tanks
cylindricalcross-section 151
hemisphericalcross-section 147
uniformcross-section 147
ducts 118
non-circular 223
dutypoint 240,242,254
dynamicalsimilarity,definition 276
dynamicpumps 235
dynamicviscosity,definition 276
E
13
263
economicpipediameter
economicaldepthof flow
eddies
elbows,definition
energy
enlargements
entrancelosses,definition
entrancetransitionlength
definition
equations
Bernoulli
153
196
175
44
276
35-59
44
276
65
276
42,43,45,50,115,119,
126,129,131,202,
207,221,240,247
200
198,199
198,199
198
68,70,71,73
202
141,143,145
214
224,225
276
Blasius
Darcy
Darcy-Weisbach
Fanning
Hagen-Poiseuille
Prandtl
Torricelli
von Karman
equivalenthydraulicdiameter
definition
295
FL DID MECHANICS
equivalentlength,definition
equivalentlengthmethod
EVLER,L.
exitheadlosses
definition
276
214,217
35
46,206,207,213,221
276
F
FANNING, IT.
Fanningequation
Fanningfrictionfactor
film thicknessin a channel
flow
chartsfor
downaninclinedplate
downa verticalwire
in branchedpipes
in circularchannels
in rectangularchannels
in taperingpipes/tubes
laminar
multiphasefluid
openchannel
overnotches
pulsed
radial
througha gap
throughanorifice
throughnon-circularducts
turbulent
velocitythrougha gap
verticalin pipes
flow measurement
by differentialhead 113-139
flowmeters 113
calibration 124
calibrationcurve 132
nozzle 113
orifice 113,114,122,128,278
Pitottubes 113,115,117,279
rotametercalibration 134
variablearea 130
venturi 113,119,122,124,
126,134,281
296
293
81
82
36
177,179
171,173,175
71,226
61-97
103
103
102
235
51
84,88
100
223
106,223
86,87
78
195
198
195
79
t1uids
definition
pressureof
fluid statics
flumes
force
centrifugal
inertial
onpipereducers
onV-bends
radial
viscous
withinahydraulicram
forcedvortices
definition
freejets
definition
freesurfaces,definition
freevortices
definition
frictionalhead,definition
frictionfactor,definition
G
gas-liquidinterface
gatevalves,definitiongaugepressure
definition
geometricalsimilarity,definition
gyroscopiceffect
H
HAGEN,G.H.L.
Hagen-Poiseuilleequation
head
definition
entranceloss
inertia
lossduetofriction
pressure
static
headcoefficient
276
1,2
1-33
167
HERSHELS, C.
hydraulicrams
hydrometers
51
64
49
38
51
64
7
52,53
277
143
277
277
52,53
277
277
276
impactpressure
impellers
inclinedlegmanometers
inertiahead
inertialforce
interface
gas-liquid
liquid-liquid
invertedmanometers
J
JOVKOWSKY, N.
K
kinematicviscosity
79
277
1,2,268
277
277
130
61
68,70,71,73
277
47
229
197
43
43
104,105,249,
252,258,260
M
MANNING,R.
Manningcoefficient
Manningformula
manometers
167
292
170,171,173
2, 17,115
netpositivesuctionhead(NPSH) 245,
247,251
definition 278
Newtoniant1uids,definition 278
NIKVRADSE, J. 193,194
non-circularducts 223
non-Newtoniant1uids,definition 278
no-slipcondition 65,67,79,82
definition 278
notches 167-191
definition 278
flow over 102
rectangular 180
trapezoidal 189
triangular 186
V 188
nozzleflowmeters I 13
NPSH (seenetpositivesuctionhead)
0
openchannels
definition
flow
278
103
L
laminarboundarylayer 65
laminarflow 61-97
definition 277
downa wire 82
generalequationfor pipefriction 199
localvelocity 68,75,86
maximumpipediilmeter 77
plate 81
throughtaperedtubes 71
throughagap 84
LANGHAAR, H. 65
liquid-liquidinterfaces 9, 11
lubrication 61-97,235,268
collarbearings 93
shafts 91
297
INDEX
119 differential 17,20,119
7 inclinedleg 23
27 inverted 19
singleleg 21
V-tube 17
115 masst1owrate,definition 277
237,256 massloading 63
23 definition 278
229 meanhydraulicdepth 168,172,173,
64 175,177
definition 278
79 memsci 2
9,11 Metric system 1
19 momentum 35-59
Mono pumps 268
MOODY, L.F. 194
41 Moody plot 205,218,223,225,293
multiphaset1uidt1ow 103
63 N
FLUID MECHANICS
orificeflowunderconstanthead 143
orificemeters 113,114,122,128
~fi~tioo n8
orifices
coefficientofdischarge
definition
dimensionalanalysis
flowthrough
P
PANNELL, J.R.
PASCAL, B.
Pascal'slaw
piezometers
definition
pipediameter
for laminart10w
pipeenlargement
pipefriction
pipelines,definition
pipereducers
pipes
accelerationin
in parallel
III serIes
rough-walled
smooth-walled
pistons,reciprocating
PITOT, H.
Pitottraverse
Pitottubes
definition
PLIMSOLL, S.
POISEUILLE, J.L.M.
positivedisplacementpumps
power,definition
powercoefficient
powerfor pumping
PRANDTL, L.
Prandtlequation
pressure
absolute
298
122,128
278
100
100
193
I
3
2
279
196
70
44
193-233
279
49
228
209
21I, 214
204,214
202
264
115
117
113,115,117
279
30
61
235,268
279
104,105
264
193
202
1,2,5,6
atmospheric 1,2
ata point 3
definition 279
differential 2,11,13,17,23
gauge 1,2,268,277
head 43
impact 115
static 115,245
vacuum 1
vapour 206,244,245,251
withina closedvessel 5
withina gasbubble 15
pressurecoefficient 108
pressuredifferential,definition 279
pressuredrop
duetoenlargements
in apipeline,scalemodel
pressurehead
definition
reciprocating
rotary
scale-upof centrifugal
specificspeed
suctionspecificspeed
235,246,262,265
235,246,268
104
249
260
51
263
99, 100
107
264
235,246,262
265
171-176
180
182,184
279
61
63,70,76,77,
80,105,108,199
279
219
114,130,134
279
235,246,268
204,214
218
R
radialforce
rampumps
RA YLEIGH, Lord
Rayleighmethod
reciprocatingpistons
reciprocatingpumps
dischargefrom
rectangularchannels
rectangularnotches
rectangularweirs
relativeroughness,definition
REYNOLDS, O.
Reynoldsnumber
44
108
43
279
pressuremeasurement
differentialmanometers 17,20, 119
inclinedlegmanometers 23
invertedmanometers 19
singlelegmanometers 21
U-tubemanometers 17
pressurerisebyvalveclosure 40
pulsationdampers 263
pulsedflow 235
pumpcharacteristic 243,253,255
pumps 235-273
axial 235,250
centrifugal 237,239,243,244,245,
249,252,254,256,258,275
characteristiccurve 243,253,255
cylinder 263
diaphragm 263
dutypoint 242,254
dynamic 235
efficiency 250,258
Mono 268
positivedisplacement 235,268
power 256,258
ram 263
definition
ringmains
rotameters
definition
rotarypumps
rough-walledpipes
ROUSE, H.
S
safetyreliefvalves
scalemodels
scale-upof centrifugalpumps
separation
definition
separators
definition
seweragepipes
shearrate,definition
shearstress
definition
for flow througha gap
similarity
dynamical,definition
geometrical,definition
singlelegmanometers
siphons
definition
sluices,definition
smooth-walledpipes
soundtransmission
INDEX
276
277
21
206
280
280
202
41
specificgravity
definition 280
measurementby hydrometer 27
specificspeed 249,251,259,260
definition 280
STANTON, T.E. 193
statichead 43
definition 280
staticpressure 115,245
steadyflow, definition 280
STEVIN, S. 1
streamlines 43,51,52,54,61,143
definition 280
suctionspecificspeed 260
surfacetension 15,102,181,185
SystemeInternational 1
T
tankdrainage
cylindricalcross-section
hemisphericalcross-section
instantaneous
268
108
104
244
279
9
n9
167
280
93
280
88
througha pipe
throughV-notches
uniformcross-section
with laminarflow
tankinflow andoutflow
taperedsections,headloss
taperedtubes,laminarflow
tension
torque
TORRICELLI, E.
Torricelli equation
tracerdilution
transitionflow, definition
141-165
151
147
157
221
188
147
161
155,159
226
71
26
89,91,94
141,146
141,143,145
124
280
299
transmissionvelocity
trapezoidalnotches
triangularnotches
turbulentflow
definition
41
189
186
193-233
280
U
U-bends
uniformflow, definition
U-tubemanometers
v
vacuum
valves
closure 40
pressurerelief 41,268
vapourbubbles 244,246
vapourpressure 206,244,245,251
variableareaflowmeters 130
VASCHY, A. 99
velocity
average 275
of flow througha gap 86,87
of fluid in apipe 73,75
velocitycoefficient 141,143,145,276
velocitygradient 75,88
definition 281
velocityhead 43
definition 281
velocityheadmethod 211,217
velocityprofile in apipe 66
venacontracta 47, 141,143,145
definition 281
300
38
280
17
VENTURI, G.B.
venturimeters
calibration
coefficientof discharge
definition
horizontal
vertical
vibration
1,2
viscometers
viscosity,definition
viscousdrag
viscousflow, definition
viscousforce
viscousshearstress
V -notches
volutechamber,definition
vonKARMAN, T.
vonKarmanequation
vortexmotion
vortices
forced
free
W
wallshearstress
water
physicalproperties
weircrest,definition
weirs
definition
rectangular
WEISBACH, J.
whirlwinds
WHITE, c.M.
119
124,134
114
281
119,122
126
238,244
89
281
64,91
281
64
75,88
188
281
194
214
51
52,53
52,53
88,169,197,223
288
281
103,167-191
281
182,184
193,198
52
194
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