Fluid Mechanics & Hydraulic Digital Material.pdf

DEPARTMENT OF MECHANICAL ENGINEERING
FLUID MECHANICS
& HYDRAULIC
MACHINES
(R18A0304)
2
NDYEAR,B.TECHI-SEM,MECHANICAL ENGINEERING

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
COURSEOBJECTIVES
UNIT -1 CO1:Togiveinsightknowledgeonfluidstatics.
UNIT -2 CO2:Togainknowledgeonfluidkinematicsand
dynamics.
UNIT -3 CO3:Togivebasicunderstandingofboundarylayer
conceptandanalyzedifferenttypesoflossesand
measurementofflow.
UNIT -4 CO4:Tobecomefamiliaraboutdifferenttypesof
turbines&abletoanalyzetheirperformance
characteristicsofvariousturbines.
.
UNIT -5 CO5:Tobeabletounderstandtheworkingofpower
absorbingdeviceslikepumps&abletoanalyzetheir
performancecharacteristics.

DEPARTMENT OF MECHANICAL ENGINEERING
UNIT 1
CO1:Togiveinsightknow ledge onflui d
statics.
FLUID STATICS

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
UNIT –I
FluidStatics:
•DimensionsandUnits.
•PhysicalPropertiesoffluids-Specificgravity,viscosity,
surfacetension.
•Vapourpressureandtheirinfluenceonfluidmotion.
•Atmospheric,gaugeandvaccumpressure.
•Measurementofpressure-Piezometer,U-tubeand
Differentialmanometers.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
COURSEOUTLINE
LECTURE LECTURETOPIC KEYELEMENTS Learningobjectives
1 Introduction to FMHM Units and
dimensions.
Remember the units
(B1)
2 Physical Properties of fluid
Definitionof
Specific gravity,
Viscosity,Surface
tension
•Analyze Law of
viscosity (B4)
•Apply Surface
tension concepts in
droplet (B3)
3 Vapour pressure and their
influence in fluid motion
Concepts of
Vapour pressure
Understanding
influence of vapour
pressure due change
in temperature (B2)
4 Atmospheric, gauge and vaccum
pressure
Classification of
Pressure
Evaluate pressure
(B5)
5 Measurement of pressure Derivation of
Piezometer,U-
Tube Manometer
Apply to measure
pressure (B3)
6 Measurement of pressure Derivation of
Differential
manometer
Apply to measure
pressure (B3)
UNIT-1

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
COURSEOUTLINE
LECTURE LECTURETOPIC KEYELEMENTS Learningobjectives
7 Example Problems (2/3 classes)
UNIT-1

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 1
•Introduction to Fluid Mechanics
•Application of Fluid Mechanics
•Units and Dimensions
•Definitions
•Problems
Introduction –Units and
Dimensions
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
INTRODUCTION
•FluidMechanicsisbasicallyastudyof:
Physicalbehavioroffluidsandfluidsystemsandlaws
governingtheirbehavior.
Actionofforcesonfluidsandtheresultingflowpattern.
•Fluidisfurthersub-dividedintoliquidandgas.
•Theliquidsandgasesexhibitdifferentcharacteristicsonaccountof
theirdifferentmolecularstructure.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
FLUID MECHANICS COVER MANY AREAS LIKE:
•Designofwiderangeofhydraulicstructures(dams,canals,weirs
etc)andmachinery(Pumps,Turbinesetc).
•Designofcomplexnetworkofpumpingandpipelinesfor
transportingliquids.Flowofwaterthroughpipesanditsdistribution
toservicelines.
•Fluidcontroldevicesbothpneumaticandhydraulic.
•Designandanalysisofgasturbinesandrocketenginesandair–
craft.
•Powergenerationfromhydraulic,streamandGasturbines.
•Methodsanddevicesformeasurementofpressureandvelocityof
afluidinmotion.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
UNITS AND DIMENSIONS :
•Adimensionisanamewhichdescribesthemeasurable
characteristicsofanobjectsuchasmass,lengthandtemperature
etc.aunitisacceptedstandardformeasuringthedimension.The
dimensionsusedareexpressedinfourfundamentaldimensions
namelyMass,Length,TimeandTemperature.
•Mass(M)–Kg
•Length(L)–m
•Time(T)–S
•Temperature(t)–
0
CorK(Kelvin)

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
UNITS AND DIMENSIONS:
•Density:Massperunitvolume=kg/m
3
•Newton:Unitofforceexpressedintermsofmassand
acceleration,accordingtoNewton’s2
nd
lawmotion.Newtonisthat
forcewhichwhenappliedtoamassof1kggivesanacceleration
1m/Sec
2
.F=MassxAcceleration=kg–m/sec
2
=N.
•Pascal:APascalisthepressureproducedbyaforceofNewton
uniformlyappliedoveranareaof1m
2
.Pressure=Forceperunit
area=N/m
2
=PascalorP
a.
•Joule:Ajouleistheworkdonewhenthepointofapplicationof
forceof1NewtonisdisplacedWork=Forceperunit=N/m=Jor
Joule.
•Watt:AWattrepresentsaworkequivalentofaJouledoneper
second.
•Power=Workdoneperunittime=J/Sec=WorWatt.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
DEFINITIONS
DensityorMassDensity:
•Thedensityormassdensityofafluidisdefinedastheratioofthe
massofthefluidtoitsvolume.Thusthemassperunitvolumeof
thefluidiscalleddensity.
•Itisdenotedbyρ.
•TheunitofmassdensityisKg/m
3
&#3627409164;=
Massoffluid
Volumeoffluid
•Thevalueofdensityofwateris1000Kg/m
3
.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
DEFINITIONS
•SpecificweightorSpecificdensity:Itistheratiobetweenthe
weightsofthefluidtoitsvolume.Theweightperunitvolumeofthe
fluidiscalledweightdensityanditisdenotedbyw.
w=
Weightoffluid
Volumeoffluid
=
Massoffluid×Accelerationduetogravity
Volumeoffluid
=
Massoffluid×g
Volumeoffluid
=ρ×g
•Specificvolume:Itisdefinedasthevolumeofthefluidoccupiedby
aunitmassorvolumeperunitmassoffluidiscalledSpecificvolume.
Specificvolume=
Volumeofthefluid
Massoffluid
=
1
Massoffluid
Volumeofthefluid
=
1
ρ
•ThustheSpecificvolumeisthereciprocalofMassdensity.Itis
expressedasm
3
/
/
kg
/
andiscommonlyappliedtogases.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
DEFINITIONS
•SpecificGravity:ItisdefinedastheratiooftheWeightdensity(or
density)ofafluidtotheWeightdensity(ordensity)ofastandard
fluid.Forliquidsthestandardfluidtakeniswaterandforgasesthe
standardliquidtakenisair.TheSpecificgravityisalsocalled
relativedensity.Itisadimensionlessquantityanditisdenotedby
ѕ.
•S(forliquids)=
weightdensityofliquid
weightdensityofwater
•S(forgases)=
weightdensityofgas
weightdensityofair

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEMS
1.Calculatethedensity,specificweightandweightofoneliterof
petrolofspecificgravity=0.7
Sol: i)Densityofaliquid=S
&#3627409164;=0.7×1000=700kg/m
3
ii)Specificweightw=&#3627409164;×g=700×9.81=6867N/m
3
iii)Weight(w) Volume=1liter=0.001m
3
Weknowthat,specificweightw=
weightoffluid
volumeofthefluid
Weightofpetrol=w×volumeofpetrol
=6867×0.001
=6.867N

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 2
•Properties of fluids
•Viscosity
•Kinematic Viscosity
•Newton’s Law of viscosity
•Types of Fluids
•Surface tension
Properties of Fluids
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROPERTIES OF FLUID
•Viscosity
•Surface tension
on liquid droplet
on hollow bubble
on liquid jet
•Capillarity
Capillary rise
Capillary fall

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
VISCOSITY
DEFINITION
•Itisdefinedasthepropertyofafluid
whichoffersresistancetothe
movementofonelayerofthefluid
overanotheradjacentlayerofthe
fluid.
•Whenthetwolayersofafluid,ata
distance‘dy’apart,moveoneover
theotheratdifferentvelocities,sayu
andu+du.
•Theviscositytogetherwithrelative
velocitiescausesashearstress
actingbetweenthefluidlayers.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
VISCOSITY
•Thetoplayercausesashearstressontheadjacentlowerlayerwhile
thelowerlayercausesashearstressontheadjacenttoplayer.
•Thisshearstressisproportionaltotherateofchangeofvelocitywith
respecttoy.Itisdenotedbysymbolτ(tau).
τ∝
du
dy
τ=μ
du
dy
•Whereµistheconstantofproportionalityandisknownastheco-
efficientofdynamicviscosityoronlyviscosity.
du
dy
representstherateof
shearstrainorrateofsheardeformationorvelocitygradient.
•Fromtheaboveequation,wehaveμ=
τ
du
dy
•Thus,viscosityisalsodefinedastheshearstressrequiredproducing
unitrateofshearstrain.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
KINEMATIC VISCOSITY
•Itisdefinedastheratiobetweendynamicviscosityanddensityof
fluid.
•Itisdenotedbysymbol??????(nu)
??????=
viscosity
density
=
μ
ρ
•TheunitofviscosityinCGSiscalledpoiseandisequaltodyne-
see/cm
2
•Theunitofkinematicviscosityism
2
/sec
•Thusonestoke=cm
2
/sec=
1
100
2
m
2
/sec=10
-4
m
2
/sec

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
NEWTON’S LAW OF VISCOSITY
•Itstatesthattheshearstress(τ)onafluidelementlayerisdirectly
proportionaltotherateofshearstrain.
•Theconstantofproportionalityiscalledtheco-efficientofviscosity.
•It is expressed as:
•FluidswhichobeyaboverelationareknownasNEWTONIANfluids
andfluidswhichdonotobeytheaboverelationarecalledNON-
NEWTONIANfluids.
τ=μ
du
dy

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
TYPES OF FLUIDS
•The fluids may be classified in to
the following five types.
Ideal fluid
Real fluid
Newtonion fluid
Non-Newtonionfluid
Ideal plastic fluid

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
TYPES OF FLUIDS
•Idealfluid:Afluidwhichiscompressibleandishavingno
viscosityisknownasidealfluid.Itisonlyanimaginaryfluidasall
fluidshavesomeviscosity.
•Realfluid:Afluidpossessingaviscosityisknownasrealfluid.All
fluidsinactualpracticearerealfluids.
•Newtonianfluid:Arealfluid,inwhichthestressisdirectly
proportionaltotherateofshearstrain,isknownasNewtonian
fluid.
•Non-Newtonianfluid:Arealfluidinwhichshearstressisnot
proportionaltotherateofshearstrainisknownasNon-Newtonian
fluid.
•Idealplasticfluid:Afluid,inwhichshearstressismorethanthe
yieldvalueandshearstressisproportionaltotherateofshear
strainisknownasidealplasticfluid.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
SURFACE TENSION
•Surfacetensionisdefinedasthetensileforceactingonthesurface
ofaliquidiscontactwithagasoronthesurfacebehaveslikea
membraneundertension.
•Themagnitudeofthisforceperunitlengthoffreesurfacewillhave
thesamevalueasthesurfaceenergyperunitarea.
•Itisdenotedby??????(sigma).
•InMKSunitsitisexpressedasKgf/mwhileinSIunitsasN/m

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
SURFACE TENSION ON LIQUID DROPLET
•Considerasmallsphericaldroplet
ofaliquidofradius‘r’ontheentire
surfaceofthedroplet,thetensile
forceduetosurfacetensionwillbe
acting
•Letσ=surfacetensionofthe
liquid
p=pressureintensityinside
thedroplet(Inexcessofoutside
pressureintensity)
d=Diameterofdroplet
•Let,thedropletiscutintotwo
halves.Theforcesactingonone
half(saylefthalf)willbe
FORCES ON DROPLET

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
SURFACE TENSION ON LIQUID DROPLET
•Tensileforceduetosurfacetensionactingaroundthe
circumferenceofthecutportion=σ×circumference=σ×&#3627409163;d
•Pressureforceonthearea
??????
4
d
2
=p×
??????
4
d
2
•These two forces will be equal to and opposite under equilibrium
conditions i.e.
p ×
??????
4
d
2
= σ &#3627409163;d
p =
σ??????d
??????
4
d
2
p=
4??????
??????

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
SURFACE TENSION ON HOLLOW BUBBLE
•Ahallowsbubblelikesoapinairhastwosurfacesincontactwith
air,oneinsideandotheroutside.
•Thus,twosurfacesaresubjectedtosurfacetension.
??????×
??????
4
d
2
=2(σ&#3627409163;??????)
p=
8??????
??????

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
SURFACE TENSION ON LIQUID JET
•Consideraliquidjetofdiameter‘d’length‘L’
•Let,p=pressureintensityinsidetheliquidjet
abovetheoutsidepressure
σ=surfacetensionoftheliquid
•Considertheequilibriumofthesemi-jet.
•Forceduetopressure=p×areaofthesemi-
jet=p×L×d
•Forceduetosurfacetension=σ×2L
Equatingaboveforcesweget,
p×L×d=σ×2L
p=
2??????
??????

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 3
•Capillarity
•Expression for Capillary rise
•Expression for Capillary fall
•Vapor pressure
•Variation of vapor pressure with
temperature
Concept of Vapor pressure
& its variation
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
CAPILLARITY
•Capillarityisdefinedasaphenomenonofriseorfallofaliquid
surfaceinasmalltuberelativetotheadjacentgenerallevelof
liquidwhenthetubeisheldverticallyintheliquid.
•Theriseofliquidsurfaceisknownascapillaryrise,whilethefallof
theliquidsurfaceisknownascapillarydepression.
•Itisexpressedintermsof‘cm’or‘mm’ofliquid.
•Itsvaluedependsuponthespecificweightoftheliquid,diameterof
thetubeandsurfacetensionoftheliquid.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
EXPRESSION FOR CAPILLARY RISE
•Consideraglasstubeofsmall
diameter‘d’openedatbothends
andisinsertedinaliquid.
•Theliquidwillriseinthetubeabove
theleveloftheliquidoutsidethe
tube.
•Let‘h’betheheightoftheliquidin
thetube.
•Underastateofequilibrium,the
weightoftheliquidofheight‘h’is
balancedbytheforceatthesurface
oftheliquidinthetube.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
EXPRESSION FOR CAPILLARY RISE
•But,theforceatthesurfaceoftheliquidinthetubeisdueto
surfacetension.
•Letσ=surfacetensionofliquid
θ=Angleofcontactbetweentheliquidandglasstube
•Theweightoftheliquidofheight‘h’inthetube
=(areaofthetube×h)×&#3627409164;×g=
??????
4
d
2
×h×&#3627409164;×g
Where‘&#3627409164;′isthedensityoftheliquid.
•Theverticalcomponentofthesurfacetensileforce=(σ×
circumference)×cosθ=σ×&#3627409163;d×cosθ

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
EXPRESSION FOR CAPILLARY RISE
Forequilibrium,
??????
??????
d
2
×h×&#3627409164;×g=σ&#3627409163;dcosθ,
h=
σπdcosθ
??????
??????
d
2
×ρ×g
Thevalueofθisequalto‘0’betweenwaterandcleanglasstube,
thencosθ=1,then
h =
4σcosθ
ρg??????
h =
4σ
ρg??????

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
EXPRESSION FOR CAPILLARY FALL
•Iftheglasstubeisdippedinmercury,theLevelofmercuryinthe
tubewillbelowerthanthegeneralleveloftheoutsideliquid.
•Thevalueθofforglass&mercury128
0
h =
4σcosθ
ρg??????

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
VAPOURPRESSURE
•Achangefromtheliquidstatetothegaseousstateisknownas
Vaporizations.
•Thevaporization(whichdependsupontheprevailingpressureand
temperaturecondition)occursbecauseofcontinuousescapingof
themoleculesthroughthefreeliquidsurface.
•Consideraliquidatatemp.of20°Candpressureisatmosphericis
confinedinaclosedvessel.
•Thisliquidwillvaporizeat100°C,themoleculesescapefromthe
freesurfaceoftheliquidandgetaccumulatedinthespace
betweenthefreeliquidsurfaceandtopofthevessel.
•Theseaccumulatedvapoursexertapressureontheliquidsurface.
Thispressureisknownasvapourpressureoftheliquidor
pressureatwhichtheliquidisconvertedintovapours.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
VAPOURPRESSURE
•Considerthesameliquidat20°Catatmosphericpressureinthe
closedvesselandthepressureabovetheliquidsurfaceisreduced
bysomemeans;theboilingtemperaturewillalsoreduce.
•Ifthepressureisreducedtosuchanextentthatitbecomesequal
toorlessthanthevapourpressure,theboilingoftheliquidwill
start,thoughthetemperatureoftheliquidis20°C.
•Thus,theliquidmayboilattheordinarytemperature,ifthe
pressureabovetheliquidsurfaceisreducedsoastobeequalor
lessthanthevapourpressureoftheliquidatthattemperature.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
VARIATION OF VAPOURPRESSURE WITH
TEMPERATURE
•Thevapourpressureofaliquidvarieswithitstemperature.
•Asthetemperatureofaliquidorsolidincreasesitsvapour
pressurealsoincreases.
•Conversely, vapour pressure decreases as
thetemperaturedecreases.

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 4
•Pressure Measuring System
•Atmospheric, Gauge & Vaccum
pressure
•Measurement of pressure
•Piezometer
•U Tube manometer
Pressure Measuring System
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PRESSURE MEASURING SYSTEM
•Thepressureonafluidis
measureintwodifferent
systems.
•Inonesystem,itismeasured
abovetheabsolutezeroor
completevacuumanditis
calledtheAbsolutepressure.
•Inothersystem,pressureis
measured above the
atmosphericpressureandis
calledGaugepressure.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
DEFINITIONS
•ABSOLUTE PRESSURE:Itisdefinedasthepressurewhichis
measuredwithreferencetoabsolutevacuumpressure.
•GAUGEPRESSURE:Itisdefinedasthepressure,whichismeasured
withthehelpofapressuremeasuringinstrument,inwhichthe
atmosphericpressureistakenasdatum.Theatmosphericonthe
scaleismarkedaszero.
•VACUUM PRESSURE:Itisdefinedasthepressurebelowthe
atmosphericpressure
i)Absolutepressure=Atmosphericpressure+gaugepressure
p
ab=p
atm+p
guage
Vaccumpressure=atmosphericpressure-Absolutepressure
•Theatmosphericpressureatsealevelat15
0
Cis10.13N/cm
2
or
101.3KN/m
2
inSIUnitsand1.033Kgf/cm
2
inMKSSystem.
•Theatmosphericpressureheadis760mmofmercuryor10.33mof
water.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
MEASUREMENT OF PRESSURE
•Thepressureofafluidismeasuredbythefallowingdevices.
Manometers
Mechanicalgauges.
•Manometers:Manometersaredefinedasthedevicesusedfor
measuringthepressureatapointinafluidbybalancingthe
columnoffluidbythesameoranothercolumnoffluid.Theyare
classifiedas:
SimpleManometers
DifferentialManometers

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
MEASUREMENT OF PRESSURE
•MechanicalGauges:Thesearedefinedasthedevicesusedfor
measuringthepressurebybalancingthefluidcolumnbythespring
ordeadweight.
•ThecommonlyusedMechanicalpressuregaugesare:
Diaphragmpressuregauge
Bourdontubepressuregauge
Dead–Weightpressuregauge
Bellowspressuregauge.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
SIMPLE MANOMETER
•Asimplemanometerconsistsofaglasstubehavingoneofits
endsconnectedtoapointwherepressureistobemeasuredand
theotherendremainsopentotheatmosphere.
•Thecommontypesofsimplemanometersare:
Piezometer.
U-tubemanometer(forgauge&vaccumpressure)
Singlecolumnmanometer
VerticalSinglecolumnmanometer
InclinedSinglecolumnmanometer

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PIEZOMETER
•Itisasimplestformofmanometerused
formeasuringgaugepressure.
•Oneendofthismanometerisconnected
tothepointwherepressureistobe
measuredandotherendisopentothe
atmosphere.
•TheriseofliquidinthePiezometergives
pressureheadatthatpointA.
•Theheightofliquidsaywateris‘h’in
piezometertube,then
PressureatA=??????gh
??????
??????
&#3627409360;

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
U-TUBE MANOMETER
•Itconsistsofaglasstubebentinu-shape,oneendofwhichis
connectedtoapointatwhichpressureistobemeasuredand
otherendremainsopentotheatmosphere.
•Thetubegenerallycontainsmercuryoranyotherliquidwhose
specificgravityisgreaterthanthespecificgravityoftheliquid
whosepressureistobemeasured.
•ForGaugePressure:LetBisthepointatwhichpressureistobe
measured,whosevalueisp.ThedatumlineA–A

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
U-TUBE MANOMETER
GAUGE PRESSURE
•Let
h
1 = height of light liquid above
datum line
h
2=height of heavy liquid above
datum line
S
1= sp. gravity of light liquid
ρ
1= density of light liquid
= 1000 S
1
S
2= sp. gravity of heavy liquid
ρ
2= density of heavy liquid
= 1000 S
2

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
U-TUBE MANOMETER
•Asthepressureisthesameforthehorizontalsurface.Hencethe
pressureabovethehorizontaldatumlineA–Aintheleftcolumn
andtherightcolumnofU–tubemanometershouldbesame.
•PressureaboveA—Aiontheleftcolumn=p+&#3627409164;
1gh
1
•PressureaboveA–Aintheleftcolumn=
&#3627409164;
2
gh
2
•Henceequatingthetwopressuresp+&#3627409164;
1gh
1=
&#3627409164;
2
gh
2
p=??????
&#3627409360;gh
2-??????
&#3627409359;gh
1

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
U-TUBE MANOMETER
VACCUM PRESSURE
•Formeasuringvacuumpressure,
thelevelofheavyfluidinthe
manometerwillbeasshownin
fig.
•PressureaboveAAintheleft
column=&#3627409164;
2gh
2+&#3627409164;
1gh
1+P
•Pressureheadintheright
columnaboveAA=0
•Therefore,equating two
pressuresweget,
&#3627409164;
1gh
2+&#3627409164;
1gh
1+P=0
p=−(??????
&#3627409360;gh
2+??????
&#3627409359;gh
1)

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 5
•Single column Manometer
•Vertical single column manometer
•Inclined single column manometer
•Differential manometer
•U-Tube differential manometer
•Inverted U-Tube differential
manometer
Measurement of pressure-
Types of Manometers
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
SINGLE COLUMN MANOMETER
•SinglecolumnmanometerisamodifiedformofaU-tube
manometerinwhichareservoir,havingalargecrosssectional
area(about.100times)ascomparedtotheareaoftubeis
connectedtooneofthelimbs(sayleftlimb)ofthemanometer.
•Duetolargecrosssectionalareaofthereservoirforanyvariation
inpressure,thechangeintheliquidlevelinthereservoirwillbe
verysmallwhichmaybeneglectedandhencethepressureis
givenbytheheightoftheliquidintheotherlimb.
•Theotherlimbmaybeverticalorinclined.
•Thus,therearetwotypesofsinglecolumnmanometer
VerticalSingleColumnManometer
InclinedSingleColumnManometer

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
VERTICAL SINGLE COLUMN MANOMETER
•LetX–Xbethedatumlineinthe
reservoirandintherightlimbofthe
manometer,whenitisconnectedtothe
pipe,whentheManometeris
connectedtothepipe,duetohigh
pressureatA.
•Theheavypressureinthereservoirwill
bepusheddownwardsandwillrisein
therightlimb.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
DERIVATION
•Let,∆ h= fall of heavy liquid in the reservoir
h
2= rise of heavy liquid in the right limb
h
1= height of the centreof the pipe above X –X
p
A= Pressure at A, which is to be measured.
A = Cross-sectional area of the reservoir
a = cross sectional area of the right limb
S
1= Specific. Gravity of liquid in pipe
S
2= sp. Gravity of heavy liquid in the reservoir and right limb
&#3627409164;
1= density of liquid in pipe
&#3627409164;
2= density of liquid in reservoir

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Fallofheavyliquidreservoirwillcauseariseofheavyliquidlevel
intherightlimb
A×∆h=a×h
2
∆h=
??????×ℎ
2
??????
………………… .(1)
•NowconsiderthedatumlineY–Y
•ThenthepressureintherightlimbaboveY–Y=&#3627409164;
2×g×(∆h+h
2)
•PressureintheleftlimbaboveY—Y=&#3627409164;
1×g×(∆h+h
1)+P
A
•Equatingthepressures,wehave
&#3627409164;
2g×(∆h+h
2)=&#3627409164;
1×g×(∆h+h
1)+p
A
p
A=&#3627409164;
2×g×(∆h+h
2)-&#3627409164;
1×g×(∆h+h
1)
=∆h(&#3627409164;
2g&#3627409164;
1g)+h
2&#3627409164;
2g-h
1&#3627409164;
1g

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•But,fromeq.(1) ∆h=
??????×ℎ
2
??????
•P
A=
??????×ℎ
2
??????
(&#3627409164;
2g-&#3627409164;
1g)+h
2&#3627409164;
2g-h
1&#3627409164;
1g
•AstheareaAisverylargeascomparedtoa,hencetheratio
a
A
becomesverysmallandcanbeneglectedThen,
……………… (2)
p
A=h
2??????
&#3627409360;g-h
1??????
&#3627409359;g

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
INCLINED SINGLE COLUMN MANOMETER
•Themanometerismoresensitive.
•Duetoinclinationthedistancemovedby
heavyliquidintherightlimbwillbemore.
•LetL=lengthofheavyliquidmovedin
theritelimb
θ=inclinationofrightLimbwith
horizontal.
H
2=verticalriseofheavyliquidin
therightlimbaboveX–X=Lsinθ
•Fromaboveeq.(2),thepressureatAis
p
A=h
2&#3627409164;
2g-h
1&#3627409164;
1g
•Substitutingthevalueofh
2
p
A=Lsinθ??????
&#3627409360;g-h
1??????
&#3627409359;g

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
DIFFERENTIAL MANOMETERS
•Differentialmanometersarethedevicesusedformeasuringthe
differenceofpressurebetweentwopointsinapipeorintwo
differentpipes.
•AdifferentialmanometerconsistsofaU-tube,containingheavy
liquid,whosetwoendsareconnectedtothepoints,whose
differenceofpressureistobemeasured.
•ThecommontypesofU-tubedifferentialmanometersare:
U-Tubedifferentialmanometer
InvertedU-tubedifferentialmanometer

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
U-TUBE DIFFERENTIAL MANOMETER
TWO POINTS A AND B ARE AT DIFFERENT LEVELS AND
ALSO CONTAINS LIQUIDS OF DIFFERENT SP.GR.
•ThesepointsareconnectedtotheU–
Tubedifferentialmanometer.
•LetthepressureatAandBarep
Aandp
B
•Leth=Differenceofmercurylevelsin
theu–tube
y=DistanceofcentreofBfromthe
mercurylevelintherightlimb
x=DistanceofcentreofAfromthe
mercurylevelintheleftlimb
&#3627409164;
1=DensityofliquidA
&#3627409164;
2=DensityofliquidB
&#3627409164;
??????=Densityofheavyliquidormercury

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Taking datum line at X –X
•Pressure above X –X in the left limb =&#3627409164;
1g(h + x) + p
A
(where p
A= Pressure at A)
•Pressure above X –X in the right limb = &#3627409164;
??????g h + &#3627409164;
2g y + P
B
(where p
B= Pressure at B)
•Equating the above two pressures, we have
&#3627409164;
1g(h + x) + p
A= &#3627409164;
??????g h + &#3627409164;
2g y + p
B
p
A-p
B= &#3627409164;
??????g h + &#3627409164;
2g y -&#3627409164;
1g ( h + x)
= h g (&#3627409164;
??????-&#3627409164;
1) + &#3627409164;
2g y -&#3627409164;
1g x
•Therefore, Difference of Pressures at A and B is
h g (&#3627409164;
??????-ℓ
1) + &#3627409164;
2g y -&#3627409164;
1g x

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
U-TUBE DIFFERENTIAL MANOMETER
TWO POINTS A AND B ARE AT SAME LEVELS AND ALSO
CONTAINS SAME LIQUIDS OF SP.GR.
•ThenpressureaboveX–Xintheright
limb=&#3627409164;
??????gh+&#3627409164;
1gx+P
B
•PressureaboveX—Xintheleftlimb=
&#3627409164;
1g(h+x)+P
A
•Equatingthetwopressures
&#3627409164;
??????gh+&#3627409164;
1gx+p
B=&#3627409164;
1g(h+x)+p
A
P
A-P
B=&#3627409164;
??????gh+&#3627409164;
1gx-&#3627409164;
1g(h+x)
=gh(&#3627409164;
??????-&#3627409164;
1)
•Therefore,DifferenceofpressureatA
andB=gh(&#3627409164;
??????-&#3627409164;
1)

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 6
•Applications
•Problems on Viscosity
•Problems on manometers
Applications & Problems
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
APPLICATIONS
•Specificpressuremonitoringapplications
•Visualmonitoringofairandgaspressureforcompressors.
•Vacuumequipmentandspecialtytankapplicationssuchas
medicalgascylinders,fireextinguishers.
•Inpowerplants,mercuryabsolutemanometerhavebeenusedto
checkcondenserefficiencybymonitoringvacuumatseveralpoints
ofthecondenser
•Usedfortheresearchofatmosphereofotherplanets.
•Andmanymoreapplicationssuchasinwhetherstudies,research
labs,gasanalysisandinmedicalequipment's.
•Neveroperatedamageequipment.
•Meteranditstubingshouldbefreefromanybreakingand
blockage.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Inaddition,sincethetubesinmanymanometersaremadeofglass
andcanbeeasilybroken,itisimportanttousecareinhandling
thesemanometers.
•Electronicmanometersdonotmeasurewaterpressures;under
theseconditionstheywillfail.Donotexceed10PSIinput
pressure.
•Sometypesofliquidsusedinmanometersaretoxicandcanbe
damagingtotheenvironment.Therefore,whenusingmanometers
tomeasureorindicatepressure,donotconnectanymanometerto
apressurethathasthepotentialtoexceedtherangeofthe
manometer.Thiscouldcausetheliquidtobeforcedoutofthetube

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM-1
•Twohorizontalplatesareplaced1.25cmapart,thespacebetweenthem
beingfilledwithoilofviscosity14poises.Calculateshearstressinoil,if
theupperplateismovedvelocityof2.5m/sec.
Sol:GivenData
Distancebetweentheplates,dy=1.25cm=0.0125m
Viscosity,µ=14poise=
14
10
Ns/m
2
Velocityofupperplate,u=2.5m/sec
Shearstress,τ=µ
du
dy
Wheredu=changeofvelocitybetweenplates=u–0
=u=2.5m/sec
τ=
14
10
×
2.5
0.0125
=280N/m
2

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM-2
•Thedynamicviscosityofoilusedforlubricationbetweenashaftand
sleeveis6poise.Theshaftdia.is0.4mandrotatesat190rpm.Calculate
thepowerlostinthebearingforasleevelengthof90mm.Thethickness
foilfilmis1.5mm
Sol:GivenData,
Viscosity,µ=6poise=
6
10
Ns
m
2
=0.6
Ns
m
2
Dia.ofshaft,D=0.4M
Speedofshaft,N=190rpm
Sleevelength,L=90mm=90
Thicknessofafilm,t=1.5mm
=1.5×10
-3
m

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
Tangentialvelocityofshaft=u=
πDN
60
=
π×0.4×190
60
=3.98m/sec
Usingtherelationτ=µ
du
dy
Wheredu=changeofvelocity=u–0=u=3.98m/sec
dy=changeofdistance=t=1.5
Then,Shearstressontheshaft
τ=0.6×
3.98
1.5×10
−3
=1592N/m
2
We know that, Shear force on the shaft, F = shear stress ×Area
=1592 ×πDL= 180.05 N
Torque on the shaft, T = Force×
D
2
= 36.01 Nm
Therefore, Power lost, P =
2πNT
60
=716.48 W

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM 3

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM 4

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEMS ON SURFACE TENSION

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM-6

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM 7

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM-8
•TherightlimbofasimpleU–tubemanometercontainingmercuryisopen
totheatmosphere,whiletheleftlimbisconnectedtoapipeinwhicha
fluidofsp.gr.0.9isflowing.Thecentreofpipeis12cmbelowthelevelof
mercuryintherightlimb.Findthepressureoffluidinthepipe,ifthe
differenceofmercurylevelinthetwolimbsis20cm.
Sol:Givendata,
Sp.gr.ofliquid,S
1=0.9
Densityoffluid,&#3627409164;
1=S
1×1000=0.9×1000=900kg/m
3
Sp.gr. of mercury, S
2= 13.6
Density of mercury, &#3627409164;
2=13.6×1000 = 13600 kg/m
3
Difference of mercury level, h
2= 20cm = 0.2m
Height of the fluid from A –A, h
1 = 20 –12 = 8cm = 0.08 m

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
Let ‘P’ be the pressure of fluid in pipe
Equating pressure at A –A, we get
p + &#3627409164;
1gh
1= &#3627409164;
2gh
2
p + 900 ×9.81 ×0.08 = 13.6 ×1000 ×9.81 ×0.2
p = 13.6 ×1000 ×9.81 ×0.2 –900 ×9.81 ×0.08
p = 26683 –706
p = 25977 N/m
2
p = 2.597 N/cm
2
Therefore, Pressure of fluid P=2.597 N/ cm
2

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM-9
•AsimpleU–tubemanometercontainingmercuryisconnectedtoapipein
whichafluidofsp.gr.0.8Andhavingvacuumpressureisflowing.The
otherendofthemanometerisopentoatmosphere.Findthevacuum
pressureinpipe,ifthedifferenceofmercurylevelinthetwolimbsis40cm.
andtheheightofthefluidinthelefttubefromthecentreofpipeis15cm
below.
Sol:Givendata,
Sp.gr of fluid, S
1 = 0.8
Sp.gr. of mercury, S
2= 13.6
Density of the fluid &#3627409164;
1= S
1 ×1000 = 0.8 ×1000
=800kg/m
3
Density of mercury &#3627409164;
2= 13.6 ×1000
Difference of mercury level h
2= 40cm = 0.4m
Height of the liquid in the left limb = 15cm =0.15m

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
Let the pressure in the pipe = p
Equating pressures above datum line A–A
&#3627409164;
2gh
2 + &#3627409164;
1gh
1 + P = 0
P = -[&#3627409164;
2gh
2 + &#3627409164;
1gh
1]
= -[13.6 ×1000 ×9.81 ×0.4 + 800 ×9.81 ×0.15]
= 53366.4 + 1177.2
= -54543.6 N/m
2
Therefore, the vacuum pressure in pipe, P = -5.454 N/cm
2

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM-10
•Asinglecolumnmanometerisconnectedtothepipecontainingliquidof
sp.gr.0.9.Findthepressureinthepipeiftheareaofthereservoiris100
timestheareaofthetubeofmanometer.sp.gr.ofmercuryis13.6.Height
oftheliquidfromthecentreofpipeis20cmanddifferenceinlevelof
mercuryis40cm.
Sol:Givendata,
Sp.gr.ofliquidinpipe,S
1=0.9
Density, &#3627409164;
1= 900 kg/ m
3
Sp.gr. of heavy liquid, S
2 = 13.6
Density, &#3627409164;
2= 13600
Areaofreservoir
Areaofrightlimb
=
A
a
= 100
Height of the liquid, h
1 = 20cm = 0.2m
Rise of mercury in the right limb, h
2= 40cm = 0.4m

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
Pressure in pipe A
p
A=
A
a
×h
2[&#3627409164;
2g -&#3627409164;
1g]
+ h
2&#3627409164;
2g -h
1&#3627409164;
1g
=
1
100
×0.4 [13600 ×9.81 –900 x 9.81] + 0.4 ×13600 ×
9.81 –0.2 ×900 ×9.81
=
0.4
100
[133416 –8829] + 53366.4 –1765.8
= 533.664 + 53366.4 –1765.8
= 52134 N/m
Therefore, Pressure in pipe A=5.21 N/ cm
2

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM-11
•Apipecontainsanoilofsp.gr.0.9.Adifferentialmanometerisconnectedat
thetwopointsAandBshowsadifferenceinmercurylevelat15cm.find
thedifferenceofpressureatthetwopoints.
Sol:Givendata,
Sp.gr.ofoilS
1=0.9:density&#3627409164;
1=0.9x1000=900kg/m
3
Differenceoflevelinthemercuryh=15cm=0.15m
Sp.gr.ofmercury=13.6,Density=13.6×1000=13600kg/m
3
Thedifferenceofpressurep
A–p
B=g×h×(&#3627409164;
??????-&#3627409164;
1)
=9.81x0.15(13600–900)
=18688N/m
2
Therefore,p
A–p
B= 18688 N/ m
2

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM-12
•AdifferentialmanometerisconnectedattwopointsAandB.AtBair
pressureis9.81N/cm
2
.FindabsolutepressureatA.
Sol:Givendata,
Density of air = 0.9 ×1000 = 900 kg/m
3
Density of mercury = 13.6 ×10
3
kg/ m
3
Pressure at B= 9.81 N/cm
2
= 98100 N/m
2
Let pressure at A is p
A
Taking datum as X –X
Pressure above X –X in the right limb
= 1000 ×9.81 ×0.6 + p
B= 5886 + 98100 = 103986
Pressure above X –X in the left limb
= 13.6 ×10
3
×9.81 ×0.1 +0900 ×9.81 ×0.2 + p
A
= 13341.6 +1765.8 +p
A

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
Equatingthetwopressuresheads,weget
103986=13341.6+1765.8+p
A
103986=15107.4+p
A
p
A=103986–15107.4
=88878.6N/m
2
Therefore,PressureatA,p
A= 8.887 N/cm
2

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM-8
•Waterisflowingthroughtwodifferentpipestowhichaninverted
differentialmanometerhavinganoilofsp.gr.0.8isconnected.The
pressureheadinthepipeAis2mofwater.FindthepressureinthepipeB
forthemanometerreadingsshowninfig.
Sol:Givendata,
Pressure head at A =
pA
ρg
= 2m of water
p
A= ρ×g ×2 = 1000 ×9.81 ×2 = 19620 N/m
2
Pressure below X –X in the left limb
= p
A-&#3627409164;
1gh
1
= 19620 –1000 ×9.81 ×0.3 = 16677 N/m
2

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
Pressure below X –X in the right limb
= p
B–1000 ×9.81 ×0.1 –800 ×9.81 ×0.12
= p
B–981 –941.76 = p
B–1922.76
Equating the two pressures, we get,
16677 = p
B-1922.76
p
B= 16677 + 1922.76
p
B= 18599.76 N/m
2
Therefore, Pressure at B, p
B= 1.859 N/cm
2

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
ASSIGNMENT QUESTIONS
•Twolargeplanesareparalleltoeachotherandareinclinedat30°tothe
horizontalwiththespacebetweenthemfilledwithafluidofviscosity20
cp.Asmallthinplateof0.125msquareslidesparallelandmidway
betweentheplanesandreachesaconstantvelocityof2m/s.Theweight
oftheplateis1N.Determinethedistancebetweentheplates.
•AU-tubemercurymanometerisusedtomeasurethepressureofoil
flowingthroughapipewhosespecificgravityis0.85.Thecenterofthe
pipeis15cmbelowthelevelofmercury.Themercuryleveldifferencein
themanometeris25cm,determinetheabsolutepressureoftheoil
flowingthroughthepipe.Atmosphericpressureis750mmofHg.
•Asinglecolumnverticalmanometerisconnectedtoapipecontainingoilof
specificgravity0.9.Theareaofthereservoiris80timestheareaofthe
manometertube.Thereservoircontainsmercuryofsp.gr.13.6.Thelevel
ofmercuryinthereservoirisataheightof30cmbelowthecenterofthe
pipeanddifferenceofmercurylevelsinthereservoirintherightlimbis50
cm.findthepressureinthepipe.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Findtheheightthroughwhichwaterrisesbycapillaryactioninaglass
tubeof2mmboreifthesurfacetensionatetheprevailingtemperatureis
0.075N/m.
•Thespacebetweentwoparallelsquareplateseachofside0.8misfilled
withanoilofspecificgravity0.8.Ifthespacebetweentheplatesis
12.5mmandtheupperplatewhichmoveswithvelocityof1.25m/s
requiresaforceof51.2N.Determine(i)Dynamicviscosityofoilinpoise
(ii)Kinematicviscosityinstokes.
•ListallthepropertiesoffluidandderiveNewton’slawofviscosity.
•Explainatmospheric,gaugeandvaccumpressurewiththehelpofaneat
sketch.

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
UNIT 2
CO2:To gain knowledge on fluid
kinematics anddynamics .
FLUID KINEMATICS
& DYNAMICS

DEPARTMENT OF MEC HANIC AL ENGINEERING
UNIT –II
Fluid Kinematics:
•Stream Line, Path line, streak lines and stream tubes.
•Classification of flows-steady & unsteady, uniform &
non-uniform, laminar & turbulent, rotational &
irrotationalflows.
•Equation of continuity-1D flow.
Fluid Dynamics:
•Surface & body forces-Euler’s and Bernouli’s equation
for flow along a stream line.
•Momentum equation & its application on force on pipe
bend.

DEPARTMENT OF MEC HANIC AL ENGINEERING
COURSEOUTLINE
LECTURE LECTURETOPIC KEYELEMENTS Learningobjectives
1 Introduction Stream line, stream
tube
Understanding the
basics (B2)
2 Classification of flows Uniform & Non-
uniform
Laminar & Turbulent
Understanding
different types of
flows (B2)
3 Equation of Continuity 1 D Flow •Analyze equation
of continuity (B4)
•Apply 1D flow
equation (B3)
4 Surface & body forces Euler’s Equation Evaluate Velocity in a
flow (B5)
5 Bernouli’s Equation of motion For a stream lineEvaluate velocity&
Pressure (B5)
6 Momentum equation Application on pipe
bend
Evaluate force exerted
on pipe bend (B5)
7 Example Problems on Bernouli’s & Momentum Equations.
UNIT-2

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 1
•Introduction to Fluid Kinematics
•Stream line
•Path line
•Streak Line
•Stream tube
Classification of flows
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
FLUID KINEMATICS
•Kinematicsisdefinedasabranchofsciencewhichdealswith
motionofparticleswithoutconsideringtheforcescausingthe
motion.
•Thevelocityatanypointinaflowfieldatanytimeisstudiedinthis.
•Oncethevelocityisknown,thenthepressuredistributionand
hencetheforcesactingonthefluidcanbedetermined.

DEPARTMENT OF MEC HANIC AL ENGINEERING
STREAM LINE
•Astreamlineisanimaginarylinedrawnina
flowfieldsuchthatthetangentdrawnatany
pointonthislinerepresentsthedirectionof
velocityvector.
•Fromthedefinitionitisclearthattherecan
benoflowacrossstreamline.
•Consideringaparticlemovingalonga
streamlineforaveryshortdistance‘ds’
havingitscomponentsdx,dyanddz,along
threemutuallyperpendicularco-ordinate
axes.
•LetthecomponentsofvelocityvectorVs
alongx,yandzdirectionsbeu,vandw
respectively.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Thetimetakenbythefluidparticletomoveadistance‘ds’along
thestreamlinewithavelocityVsis:
&#3627408481;=
ds
Vs
whichissameas&#3627408481;=
dx
u
=
dy
v
=
ds
w
•Hencethedifferentialequationofthesteamlinemaybewritten
as:
dx
u
=
dy
v
=
ds
w

DEPARTMENT OF MEC HANIC AL ENGINEERING
PATH LINE
•Apathlineislocusofafluidparticle
asitmovesalong.
•Inotherwordsapathlineisacurve
tracedbyasinglefluidparticle
duringitsmotion.
•Astreamlineattimet
1indicatingthe
velocityvectorsforparticlesAand
B.
•Attimest
2andt
3theparticleA
occupiesthesuccessivepositions.
•Thelinecontainingthesevarious
positionsofArepresentsitsPath
line

DEPARTMENT OF MEC HANIC AL ENGINEERING
STREAK LINE
•Whenadyeisinjectedinaliquidorsmokeinagas,soastotrace
thesubsequentmotionoffluidparticlespassingafixedpoint,the
pathfallowedbydyeorsmokeiscalledthestreakline.
•Thusthestreaklineconnectsallparticlespassingthroughagiven
point.
•Insteadyflow,thestreamlineremainsfixedwithrespecttoco-
ordinateaxes.
•Streamlinesinsteadyflowalsorepresentthepathlinesandstreak
lines.
•Inunsteadyflow,afluidparticlewillnot,ingeneral,remainonthe
samestreamline(exceptforunsteadyuniformflow).
•Hencethestreamlinesandpathlinesdonotcoincideinunsteady
non-uniformflow.

DEPARTMENT OF MEC HANIC AL ENGINEERING
STREAM TUBE
•Ifstreamlinesaredrawnthroughaclosed
curve,theyformaboundarysurfaceacross
whichfluidcannotpenetrate.
•Suchasurfaceboundedbystreamlinesis
knownasStreamtube.
•Fromthedefinitionofstreamtube,itisevident
thatnofluidcancrosstheboundingsurfaceof
thestreamtube.
•Thisimpliesthatthequantityoffluidentering
thestreamtubeatoneendmustbethesame
asthequantityleavingattheotherend.
•TheStreamtubeisassumedtobeasmall
cross-sectionalarea,sothatthevelocityoverit
couldbeconsidereduniform.

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 2
•Classification of flows
•Steady and unsteady flows
•Uniform & non-uniform flow
•Laminar & turbulent flow
•Compressible & incompressible flow
•Rotational & irrotationalflow
•One dimensional flow
•Two & three dinemsionalflow
•Rate of flow
Classification of flows
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
CLASSIFICATION OF FLOWS
•Thefluidflowisclassifiedas:
Steadyandunsteadyflows.
UniformandNon-uniformflows.
LaminarandTurbulentflows.
Compressibleandincompressibleflows.
RotationalandIrrotationalflows.
One,twoandthreedimensionalflows.

DEPARTMENT OF MEC HANIC AL ENGINEERING
STEADY & UNSTEADY FLOW
•Steadyflowisdefinedastheflowinwhichthefluidcharacteristics
likevelocity,pressure,densityetc.atapointdonotchangewith
time.
•Thusforasteadyflow,wehave
??????V
??????t
x,y,z
=0,
??????p
??????t
x,y,z
=0,
??????ρ
??????t
x,y,z
=0
•Un-Steadyflowistheflowinwhichthevelocity,pressure,density
atapointchangeswithrespecttotime.
•Thusforun-steadyflow,wehave
??????V
??????t
x,y,z
≠0,
??????p
??????t
x,y,z
≠0,
??????ρ
??????t
x,y,z
≠0

DEPARTMENT OF MEC HANIC AL ENGINEERING
UNIFORM & NON -UNIFORM FLOW
•Uniformflowisdefinedastheflowinwhichthevelocityatany
giventimedoesnotchangewithrespecttospace.(i.e.thelength
ofdirectionofflow)
•Foruniformflow
??????V
??????s
t=const
=0
Where,????????????=Changeofvelocity
??????s=Lengthofflowinthedirectionof–S
•Non-uniformistheflowinwhichthevelocityatanygiventime
changeswithrespecttospace.
•ForNon-uniformflow
??????V
??????s
t=const
≠0

DEPARTMENT OF MEC HANIC AL ENGINEERING
LAMINAR & TURBULENT FLOW
•Laminarflowisdefinedastheflowinwhichthefluidparticlesmove
alongwell-definedpathsorstreamlineandallthestreamlinesare
straightandparallel.
•Thustheparticlesmoveinlaminasorlayersglidingsmoothlyover
theadjacentlayer.Thistypeofflowisalsocalledstreamlineflowor
viscousflow.
•Turbulentflowistheflowinwhichthefluidparticlesmoveina
zigzagway.
•Duetothemovementoffluidparticlesinazigzagway,theeddies
formationtakesplace,whichareresponsibleforhighenergyloss.

DEPARTMENT OF MEC HANIC AL ENGINEERING
LAMINAR & TURBULENT FLOW
•Forapipeflow,thetypeofflowisdeterminedbyanon-
dimensionalnumber
VD
??????
calledtheReynoldsnumber.
WhereD=Diameterofpipe.
V=Meanvelocityofflowinpipe.
??????=Kinematicviscosityoffluid.
•IftheReynoldsnumberislessthan2000,theflowiscalledLaminar
flow.
•IftheReynoldsnumberismorethan4000,itiscalledTurbulent
flow.
•IftheReynoldsnumberisbetween2000and4000theflowmaybe
LaminarorTurbulentflow.

DEPARTMENT OF MEC HANIC AL ENGINEERING
COMPRESSIBLE & INCOMPRESSIBLE FLOW
•Compressibleflowistheflowinwhichthedensityoffluidchanges
frompointtopointorinotherwordsthedensityisnotconstantfor
thefluid.
•Forcompressibleflow,ρ≠Constant.
•Incompressibleflowistheflowinwhichthedensityisconstantfor
thefluidflow.
•Liquidsaregenerallyincompressible,whilethegasesare
compressible.
•Forincompressibleflow,ρ=Constant

DEPARTMENT OF MEC HANIC AL ENGINEERING
ROTATIONAL & IRROTATIONAL FLOW
•Rotationalflowisatypeofflowinwhichthefluidparticleswhile
flowingalongstreamlinesalsorotateabouttheirownaxis.
•Andifthefluidparticles,whileflowingalongstreamlines,donot
rotateabouttheirownaxis,theflowiscalledIrrotationalflow.

DEPARTMENT OF MEC HANIC AL ENGINEERING
ONE DIMENSION FLOW
•1Dflowisatypeofflowinwhichflowparametersuchasvelocity
isafunctionoftimeandonespaceco-ordinateonly,say‘x’.
•Forasteadyone-dimensionalflow,thevelocityisafunctionofone
spaceco-ordinateonly.
•Thevariationofvelocitiesinothertwomutuallyperpendicular
directionsisassumednegligible.
•Henceforonedimensionalflowu=f(x),v=0andw=0
Whereu,vandwarevelocitycomponentsinx,yandz
directionsrespectively.

DEPARTMENT OF MEC HANIC AL ENGINEERING
TWO DIMENSION FLOW
•2Dflowisthetypeofflowinwhichthevelocityisafunctionoftime
andtwospaceco-ordinates,sayxandy.
•Forasteadytwo-dimensionalflowthevelocityisafunctionoftwo
spaceco-ordinatesonly.
•Thevariationofvelocityinthethirddirectionisnegligible.
•Thusfortwodimensionalflowu=f
1(x,y),v=f
2(x,y)andw=0.
•3Dflowisthetypeofflowinwhichthevelocityisafunctionoftime
andthreemutuallyperpendiculardirections.
•Butforasteadythree-dimensionalflow,thefluidparametersare
functionsofthreespaceco-ordinates(x,y,andz)only.
•Thusforthree-dimensionalflowu=f
1(x,y,z),v=f
2(x,y,z),z
=f
3(x,y,z).

DEPARTMENT OF MEC HANIC AL ENGINEERING
RATE OF FLOW OR DISCHARGE (Q)
•Itisdefinedasthequantityofafluidflowingpersecondthrougha
sectionofpipeorchannel.
•Foranincompressiblefluid(orliquid)therateofflowordischarge
isexpressedasthevolumeoftheliquidflowingcrossthesection
persecond.orcompressiblefluids.
•Therateofflowisusuallyexpressedastheweightoffluidflowing
acrossthesection.
•Thus(i)ForliquidstheunitofQism
3
/secorLitres/sec.
(ii)ForgasestheunitofQisKgf/secorNewton/sec.
•ThedischargeQ=A×V
Where,A=Areaofcross-sectionofpipe.
V=Averagevelocityoffluidacrossthesection.

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 3
•Continuity equation
•Continuity equation for 3D flow
•Example Problem
Equation of Continuity
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
EQUATION OF CONTINUITY
•Theequationbasedontheprinciple
ofconservationofmassiscalled
Continuityequation.
•Thusforafluidflowingthroughthe
pipeatallcross-sections,the
quantityoffluidpersecondis
constant.
•Considertwocross-sectionsofa
pipe.
•LetV
1=Averagevelocityatcross-
section1-1
ρ
1=Densityoffluidatsection1-1
A
1=Areaofpipeatsection1-1
AndV
2,ρ
2,A
2arethecorresponding
valuesatsection2-2

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Thentherateflowatsection1-1=ρ
1A
1V
1
•Rateofflowatsection2-2=ρ
2A
2V
2
•Accordingtolawofconservationofmass,Rateofflowatsection1-
1=Rateofflowatsection2-2
ρ
1A
1V
1=ρ
2A
2V
2
•Thisequationisapplicabletothecompressibleaswellas
incompressiblefluidsandiscalled“Continuityequation”.
•Ifthefluidisincompressible,thenρ
1=ρ
2andthecontinuityequation
reducesto
A
1V
1= A
2V
2

DEPARTMENT OF MEC HANIC AL ENGINEERING
EQUATION OF CONTINUITY FOR 3D FLOW
•Considerafluidelementoflengths
dx,dyanddzinthedirectionofx,y
andz.
•Letu,vandwaretheinletvelocity
componentsinx,yandzdirections
respectively.
•Massoffluidenteringtheface
ABCDpersecond=ρ×velocityin
x–direction×AreaofABCD
=ρ×u×(dy×dz)

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Then the mass of fluid leaving the face EFGH per second
= ρ ×u ×(dy×dz)+
??????
??????&#3627408485;
ρudydzdx
•Gain of mass in x-direction = Mass through ABCD –Mass through
EFGH per second= ρ u dydz-ρu dydz-
??????
??????&#3627408485;
ρudydzdx
= -
??????
??????&#3627408485;
ρudydzdx
= -
??????
??????&#3627408485;
ρudxdydz___________ (1)
•Similarly gain of mass in y-direction
= −
??????
??????&#3627408486;
&#3627409164;??????dx dydz_________(2)
•Similarly gain of mass in z-direction
= −
??????
??????z
(ρw)dx dydz__________(3)

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Netgainofmass=−
??????
??????x
ρu+
??????
??????y
ρv+
??????
??????z
ρwdxdydz(4)
•Sincemassisneithercreatednordestroyedinthefluidelement,
thenetincreaseofmassperunittimeinthefluidelementmustbe
equaltotherateofincreaseofmassoffluidintheelement.
•Butthemassoffluidintheelementisρdxdydzanditsrateof
increasewithtimeis
??????
??????&#3627408481;
(ρ.dx.dy.dz.)or
??????ρ
??????&#3627408481;
.dx.dy.dz.(5)
•Equatingthetwoexpressions(4)&(5)
-(
??????
??????&#3627408485;
ρu+
??????
??????&#3627408486;
ρv+
??????
??????&#3627408487;
ρw)dxdydz=
??????ρ
??????&#3627408481;
.dx.dy.dz.
_____(6)????????????
??????&#3627408533;
+
??????
??????&#3627408537;
??????&#3627408430;+
??????
??????&#3627408538;
??????&#3627408431;+
??????
??????&#3627408539;
??????&#3627408432;= o

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Thisequationisapplicableto
 Steadyandunsteadyflow
 Uniformandnon-uniformflow,and
 Compressibleandincompressibleflow.
•Forsteadyflow
????????????
??????&#3627408533;
=0andhenceequation(6)becomes
_____(7)
•Ifthefluidisincompressible,thenρisconstantandtheabove
equationbecomes
_____ (8)
•This is the continuity equation in three -dimensional flow.
??????
??????&#3627408537;
??????&#3627408430;+
??????
??????&#3627408538;
??????&#3627408431;+
??????
??????&#3627408539;
??????&#3627408432;= 0
??????&#3627408534;
??????&#3627408537;
+
??????&#3627408535;
??????&#3627408538;
+
??????&#3627408536;
??????&#3627408539;
= o

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 1
•Thediameterofapipeatsections1and2are10cmand15cms
respectively.Findthedischargethroughpipe,ifthevelocityofwater
flowingthroughthepipeatsection1is5m/sec.determinethevelocity
atsection2.
Sol:Givendata
At section 1, D
1= 10cms = 0.1m
&#3627408436;
1=
&#3627409163;
4
&#3627408439;
1
2
=
&#3627409163;
4
0.1
2
=0.007254&#3627408474;
2
V
1= 5m/sec
At section 2, D
2= 15cms =0.15
&#3627408436;
2=
&#3627409163;
4
0.15
2
=0.01767&#3627408474;
2
Discharge through pipe, Q = A
1×V
1= 0.007854 ×5 = 0.03927 m
3
/sec
We have, A
1V
1= A
2V
2
??????
2=
&#3627408436;
1??????
1
&#3627408436;
2
=
0.007854
0.01767
×5.0=2.22&#3627408474;/&#3627408480;

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 4
•Fluid Dynamics
•Surface, Line & Body forces
•Forces acting on fluid in motion
•Euler’s equation of motion
Fluid Dynamics-Surface &
Body forces
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
FLUID DYNAMICS
•Afluidinmotionissubjectedtoseveralforces,whichresultsinthe
variationoftheaccelerationandtheenergiesinvolvedintheflow
ofthefluid.
•Thestudyoftheforcesandenergiesthatareinvolvedinthefluid
flowisknownasDynamicsoffluidflow.
•Thevariousforcesactingonafluidmassmaybeclassifiedas:
Bodyorvolumeforces
Surfaceforces
Lineforces.

DEPARTMENT OF MEC HANIC AL ENGINEERING
FORCES
•Bodyforces:Thebodyforcesaretheforceswhichare
proportionaltothevolumeofthebody.
Examples:Weight,Centrifugalforce,magneticforce,
Electromotiveforceetc.
•Surfaceforces:Thesurfaceforcesaretheforceswhichare
proportionaltothesurfaceareawhichmayincludepressureforce,
shearortangentialforce,forceofcompressibilityandforcedueto
turbulenceetc.
•Lineforces:Thelineforcesaretheforceswhichareproportional
tothelength.
Example:surfacetension.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•ThedynamicsoffluidflowisgovernedbyNewton’ssecondlawof
motionwhichstatesthattheresultantforceonanyfluidelement
mustbeequaltotheproductofthemassandaccelerationofthe
elementandtheaccelerationvectorhasthedirectionofthe
resultantvector.
•Thefluidisassumedtobeincompressibleandnon-viscous.
∑F
x=M.a
Where∑Frepresentstheresultantexternalforceactingonthe
fluidelementofmassMandaistotalacceleration.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Boththeaccelerationandtheresultantexternalforcemustbe
alongsamelineofaction.
•Theforceandaccelerationvectorscanberesolvedalongthethree
referencedirectionsx,yandzandthecorrespondingequations
maybeexpressedas;
∑F
x=M.a
x
∑F
y=M.a
y
∑F
z=M.a
z
Where∑F
x,∑F
yand∑F
zarethecomponentsofthe
resultantforceinthex,yandzdirectionsrespectively,anda
x,a
y
anda
zarethecomponentsofthetotalaccelerationinx,yandz
directionsrespectively.

DEPARTMENT OF MEC HANIC AL ENGINEERING
FORCES ACTING ON FLUID IN MOTION
•Thevariousforcesthatinfluencethemotionoffluidaredueto
gravity,pressure,viscosity,turbulenceandcompressibility.
•Thegravityforce‘Fg’isduetotheweightofthefluidandisequal
toMg.Thegravityforceperunitvolumeisequalto“ρg”.
•Thepressureforce‘Fp’isexertedonthefluidmass,ifthereexists
apressuregradientbetweenthetwopointsinthedirectionofthe
flow.
•Theviscousforce‘Fv’isduetotheviscosityoftheflowingfluidand
thusexistsincaseofallrealfluids.
•Theturbulentflow‘Ft’isduetotheturbulenceofthefluidflow.
•Thecompressibilityforce‘Fc’isduetotheelasticpropertyofthe
fluidanditisimportantonlyforcompressiblefluids.

DEPARTMENT OF MEC HANIC AL ENGINEERING
FORCES ACTING ON FLUID IN MOTION
•Ifacertainmassoffluidinmotionisinfluencedbyalltheabove
forces,thenaccordingtoNewton’ssecondlawofmotion
•ThenetforceF
x=M.a
x=(F
g)
x+(F
p)
x+(F
v)
x+(F
t)
x+(F
c)
x
•Ifthenetforceduetocompressibility(F
c)isnegligible,theresulting
netforce
F
x=(F
g)x+(F
p)
x+(F
v)
x+(F
t)
x
andtheequationofmotionsarecalledReynolds’sequationsof
motion.
•Forflowwhere(Ft)isnegligible,theresultingequationsofmotion
areknownasNavier–Stokesequation.
•Iftheflowisassumedtobeideal,viscousforce(F
v)iszeroandthe
equationsofmotionareknownasEuler’sequationofmotion.

DEPARTMENT OF MEC HANIC AL ENGINEERING
EULER’S EQUATION OF MOTION
•Inthisequationofmotionthe
forcesduetogravityand
pressurearetakeninto
consideration.
•Thisisderivedbyconsidering
themotionofthefluidelement
alongastream-lineas:
•Considerastream-linein
whichflowistakingplaceins-
direction.
•Consideracylindricalelement
ofcross-sectiondAandlength
ds.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Theforcesactingonthecylindricalelementare:
PressureforcepdAinthedirectionofflow.
Pressureforce&#3627408477;+
??????p
??????s
ds&#3627408465;&#3627408436;
Weightofelement&#3627409164;gdA.ds
•Let??????istheanglebetweenthedirectionofflowandthelineof
actionoftheweightoftheelement.
•TheresultantforceonthefluidelementinthedirectionofSmust
beequaltothemassoffluidelement×accelerationinthedirection
ofs.
&#3627408477;&#3627408465;&#3627408436;−&#3627408477;+
??????p
??????s
&#3627408465;&#3627408480;&#3627408465;&#3627408436;−&#3627409164;??????&#3627408465;&#3627408436;&#3627408465;&#3627408480;cosθ=&#3627409164;&#3627408465;&#3627408436;&#3627408465;&#3627408480;×&#3627408462;
&#3627408480;___(1)
Whereasistheaccelerationinthedirectionofs.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Now, a
s=
dv
dt
where ‘v’ is a function of s and t.
a
s=
????????????
??????&#3627408480;
??????&#3627408480;
??????&#3627408481;
+
????????????
??????&#3627408481;
=
??????????????????
??????&#3627408480;
+
????????????
??????&#3627408481;
•If the flow is steady, then
????????????
??????&#3627408481;
=0.So, a
s=
??????????????????
??????&#3627408480;
•Substituting the value of a
sin equation (1) and simplifying, we get
−
??????p
??????s
dsdA−ρgdAdscosθ=ρdAds×
??????v
??????s
•Dividing by &#3627409164;dA.ds, −
1
ρ
×
??????p
??????s
−gcosθ=
v??????v
??????s
1
ρ
×
??????p
??????s
+gcosθ+
v??????v
??????s
=0 But we have cosθ=
dz
ds
1
ρ
×
??????p
??????s
+g
dz
ds
+
v??????v
??????s
=0
∴This equation is known as Euler’s equation of motion.
????????????
??????
+????????????&#3627408435;+&#3627408431;??????&#3627408431;=&#3627409358;

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 5
•Bernoulli’s equation
•Assumptions of Bernoulli’s equation
•Momentum equation
Bernoulli’s Equation
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
BERNOULLI’S EQUATION
•Bernoulli’sequationisobtainedbyintegratingtheEuler’sequation
ofmotionas
න
&#3627408465;&#3627408477;
&#3627409164;
+න??????&#3627408465;??????+න??????&#3627408465;??????=&#3627408438;&#3627408476;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;
•Iftheflowisincompressible,&#3627409164;isconstantand
p
ρ
+gz+
v
2
2
=constant
p
ρg
+z+
v
2
2g
=constant
p
ρg
+
v
2
2g
+z=constant

DEPARTMENT OF MEC HANIC AL ENGINEERING
•TheaboveequationisBernoulli’sequationinwhich
p
ρg
=Pressureenergyperunitweightoffluidorpressurehead.
v
2
2g
=KineticenergyperunitweightoffluidorKinetichead.
z=PotentialenergyperunitweightoffluidorPotentialhead.

DEPARTMENT OF MEC HANIC AL ENGINEERING
ASSUMPTIONS OF BERNOULLI’S EQUATION
•Thefollowingaretheassumptionsmadeinthederivationof
Bernoulli’sequation.
Thefluidisideali.e.Viscosityiszero.
Theflowissteady.
Theflowisincompressible.
TheflowisIrrotational.

DEPARTMENT OF MEC HANIC AL ENGINEERING
MOMENTUM EQUATION
•Itisbasedonthelawofconservationofmomentumoronthe
momentumprinciple,whichstatesthatthenetforceactingona
fluidmassequaltothechangeinthemomentumoftheflowper
unittimeinthatdirection.
•Theforceactingonafluidmass‘m’isgivenbyNewton’ssecond
lawofmotion.
F=m×a
•WhereaistheaccelerationactinginthesamedirectionasforceF.
But&#3627408462;=
????????????
??????&#3627408481;
??????=&#3627408474;
????????????
??????&#3627408481;
=
??????????????????
??????&#3627408481;
(Sincemisaconstantandcanbetaken
insidedifferential)

DEPARTMENT OF MEC HANIC AL ENGINEERING
??????=
??????????????????
??????&#3627408481;
Theaboveequationisknownasthemomentumprinciple.
F.dt=d(mv)
Theaboveequationisknownastheimpulsemomentumequation.
•Itstatesthattheimpulseofaforce‘F’actingonafluidmassmina
shortintervaloftime‘dt’isequaltothechangeofmomentum
‘d(mv)’inthedirectionofforce.

DEPARTMENT OF MEC HANIC AL ENGINEERING
FORCE EXERTED BY A FLOWING FLUID ON A
PIPE-BEND:
•The impulse momentum equation
is used to determine the resultant
force exerted by a flowing fluid on
a pipe bend.
•Consider two sections (1) and (2)
as above
•Let v
1= Velocity of flow at section
(1)
•P
1= Pressure intensity at section
(1)
•A
1= Area of cross-section of pipe
at section (1)
•And V
2, P
2, A
2are corresponding
values of Velocity, Pressure, Area
at section (2)

DEPARTMENT OF MEC HANIC AL ENGINEERING
•LetF
xandF
ybethecomponentsoftheforcesexertedbythe
flowingfluidonthebendinxandydirectionsrespectively.
•Thentheforceexertedbythebendonthefluidinthedirectionsof
xandywillbeequaltoF
XandF
Ybutintheoppositedirections.
•Hencethecomponentoftheforceexertedbythebendonthefluid
inthex–direction=-F
xandinthedirectionofy=-F
y.
•Theotherexternalforcesactingonthefluidarep
1A
1andp
2A
2on
thesections(1)and(2)respectively.Thenthemomentumequation
inx-directionisgivenby
•Net force acting on the fluid in the direction of x = Rate of change
of momentum in x –direction
= p
1 A
1–p
2A
2Cos??????-F
x= (Mass per second) (Change of velocity)

DEPARTMENT OF MEC HANIC AL ENGINEERING
= &#3627409164;Q (Final velocity in x-direction –Initial velocity in x-direction)
= &#3627409164;Q (V
2Cos??????-V
1)
F
x= &#3627409164;Q (V
1-V
2Cos??????) + p
1 A
1–p
2A
2Cos?????? (1)
•Similarly the momentum equation in y-direction gives
0 -p
2A
2Sin??????-F
y= &#3627409164;Q (V
2Sin??????-0)
F
y= &#3627409164;Q (-V
2Sin??????) -p
2A
2Sin?????? (2)
•Now the resultant force (F
R) acting on the bend
F
R= ??????
&#3627408485;
2
+??????
&#3627408486;
2
•And the angle made by the resultant force with the horizontal
direction is given by
&#3627408533;??????????????????=
??????
&#3627408538;
??????
&#3627408537;

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 6
•Problems on Bernoulli’s equation
Problems
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 1
•Apipethroughwhichwaterisflowingishavingdiameters20cms
and10cmsatcross-sections1and2respectively.Thevelocityof
wateratsection1is4m/sec.Findthevelocityheadatsection1
and2andalsorateofdischarge?
Sol:Givendata
D
1= 20cms = 0.2m
A
1=
&#3627409163;
4
×0.2
2
=0.0314&#3627408474;
2
V
1= 4 m/sec
D
2= 10 cm = 0.1 m
A
2=
&#3627409163;
4
×0.1
2
=0.007854&#3627408474;
2

DEPARTMENT OF MEC HANIC AL ENGINEERING
i)Velocityheadatsection1
??????
1
2
2??????
=
4×4
2×9.81
=0.815&#3627408474;
ii)Velocity head at section 2
??????
2
2
2??????
To find V
2, apply continuity equation, A
1V
1= A
2V
2
??????
2=
??????
1??????1
??????
2
=
0.0314×4
0.00785
=16&#3627408474;/&#3627408480;&#3627408466;&#3627408464;
•Velocity head at section 2
•
??????
2
2
2??????
=
16×16
2×9.81
=13.047&#3627408474;
•iii)Rate of discharge
Q = A
1V
1= A
2V
2
= 0.0314 ×4 = 0.1256 m
3
/sec
Q = 125.6 Liters/sec

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 2
•Waterisflowingthroughapipeof5cmdia.Underapressureof
29.43N/cm
2
andwithmeanvelocityof2m/sec.findthetotalhead
ortotalenergyperunitweightofwateratacross-section,whichis
5mabovedatumline.
Sol:Givendata
Dia. of pipe, d = 5cm = 0.05m
Pressure, P = 29.43N/cm
2
= 29.43 x 10
4
N/m
2
Velocity, V = 2 m/sec
Datum head, Z = 5m
Total head = Pressure head + Kinetic head + Datum head
Pressurehead=
&#3627408477;
&#3627409164;??????
=
29.43×10
4
1000×9.81
=30&#3627408474;

DEPARTMENT OF MEC HANIC AL ENGINEERING
Kinetichead=
??????
2
2??????
=
2×2
2×9.81
=0.204&#3627408474;
Datum head = Z = 5m
&#3627408477;
&#3627409164;??????
+
??????
2
2??????
+&#3627408461;=30+0.204+5=35.204&#3627408474;
Total head = 35.204m

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 3
•Waterisflowingthroughapipehavingdiameters20cmsand10cmsat
sections1and2respectively.Therateofflowthroughpipeis35
liters/sec.Thesection1is6mabovethedatumandsection2is4m
abovethedatum.Ifthepressureatsection1is39.24n/cm
2
.Findthe
intensityofpressureatsection2?
Sol:Givendata
Atsection1,D
1=20cm=0.2m
A
1=
&#3627409163;
4
×0.2
2
=0.0314&#3627408474;
2
P
1= 39.24N/cm
2
=39.24 x 10
4
N/m
2
Z
1= 6m
At section 2, D
2=10cm = 0.1m
A
2=
&#3627409163;
4
×0.1
2
=0.0007854&#3627408474;
2
Z
2 = 4m

DEPARTMENT OF MEC HANIC AL ENGINEERING
Rate of flow Q = 35 lt/sec = (35/1000) m
3
/sec = 0.035 m
3
/sec
Q = A
1V
1= A
2V
2
??????
1=
&#3627408452;
??????
1
=
0.035
0.0314
=1.114&#3627408474;/&#3627408480;&#3627408466;&#3627408464;, ??????
2=
&#3627408452;
??????
2
=
0.035
0.007854
=4.456&#3627408474;/&#3627408480;&#3627408466;&#3627408464;
Applying Bernoulli’s equation at sections 1 and 2
&#3627408451;
1
&#3627409164;??????
+
??????
1
2
2??????
+&#3627408461;
1=
&#3627408451;
2
&#3627409164;??????
+
??????
2
2
2??????
+??????
=39.24×10
4
1000×9.81+
1.114
2
2×9.81
+6=
&#3627408451;
2
1000×9.81
+
4.456
2
2×9.81
+
4
= 40+0.063+6=
&#3627408451;
2
9810
+1.102+4
= 46.063 =
&#3627408451;
2
9810
+ 5.102
=
&#3627408451;
2
9810
= 46.063 –5.102 = 41.051
Therefore, P
2= 41.051 ×9810 =402710 N/m
2
P
2= 40.271N/cm
2

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 4
•Waterisflowingthroughapipehavingdiameter300mmand
200mmatthebottomandupperendrespectively.Theintensityof
pressureatthebottomendis24.525N/cm
2
andthepressureatthe
upperendis9.81N/cm
2
.Determinethedifferenceindatumheadif
therateofflowthroughis40lit/sec?
Sol:Givendata
Section1,D
1=300mm=0.3m
&#3627408436;
1=
&#3627409163;
4
×0.3
2
A
1=0.07065m
2
P
1=24.525N/cm
2
=24.525×10
4
N/m
2
Section2,D
2=200mm0.2m
A
2=
&#3627409163;
4
)×(0.2)
2
=0.0314m
2
P
2=9.81N/cm
2
=9.81×10
4
N/m
2

DEPARTMENT OF MEC HANIC AL ENGINEERING
Rate of flow, Q = 40 lit/Sec =40/1000 = 0.04 m
3
/sec
Q = A
1V
1= A
2V
2
??????
1=
Q
A
1
=
0.04
0.07065
=0.566m/sec
??????
2=
Q
A
2
=
0.04
00.0314
=1.274m/sec
Applying Bernoulli’s equation at sections 1 and 2
??????
1
&#3627409164;??????
+
??????
1
2
2??????
+&#3627408461;
1=
??????
2
&#3627409164;??????
+
??????
2
2
2??????
+&#3627408461;
2
=
24.525×10
4
1000×9.81
+
0.566
2
2×9.81
+&#3627408461;
1=
9.81×10
4
1000×9.81
+
1.274
2
2×9.81
+&#3627408461;
2
= 25 + 0.32 +Z
1= 10 +1.623 + Z
2
= Z
2–Z
1= 25.32 –11.623 = 13.697 or say 13.70m
The difference in datum head =
Z
2–Z
1= 13.70m

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 7
•Problems on Momentum equation
•Applications
•Assignment Questions
Problems
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 1
•Thewaterisflowingthroughataperpipeoflength100mhaving
diameters600mmattheupperendand300mmatthelowerend,at
therateof50lts/sec.thepipehasaslopeof1in30.Findthepressure
atthelowerend,ifthepressureatthehigherlevelis19.62N/cm
2
?
Sol:Givendata
Length of pipe L = 100m
Dia. At the upper end D
1= 600mm = 0.6
&#3627408436;
1=
π
4
×0.6
2
=0.2827m
2
P
1= 19.62N/cm
2
= 19.62 ×10
4
N/m
2
Dia. at the lower end D
2= 300mm = 0.3m
&#3627408436;
2=
π
4
×0.3
2
=0.07065m
2
Rate of flow Q = 50 Lts/sec =
50
1000
= 0.05 m
3
/sec

DEPARTMENT OF MEC HANIC AL ENGINEERING
Let the datum line is passing through the centreof the lower end.
Then Z
2= o
As slope is 1 in 30 means &#3627408461;
1=
1
30
×100=
10
3
m
We also know that, Q = A
1V
1= A
2V
2
??????
1=
Q
A1
=
0.05
0.2827
=0.177m/sec
??????
2=
Q
A2
=
0.05
0.07065
=0.707m/sec
Applying Bernoulli’s equation at sections 1 and 2
&#3627408451;1
&#3627409164;??????
+
??????1
2
2??????
+&#3627408461;
1=
&#3627408451;2
&#3627409164;??????
+
??????2
2
2??????
+&#3627408461;
2

DEPARTMENT OF MEC HANIC AL ENGINEERING
=
19.62×10
4
1000×9.81
+
0.177
2
2×9.81
+
10
3
=
&#3627408451;2
1000×9.81
+
0.707
2
2×9.81
+0
= 20 +0.001596 + 3.334 =
P2
9810
+ 0.0254
= 23.335 =
P2
9810
+ 0.0254
=
P2
9810
= 23.335 –0.0254 =23.31
= P
2= 23.31 ×9810 = 228573N/m
2
P
2= 22.857N/cm
2

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 2
•A45°reducingbendisconnectedtoapipeline,thediametersat
inletandoutletofthebendbeing600mmand300mmrespectively.
Findtheforceexertedbythewateronthebend,iftheintensityof
pressureattheinlettothebendis8.829N/cm
2
andrateofflowof
wateris600Lts/sec.
Sol:Givendata
Angle of bend ??????= 45°
Dia. at inlet D
1= 600mm = 0.6m
&#3627408436;
1=
π
4
×(0.6)
2
= 0.2827 m
2
Dia. at outlet D
2= 300mm = 0.3m
A
2=
π
4
×(0.3)
2
= 0.07065 m
2

DEPARTMENT OF MEC HANIC AL ENGINEERING
ApplyingBernoulli’sequationatsections1and2,weget
&#3627408451;1
&#3627409164;??????
+
??????1
2
2??????
+&#3627408461;
1=
&#3627408451;2
&#3627409164;??????
+
??????2
2
2??????
+&#3627408461;
2
ButZ
1=Z
2,then
&#3627408451;1
&#3627409164;??????
+
??????1
2
2??????
=
&#3627408451;2
&#3627409164;??????
+
??????2
2
2??????
=
8.829×10
4
1000×9.81
+
2.122
2
2×9.81
=
P2
1000×9.81
=
8.488
2
2×9.81
=9+0.2295=
P2
9810
+3.672
=
P2
9810
=9.2295-3.672=5.5575mofwater
P
2=5.5575×9810=5.45×10
4
N/m
2

DEPARTMENT OF MEC HANIC AL ENGINEERING
ForceexertedonthebendinXandY–directions
F
x=&#3627409164;Q(V
1-V
2Cos??????)+P
1A
1–P
2A
2Cos??????
=1000×0.6(2.122–8.488Cos45°)+8.829×10
4
×0.2827–
5.45×10
4
×0.07065×Cos45°
=-2327.9+24959.6–2720.3=24959.6-5048.2=19911.4N
F
y=&#3627409164;Q(-V
2Sin??????)-P
2A
2Sin??????
=1000×0.6(-8.488Sin45°)-5.45×10
4
×0.07068Sin45°
=-3601.1-2721.1=-6322.2N
(-vesignmeansFyisactinginthedownwarddirection)
F
x=19911.4 N
F
y= -6322.2N

DEPARTMENT OF MEC HANIC AL ENGINEERING
ThereforetheResultantForceF
R=??????
&#3627408485;
2
+??????
&#3627408486;
2
=(19911.4)
2
+(-
6322.2)
2
=20890.9N
TheanglemadebyresultantforcewithX–axisisTan??????=
Fx
Fy
=(6322.2/19911.4)=0.3175
F
R = 20890.9 N
??????= tan
-1
0.3175= &#3627409359;&#3627409365;
&#3627409358;
??????&#3627409364;
′

DEPARTMENT OF MEC HANIC AL ENGINEERING
APPLICATIONS
•Forsizingofpumps:Voluteinthecasingofcentrifugalpumps
convertsvelocityoffluidintopressureenergybyincreasingareaof
flow.Theconversionofkineticenergyintopressureisaccordingto
Bernoulli’sequation.
•CarburetorworksonprincipleofBernoulli’sprinciple:thefasterthe
airmoves,theloweritsstaticpressureandhigheritsdynamic
pressure.
•Application of the Momentum Equation
Force due to the flow of fluid round a pipe bend.
Force on a nozzle at the outlet of a pipe.
Impact of a jet on a plane surface.

DEPARTMENT OF MEC HANIC AL ENGINEERING
ASSIGNMENT QUESTIONS
•Apipeline300mlonghasaslopeof1in100andtapersfrom1.2m
diameteratthehighendto0.6matthelowend.Thedischarge
throughthepipeis5.4m
3
/min.Ifthepressureatthehighendis
70kPa,findthepressureatthelowerend.Neglectlosses.
•Apipe1of450mmindiameterbranchesintotwopipes(2&3)of
diameter300mmand200mm.Ifaveragevelocityin450mmdiameter
pipeis3/s.Find(i)Dischargethrough450mmdiameterpipe.(ii)
Velocityin200mmdiameterpipeifthevelocityin300mmpipeis
2.5m/s.
•ApipelineABC200mlongislaidonanupwardslope1in40.The
lengthoftheportionABis100manditsdiameteris100mm.AtBthe
pipesectionsuddenlyenlargesto200mmdiameterandremainssofor
theremainderofitslengthBC,100m.Aflowof0.0m
3
/sispumpedinto
thepipeatitslowerendAandisdischargedattheupperendCinto
closedtank.Thepressureatthesupplyendis200kN/m
2
.Whatisthe
pressureatC?

DEPARTMENT OF MEC HANIC AL ENGINEERING
•A45
0
reducingbendisconnectedinapipeline,thediametersat
theinletandoutletofthebendbeing40cmand20cm.Findthe
forceexertedbywateronthebendiftheintensityofpressureat
inletofbendis21.58N/cm
2
.Therateofflowofwateris500litres/s.
•DeriveBernoulli’sequationfromEuler’sequationofmotion.What
aretheassumptionsmadeinderivingBernoulli’stheorem?

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
UNIT3
CO3:Togivebasicunderstandingofboundarylayer
conceptandanalyzedifferenttypesoflossesand
measurementofflow.

DEPARTMENT OF MECHANICAL ENGINEERING
UNIT 3
CO3:To gi ve b asi cun der st a ndi ng of
b o un dar yl a yerc on ce pta ndan al yz edi f f er e nt
t yp e sofl o s s e sa n dm e a s u r e m e ntoff l o w.
BOUNDARY LAYER
CONCEPT &
CLOSED CONDUIT
FLOW

DEPARTMENT OF MEC HANIC AL ENGINEERING
UNIT –III
BoundaryLayerConcept:
•Definition-thickness.
•Characterizationalongthinplate.
•Laminarandturbulentboundarylayers(NoDerivation).
ClosedConduitFlow:
•Reynold’sexperiment,DarcyWeisbachequation.
•Major&minorlosses,pipesinseries&parallel.
•Totalenergylineandhydraulicgradientline.
•Measurementofflow-Pitottube,Venturimeter&
Orificemeter.

DEPARTMENT OF MEC HANIC AL ENGINEERING
COURSEOUTLINE
LECTURE LECTURETOPIC KEYELEMENTS Learningobjectives
1 Introductionto boundary layer
concept
Characterization.
Laminar & Turbulent
Understanding
concepts (B2)
2 Closed conduit flow
Reynolds experimentUnderstanding type
of flow(B2)
3 Darcy Weisbach equation
Major losses
Friction factor
Apply law of
conservation of mass
(B3)
4 Minor losses Pipes in series &
parallel
Analyzedifferent types
of losses in pipe flow
(B4)
5 Problems on major & minor losses
6 Total Energy Line and Hydraulic
Gradient Line
HGL & TEL Understanding
concepts (B2)
7 Measurement of flow Pitot tube
Venturimeter
Evaluateflow through
venturimeter (B5)
8 Measurement of flow Orificemeter Evaluateflow through
orificemeter (B5)
9 Problems on Venturimeter & orificemeter
UNIT-3

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 1
•Introduction to Boundary layer
•Boundary layer Characteristics
•Boundary layer thickness
Boundary Layer Concepts
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
BOUNDARY LAYER CHARACTERISTICS
•Boundarylayeristheregionscloseto
thesolidboundarywheretheeffects
ofviscosityareexperiencedbythe
flow.
•Intheregionsoutsidetheboundary
layer,theeffectofviscosityis
negligibleandthefluidistreatedas
inviscid.
•So,theboundarylayerisabuffer
regionbetweenthewallbelowand
theinviscidfree-streamabove.
•Thisapproachallowsthecomplete
solutionofviscousfluidflowswhich
wouldhavebeenimpossiblethrough
Navier-Stokesequation.
REPRESENTATION OF
BOUNDARY LAYER ON
A FLAT PLATE

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Thequalitativepictureoftheboundary-layergrowthoveraflat
plateisshowninFig.
•Alaminarboundarylayerisinitiatedattheleadingedgeofthe
plateforashortdistanceandextendstodownstream.
•Thetransitionoccursoveraregion,aftercertainlengthinthe
downstreamfollowedbyfullyturbulentboundarylayers.
•Forcommoncalculationpurposes,thetransitionisusually
consideredtooccuratadistancewheretheReynoldsnumberis
about500,000.Withairatstandardconditions,movingatavelocity
of30m/s,thetransitionisexpectedtooccuratadistanceofabout
250mm.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•A typical boundary layer flow is characterized by certain
parameters as given below
Boundary thickness
Free stream flow (no viscosity)
Concepts of displacement thickness

DEPARTMENT OF MEC HANIC AL ENGINEERING
BOUNDARY LAYER THICKNESS
•Itisknownthatno-slipconditionshavetobesatisfiedatthesolid
surface:thefluidmustattainthezerovelocityatthewall.
•Subsequently,abovethewall,theeffectofviscositytendsto
reduceandthefluidwithinthislayerwilltrytoapproachthefree
streamvelocity.
•Thus,thereisavelocitygradientthatdevelopswithinthefluid
layersinsidethesmallregionsneartosolidsurface.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Theboundarylayerthickness
isdefinedasthedistancefrom
thesurfacetoapointwhere
thevelocityisreaches99%of
thefreestreamvelocity.
•Thus,thevelocityprofile
merges smoothly and
asymptoticallyintothefree
streamasshowninFig.2.

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 2
•Reynold’sExperiment
•Reynoldnumber
Reynold’sExperiment
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
REYNOLD’S EXPERIMENT
REYNOLD’S APPARATUS

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Itconsistsofaconstantheadtankfilledwithwater,asmalltank
containingdye,ahorizontalglasstubeprovidedwithabell-
mouthedentranceaandregulatingvalve.
•Thewaterwasmadetoflowfromthetankthroughtheglasstube
intotheatmosphereandthevelocityifflowwasvariedby
adjustingtheregulatingvalve.
•Theliquiddyehavingthesamespecificweightasthatofwater
wasintroducedintotheflowatthebell–mouththroughasmall
tube.
•Fromtheexperimentsitwasdisclosedthatwhenthevelocityof
flowwaslow,thedyeremainedintheformofastraightlineand
stablefilamentpassingthroughtheglasstubesosteadythatit
scarcelyseemedtobeinmotionwithincreaseinthevelocityof
flowacriticalstatewasreachedatwhichthefilamentofdye
showedirregularitiesandbeganofwaver.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Furtherincreaseinthevelocityofflowthefluctuationsinthe
filamentofdyebecamemoreintenseandultimatelythedye
diffusedovertheentirecross-sectionofthetube,dueto
interminglingoftheparticlesoftheflowingfluid
•Reynolds’sdeducedfromhisexperimentsthatatlowvelocitiesthe
interminglingofthefluidparticleswasabsentandthefluidparticles
movedinparallellayersorlamina,slidingpasttheadjacentlamina
butnotmixingwiththem,whichisthelaminarflow.
•Athighervelocitiesthedyefilamentdiffusedthroughthetubeit
wasapparentthattheinterminglingoffluidparticleswasoccurring
inotherwordstheflowwasturbulent.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Thevelocityatwhichtheflowchangesfromthelaminarto
turbulentforthecaseofagivenfluidatagiventemperatureandin
agivenpipeisknownasCriticalVelocity.
•Thestateofflowinbetweenthesetypesofflowisknownas
transitionalstateorflowintransition.
•Reynoldsdiscoveredthattheoccurrenceoflaminarandturbulent
flowwasgovernedbytherelativemagnitudesoftheinertiaandthe
viscousforces.
•Atlowvelocitiestheviscousforcesbecomepredominantandflow
isviscous.
•Athighervelocitiesofflowtheinertialforcespredominanceover
viscousforces.

DEPARTMENT OF MEC HANIC AL ENGINEERING
REYNOLD’S NUMBER
•Reynolds related the inertia to viscous forces and arrived at a
dimension less parameter.
&#3627408453;
&#3627408466;&#3627408476;&#3627408479;??????
&#3627408466;=
inertiaforce
viscousforce
=
??????
??????
??????
??????
•According to Newton’s 2
nd
law of motion, the inertia force F
iis
given by
F
i= mass ×acceleration
= ρ ×volume ×acceleration (ρ = mass density)
= ρ ×L
3
×
L
T
2
= ρ L
2
V
2
----(1) (L = Linear dimension)

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Similarly viscous force F
V is given by Newton’s 2
nd
law of velocity
as
F
V = τ ×area τ = shear stress
= µ
dv
dy
×L
2
= µVL ----------------(2)
V = Average Velocity of flow
µ = Viscosity of fluid
•In case of pipes L = D
R
eor N
R =
ρL
2
V
2
μVL
=
ρVL
μ

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Incaseofflowthroughpipes
Whereµ/ρ=kinematicviscosityoftheflowingliquid??????
•TheReynoldsnumberisaveryusefulparameterinpredicting
whethertheflowislaminarorturbulent.
R
e<2000viscous/laminarflow
R
e2000to4000transientflow
R
e>4000Turbulentflow
&#3627408453;
&#3627408466;=
ρDV
μ
or
??????&#3627408439;
??????
??????

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 3
•Darcy Weisbachequation-Major
loss
•Derivation
•Minor Losses in pipes
Losses in pipes
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
FRICTIONAL LOSS IN PIPE FLOW –DARCY
WEISBACK EQUATION
•Whenaliquidisflowing
throughapipe,thevelocityof
theliquidlayeradjacenttothe
pipewalliszero.
•Thevelocityofliquidgoeson
increasingfromthewalland
thusvelocitygradientand
henceshearstressesare
producedinthewholeliquid
duetoviscosity.
•Thisviscousactioncauses
lossofenergy,whichisknown
asfrictionalloss.

DEPARTMENT OF MEC HANIC AL ENGINEERING
DERIVATION
•Considerauniformhorizontalpipehavingsteadyflow.Let1-1,2-2
aretwosectionsofpipe.
LetP
1=Pressureintensityatsection1-1
V
1=Velocityofflowatsection1-1
L=Lengthofpipebetweensection1-1and2-2
d=Diameterofpipe
&#3627408467;
′
=Fractionalresistanceforunitwettedareaperaunitvelocity
h
f=Lossofheadduetofriction
•AndP
2,V
2=arevaluesofpressureintensityandvelocityatsection
2-2

DEPARTMENT OF MEC HANIC AL ENGINEERING
•ApplyingBernoulli’sequationbetweensections1-1and2-2
Totalheadat1-1=totalheadat2-2+lossofheadduetofriction
between1-1and2-2
&#3627408451;1
&#3627409164;&#3627408468;
+
??????1
2
2&#3627408468;
+&#3627408461;
1=
&#3627408451;2
&#3627409164;&#3627408468;
+
??????2
2
2&#3627408468;
+&#3627408461;
2
Z
1=Z
2aspipeishorizontal
V
1=V
2asdia.ofpipeissameat1-1and2-2
&#3627408451;1
&#3627409164;&#3627408468;
=
&#3627408451;2
&#3627409164;&#3627408468;
+ℎ
&#3627408467;Or
ℎ
&#3627408467;=
&#3627408451;1
&#3627409164;&#3627408468;
−
&#3627408451;2
&#3627409164;&#3627408468;
(1)
•Buth
fisheadislostduetofrictionandhencetheintensityof
pressurewillbereducedinthedirectionflowbyfrictional
resistance.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Now,FrictionalResistance=Frictionalresistanceperunitwetted
areaperunitvelocity×WettedArea×(velocity)
2
??????
1=&#3627408467;
′
×&#3627409163;&#3627408465;??????×&#3627408457;
2
[∵Wettedarea=&#3627409163;&#3627408465;×??????,Velocity=V=V
1=V
2]
??????
1=&#3627408467;
′
×&#3627408477;??????&#3627408457;
2
(2)[∵&#3627409163;d=perimeter=p]
Theforcesactingonthefluidbetweensection1-1and2-2are
Pressureforceatsection1-1=P
1×A (whereA=areaofpipe)
Pressureforceatsection2-2=P
2×A
Frictionalforce=F
1
•Resolving all forces in the horizontal direction, we have
P
1 A –P
2A –F
1= 0
(P
1–P
2)A = F
1= &#3627408467;
′
×&#3627408477;×??????×&#3627408457;
2
from equation –(2)
P
1 –P
2 =
&#3627408467;
′
×&#3627408477;×??????×??????
2
A
But from equation (1) P
1-P
2= ρgh
f

DEPARTMENT OF MEC HANIC AL ENGINEERING
•EquatingthevalueofP
1-P
2,weget
ρgh
f=
&#3627408467;
′
×&#3627408477;×??????×??????
2
A
ℎ
&#3627408467;=
f
′
pg
×
P
A
×L×V
2
(3)
•Intheequation(3)
P
A
=
WettedPermiter
Area
=
πd
π
4
d
2
=
π
d
ℎ
&#3627408467;=
&#3627408467;
′
&#3627409164;&#3627408468;
×
4
&#3627408465;
×??????×&#3627408457;
2
=
&#3627408467;
′
&#3627409164;&#3627408468;
×
4????????????
2
&#3627408465;
Putting
&#3627408467;
′
&#3627409164;
=
&#3627408467;
2
Wherefisknownasco-efficientoffriction.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Equation(4)becomesasℎ
&#3627408467;=
4&#3627408467;
2&#3627408468;
×
????????????
2
&#3627408465;
•ThisEquationisknownasDarcy–Weisbachequation,commonly
usedforfindinglossofheadduetofrictioninpipes
•Thenfisknownasafrictionfactororco-efficientoffrictionwhich
isadimensionlessquantity.fisnotaconstantbut,itsvalue
dependsupontheroughnessconditionofpipesurfaceandthe
Reynoldsnumberoftheflow.
ℎ
&#3627408467;=
4&#3627408467;??????&#3627408457;
2
2&#3627408468;&#3627408465;

DEPARTMENT OF MEC HANIC AL ENGINEERING
MINOR LOSSES IN PIPES
•Thelossofenergyduetofrictionisclassifiedasamajorloss,
becauseincaseoflongpipelinesitismuchmorethanthelossof
energyincurredbyothercauses.
•Theminorlossesofenergyarecausedonaccountofthechange
inthevelocityofflowingfluids(eitherinmagnitudeordirection).
•Incaseoflongpipestheselosesarequitesmallascomparedwith
thelossofenergyduetofrictionandhencethesearetermedas
‘’minorlosses‘’
•Whichmayevenbeneglectedwithoutseriouserror,howeverin
shortpipestheselossesmaysometimesoutweighthefrictionloss.
•Someofthelossesofenergywhichmaybecausedduetothe
changeofvelocityare:

DEPARTMENT OF MEC HANIC AL ENGINEERING
MINOR LOSSES IN PIPES
•Loss of energy due to sudden enlargement, ℎ
&#3627408466;=
??????1−??????2
2
2&#3627408468;
•Loss of energy due to sudden contraction, ℎ
&#3627408464;=0.5
??????2
2
2&#3627408468;
•Loss of energy at the entrance to a pipe, ℎ
??????=0.5
??????
2
2&#3627408468;
•Loss of energy at the exit from a pipe, ℎ
&#3627408476;=
??????
2
2&#3627408468;
•Loss of energy due to gradual contraction or enlargement, ℎ
&#3627408473;=
&#3627408472;??????1−??????2
2
2&#3627408468;
•Loss of energy in the bends, ℎ
&#3627408463;=
&#3627408472;??????
2
2&#3627408468;
•Loss of energy in various pipe fittings, ℎ
&#3627408473;=
&#3627408472;??????
2
2&#3627408468;

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 4
•Loss due to sudden enlargement
•Pipes in series
•Pipes in parallel
Minor Losses in Pipe
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
LOSS OF HEAD DUE TO SUDDEN
ENLARGEMENT
•Consideraliquidflowingthrougha
pipewhichhassuddenenlargement.
•Considertwosections1-1and2-2
beforeandafterenlargement.
•Duetosuddenchangeofdiameterof
thepipefromD
1toD
2.
•Theliquidflowingfromthesmaller
pipeisnotabletofallowtheabrupt
changeoftheboundary.
•Thustheflowseparatesfromthe
boundaryandturbulenteddiesare
formed.
•Thelossofheadtakesplacedueto
theformationoftheseeddies

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Let&#3627408477;
′
=Pressureintensityoftheliquideddiesonthearea(A
2–A
1)
h
e=lossofheadduetothesuddenenlargement.
•ApplyingBernoulli’sequationatsection1-1and2-2
&#3627408477;1
&#3627409164;&#3627408468;
+
??????1
2
2&#3627408468;
+??????
1=
&#3627408477;2
&#3627409164;&#3627408468;
+
??????2
2
2&#3627408468;
+??????
2+Lossofheadduetosudden
enlargement
•But,z
1=z
2aspipeishorizontal
&#3627408477;1
&#3627409164;&#3627408468;
+
??????1
2
2&#3627408468;
=
&#3627408477;2
&#3627409164;&#3627408468;
+
??????2
2
2&#3627408468;
+ℎ
&#3627408466;
Orℎ
&#3627408466;=
&#3627408477;1
&#3627409164;&#3627408468;
−
&#3627408477;2
&#3627409164;&#3627408468;
+
??????1
2
2&#3627408468;
−
??????2
2
2&#3627408468;
(1)
•Theforceactingontheliquidinthecontrolvolumeinthedirection
offlow
??????
??????=&#3627408477;
1??????
1+&#3627408477;
′
??????
2−??????
1−&#3627408477;
2??????
2

DEPARTMENT OF MEC HANIC AL ENGINEERING
•But experimentally it is found that &#3627408477;
′
= &#3627408477;
1
??????
??????=&#3627408477;
1??????
1+&#3627408477;
1??????
2−??????
1−&#3627408477;
2??????
2
=&#3627408477;
1??????
2−&#3627408477;
2??????
2
=&#3627408477;
1−&#3627408477;
2??????
2 (2)
•Momentum of liquid/ second at section 1-1 = mass ×velocity
=&#3627409164;??????
1&#3627408457;
1×&#3627408457;
1
=&#3627409164;??????
1&#3627408457;
1
2
•Momentum of liquid/ second at section 2-2 &#3627409164;??????
2&#3627408457;
2×&#3627408457;
2=&#3627409164;??????
2&#3627408457;
2
2
•Change of momentum/second =&#3627409164;??????
2&#3627408457;
2
2
−&#3627409164;??????
1&#3627408457;
1
2
(3)

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Butfromcontinuityequation,wehave
A
1V
1=A
2V
2
Or ??????
1=
??????2??????2
??????1
∴Changeofmomentum/sec=&#3627409164;??????
2&#3627408457;
2
2
−&#3627409164;×
??????2??????2
??????1
×&#3627408457;
1
2
=&#3627409164;??????
2&#3627408457;
2
2
−&#3627409164;??????
1&#3627408457;
1&#3627408457;
2
=&#3627409164;??????
2&#3627408457;
2
2
−&#3627408457;
1&#3627408457;
2___________(4)
•Nowthenetforceactingonthecontrolvolumeinthedirectionof
flowmustbeequaltorateofchangeofmomentumpersecond.
Henceequatingequation(2)andequation(4)
&#3627408477;
1−&#3627408477;
2??????
2=&#3627409164;??????
2&#3627408457;
2
2
−&#3627408457;
1&#3627408457;
2
&#3627408477;1−&#3627408477;2
&#3627409164;
=&#3627408457;
2
2
−&#3627408457;
1&#3627408457;
2

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Dividing both sides by ‘g’ we have
&#3627408477;1−&#3627408477;2
&#3627409164;&#3627408468;
=
??????2
2
−??????1??????2
&#3627408468;
Or
&#3627408477;1
&#3627409164;&#3627408468;
−
&#3627408477;2
&#3627409164;&#3627408468;
=
??????2
2
−??????1??????2
&#3627408468;
•Substituting in equation (1)
ℎ
&#3627408466;=
&#3627408457;
2
2
−&#3627408457;
1&#3627408457;
2
&#3627408468;
+
&#3627408457;
1
2
2&#3627408468;
−
&#3627408457;
2
2
2&#3627408468;
=
2&#3627408457;
2
2
−2&#3627408457;
1&#3627408457;
2+&#3627408457;
1
2
−&#3627408457;
2
2
2&#3627408468;
=
&#3627408457;
1
2
+&#3627408457;
2
2
−2&#3627408457;
1&#3627408457;
2
2&#3627408468;
ℎ
&#3627408466;=
&#3627408457;
1−&#3627408457;
2
2
2&#3627408468;

DEPARTMENT OF MEC HANIC AL ENGINEERING
PIPES IN SERIES
•Ifapipelineconnectingtwo
reservoirsismadeupof
severalpipesofdifferent
diametersd
1,d
2,d
3,etc.and
lengthsL
1,L
2,L
3etc.all
connectedinseries(i.e.endto
end),thenthedifferenceinthe
liquidsurfacelevelsisequalto
thesumoftheheadlossesin
allthesections.
•Furtherthedischargethrough
eachpipewillbesame.

DEPARTMENT OF MEC HANIC AL ENGINEERING
??????=
0.5&#3627408457;
1
2
2&#3627408468;
+
4&#3627408467;
1??????
1&#3627408457;
1
2
2&#3627408468;&#3627408465;
1
+
0.5&#3627408457;
2
2
2&#3627408468;
+
4&#3627408467;
2??????
2&#3627408457;
2
2
2&#3627408468;&#3627408465;
2
+
0.5&#3627408457;
3
2
2&#3627408468;
+
4&#3627408467;
3??????
3&#3627408457;
3
2
2&#3627408468;&#3627408465;
3
•Also,&#3627408452;=
&#3627409163;×&#3627408465;1
2
4
×&#3627408457;
1=
&#3627409163;×&#3627408465;2
2
4
×&#3627408457;
2=
&#3627409163;×&#3627408465;3
2
4
×&#3627408457;
3
•Howeveriftheminorlossesareneglectedascomparedwiththe
lossofheadduetofrictionineachpipe,then
•Theaboveequationmaybeusedtosolvetheproblemsofpipe
linesinseries.
??????=
4&#3627408467;
1??????
1&#3627408457;
1
2
2&#3627408468;&#3627408465;
1
+
4&#3627408467;
2??????
2&#3627408457;
2
2
2&#3627408468;&#3627408465;
2
+
4&#3627408467;
3??????
3&#3627408457;
3
2
2&#3627408468;&#3627408465;
3

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Therearetwotypesofproblemswhichmayariseforthepipelines
inseries.Viz.
GivenadischargeQtodeterminetheheadHand
GivenHtodeterminedischargeQ.
•Iftheco-efficientoffrictionissameforallthepipesi.e.f
1=f
2=f
3,
then
??????=
4&#3627408467;
1
2&#3627408468;
??????
1&#3627408457;
1
2
&#3627408465;
1
+
??????
2&#3627408457;
2
2
&#3627408465;
2
+
??????
3&#3627408457;
3
2
&#3627408465;
3

DEPARTMENT OF MEC HANIC AL ENGINEERING
PIPES IN PARALLEL
•Whenamainpipelinedividesinto
twoormoreparallelpipes,whichmay
againjointogetherdownstreamand
continueasmainline,thepipesare
saidtobeinparallel.
•Thepipesareconnectedinparallelin
ordertoincreasethedischarge
passingthroughthemain.
•Itisanalogoustoparallelelectric
currentinwhichthedropinpotential
andflowofelectriccurrentcanbe
comparedtoheadlossandrateof
dischargeinafluidflowrespectively.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Therateofdischargeinthemainlineisequaltothesumofthe
dischargesineachoftheparallelpipes.
•Thus,Q=Q
1+Q
2
•Theflowofliquidinpipes(1)and(2)takesplaceunderthedifference
ofheadbetweenthesectionsAandBandhencethelossofhead
betweenthesectionsAandBwillbethesamewhethertheliquid
flowsthroughpipe(1)orpipe(2).
•ThusifD
1,D
2andL
1,L
2arethediametersandlengthsofthepipes(1)
and(2)respectively,thenthevelocitiesofflowV
1andV
2inthetwo
pipesmustbesuchastogive
•Assumingsamevalueoffforeachparallelpipe
??????
1??????
1
2
2&#3627408468;&#3627408465;
1
=
??????
2??????
2
2
2&#3627408468;&#3627408465;
2
ℎ
&#3627408467;=
&#3627408467;??????
1&#3627408457;
1
2
2&#3627408468;&#3627408465;
1
=
&#3627408467;??????
2&#3627408457;
2
2
2&#3627408468;&#3627408465;
2

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 5&6
•Problem on Darcy Wiesbach
equation
•Problem on Minor losses
•Problems on pipes in series &
parallel
Example Problems
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 1
•Inapipeofdiameter350mmandlength75mwaterisflowingata
velocityof2.8m/s.FindtheheadlostduetofrictionbyusingDarcy
Weisbachequation?

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM-2

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 3

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 4

DEPARTMENT OF MEC HANIC AL ENGINEERING

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM-5

DEPARTMENT OF MEC HANIC AL ENGINEERING

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM-6

DEPARTMENT OF MEC HANIC AL ENGINEERING

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 7
•Hydraulic Gradient line
•Total energy line
•Pitottube
HGL & TEL
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
HYDRAULIC GRADIENT LINE AND TOTAL
ENERGY LINE
•Consideralongpipeline
carryingliquidfromareservoirA
toreservoirB.
•Atseveralpointsalongthe
pipelineletpiezometersbe
installed.
•Theliquidwillriseinthe
piezometerstocertainheights
correspondingtothepressure
intensityateachsection.
•Theheightoftheliquidsurface
abovetheaxisofthepipeinthe
piezometeratanysectionwillbe
equaltothepressurehead(p/w)
atthatsection.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Onaccountoflossofenergyduetofriction,thepressureheadwill
decreasegraduallyfromsectiontosectionofpipeinthedirection
offlow.
•Ifthepressureheadsatthedifferentsectionsofthepipeare
plottedtoscaleasverticalordinatesabovetheaxisofthepipeand
allthesepointsarejoinedbyastraightline,aslopinglineis
obtained,whichisknownasHydraulicGradientLine(H.G.L).
•Sinceatanysectionofpipetheverticaldistancebetweenthepipe
axisandHydraulicgradientlineisequaltothepressureheadat
thatsection,itisalsoknownaspressureline.
•MoreoverifZistheheightofthepipeaxisatanysectionabovean
arbitrarydatum,thentheverticalheightoftheHydraulicgradient
lineabovethedatumatthatsectionofpiperepresentsthe
piezometricheadequalto(p/w+z)..

DEPARTMENT OF MEC HANIC AL ENGINEERING
•SometimestheHydraulicgradientlineisalsoknownas
piezometricheadline.
•AttheentrancesectionofthepipeforsomedistancetheHydraulic
gradientlineisnotverywelldefined.
•Thisisbecauseasliquidfromthereservoirentersthepipe,a
suddendropinpressureheadtakesplaceinthisportionofpipe.
•Furthertheexitsectionofpipebeingsubmerged,thepressure
headatthissectionisequaltotheheightoftheliquidsurfacein
thereservoirBandhencethehydraulicgradientlineattheexit
sectionofpipewillmeettheliquidsurfaceinthereservoirB.

DEPARTMENT OF MEC HANIC AL ENGINEERING
TOTAL ENERGY LINE
•Ifatdifferentsectionsofpipethetotalenergy(intermsofhead)is
plottedtoscaleasverticalordinateabovetheassumeddatumand
allthesepointsarejoined,thenastraightslopinglinewillbe
obtainedandisknownasenergygradelineorTotalenergyline
(T.E.L).
•Sincetotalenergyatanysectionisthesumofthepressurehead
(p/w),datumheadzandvelocityhead
??????
2
2&#3627408468;
andthevertical
distancebetweenthedatumandhydraulicgradelineisequalto
thepiezometrichead(p/w+z),theenergygradelinewillbe
paralleltothehydraulicgradeline,withaverticaldistancebetween
themequalto
??????
2
2&#3627408468;
.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Attheentrancesectionofthepipethereoccurssomelossof
energycalled“Entranceloss”equaltoh
L=0.5
??????
2
2&#3627408468;
andhencethe
energygradelineatthissectionwilllieataverticaldepthequalto
0.5
??????
2
2&#3627408468;
belowtheliquidsurfaceinthereservoirA.

DEPARTMENT OF MEC HANIC AL ENGINEERING
PITOTTUBE
•APitottubeisasimpledevice
usedformeasuringthevelocity
offlow.
•Thebasicprincipleusedinthisis
thatifthevelocityofflowata
particularpointisreducedto
zero,whichisknownas
stagnationpoint,thepressure
thereisincreaseddueto
conversionofthekineticenergy
intopressureenergyandby
measuringtheincreasein
pressureenergyatthispoint,the
velocityofflowmaybe
determined.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Simplestformofapitottubeconsistsofaglasstube,largeenough
forcapillaryeffectstobenegligibleandbentatrightangles.
•Asingletubeofthistypeisusedformeasuringthevelocityofflow
inanopenchannel.
•Thetubeisdippedverticallyintheflowingstreamoffluidwithits
openendAdirectedtofacetheflowandotheropenendprojecting
abovethefluidsurfaceinthestream.
•Thefluidentersthetubeandthelevelofthefluidinthetube
exceedsthatofthefluidsurfaceinthesurroundingstream.Thisis
sobecausetheendAofthetubeisastagnationpoint,wherethe
fluidisatrest,andthefluidapproachingendAdividesatthispoint
andpassesaroundtube.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Sinceatstagnationpointthekineticenergyisconvertedinto
pressureenergy,thefluidinthetuberisesabovethesurrounding
fluidsurfacebyaheight,whichcorrespondstothevelocityofflow
offluidapproachingendAofthetube.
•Thepressureatthestagnationpointisknownasstagnation
pressure.
•Considerapoint1slightlyupstreamofendAandlyingalongthe
samehorizontalplaneintheflowingstreamofvelocityV.
•Nowifthepoint1andAareataverticaldepthofh
Ofromthefree
surfaceoffluidandhistheheightofthefluidraisedinthepitot
tubeabovethefreesurfaceoftheliquid.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•ThenbyapplyingBernoulli’sequationbetweenthepoint1andA,
neglectinglossofenergy,
weget ℎ
&#3627408476;=
??????
2
2&#3627408468;
=ℎ
&#3627408476;+ℎ
•(h
o+h)isthestagnationpressureheadatapointA,whichconsistsof
staticpressureheadh
Oanddynamicpressureheadh.
•Simplifyingtheexpression,
??????
2
2&#3627408468;
=ℎOr&#3627408483;=2&#3627408468;ℎ(1)
•Thisequationindicatesthatthedynamicpressureheadhis
proportionaltothesquareofthevelocityofflowclosetoendA.
•Thusthevelocityofflowatanypointintheflowingstreammaybe
determinedbydippingthePitottubetotherequiredpointand
measuringtheheight‘h’ofthefluidraisedinthetubeabovethefree
surface.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Thevelocityofflowgivenbytheaboveequation(1)ismorethan
actualvelocityofflowasnolossofenergyisconsideredinderiving
theaboveequation.
•WhentheflowishighlyturbulentthePitottuberecordsahigher
valueofh,whichishigherthanthemeanvelocityofflow.
•Inordertotakeintoaccounttheerrorsduetotheabovefactors,
theactualvelocityofflowmaybeobtainedbyintroducingaco-
efficientCorC
VcalledPitottubeco-efficient.
•Sotheactualvelocityisgivenby
(ProbablevalueofCis0.98)&#3627408483;=&#3627408438;2&#3627408468;ℎ

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 8
•Venturimeter
•Derivation of co-efficient of
discharge for venturimeter
•Orificmeter
Measurement of flow
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
VENTURIMETER
•Aventurimeterisadeviceused
formeasuringtherateofflow
offluidthroughapipe.
•Thebasicprincipleonwhich
venturimeterworksisthatby
reducingthecross-sectional
areaoftheflowpassage,a
pressuredifferenceiscreated
andthemeasurementofthe
pressuredifferenceenables
thedeterminationofthe
dischargethroughthepipe.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Aventuremeterconsistsof(1)aninletsection,followedbya
convergingcone(2)acylindricalthroatand(3)agradually
divergentcone.
•Theinletsectionofventuremeteristhesamediameterasthatof
thepipewhichisfollowedbyaconvergentcone.
•Theconvergentconeisashortpipe,whichtapersfromtheoriginal
sizeofthepipetothatofthethroatoftheventuremeter.
•Thethroatoftheventuremeterisashortparallel–sidedtube
havingitscross-sectionalareasmallerthanthatofthepipe.
•Thedivergentconeoftheventuremeterisagraduallydiverging
pipewithitscross-sectionalareaincreasingfromthatofthethroat
totheoriginalsizeofthepipe.
•Attheinletsectionandthethroati.esections1and2ofthe
venturemeterpressuregaugesareprovided.

DEPARTMENT OF MEC HANIC AL ENGINEERING
DERIVATION
•Leta
1anda
2bethecross-sectionareasatinletandthroat
sections,atwhichP
1andP
2thepressuresandvelocitiesV
1andV
2
respectively.
•Assumingtheflowingfluidisincompressibleandthereisnolossof
energybetweensection1and2andapplyingBernoulli’sequation
betweensections1and2,weget,
&#3627408451;
1
??????
+
&#3627408483;
1
2
2&#3627408468;
+??????
1=
&#3627408451;
2
??????
+
&#3627408483;
2
2
2&#3627408468;
+??????
2
Where??????isthespecificweightofflowingfluid.
•If the venturimeter is connected in a horizontal pipe, then Z
1= Z
2,
then
&#3627408451;1
??????
+
??????1
2
2&#3627408468;
=
&#3627408451;2
??????
+
??????2
2
2&#3627408468;
&#3627408451;
1
??????
−
&#3627408451;
2
??????
=
&#3627408483;
2
2
2&#3627408468;
−
&#3627408483;
1
2
2&#3627408468;

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Intheaboveexpression
&#3627408451;1
??????
−
&#3627408451;2
??????
??????&#3627408480;&#3627408481;ℎ&#3627408466;&#3627408477;&#3627408479;&#3627408466;&#3627408480;&#3627408480;&#3627408482;&#3627408479;&#3627408466;&#3627408465;??????&#3627408467;&#3627408467;&#3627408466;&#3627408479;&#3627408466;&#3627408475;&#3627408464;&#3627408466;
betweenthepressureheadsatsection1and2,isknownas
ventureheadandisdenotedbyh
ℎ=
??????2
2
2&#3627408468;
−
??????1
2
2&#3627408468;
&#3627408452;
&#3627408481;ℎ=&#3627408462;
1&#3627408483;
1=&#3627408462;
2&#3627408483;
2,
&#3627408483;
1=
&#3627408452;??????ℎ
&#3627408462;1
,&#3627408483;
2=
&#3627408452;??????ℎ
&#3627408462;2
ℎ=
&#3627408452;??????ℎ
2
2&#3627408468;
1
&#3627408462;2
2
−
1
&#3627408462;1
2
&#3627408452;
&#3627408481;ℎ=
&#3627408462;1&#3627408462;22&#3627408468;ℎ
&#3627408462;1
2
−&#3627408462;2
2

DEPARTMENT OF MEC HANIC AL ENGINEERING
&#3627408452;=&#3627408438;
&#3627408465;&#3627408452;
&#3627408481;ℎ=
&#3627408438;&#3627408465;&#3627408462;1&#3627408462;22&#3627408468;ℎ
&#3627408462;1
2
−&#3627408462;2
2
=&#3627408438;
&#3627408465;&#3627408438;ℎ∵&#3627408438;=
&#3627408462;1&#3627408462;22&#3627408468;
&#3627408462;1
2
−&#3627408462;2
2
C
d=Co-efficientofdischarge<1
??????
??????&#3627408516;&#3627408533;&#3627408534;????????????=??????
&#3627408517;????????????

DEPARTMENT OF MEC HANIC AL ENGINEERING
ORIFICEMETER
•Anorificemeterisasimple
deviceformeasuringthe
dischargethroughpipes.
•Orificemeteralsoworksonthe
sameprincipleasthatof
venturemeteri.ebyreducing
cross-sectionalareaoftheflow
passage,apressuredifference
betweenthetwosectionsis
developed and the
measurementofthepressure
differenceenables the
determinationofthedischarge
throughthepipe.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Orificemeterisacheaperarrangementandrequiressmallerlength
andcanbeusedwherespaceislimited.
•Thisgivesthedischargethroughanorificemeterandissimilartothe
dischargethroughventuremeter.
•Theco-efficientCmaybeconsideredastheco-efficientofdischarge
ofanorificemeter.
•Theco-efficientofdischargeforanorificemeterissmallerthanthat
foraventuremeter.
•Thisisbecausetherearenogradualconverginganddivergingflow
passagesasinthecaseofventuremeter,whichresultsinagreater
lossofenergyandconsequentreductionoftheco-efficientof
dischargeforanorificemeter.
&#3627408452;=&#3627408438;&#3627408462;
0&#3627408462;
1
2&#3627408468;ℎ
&#3627408462;
1
2
−&#3627408462;
0
2

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 9&10
•Example problems on
 Pitottube
 Venturimeter
 Orificemeter
•Applications
•Assignment questions
Example Problems
TOPICS TO BE COVERED

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 1
•Atasuddenenlargementofawatermainfrom240mmto480mm
diameter,thehydraulicgradientrisesby10mm.Estimatetherateofflow.
Sol:GivenData
Dia.ofsmallerpipeD
1=240mm=0.24m
AreaA
1=
&#3627409163;
4
D
1
2
=
&#3627409163;
4
(0.24)
2
Dia.oflargerpipeD
2=480mm=0.48m
AreaA
2=
&#3627409163;
4
&#3627408439;
2
2
=
&#3627409163;
4
(0.48)
2
Riseofhydraulicgradienti.e.&#3627408461;
2+
&#3627408451;2
&#3627409164;&#3627408468;
−&#3627408461;
1+
&#3627408451;1
&#3627409164;&#3627408468;
=10mm=
10
1000
&#3627408474;=
1
100
&#3627408474;
Lettherateofflow=Q
ApplyingBernoulli’sequationtobothsectionsi.esmallerandlargersections
P1
ρg
+
V1
2
2g
+Z
1=
P2
ρg
+
V2
2
2g
+Z
2+Headlossduetoenlargement (1)

DEPARTMENT OF MEC HANIC AL ENGINEERING
But head loss due to enlargement,ℎ
&#3627408466;=
V
1−V
2
2
2g
(2)
From continuity equation, we have A
1V
1= A
2V
2&#3627408457;
1=
??????
2??????
2
??????
1
&#3627408457;
1=
&#3627409163;
4
&#3627408439;
2
2
&#3627408457;
2
&#3627409163;
4
&#3627408439;
1
2
=
&#3627408439;
2
&#3627408439;
1
2
×&#3627408457;
2=
0.48
0.24
2
&#3627408457;
2=2
2
&#3627408457;
2=4&#3627408457;
2
Substituting this value in equation (2), we get
ℎ
&#3627408466;=
4&#3627408457;
2−&#3627408457;
2
2
2&#3627408468;
=
3&#3627408457;
2
2
2&#3627408468;
=
9&#3627408457;
2
2
2&#3627408468;
Now substituting the value of h
eand V
1 in equation (1)
P
1
ρg
+
4&#3627408457;
2
2
2g
+&#3627408461;
1=
P
2
ρg
+
V
2
2
2g
+Z
2+
9&#3627408457;
2
2
2&#3627408468;
16V
2
2
2g
−
V
2
2
2g
−
9&#3627408457;
2
2
2&#3627408468;
=
P
2
ρg
+Z
2−
P
1
ρg
+&#3627408461;
1

DEPARTMENT OF MEC HANIC AL ENGINEERING
But Hydraulic gradient rise =
P2
ρg
+Z
2−
P1
ρg
+&#3627408461;
1=
1
100
m
6V2
2
2g
=
1
100
m &#3627408457;
2=
2×9.81
6×100
=0.1808=0.181&#3627408474;/&#3627408480;&#3627408466;&#3627408464;
Discharge, Q= A
2V
2=
&#3627409163;
4
D
2
2
V
2
=
&#3627409163;
4
(0.48)
2
×0.181 = 0.03275m
3
/sec
Q = 32.75Lts/sec

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 2
•A150mmdia.pipereducesindia.abruptlyto100mmdia.Ifthe
pipecarrieswaterat30lts/sec,calculatethepressurelossacross
thecontraction.Takeco-efficientofcontractionas0.6
Sol:GivenData
Dia. of larger pipe,D
1= 150mm = 0.15m
Area of larger pipe, A
1=
&#3627409163;
4
(0.15)
2
= 0.01767m
2
Dia. of smaller pipe, D
2= 100mm = 0.10m
Area of smaller pipe, A
2=
&#3627409163;
4
(0.10)
2
= 0.007854m
2
Discharge, Q = 30 lts/sec = 0.03m
3
/sec
Co-efficient of contraction, C
C= 0.6

DEPARTMENT OF MEC HANIC AL ENGINEERING
From continuity equation, we have Q = A
1V
1= A
2V
2
&#3627408457;
1=
&#3627408452;
??????1
=
0.03
0.01767
=1.697&#3627408474;/&#3627408480;&#3627408466;&#3627408464;
&#3627408457;
2=
&#3627408452;
??????2
=
0.03
0.007854
=3.82&#3627408474;/&#3627408480;&#3627408466;&#3627408464;
Applying Bernoulli’s equation before and after contraction
P1
ρg
+
V1
2
2g
+Z
1=
P2
ρg
+
V2
2
2g
+Z
2+h
c (1)
But Z
1= Z
2and h
cthe head loss due to contraction is given by the
equation
ℎ
&#3627408464;=
??????2
2
2&#3627408468;
1
&#3627408438;&#3627408464;
−1
2
=
3.82
2
2×9.81
1
0.6
−1
2
=0.33

DEPARTMENT OF MEC HANIC AL ENGINEERING
Substituting these values in equation (1), we get
&#3627408451;1
&#3627409164;&#3627408468;
+
1.697
2
2×9.81
=
&#3627408451;2
&#3627409164;&#3627408468;
+
3.82
2
2×9.81
+0.33
&#3627408451;1
&#3627409164;&#3627408468;
+0.1467=
&#3627408451;2
&#3627409164;&#3627408468;
+0.7438+0.33
&#3627408451;1
&#3627409164;&#3627408468;
−
&#3627408451;2
&#3627409164;&#3627408468;
=0.7438+0.33−0.1467=0.9271&#3627408474;&#3627408476;&#3627408467;&#3627408458;&#3627408462;&#3627408481;&#3627408466;&#3627408479;
&#3627408451;
1−&#3627408451;
2=&#3627409164;&#3627408468;×0.9271=1000×9.81×0.9271=0.909×10
4
??????/&#3627408474;
2
= 0.909 N/cm
2
Pressure loss across contraction =
P
1-P
2= 0.909N/cm
2

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 3
•Apitottubeisplacedinthecentreofa300mmpipelinehasoneend
pointingupstreamandotherperpendiculartoit.Themeanvelocityin
thepipeis0.80ofthecentralvelocity.Findthedischargethroughthe
pipe,ifthepressuredifferencebetweenthetwoorificesis60mmof
water.Co-efficientofPitottubeC
V=0.98
Sol:GivenData
Diameterofpipe=300mm=0.3m
Differenceofpressureheadh=60mmofwater=0.06mofwater
Meanvelocityത&#3627408457;=0.80×centralvelocity
Centralvelocity=&#3627408438;
??????2&#3627408468;ℎ=0.98×2×9.81×0.06=1.063
&#3627408474;
&#3627408480;&#3627408466;&#3627408464;
Meanvelocity=0.8×1.063=0.8504m/sec
Discharge,Q=Areaofpipe×Meanvelocity=A×ത&#3627408457;
=
&#3627409163;
4
(0.3)
2
×0.8504
Q= 0.06m
3
/sec

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 4
•Findthevelocityofflowofanoilthroughapipe,whenthe
differenceofmercurylevelinadifferentialU-tubemanometer
connectedtothetwotappingsofthepitottubeis100mm.Co-
efficientofpitottubeC=0.98andsp.gr.ofoil=0.8.
Sol:GivenData
Differenceofmercurylevelx=100mm=0.1m
Sp.gr.ofoil=0.8,C
V=0.98
Differenceofpressureheadℎ=??????
????????????
??????
−1=0.1
13.6
0.8
−1
=1.6mofoil
Velocityofflow=C
V2×&#3627408468;×ℎ=0.982×9.81×1.6
V= 5.49m/sec

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 5
•Apitottubeisusedtomeasurethevelocityofwaterinapipe.The
stagnaturepressureheadis6mandstaticpressureheadis5m.
Calculatethevelocityofflow.Co-efficientofpitottubeis0.98.
Sol:GivenData
Stagnaturepressurehead,h
S=6m
Staticpressurehead,h
t=5m
h=h
S–h
t=6–5=1m
Velocityofflow,V=C
V2×&#3627408468;×ℎ
=0.982×9.81×1
V=4.34m/sec

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 6
•Apitottubeisinsertedinapipeof300mmdiameter.Thestatic
pressureinthepipeis100mmofmercury(Vacuum).The
stagnationpressureatthecentreofthepipeisrecordedbyPitot
tubeis0.981N/cm
2
.Calculatetherateofflowofwaterthroughthe
pipe.Themeanvelocityofflowis0.85timesthecentralvelocityC
V
=0.98
•Sol:GivenData
Diameterofpiped=0.3m
Areaofpipea=
&#3627409163;
4
(0.3)
2
=0.07068m
2
Staticpressurehead=100mmofmercury=
100
1000
×13.6=1.36mof
water
Stagnationpressurehead=
0.981×10
4
1000
×9.81=1&#3627408474;

DEPARTMENT OF MEC HANIC AL ENGINEERING
h=Stagnationpressurehead–staticpressurehead=1–(-1.36)=
2.36m
Velocityatcentre=C
V2&#3627408468;ℎ=0.982×9.81×2.36=6.668m/sec
Meanvelocityത&#3627408457;=0.85×6.668=5.6678m/sec
Rateofflowofwater=ത&#3627408457;×Areaofpipe
=5.6678×0.07068
Q=0.4006m
3
/sec

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 7

DEPARTMENT OF MEC HANIC AL ENGINEERING
PROBLEM 8

DEPARTMENT OF MEC HANIC AL ENGINEERING
APPLICATIONS
•TheVenturimeterwhichhaslongbeenusedinhydraulicsishere
appliedtothemeasurementofvolumeflowofbloodthrough
vessels.
•Influidflow,frictionloss(orskinfriction)isthelossofpressureor
“head”thatoccursinpipeorductflowduetotheeffectofthe
fluid'sviscositynearthesurfaceofthepipeorduct.
•Inmechanicalsystemssuchasinternalcombustionengines,the
termreferstothepowerlostinovercomingthefrictionbetweentwo
movingsurfaces,adifferentphenomenon.
•Rigorouscalculationofthepressurelossforflowinggases,based
onasproperties,flow,andpipingconfiguration(pipelength,
fittings,andvalves).Resultscanbeprintedoutor"cutandpaste"
intootherapplications.

DEPARTMENT OF MEC HANIC AL ENGINEERING
ASSIGNMENT QUESTIONS
•Tworeservoirsareconnectedbyapipelineconsistingoftwo
pipes,oneof15cmdiameterandlength6mandtheotherdiameter
22.5cmand16mlength.Ifthedifferenceofwaterlevelsinthetwo
reservoirsis6m.Calculatethedischargeanddrawtheenergy
gradientline.Takef=0.004
•Aventurimeterhasanarearatioof9to1,thelargerdiameter
being300mm.Duringtheflow,therecordedpressureheadinthe
largersectionis6.5mandthatatthethroat4.25m.TakeCd=
0.99,computethedischargethroughthemeter.
•Acrudeoilofkinematicviscosity0.4stokeisflowingthroughapipe
ofdiameter300mmattherateof300litres/sec.Findtheheadlost
duetofrictionforalengthof50mofthepipe.
•DeriveDarcyWeisbachequation.

DEPARTMENT OF MEC HANIC AL ENGINEERING
•Therateofflowofwaterthroughahorizontalpipeis0.254m
3
/s.
Thediameterofthepipewhichis200mmissuddenlyenlargedto
400mm.Thepressureintensityinthesmallerpipeis11.772N/cm
2
.
Determinei)Lossofheadduetosuddencontractionii)pressure
intensityinthelargerpipeiii)powerlostduetofriction
•Explainboundarylayerconceptindetail.

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

www.mrcet.ac.in
UNIT 4
CO1:ExplaintheresolutionofAsystemofforces,computetheir
resultantandsolveproblemsusingequationsofequilibrium.
BASICS OF TURBO
MACHINERY&
HYDRAULIC TURBINES

DEPARTMENT OF MECHANICAL ENGINEERING
UNIT –IV (SYLLABUS)
Basics of Turbo Machinery:
•Hydrodynamic force of jets on stationary
•Jet on moving flat,
•jet on inclined, and curved vanes.
Hydraulic Turbines:
•Classification of turbines
•impulse and reaction turbines,
•Peltonwheel turbine,
•Francis turbine and Kaplan turbine
•draft tube.
•Performance of hydraulic turbines: characteristic curves,
cavitation, surge tank, water hammer

DEPARTMENT OF MECHANICAL ENGINEERING
COURSE OUTLINE
LECTURELECTURE TOPIC KEY ELEMENTS LEARNING OBJECTIVES
1
Introduction to hydrodynamic force
on jets
Derivation of force on Stationary-
Flat, inclined & curved plate
Evaluate force exerted (B5)
2Hydrodynamic force on jets
Derivation of force on Moving-
Flat, inclined & curved plates
Evaluate force exerted (B5)
3Example Problems on force on jets for stationary & moving plates
4Classification of turbinesImpulse & Reaction turbines
Understanding the types of
turbines (B2)
5Peltonwheel Turbine
Working principle, derivation of
work done & η
Understand working principle (B2)
Evaluate the efficiency (B5)
6Francis & Kaplan turbine
Working principle, derivation of
work done & η
Understand working principle (B2)
Evaluate the efficiency (B5)
7Hydraulic Design-Draft Tube theoryFunctions & η Evaluate the efficiency (B5)
8Example Problems on turbines
9Geometric similarity Derivation of Unit & specificquantitiesEvaluate the unit quantities (B5)
UNIT -4

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 1
•Derivation of force on Stationary-
Flat, inclined & curved plate
Introduction to hydrodynamic
force on jets
TOPICS TO BE COVERED

DEPARTMENT OF MECHANICAL ENGINEERING
HYDRO-DYNAMIC FORCE OF JETS:
•Theliquidcomesoutintheformofajetfromtheoutletofanozzle,
theliquidisflowingunderpressure.
•Ifsomeplate,whichmaybefixedormoving,isplacedinthepath
ofthejet,aforceisexertedbythejetontheplate.
•ThisforceisobtainedbyNewton’ssecondlawofmotionorfrom
Impulse–Momentumequation.
F=ma

DEPARTMENT OF MECHANICAL ENGINEERING
Thefollowingcasesofimpactofjeti.e.theforceexertedbythejet
onaplatewillbeconsidered.
1)Forceexertedbythejetonastationaryplate,when
a)Plateisverticaltothejet.
b)Plateisinclinedtothejetand
c)Plateiscurved
2)Forceexertedbythejetonamovingplate,when
a)Plateisverticaltothejet.
b)Plateisinclinedtothejet.
c)Plateiscurved.

DEPARTMENT OF MECHANICAL ENGINEERING
JET ON A STATIONARY VERTICAL PLATE
Considerajetofwatercomingoutfromthenozzle,strikesaflatverticalplate.
LetV=Velocityofjet.
d=Diameterofjet.
a=Areaofcross-sectionofjet.=d
2
Thejetofwaterafterstrikingtheplatewillmovealongtheplate.Buttheplateisat
rightanglestothejet.Hencethejetafterstrikingwillbedeflectedthrough90.

DEPARTMENT OF MECHANICAL ENGINEERING
•Theforceexertedbythejetontheplateinthedirectionofjet,
F
x=Rateofchangeofmomentuminthedirectionofforce.
•Forderivingtheaboveequation,wehavetakeninitialvelocityminusfinalvelocityand
notfinalvelocityminusinitialvelocity.Iftheforceexertedonthejetistobe
calculated,thenfinalvelocityminusinitialvelocityistobetaken =
Initial momentum−Final momentum
Time

=
Mass × Initial velocity − Mass × Final velocity
Time

=
&#3627408448;??????&#3627408480;&#3627408480;
&#3627408455;&#3627408470;&#3627408474;&#3627408466;
(Initial velocity – Final velocity)
= &#3627408448;??????&#3627408480;&#3627408480;/&#3627408480;&#3627408466;&#3627408464; × (Velocity of jet before striking – Final velocity of jet after striking)
= &#3627409164; a V (V – o)
&#3627408441;
&#3627408485;=&#3627409164;??????&#3627408457;
2

DEPARTMENT OF MECHANICAL ENGINEERING
JET ON A STATIONARY INCLINED FLAT
PLATE
Letajetofwatercomingoutfromthe
nozzle,strikesaninclinedflatplate.
V=VelocityofjetinthedirectionofX
=Anglebetweenthejetandplate.
a=Areaofcross-sectionofjet.
Massofwaterpersecondstrikingthe
plate=pav

DEPARTMENT OF MECHANICAL ENGINEERING
Letusfindtheforceexertedbythejetontheplateinthedirection
normaltotheplate.LetthisforceisrepresentedbyF
n.
Then
F
n=Massofjetstrikingpersecond*(Initialvelocityofjetbefore
strikinginthedirectionofn–Finalvelocityofjetafterstriking
inthedirectionofn)

DEPARTMENT OF MECHANICAL ENGINEERING= &#3627409164; a V (Vsin &#3627409155; – 0) = &#3627409164; a V
2
sin &#3627409155; -------------- (1)
This force can be resolved in two components, one in the direction of the jet and the other
perpendicular to the direction of flow.
Then we have Fx = Component of Fn in the direction of flow.
&#3627408441;
&#3627408485;=&#3627408441;
&#3627408475;cos 90−&#3627409155; =&#3627408441;
&#3627408475;sin&#3627409155;−&#3627409164;??????&#3627408457;
2
sin&#3627409155;×sin&#3627409155;
&#3627408441;
&#3627408485;=&#3627409164;??????&#3627408457;
2
&#3627408480;&#3627408470;&#3627408475;
2
&#3627409155; (1)
And F y = Component of Fn in the direction perpendicular to the flow.
&#3627408441;
&#3627408486;=&#3627408441;
&#3627408475;sin 90−&#3627409155; =&#3627408441;
&#3627408475;cos&#3627409155;=&#3627409164;??????&#3627408483;
2
sin&#3627409155;×cos&#3627409155;
&#3627408441;
&#3627408486;=&#3627409164;??????&#3627408483;
2
&#3627408480;&#3627408470;&#3627408475;&#3627409155;&#3627408464;&#3627408476;&#3627408480;&#3627409155; (2)

DEPARTMENT OF MECHANICAL ENGINEERING
JET STRIKES THE CURVED PLATE AT
THE CENTREComponent of velocity in the direction of jet = −&#3627408457; &#3627408438;&#3627408476;&#3627408480;&#3627409155;
(-ve sign is taken as the velocity at out let is in the opposite direction of the jet of water coming
out from nozzle.)
Component of velocity perpendicular to the jet = V sin&#3627409155;
Force exerted by the jet in the direction of the jet
Fx = Mass per sec (V1x – V2x)
Where V1x = Initial velocity in the direction of jet = V
V2x = Final velocity in the direction of jet = −&#3627408457; &#3627408438;&#3627408476;&#3627408480;&#3627409155;
Fx = &#3627409164;aV [V – (−&#3627408457; &#3627408438;&#3627408476;&#3627408480;&#3627409155;)] = &#3627409164;aV [V+ V Cos&#3627409155;] = &#3627409164;aV
2
(1 + Cos&#3627409155;) ---------- (1)
Similarly Fy = Mass per second (V1y – V2y)
Where V1y = Initial velocity in the direction of y = 0
V2y = Final velocity in the direction of y = V sin&#3627409155;
Fy = &#3627409164;aV [0 – V Sin&#3627409155;] = −&#3627409164;aV
2
Sin&#3627409155; ------------- (2)

DEPARTMENT OF MECHANICAL ENGINEERING
JET STRIKES THE CURVED PLATE AT ONE END TANGENTIALLY
WHEN THE PLATE IS SYMMETRICALLet V = Velocity of jet of water.
&#3627409155; = Angle made by the jet with x–axis at the inlet tip of the curved plate.

If the plate is smooth and loss of energy due to impact is zero, then the velocity of water at the out
let tip of the curved plate will be equal to V. The force exerted by the jet of water in the direction
of x
and y are
Fx = (mass/sec) × (V1x – V2x)
= &#3627409164;aV [V Cos&#3627409155;−(−&#3627408457; &#3627408438;&#3627408476;&#3627408480;&#3627409155;)]
= 2 ??????aV
2
Cos??????
Fy = &#3627409164;aV (V1y – V2y) = &#3627409164;aV [V sin&#3627409155; −&#3627408457;&#3627408454;&#3627408470;&#3627408475;&#3627409155;] = 0

DEPARTMENT OF MECHANICAL ENGINEERING
JET STRIKES THE CURVED PLATE AT ONE END
TANGENTIALLY WHEN THE PLATE IS UN -SYMMETRICALWhen the curved plate is unsymmetrical about x- axis, then the angles made by tangents drawn at
inlet and outlet tips of the plate with x- axis will be different.
Let &#3627409155; = Angle made by tangent at the inlet tip with x-axis.
∅ = Angle made by tangent at the outlet tip with x-axis
The two components of velocity at inlet are
V1x = V cos&#3627409155; and V1y = V sin&#3627409155;
The two components of velocity at outlet are
V2x = −&#3627408457;cos∅ and V2y = V sin ∅
The forces exerted by the jet of water in the directions of x and y are:
Fx = &#3627409164;aV (V1x – V2x) = &#3627409164;aV (V Cos&#3627409155; + V Cos ∅)
=&#3627409164;??????&#3627408457;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;+&#3627408464;&#3627408476;&#3627408480;∅
Fy = &#3627409164;aV (V1y – V2y) = &#3627409164;aV [V sin&#3627409155;−V sin ∅] = ?????? a V
2
(sin?????? - Sin ∅)

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DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 2
•Flat vertical plate moving in the
direction of jet
•Flat vertical plate moving in away
from the jet
•Flat vertical plate moving in the
direction of jet
•Curved plate moving in the direction
of jet
Hydrodynamic force on jets
on Moving Plates.
TOPICS TO BE COVERED

DEPARTMENT OF MECHANICAL ENGINEERING
FORCE ON FLAT VERTICAL PLATE
MOVING IN THE DIRECTION OF JET
LetV=Velocityofjet.
a=Areaofcross-sectionofjet.
u=Velocityofflatplate.
•Inthiscase,thejetdoesnotstriketheplatewith
avelocityv,butitstrikeswitharelativevelocity,
whichisequaltotheabsolutevelocityofjetof
waterminusvelocityoftheplate.
Hencerelativevelocityofthejetwithrespectto
plate=V–u
Massofwaterstrikingtheplatepersecond=
=a(V–u)
Forceexertedbythejetonthemovinginthe
directionoftheplate
F
x=massofwaterstrikingpersecond(Initial
velocitywithwhichwaterstrikes– Final
velocity)
=a(V–u)[(V–u)–0]=a(V–u)
2

DEPARTMENT OF MECHANICAL ENGINEERING
Sincefinalvelocityinthedirectionofjetiszero.
Inthis,casetheworkwillbedonebythejetontheplate,astheplate
ismoving.Forstationaryplates,theworkdoneiszero.
Theworkdonepersecondbythejetontheplate=
=F
xu=a(v–u)
2
u------------(2)
Intheaboveequation(2),ifthevalueofforwateristakeninS.I
units(i.e.1000kg/m
3
)theworkdonewillbeinNm/s.Thetermis
equaltoWatt(W).

DEPARTMENT OF MECHANICAL ENGINEERING
FORCE ON INCLINED PLATE MOVING IN THE
DIRECTION OF JET
LetV=Absolutevelocityofwater.
u=Velocityofplateinthedirectionofjet.
a=Cross-sectionalareaofjet

DEPARTMENT OF MECHANICAL ENGINEERING &#3627409155; = Angle between jet and plate.
Relative velocity of jet of water = &#3627408457;−&#3627408482;
The velocity with which jet strikes = &#3627408457;−&#3627408482;
Mass of water striking per second = &#3627409164; a (V – u)
If the plate is smooth and loss of energy due to impact of the jet is assumed zero, the jet of water
will leave the inclined plate with a velocity equal to &#3627408457;−&#3627408482; .
The force exerted by the jet of water on the plate in the direction normal to the plate is given as
Fn = Mass striking per sec x (Initial velocity in the normal direction with which jet strikes – final velocity)
= &#3627409164; a (V – u) [(V – u) sin&#3627409155;− 0]
= ?????? a (V– u)
2
sin??????
This normal force Fn is resolved in to two components, namely Fx and Fy in the direction of jet
and perpendicular to the direction of jet respectively.
&#3627408441;
&#3627408485;=&#3627408441;
&#3627408475; &#3627408480;&#3627408470;&#3627408475;&#3627409155;=&#3627409164;?????? &#3627408457;−&#3627408482;
2
&#3627408480;&#3627408470;&#3627408475;
2
&#3627409155;
&#3627408441;
&#3627408486;=&#3627408441;
&#3627408475; &#3627408464;&#3627408476;&#3627408480;&#3627409155;=&#3627409164;?????? &#3627408457;−&#3627408482;
2
&#3627408480;&#3627408470;&#3627408475;&#3627409155;&#3627408464;&#3627408476;&#3627408480;&#3627409155;
Work done per second by the jet on the plate
= Fx × Distance per second in the direction of x
= Fx × u = &#3627409164;a(V – u)
2
sin
2
&#3627409155; × u
= ?????? a (V – u)
2
u sin
2
?????? N m/sec

DEPARTMENT OF MECHANICAL ENGINEERING
FORCE ON THE CURVED PLATE WHEN THE PLATE IS
MOVING IN THE DIRECTION OF JET Let V = absolute velocity of jet.
a = area of jet.
u = Velocity of plate in the direction of jet.
Relative velocity of jet of water or the velocity with which jet strikes the curved plate = V – u
If the plate is smooth and the loss of energy due to impact of jet is zero, then the velocity with
which the
jet will be leaving the curved vane = (V – u)

DEPARTMENT OF MECHANICAL ENGINEERINGThis velocity can be resolved in to two components, one in the direction of jet and the other
perpendicular to the direction of jet.
Component of the velocity in the direction of jet = - (V – u) Cos&#3627409155;
(-ve sign is taken as at the out let, the component is in the opposite direction of the jet).
Component of velocity in the direction perpendicular to the direction of jet = (V – u) sin&#3627409155;
Mass of water striking the plate = &#3627409164;?????? × velocity with which jet strikes the plate.
= &#3627409164;?????? (V – u)
∴Force exerted by the jet of water on the curved plate in the direction of jet Fx
Fx = Mass striking per sec [Initial velocity with which jet strikes the plate in
the direction of jet −Final velocity]
=&#3627409164;?????? &#3627408457;−&#3627408482; &#3627408457;−&#3627408482; − − &#3627408457;−&#3627408482; cos&#3627409155;
=&#3627409164;?????? &#3627408457;−&#3627408482;
2
1+cos&#3627409155; (1)
Work done by the jet on the plate per second
= Fx ×Distance travelled per second in the direction of x
=&#3627408441;
&#3627408485;×&#3627408482;=&#3627409164;?????? &#3627408457;−&#3627408482;
2
1+cos&#3627409155; ×&#3627408482;
=&#3627409164;?????? &#3627408457;−&#3627408482;
2
×&#3627408482; 1+cos&#3627409155;

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DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 3
•PROBLEMS ON STATIONARY
PLATES
•PROBLEMS ON MOVING PLATES
Example Problems on force
on jets for stationary &
moving plates
TOPICS TO BE COVERED

DEPARTMENT OF MECHANICAL ENGINEERING
1.Waterisflowingthroughapipeattheendofwhichanozzleis
fitted.Thediameterofthenozzleis100mmandtheheadofthe
wateratthecentreofthenozzleis100m.Findtheforceexertedby
thejetofwateronafixedverticalplate.Theco-efficientofvelocity
isgivenas0.95.
ANS:GIVEN:Diameterofnozzled=100mm=0.1m
Areaofnozzle=
HeadofwaterH=100m
Co-efficientofvelocity=

DEPARTMENT OF MECHANICAL ENGINEERINGTheoretical velocity of jet of water Vth = 2&#3627408468;ℎ = 2×9.81×100 = 44.294m/sec
But Cv =
&#3627408436;&#3627408464;&#3627408481;&#3627408482;??????&#3627408473; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486;
&#3627408455;ℎ&#3627408466;&#3627408476;&#3627408479;&#3627408470;&#3627408481;&#3627408470;&#3627408464;??????&#3627408473; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486;

∴ Actual velocity of jet of water = &#3627408438;
&#3627408483; × Vth = 0.95 × 44.294 = 42.08m/sec
Force exerted on a fixed vertical plate
F = &#3627409164;??????&#3627408457;
2
=1000×0.007854× 42.08
2

F= 13907.2 N = 13.9 kN

DEPARTMENT OF MECHANICAL ENGINEERING
2.Ajetofwaterofdiameter75mmmovingwithavelocityof25m/sec
strikesafixedplateinsuchawaythattheanglebetweenthejet
andplateis60.Findtheforceexertedbythejetontheplate
I.Inthedirectionnormaltotheplateand
II.Inthedirectionofthejet.
ANS:Given:Diameter of the jet d =75mm = 0.075m
Area of the jet
&#3627409163;
4
× 0.075
2
=0.004417&#3627408474;
2

Velocity of jet V = 25m/sec
Angle between jet and plate &#3627409155; = 60°

DEPARTMENT OF MECHANICAL ENGINEERINGi) The force exerted by the jet of water in the direction normal to the plate
&#3627408441;
&#3627408475;=&#3627409164;??????&#3627408457;
2
&#3627408480;&#3627408470;&#3627408475;&#3627409155;=1000×0.004417× 25
2
&#3627408480;&#3627408470;&#3627408475; 60
0

??????
&#3627408527;=&#3627409360;&#3627409361;&#3627409367;&#3627409358;.&#3627409365; ??????
ii) The force in the direction of jet
&#3627408441;
&#3627408485;=&#3627409164;??????&#3627408457;
2
&#3627408480;&#3627408470;&#3627408475;
2
&#3627409155;=1000×0.004417× 25
2
×&#3627408480;&#3627408470;&#3627408475;
2
60
0

??????
&#3627408537;=&#3627409360;&#3627409358;&#3627409365;&#3627409358;.&#3627409362; ??????

DEPARTMENT OF MECHANICAL ENGINEERING
3.Ajetofwaterofdia.50mmmovingwithavelocityof40m/sec
strikesacurvedfixedsymmetricalplateatthecentre.Findthe
forceexertedbythejetofwaterinthedirectionofjet,ifthe
directionofjetisdeflectedthroughanangleof120attheoutletof
curvedplate.
Ans)Given: Dia. of jet d = 0.05m
Area of jet a =
&#3627409163;
4
× 0.05
2
=0.001963&#3627408474;
2

Velocity of jet V = 40m/sec
Angle of deflection = 180 – &#3627409155; = 180 – 120 = 60°
Force exerted by the jet on the curved plate in the direction of jet
&#3627408441;
&#3627408485;=&#3627409164;??????&#3627408457;
2
1+&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408441;
&#3627408485;=1000×0.001963× 40
2
× 1+cos60
0

??????
&#3627408537;=&#3627409362;&#3627409365;&#3627409359;&#3627409359;.&#3627409359;&#3627409363; ??????

DEPARTMENT OF MECHANICAL ENGINEERING
4.Ajetofwaterofdia.75mmmovingwithavelocityof30m/secstrikesacurved
fixedplatetangentiallyatoneendatanangleof30tothehorizontal.Thejet
leavestheplateatanangleof20
0
tothehorizontal.Findtheforceexertedby
thejetontheplateinthehorizontalandverticaldirection.Given: Dia. of jet d = 75mm = 0.075m,
Area of jet ??????=
&#3627409163;
4
× 0.075
2
=0.004417 &#3627408474;
2

Velocity of jet V = 30m/sec
Angle made by the jet at inlet tip with the horizontal &#3627409155; =30°
Angle made by the jet at out let tip with the horizontal ∅ = 20°
The force exerted by the jet of water on the plate in horizontal direction Fx
&#3627408441;
&#3627408485;=&#3627409164;??????&#3627408457;
2
cos&#3627409155;+cos∅
=1000×0.004417 cos30
0
+cos20
0
× 30
2

??????
&#3627408537;=&#3627409365;&#3627409359;&#3627409365;&#3627409366;.&#3627409360; ??????
The force exerted by the jet of water on the plate in vertical direction Fy
&#3627408441;
&#3627408486;=&#3627409164;??????&#3627408457;
2
&#3627408480;&#3627408470;&#3627408475; &#3627409155;−&#3627408480;&#3627408470;&#3627408475; ∅
=1000×0.004417 sin30
0
−sin20
0
× 30
2

??????
&#3627408538;=&#3627409364;&#3627409360;&#3627409366;.&#3627409359;&#3627409361; ??????

DEPARTMENT OF MECHANICAL ENGINEERING
5.Anozzleof50mmdia.deliversastreamofwaterat20m/secperpendicularto
theplatethatmovesawayfromtheplateat5m/sec.Find:
i)Theforceontheplate.
ii)Theworkdoneand
iii)Theefficiencyofthejet.
Ans)Given: Dia. of jet d = 50mm = 0.05m,
Area of jet a =
&#3627409163;
4
(0.05)
2
= 0.0019635 m
2

Velocity of jet V = 20m/sec,
Velocity of plate u = 5m/sec
i) The force on the plate &#3627408441;
&#3627408485;=&#3627409164;?????? &#3627408457;−&#3627408482;
2

&#3627408441;
&#3627408485;=1000×0.0019635× 20−5
2

??????
&#3627408537;=&#3627409362;&#3627409362;&#3627409359;.&#3627409365;&#3627409366; ??????

DEPARTMENT OF MECHANICAL ENGINEERINGii) The work done by the jet =&#3627408441;
&#3627408485;×&#3627408482;
=441.78×5
=&#3627409360;&#3627409360;&#3627409358;&#3627409366;.&#3627409367; ??????&#3627408526; /&#3627408532;
iii) The efficiency of the jet &#3627409154;=
&#3627408450;&#3627408482;&#3627408481; &#3627408477;&#3627408482;&#3627408481; &#3627408476;&#3627408467; &#3627408471;&#3627408466;&#3627408481;
??????&#3627408475;&#3627408477;&#3627408482;&#3627408481; &#3627408476;&#3627408467; &#3627408471;&#3627408466;&#3627408481;

=
&#3627408458;&#3627408476;&#3627408479;&#3627408472; &#3627408465;&#3627408476;&#3627408475;&#3627408466;/&#3627408480;&#3627408466;&#3627408464;
??????.&#3627408440; &#3627408476;&#3627408467; &#3627408471;&#3627408466;&#3627408481;/&#3627408480;&#3627408466;&#3627408464;
=
&#3627408441;
&#3627408485; ×&#3627408482;
1
2
&#3627408474;&#3627408483;
2

=
&#3627408441;
&#3627408485; × &#3627408482;
1
2
&#3627409164;??????&#3627408457; &#3627408457;
2

=
2208.9
1
2
1000×0.0019635×20 ×20
2
=
2208.9
6540

= 0.3377 = 33.77%

DEPARTMENT OF MECHANICAL ENGINEERING
6.A7.5cmdia.jethavingavelocityof30m/secstrikesaflatplate,thenormalof
whichisinclinedat45totheaxisofthejet.Findthenormalpressureonthe
plate:(i)Whentheplateisstationaryand(ii)Whentheplateismovingwitha
velocityof15m/secawayfromthejet.Alsodeterminethepowerandefficiency
ofthejetwhentheplateismoving.
Given: Dia.ofthejetd=7.5cm=0.075mArea of jet a =
&#3627409163;
4
(0.075)
2
= 0.004417m
2

Angle between jet and plate &#3627409155; = 90° – 45° = 45°
Velocity of jet V = 30m/sec
i) When the plate is stationary, the normal force &#3627408441;
&#3627408475; on the plate is
&#3627408441;
&#3627408475;=&#3627409164;??????&#3627408457;
2
sin&#3627409155;=1000×0.004417× 30
2
×sin45
0

=&#3627409360;&#3627409366;&#3627409359;&#3627409358;.&#3627409367;&#3627409364; ??????
ii) When the plate is moving with a velocity of 15m/sec away from the jet, the normal force on
the plate &#3627408441;
&#3627408475;
&#3627408441;
&#3627408475;= &#3627409164;?????? &#3627408457;−&#3627408482;
2
sin&#3627409155;=1000×0.004417× 30−15
2
×sin45
0

= 702.74 N

DEPARTMENT OF MECHANICAL ENGINEERINGiii) The power and efficiency of the jet, when the plate is moving is obtained as
Work done /sec by the jet
= Force in the direction of jet x Distance moved by plate in the direction of jet/sec
=&#3627408441;
&#3627408485;×&#3627408482; Where &#3627408441;
&#3627408485;=&#3627408441;
&#3627408475;sin&#3627409155;=702.74 ×sin45
0
=496.9 &#3627408449;
Work done/ sec = 496.9 × 15 = 7453.5Nm/s
∴ Power in kW =
&#3627408458;&#3627408476;&#3627408479;&#3627408472; &#3627408465;&#3627408476;&#3627408475;&#3627408466; /&#3627408480;&#3627408466;&#3627408464;
1000
=
7453.5
1000
= 7.453 kW
Efficiency of jet =
&#3627408450;&#3627408482;&#3627408481; &#3627408477;&#3627408482;&#3627408481;
??????&#3627408475;&#3627408477;&#3627408482;&#3627408481;
=
&#3627408458;&#3627408476;&#3627408479;&#3627408472; &#3627408465;&#3627408476;&#3627408475;&#3627408466; &#3627408477;&#3627408466;&#3627408479; &#3627408480;&#3627408466;&#3627408464;
??????.&#3627408440; &#3627408476;&#3627408467; &#3627408471;&#3627408466;&#3627408481; &#3627408477;&#3627408466;&#3627408479; &#3627408480;&#3627408466;&#3627408464;

=
7453.5
1
2
&#3627409164;??????&#3627408457; ×&#3627408457;
2=
7453.5
1
2
&#3627409164;??????&#3627408457;
3

=
7453.5
1
2
× 1000 ×0.004417 ×30
3

= 0.1249 ≃ 0.125 =12.5%

DEPARTMENT OF MECHANICAL ENGINEERING
7.Ajetofwaterofdia.7.5cmstrikesacurvedplateatitscentrewithavelocity
of20m/sec.thecurvedplateismovingwithavelocityof8m/secinthedirection
ofthejet.Thejetisdeflectedthroughanangleof165.Assumingplateis
smooth,find
1.Forceexertedontheplateinthedirectionofjet.
2.Powerofjet.
3.Efficiencyofjet.Given: Dia. of jet d = 7.5cm = 0.075m
Area of jet a =
&#3627409163;
4
(0.075)
2
= 0.004417m
2

Velocity of jet V = 20m/sec
Velocity of plate u = 8m/sec
Angle made by the relative velocity at the out let of the plate &#3627409155; = 180 °-165° =15°

DEPARTMENT OF MECHANICAL ENGINEERINGi) Force exerted by the jet on the plate in the direction of jet
&#3627408441;
&#3627408485;= &#3627409164;?????? &#3627408457;−&#3627408482;
2
1+cos&#3627409155;
&#3627408441;
&#3627408485;=1000×0.004417× 20−8
2
1+cos15
0

=&#3627409359;&#3627409360;&#3627409363;&#3627409358;.&#3627409361;&#3627409366;??????
ii) Work done by the jet on the plate per second
=&#3627408441;
&#3627408485;×&#3627408482;
=1250.38×8
=&#3627409359;&#3627409358;&#3627409358;&#3627409358;&#3627409361;.&#3627409358;&#3627409362; ??????&#3627408526;/&#3627408532;
∴ Power of jet =
10003.04
1000
= 10kW
iii)Efficiency of the jet =
&#3627408450;&#3627408482;&#3627408481; &#3627408477;&#3627408482;&#3627408481;
??????&#3627408475;&#3627408477;&#3627408482;&#3627408481;
=
&#3627408458;&#3627408476;&#3627408479;&#3627408472; &#3627408465;&#3627408476;&#3627408475;&#3627408466; &#3627408477;&#3627408466;&#3627408479; &#3627408480;&#3627408466;&#3627408464;
??????.&#3627408440; &#3627408476;&#3627408467; &#3627408471;&#3627408466;&#3627408481; &#3627408477;&#3627408466;&#3627408479; &#3627408480;&#3627408466;&#3627408464;

=
1250.38 × 8
1
2
&#3627409164;??????&#3627408457;.&#3627408457;
2
=
1250.38 × 8
1
2
×1000 ×0.004417 × (20)20
3

= 0.564 = 56.4%

DEPARTMENT OF MECHANICAL ENGINEERING
8.Ajetofwaterhavingavelocityof40m/secstrikesacurvedvane,whichis
movingwithavelocityof20m/sec.Thejetmakesanangleof30withthe
directionofmotionofvaneatinletandleavesatanangleof90tothedirectionof
motionofvaneatoutlet.Drawvelocitytrianglesatinletandoutletanddetermine
vaneanglesatinletandoutlet,sothatthewaterentersandleavesthevanes
withoutshock.
Ans)Given: Velocity of jet &#3627408457;
1= 40m/sec
Velocity of vane u1 = 20m/sec
Angle made by jet at inlet &#3627409148; =30°
Angle made by leaving jet = 90°
∴&#3627409149; = 180° - 90° = 90°
u1 = u2 = u = 20m/sec

DEPARTMENT OF MECHANICAL ENGINEERINGVane angles at inlet and outlet are &#3627409155; and ∅
From ∆ BCD we have tan &#3627409155;=
&#3627408437;&#3627408439;
&#3627408438;&#3627408439;
=
&#3627408437;&#3627408439;
&#3627408436;&#3627408439;−&#3627408436;&#3627408438;
=
&#3627408457;
&#3627408467;1
&#3627408457;
&#3627408484;1
−&#3627408482;
1

Where &#3627408457;
&#3627408467;1
=&#3627408457;
1sin&#3627409148;=40×sin30
0
=20&#3627408474;/&#3627408480;
&#3627408457;
&#3627408484;1
=&#3627408457;
1cos&#3627409148;=40×cos30
0
=34.64 &#3627408474;/&#3627408480;
&#3627408482;
1=20&#3627408474;/&#3627408480;
∴ tan&#3627409155;=
20
34.64−20
=
20
14.64
=1.366=tan53.79
0

∴ ??????=&#3627409363;&#3627409361;.&#3627409365;&#3627409367;
&#3627409358;
&#3627408528;&#3627408531; &#3627409363;&#3627409361;
&#3627409358;
&#3627409362;&#3627409365;.&#3627409362;
′

Also from ∆ BCD we have sin&#3627409155;=
&#3627408457;
&#3627408467;1
&#3627408457;
&#3627408479;1
&#3627408476;&#3627408479; &#3627408457;
&#3627408479;
1
=
&#3627408457;
&#3627408467;1
sin&#3627409155;
=
20
sin53.79
0
=24.78 &#3627408474;/&#3627408480;
∴ &#3627408457;
&#3627408479;1
=24.78&#3627408474;/&#3627408480;
But &#3627408457;
&#3627408479;2
=&#3627408457;
&#3627408479;1
=24.78
Hence, From ∆ EFG, cos∅=
&#3627408482;
2
&#3627408457;
&#3627408479;2
=
20
24.78
=0.8071=cos36.18
0

∅=&#3627409361;&#3627409364;.&#3627409359;&#3627409366;
&#3627409358;
&#3627408528;&#3627408531; &#3627409361;&#3627409364;
&#3627409358;
&#3627409359;&#3627409358;.&#3627409366;
′

DEPARTMENT OF MECHANICAL ENGINEERING
9.Astationaryvanehavinganinletangleofzerodegreeandanoutletangleof
25,receiveswateratavelocityof50m/sec.Determinethecomponentsofforce
actingonitinthedirectionofjetvelocityandnormaltoit.Alsofindtheresultant
forceinmagnitudeanddirectionperunitweightoftheflowGiven: Velocity of jet V = 50m/sec
Angle at outlet =25°
For the stationary vane, the force in the direction of jet.
&#3627408441;
&#3627408485;=&#3627408448;??????&#3627408480;&#3627408480; &#3627408477;&#3627408466;&#3627408479; &#3627408480;&#3627408466;&#3627408464; × &#3627408457;
1&#3627408485;−&#3627408457;
2&#3627408485;
Where &#3627408457;
1&#3627408485;= 50m/sec , &#3627408457;
2&#3627408485;=−50cos25
0
= −45.315

DEPARTMENT OF MECHANICAL ENGINEERING ∴ Force in the direction of jet per unit weight of water &#3627408441;
&#3627408485;
&#3627408441;
&#3627408485;=
&#3627408448;??????&#3627408480;&#3627408480;/sec[50− −45.315 ]
&#3627408458;&#3627408466;&#3627408470;&#3627408468;ℎ&#3627408481; &#3627408476;&#3627408467; &#3627408484;??????&#3627408481;&#3627408466;&#3627408479; /&#3627408480;&#3627408466;&#3627408464;
=
&#3627408448;??????&#3627408480;&#3627408480;/sec⁡[50+45.315]
&#3627408474;??????&#3627408480;&#3627408480;/&#3627408480;&#3627408466;&#3627408464; ×&#3627408468;

=
95.315
9.81
=9.716 &#3627408449;/&#3627408449;
Force exerted by the jet in perpendicular direction to the jet per unit weight of flow
&#3627408457;
1&#3627408486;=0 &#3627408457;
2&#3627408486;=50sin25
0

&#3627408441;
&#3627408486;=
&#3627408448;??????&#3627408480;&#3627408480;/sec(&#3627408457;
1&#3627408486;− &#3627408457;2&#3627408486;
)
&#3627408468;× &#3627408474;??????&#3627408480;&#3627408480; &#3627408477;&#3627408466;&#3627408479; &#3627408480;&#3627408466;&#3627408464;

=
(0−50 &#3627408454;&#3627408470;&#3627408475;25°)
&#3627408468;
=
−50 &#3627408454;&#3627408470;&#3627408475;25°
9.81

= - 2.154 N
- ve sign means the force Fy is acting in the downward direction.
∴ Resultant Force per unit weight of water FR = &#3627408441;
&#3627408485;
2
+&#3627408441;
&#3627408486;
2

&#3627408441;
&#3627408453;= (9.716)
2
+(2.154)
2
= 9.952N
The angle made by the Resultant Force with x- axis
tan &#3627409155; =
Fy
Fx
=
2.154
9.716
=0.2217
?????? = &#3627408533;??????&#3627408527;
−&#3627409359;
0.2217 = 12.50°

DEPARTMENT OF MECHANICAL ENGINEERING
10.Ajetofwaterdiameter50mmmovingwithavelocityof25m/sec
impingesonafixedcurvedplatetangentiallyatoneendatanangleof30to
thehorizontal.Calculatetheresultantforceofthejetontheplate,ifthejetis
deflectedthroughanangleof50.Takeg=10m/sec
2
.Given:
Dia. of jet d =50mm = 0.05m,
Area of jet ??????=
&#3627409163;
4
(0.05)
2
= 0.0019635 m
2

Velocity of jet V = 25m/sec,
Angle made by the jet at inlet with horizontal &#3627409155; = 30°
Angle of deflection = 50°
Angle made by the jet at the outlet with horizontal ∅
∅=&#3627409155;+??????&#3627408475;&#3627408468;&#3627408473;&#3627408466; &#3627408476;&#3627408467; &#3627408465;&#3627408466;&#3627408467;&#3627408473;&#3627408466;&#3627408464;&#3627408481;&#3627408470;&#3627408476;&#3627408475; = 30°+50° = 80°

DEPARTMENT OF MECHANICAL ENGINEERINGThe Force exerted by the jet of water in the direction of x
&#3627408441;
&#3627408485;=&#3627409164;??????&#3627408457; &#3627408457;
1&#3627408485;−&#3627408457;
2&#3627408485;
Where &#3627409164;=1000
??????=
&#3627409163;
4
0.05
2
V = 25 m/s
&#3627408457;
1&#3627408485;=&#3627408457;cos30
0
=25cos30
0

&#3627408457;
2&#3627408485;=&#3627408457;cos80
0
=25cos80
0

&#3627408441;
&#3627408485;=1000×
&#3627409163;
4
0.05
2
×25 25cos30
0
−25cos80
0
=&#3627409366;&#3627409362;&#3627409367;.&#3627409365; ??????
The Force exerted by the jet of water in the direction of y
&#3627408441;
&#3627408486;=&#3627409164;??????&#3627408457; &#3627408457;
1&#3627408486;−&#3627408457;
2&#3627408486;
&#3627408441;
&#3627408486;=1000×
&#3627409163;
4
0.05
2
×25 25sin30
0
−25sin80
0
=−&#3627409363;&#3627409367;&#3627409362;.&#3627409367; ??????
- ve sign shows that Force Fy is acting in the downward direction.
The Resultant force FR = &#3627408441;
&#3627408485;
2
+&#3627408441;
&#3627408486;
2

= (849.7)
2
+(594.9)
2

= 1037N
Angle made by the Resultant Force with the Horizontal
tan&#3627409148;=
&#3627408441;
&#3627408486;
&#3627408441;
&#3627408485;
=
594.9
849.7
=0.7
?????? = &#3627408533;??????&#3627408527;
−&#3627409359;
0.7 = 35°

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DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 4
oImpulse turbine and
oReaction turbine
CLASSIFICATION OF
HYDRAULIC TURBINES:
TOPICS TO BE COVERED

DEPARTMENT OF MECHANICAL ENGINEERING
CLASSIFICATION OF HYDRAULIC TURBINES:
TheHydraulicturbinesareclassifiedaccordingtothetypeofenergy
availableattheinletoftheturbine,directionofflowthroughthevanes,head
attheinletoftheturbineandspecificspeedoftheturbine.Thefollowingare
theimportantclassificationoftheturbines.
1.Accordingtothetypeofenergyatinlet:
(a)Impulseturbineand
(b)Reactionturbine
2.Accordingtothedirectionofflowthroughtherunner:
(a)Tangentialflowturbine
(b)Radialflowturbine.
(c)Axialflowturbine
(d)Mixedflowturbine.

DEPARTMENT OF MECHANICAL ENGINEERING
3. According to the head at inlet of the turbine:
(a)High head turbine
(b)Medium head turbine and
(c)Low head turbines.
4. According to the specific speed of the turbine:
(a)Low specific speed turbine
(b)Medium specific speed turbine
(c)High specific speed turbine.

DEPARTMENT OF MECHANICAL ENGINEERING
•Ifattheinletofturbine,theenergyavailableisonlykineticenergy,
theturbineisknownasImpulseturbine.
•Asthewaterflowsoverthevanes,thepressureisatmospheric
frominlettooutletoftheturbine.Ifattheinletoftheturbine,the
waterpossesseskineticenergyaswellaspressureenergy,the
turbineisknownasReactionturbine.
•Asthewaterflowsthroughrunner,thewaterisunderpressureand
thepressureenergygoesonchangingintokineticenergy.The
runneriscompletelyenclosedinanair-tightcasingandtherunner
andcasingiscompletelyfullofwater.

DEPARTMENT OF MECHANICAL ENGINEERING
•Ifthewaterflowsalongthetangentofrunner,theturbineisknown
asTangentialflowturbine.
•Ifthewaterflowsintheradialdirectionthroughtherunner,the
turbineiscalledRadialflowturbine.
•Ifthewaterflowsfromoutwardtoinwardsradially,theturbineis
knownasInwardradialflowturbine,ontheotherhand

DEPARTMENT OF MECHANICAL ENGINEERING
•ifthewaterflowsradiallyfrominwardtooutwards,theturbineis
knownasoutwardradialflowturbine.
•Ifthewaterflowsthroughtherunneralongthedirectionparallelto
theaxisofrotationoftherunner,theturbineiscalledaxialflow
turbine
•Ifthewaterflowsthroughtherunnerintheradialdirectionbut
leavesinthedirectionparalleltotheaxisofrotationoftherunner,
theturbineiscalledmixedflowturbine.

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DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 5
•Working principle
•Derivation of work done & η
Peltonwheel Turbine:
TOPICS TO BE COVERED

DEPARTMENT OF MECHANICAL ENGINEERING
PELTON WHEEL (TURBINE)
•Itisatangentialflowimpulseturbine.Thewaterstrikesthebucket
alongthetangentoftherunner.Theenergyavailableattheinletof
theturbineisonlykineticenergy.Thepressureattheinletandout
letofturbineisatmospheric.Thisturbineisusedforhighheads
andisnamedafterL.A.PeltonanAmericanengineer.
•Thewaterfromthereservoirflowsthroughthepenstocksattheout
letofwhichanozzleisfitted.Thenozzleincreasesthekinetic
energyofthewaterflowingthroughthepenstock.Attheoutletof
thenozzle,thewatercomesoutintheformofajetandstrikesthe
buckets(vanes)oftherunner.

DEPARTMENT OF MECHANICAL ENGINEERING
•Nozzle and flow regulating arrangement (spear)
•Runner and Buckets.
•Casing and
•Breaking jet
MAIN PARTS OF PELTON WHEEL TURBINE

DEPARTMENT OF MECHANICAL ENGINEERING
1.Nozzleandflowregulatingarrangement:Theamountofwater
strikingthebuckets(vanes)oftherunneriscontrolledbyproviding
aspearinthenozzle.Thespearisaconicalneedlewhichis
operatedeitherbyhandwheelorautomaticallyinanaxialdirection
dependinguponthesizeoftheunit.Whenthespearispushed
forwardintothenozzle,theamountofwaterstrikingtherunneris
reduced.Ontheotherhand,ifthespearispushedback,the
amountofwaterstrikingtherunnerincreases.
2.Runnerwithbuckets:Itconsistsofacirculardisconthe
peripheryofwhichanumberofbucketsevenlyspacedarefixed.
Theshapeofthebucketsisofadoublehemisphericalcuporbowl.
Eachbucketisdividedintotwosymmetricalpartsbyadividing
wall,whichisknownassplitter.

DEPARTMENT OF MECHANICAL ENGINEERING
3.Casing:Thefunctionofcasingistopreventthesplashingofthe
waterandtodischargethewatertotailrace.Italsoactsas
safeguardagainstaccidents.ItismadeofCastIronorfabricated
steelplates.ThecasingofthePeltonwheeldoesnotperformany
hydraulicfunction.
4.Breakingjet:Whenthenozzleiscompletelyclosedbymovingthe
spearintheforwarddirection,theamountofwaterstrikingthe
runnerreducestozero.Buttherunnerduetoinertiagoeson
revolvingforalongtime.Tostoptherunnerinashorttime,asmall
nozzleisprovided,whichdirectsthejetofwateronthebackofthe
vanes.ThisjetofwateriscalledBreakingjet.

DEPARTMENT OF MECHANICAL ENGINEERING
VELOCITY TRIANGLES AND WORK DONE FOR
PELTONWHEEL
•Thejetofwaterfromthenozzlestrikesthebucketatthesplitter,
whichsplitsupthejetintotwoparts.
•Thesepartsofthejet,glidesovertheinnersurfacesandcomes
outattheouteredge.
•Thesplitteristheinlettipandouteredgeofthebucketistheoutlet
tipofthebucket.Theinletvelocitytriangleisdrawnatthesplitter
andoutervelocitytriangleisdrawnattheouteredgeofthebucket.

DEPARTMENT OF MECHANICAL ENGINEERINGLet
H = Net head acting on the Pelton Wheel

Where = Gross Head

Where =diameter of penstock,
D = Diameter of wheel,
d = Diameter of Jet,
N = Speed of the wheel in r.p.m
Then V1 = Velocity of jet at inlet

DEPARTMENT OF MECHANICAL ENGINEERINGThe Velocity Triangle at inlet will be a straight line where

From the velocity triangle at outlet, we have

The force exerted by the Jet of water in the direction of motion is
(1)
As the angle β is an acute angle, +ve sign should be taken. Also this is the case of
series of vanes, the mass of water striking is and not. In equation (1) ‘a’ is the area
of the jet =
Now work done by the jet on the runner per second

Power given to the runner by the jet

DEPARTMENT OF MECHANICAL ENGINEERINGWork done/s per unit weight of water striking/s
(3)
The energy supplied to the jet at inlet is in the form of kinetic energy
K.E. of jet per second
Hydraulic efficiency,

(4)
Now

And

DEPARTMENT OF MECHANICAL ENGINEERINGSubstituting the values of and in equation (4)

(5)
The efficiency will be maximum for a given value of when

Or
Or
Or Or (6)
Equation (6) states that hydraulic efficiency of a Pelton wheel will be maximum when the
velocity of the wheel is half the velocity of the jet water at inlet. The expression for
maximum efficiency will be obtained by substituting the value of in equation (5)
Max.

DEPARTMENT OF MECHANICAL ENGINEERING
RADIAL FLOW REACTION TURBINE
•IntheRadialflowturbineswaterflowsintheradialdirection.
•Thewatermayflowradiallyfromoutwardstoinwards(i.e.towards
theaxisofrotation)orfrominwardstooutwards.
•Ifthewaterflowsfromoutwardstoinwardsthroughtherunner,the
turbineisknownasinwardsradialflowturbine.Andifthewaterflows
frominwardstooutwards,theturbineisknownasoutwardradialflow
turbine.
MainpartsofaRadialflow
Reactionturbine:
Casing
Guidemechanism
Runnerand
Drafttube

DEPARTMENT OF MECHANICAL ENGINEERING
1.Casing:incaseofreactionturbine,casingandrunnerarealwaysfull
ofwater.Thewaterfromthepenstocksentersthecasingwhichisof
spiralshapeinwhichareaofcross-sectiononeofthecasinggoeson
decreasinggradually.Thecasingcompletelysurroundstherunnerof
theturbine.Thewaterenterstherunneratconstantvelocity
throughoutthecircumferenceoftherunner.
2.GuideMechanism:Itconsistsofastationarycircularwheelall
aroundtherunneroftheturbine.Thestationaryguidevanesarefixed
ontheguidemechanism.Theguidevanesallowthewatertostrike
thevanesfixedontherunnerwithoutshockatinlet.Alsobysuitable
arrangement,thewidthbetweentwoadjacentvanesofguide
mechanismcanbealteredsothattheamountofwaterstrikingthe
runnercanbevaried.
3.Runner:Itisacircularwheelonwhichaseriesofradialcurvedvanes
arefixed.Thesurfacesofthevanesaremadeverysmooth.Theradial
curvedvanesaresoshapedthatthewaterentersandleavesthe
runnerwithoutshock.Therunnersaremadeofcaststeel,castironor
stainlesssteel.Theyarekeyedtotheshaft.

DEPARTMENT OF MECHANICAL ENGINEERING
4.Draft-Tube:Thepressureattheexitoftherunnerofareaction
turbineisgenerallylessthanatmosphericpressure.Thewaterat
exitcan’tbedirectlydischargedtothetailrace.Atubeorpipeof
graduallyincreasingareaisusedfordischargingthewaterfrom
theexitoftheturbinetothetailrace.Thistubeofincreasingarea
iscalleddraft-tube.
5.InwardRadialFlowTurbine:Intheinwardradialflowturbine,in
whichcasethewaterfromthecasingentersthestationaryguiding
wheel.Theguidingwheelconsistsofguidevaneswhichdirectthe
watertoentertherunnerwhichconsistsofmovingvanes.The
waterflowsoverthemovingvanesintheinwardradialdirection
andisdischargedattheinnerdiameteroftherunner.Theouter
diameteroftherunneristheinletandtheinnerdiameteristhe
outlet.

DEPARTMENT OF MECHANICAL ENGINEERING
VELOCITY TRIANGLES AND WORK DONE BY
WATER ON RUNNER:Work done per second on the runner by water

(1)
The equation represents the energy transfer per second to the runner.
Where Velocity of whirl at inlet
Velocity of whirl at outlet
Tangential velocity at inlet
, Where Outer dia. Of runner,
Tangential velocity at outlet

DEPARTMENT OF MECHANICAL ENGINEERINGWhere Inner dia. Of runner,
N = Speed of the turbine in r.p.m.
The work done per second per unit weight of water per second

(2)
Equation (2) represents the energy transfer per unit weight/s to
the runner. This equation is known by Euler’s equation.

DEPARTMENT OF MECHANICAL ENGINEERINGIn equation +ve sign is taken if β is an acute angle,
-ve sign is taken if β is an obtuse angle.
If then and work done per second per unit weight of water striking/s
Work done
Hydraulic efficiency
(3)
Where R.P. = Runner Power i.e. power delivered by water to the runner
W.P. = Water Power
If the discharge is radial at outlet, then

DEPARTMENT OF MECHANICAL ENGINEERING
1.Apeltonwheelhasameanbucketspeedof10m/swithajetof
waterflowingattherateof700lts/secunderaheadof30m.the
bucketsdeflectthejetthroughanangleof160
0
calculatethe
powergivenbythewatertotherunnerandhydraulicefficiencyof
theturbine?Assumeco-efficientofvelocity=0.98
ANS)Given:
Speed of bucket
Discharge Q = 700lt/sec=0.7m
3
/s
Head of water H = 30m
Angle deflection = 160
0

Co-efficient of velocity
The velocity of jet

DEPARTMENT OF MECHANICAL ENGINEERING

From the outlet velocity triangle

Work done by the jet/sec on the runner is given by equation

Power given to the turbine
The hydraulic efficiency of the turbine is given by equation

Or = 94.54%

DEPARTMENT OF MECHANICAL ENGINEERING
2.Areactionturbineworksat450rpmunderaheadof120m.itsdiameteratinletis
120cmandflowareais0.4m
2
.Theanglesmadebyabsoluteandrelativevelocities
atinletare20
0
and60
0
respectively,withthetangentialvelocity.Determine
i)Volumeflowrateii)thepowerdeveloped
iii)Thehydraulicefficiency.Assumewhirlatoutletiszero.
AnsGiven: Speed of turbine N = 450rpm
Head H = 120m
Diameter of inlet D1 = 120cm=1.2m
Flow area
Angle made by absolute velocity
Angle made by relative velocity
Whirl at outlet
Tangential velocity of the turbine at inlet

From inlet triangle

DEPARTMENT OF MECHANICAL ENGINEERING
(1)
Also

From equation (1)
i) Volume flow rate is given by equation as

ii) Work done per second on the turbine is given by equation

Power developed in
iii) The hydraulic efficiency is given by equation

= 85.95%

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DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 6
•Working principle
•Derivation of work done & η
FRANCIS & KAPLAN
TURBINE:
TOPICS TO BE COVERED

DEPARTMENT OF MECHANICAL ENGINEERING
FRANCIS TURBINE
•Theinwardflowreactionturbinehavingradialdischargeatoutletis
knownasFrancisTurbine.Thewaterenterstherunnerofthe
turbineintheradialdirectionatoutletandleavesintheaxial
directionattheinletoftherunner.ThustheFrancisturbineisa
mixedflowtypeturbine.
•Theworkdonebywaterontherunnerpersecondwillbe=
The work done per second per unit weight of water striking/sec =
Hydraulic efficiency

DEPARTMENT OF MECHANICAL ENGINEERINGImportant relations for Francis turbines:
1. The ratio of width of the wheel to its diameter is given as. The value of n
varies from 0.10 to 0.40
2. The flow ratio is given as
Flow ratio and varies from 0.15 to 0.30
3. The speed ratio varies from 0.6 to 0.9

DEPARTMENT OF MECHANICAL ENGINEERING
OUTWARD RADIAL FLOW REACTION TURBINE: In this case as the inlet of the runner is at the inner diameter of the runner, the tangential
velocity at inlet will be less than that of an outlet. i.e.
As
All the working conditions flow through the
runner blades without shock. As such eddy
losses which are inevitable in Francis and
propeller turbines are almost completely
eliminated in a Kaplan turbine.
The discharge through the runner is
obtained as

Where outer diameter of the runner
Diameter of the hub
Velocity of flow at inlet

DEPARTMENT OF MECHANICAL ENGINEERING
IMPORTANT POINTS FOR KAPLAN TURBINE:The peripheral velocity at inlet and outlet are equal.
Where Outer diameter of runner.
2. Velocity of flow at inlet and outlet are equal.

3. Area of flow at inlet = Area of flow at outlet

AXIAL FLOW REACTION TURBINE
1.Propeller Turbine
2.Kaplan Turbine

DEPARTMENT OF MECHANICAL ENGINEERING
KAPLAN TURBINE
•The main parts of the Kaplan turbine are:
1.Scroll casing
2.Guide vanes mechanism
3.Hub with vanes or runner of the turbine
4.Draft tube

DEPARTMENT OF MECHANICAL ENGINEERING
WORKING PROPORTIONS OF KAPLAN
TURBINEThe main dimensions of Kaplan Turbine runners are similar to Francis turbine runner.
However the following are main deviations,
i. Choose an appropriate value of the ratio, where d in hub or boss diameter and
D is runner outside diameter. The value of n varies from 0.35 to 0.6
ii. The discharge Q flowing through the runner is given by

The value of flow ratio for a Kaplan turbine is 0.7
iii. The runner blades of the Kaplan turbine are twisted, the blade angle being greater at
the outer tip than at the hub. This is because the peripheral velocity of the blades
being directly proportional to radius. It will valy from section to section along the
blade, and hence in order to have shock free entry and exit of water over the blades
with angles varying from section to section will have to be designed.

DEPARTMENT OF MECHANICAL ENGINEERING
AFrancisturbinewithanoverallefficiencyof75%isrequiredtoproduce148.25kW
power.Itisworkingunderaheadof7.62m.Theperipheralvelocity=0.26andthe
radialvelocityofflowatinletis0.96.Thewheelrunsat150rpmandthehydraulic
lossesintheturbineare22%oftheavailableenergy.Assumingradialdischarge
determine
i)Theguidebladeangle ii)Thewheelvaneangleatinlet
iii)Thediameterofthewheelatinlet,andiv)Widthofthewheelatinlet
AnsGiven: Overall efficiency
Head H=7.62rpm
Power Produced S.P. = 148.25kW
Speed N= 150rpm
Hydraulic loses =22% of energy
Peripheral velocity
Discharge at outlet = Radial

DEPARTMENT OF MECHANICAL ENGINEERING
The hydraulic efficiency

But ,
.
i) The guide blade angle i.e. From inlet velocity triangle

ii) The wheel angle at inlet

DEPARTMENT OF MECHANICAL ENGINEERINGiii) The diameter of wheel at inlet
Using relation

iv) Width of the wheel at inlet

But

Using equation

DEPARTMENT OF MECHANICAL ENGINEERING
•AKaplanturbinerunneristobedesignedtodevelop7357.5kWshaft
power.Thenetavailableheadis5.50m.Assumethatthespeedratiois
2.09andflowratiois0.68andtheoverallefficiencyis60%.Thediameter
ofbossisofthediameterofrunner.Findthediameteroftherunner,its
speedandspecificspeedGiven: Shaft power P = 7357.5kW
Head H = 5.5m
Speed ratio =

Flow ratio
Overall Efficiency
Diameter of boss
Using the relation

DEPARTMENT OF MECHANICAL ENGINEERING
Discharge
Using equation for discharge

Using the relation

The specific speed

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DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 7
•Draft Tube
•Functions & efficiency
Hydraulic Design-Draft
Tube theory:
TOPICS TO BE COVERED

DEPARTMENT OF MECHANICAL ENGINEERING
DRAFT TUBE:
•Thedrafttubeisapipeofgraduallyincreasingarea,which
connectstheoutletoftherunnertothetailrace.
•Itisusedfordischargingwaterfromtheexitoftheturbinetothe
tailrace.Thispipeofgraduallyincreasingareaiscalledadraft
tube.
•Oneendofthedrafttubeisconnectedtotheoutletoftherunner
andtheotherendissubmergedbelowthelevelofwaterinthetail
race.

DEPARTMENT OF MECHANICAL ENGINEERING
TYPES OF DRAFT TUBE
•Types of Draft Tube:
1.Conical Draft Tube
2.Simple Elbow Tubes
3.Moody Spreading tubes
4.Elbow Draft Tubes with Circular inlet and rectangular outlet

DEPARTMENT OF MECHANICAL ENGINEERING
DRAFT TUBE THEORYConsider a conical draft tube
Vertical height of draft tube above tail race
Y= Distance of bottom of draft tube from tail race.
Applying Bernoulli’s equation to inlet section 1-1 and outlet section 2-2of the draft
tube and taking section 2-2 a datum, we get

DEPARTMENT OF MECHANICAL ENGINEERINGWhere loss of energy between section 1-1 and 2-2.
But Atmospheric Pressure + y =
Substituting this value of in equation (1) we get

(2)

DEPARTMENT OF MECHANICAL ENGINEERINGEfficiency of Draft Tube: the efficiency of a draft tube is defined as the ratio of actual
conversion of kinetic head in to pressure in the draft tube to the kinetic head at the inlet of the
draft tube.

Let Velocity of water at inlet of draft tube
Velocity of water at outlet of draft tube
Loss of head in the draft tube
Theoretical conversion of Kinetic head into Pressure head in
Draft tube
Actual conversion of Kinetic head into pressure head
Now Efficiency of draft tube

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DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 8
•Derivation of Units&
•Specific quantities
•Surge tank
•Water hammer
Geometric similarity
TOPICS TO BE COVERED

DEPARTMENT OF MECHANICAL ENGINEERING
GEOMETRIC SIMILARITY
•Thegeometricsimilaritymustexistbetweenthemodelandits
prototype.theratioofallcorrespondinglineardimensionsinthe
modelanditsprototypeareequal.Let length of model
Breadth of model
Diameter of model
Area of model
Volume of model

DEPARTMENT OF MECHANICAL ENGINEERINGAnd Corresponding values of the proto type.
For geometrical similarity between model and prototype, we must have the relation,

Where is called scale ratio.
For area’s ratio and volume’s ratio the relation should be,

DEPARTMENT OF MECHANICAL ENGINEERING
PERFORMANCE OF HYDRAULIC TURBINES
•Inordertopredictthebehaviorofaturbineworkingundervarying
conditionsofhead,speed,outputandgateopening,theresults
areexpressedintermsofquantitieswhichmaybeobtainedwhen
theheadontheturbineisreducedtounity.Theconditionsofthe
turbineunderunitheadaresuchthattheefficiencyoftheturbine
remainsunaffected.
Thethreeimportantunitquantitiesare:
1.Unitspeed,
2.Unitdischarge,and
3.Unitpower

DEPARTMENT OF MECHANICAL ENGINEERING1. Unit Speed: it is defined as the speed of a turbine working under a unit head. It is denoted
by . The expression of unit speed is obtained as:
Let N = Speed of the turbine under a head H
H = Head under which a turbine is working
u = Tangential velocity.
The tangential velocity, absolute velocity of water and head on turbine are related as:
Where
(1)
Also tangential velocity (u) is given by
Where D = Diameter of turbine.

DEPARTMENT OF MECHANICAL ENGINEERINGFor a given turbine, the diameter (D) is constant
Or Or
(2) Where is constant of proportionality.
If head on the turbine becomes unity, the speed becomes unit speed or
When H = 1, N = Nu
Substituting these values in equation (2), we get

Substituting the value of K1 in equation (2)
(I)
2. Unit Discharge: It is defined as the discharge passing through a turbine, which is working
under a unit head (i.e. 1 m). It is denoted by the expression for unit discharge is given as:
Let H = head of water on the turbine
Q = Discharge passing through turbine when head is H on the turbine.
= Area of flow of water

DEPARTMENT OF MECHANICAL ENGINEERINGThe discharge passing through a given turbine under a head ‘H’ is given by,
Q=Area of flow Velocity
But for a turbine, area of flow is constant and velocity is proportional to

Or (3)
Where is constant of proportionality
If H = 1, (By definition)
Substituting these values in equation (3) we get

Substituting the value of in equation (3) we get

(II)
3. Unit Power: It is defined as the power developed by a turbine working under a unit head
(i.e. under a head of 1m). It is denoted by . The expression for unit power is obtained as:
Let H= Head of water on the turbine
P= Power developed by the turbine under a head of H
Q= Discharge through turbine under a head H

DEPARTMENT OF MECHANICAL ENGINEERINGThe overall efficiency is given as

(4) Where is a constant of proportionality
When H=1 m,

Substituting the value of in equation (4) we get

(III)

DEPARTMENT OF MECHANICAL ENGINEERING
UNIT QUANTITIESUse of Unit Quantities :
If a turbine is working under different heads, the behaviour of the turbine can be
easily known from the values of the unit quantities i.e. from the value of unit speed, unit
discharge and unit power.
Let H1, H2, H3, -------- are the heads under which a turbine works,
N1, N2, N3, --------- are the corresponding speeds,
Q1, Q2, Q3, -------- are the discharge and
P1, P2, P3, --------- are the power developed by the turbine.
Using equation I, II, III respectively,
(IV)
Hence, if the speed, discharge and power developed by a turbine under a head are known,
then by using equation (IV) the speed, discharge, power developed by the same turbine a
different head can be obtained easily.

DEPARTMENT OF MECHANICAL ENGINEERING
CHARACTERISTIC CURVES OF HYDRAULIC
TURBINES:
•Characteristiccurvesofahydraulicturbinearethecurves,withthe
helpofwhichtheexactbehaviourandperformanceoftheturbine
underdifferentworkingconditionscanbeknown.Thesecurvesare
plottedfromtheresultsofthetestsperformedontheturbineunder
differentworkingconditions.
1)Speed(N)
2)Head(H)
3)Discharge(Q)
4)Power(P)
5)OverallEfficiency()and
6)Gateopening.

DEPARTMENT OF MECHANICAL ENGINEERING
MAIN CHARACTERISTIC CURVES OR
CONSTANT HEAD CURVES

DEPARTMENT OF MECHANICAL ENGINEERING
CAVITATION:
Cavitationisdefinedasthephenomenonofformationofvapour
bubblesofaflowingliquidinaregion,wherethepressureofthe
liquidfallsbelowitsvapourpressureandthesuddencollapsingof
thesevapourbubblesinaregionofhigherpressure.
PrecautionagainstCavitation:
•Thepressureoftheflowingliquidinanypartofthehydraulic
systemshouldnotbeallowedtofallbelowitsvapourpressure.If
theflowingliquidiswater,thentheabsolutepressureheadshould
notbebelow2.5mofwater.
•ThespecialmaterialsorcoatingssuchasAluminum-bronzeand
stainlesssteel,whicharecavitationresistantmaterials,shouldbe
used.
EffectsofCavitation

DEPARTMENT OF MECHANICAL ENGINEERING
•Themetallicsurfacesaredamagedandcavitiesareformedonthe
surfaces.
Duetosuddencollapseofvapourbubbles,considerablenoiseand
vibrationsareproduced.
•Theefficiencyofaturbinedecreasesduetocavitation.Dueto
pittingaction,thesurfaceoftheturbinebladesbecomesroughand
theforceexertedbythewaterontheturbinebladesdecreases.
Hence,theworkdonebywateroroutputhorsepowerbecomesless
andefficiencydecreases.

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 9
•Surge tank
•Water hammer
•Applications
•Assignment questions
Surge Tank & Water
Hammer
TOPICS TO BE COVERED

DEPARTMENT OF MECHANICAL ENGINEERING
SURGE TANK

DEPARTMENT OF MECHANICAL ENGINEERING
WATER HAMMER
Thepressureriseduetowaterhammerdependsupon:
1.Velocityofflowofwaterinpipe.
2.Thelengthofpipe.
3.Timetakentoclosethevalve.
4.Elasticpropertiesofthematerialofthepipe.
Thefollowingcasesofwaterhammerinpipeswillbeconsidered.
1.Gradualclosureofvalve
2.Suddenclosureofvalveconsideringpipeinrigid
3.Suddencloserofvalveconsideringpipeelastic.

DEPARTMENT OF MECHANICAL ENGINEERING
APPLICATIONS
•Peltonwheelsarepreferredturbineforhydropower,whentheavailablewater
sourcehasrelativelyhighhydraulicheadatlowflowrates.
•Peltonwheelsaremadeinallsizes.Formaximumpowerandefficiency,the
wheelandturbinesystemisdesignedsuchwaythatthewaterjetvelocityis
twicethevelocityoftherotatingbuckets.
•ThereexistinmultitonPeltonwheelmountedonverticaloilpadbearingin
hydroelectricpower.
•Kaplanturbinesarewidelyusedthroughouttheworldforelectricalpower
production.
•InexpensivemicroturbinesonKaplanturbinemodelaremanufacturedfor
individualpowerproductionwithaslittleastwofeetofhead.
•LargeKaplanturbinesareindividuallydesignedforeachsitetooperateatthe
highestpossibleefficiency.,typicallyover90%.Theyareveryexpensiveto
design,manufactureandinstall,butoperatefordecades.
•Francisturbinesareusedforpumpedstorage,whereareservoirisfilledbythe
turbine(actingasapump)drivenbythegeneratoractingasalargeelectrical
motorduringperiodsoflowpowerdemand.

DEPARTMENT OF MECHANICAL ENGINEERING
ASSIGNMENT QUESTIONS
•APeltonwheelistobedesignedforthefollowingspecifications.
Power=735.75kWS.Phead=200m,Speed=800rpm,overall
efficiency=0.86andjetdiameterisnottoexceedone-tenththewheel
diameter.Determine:(i)Wheeldiameter,(ii)thenoofjetsrequired
and(iii)diameterofthejet.TakeCv=0.98andspeedratio=0.45.
•Afrancisturbinewithanoverallefficiencyof70%isrequiredto
produce147.15kW.Itisworkingunderaheadof8m.Theperipheral
velocity=0.30andtheradialvelocityofflowatinletis0.96.Thewheel
runsat200rpmandthehydrauliclossesintheturbineare20%ofthe
availableenergy.Assumeradialdischarge,determine:(i)theguide
bladeangle,(ii)thewheelvaneangleatinlet(iii)thediameterof
wheelatinletand(iv)widthofwheelatinlet.
•OneoftheKaplanturbine,installedatGanguwalpowerhouseisrated
at25000kWwhenworkingunder30mofheadat180rpm.Findthe
diameteroftherunner,iftheoverallefficiencyoftheturbineis0.91.
Assumeflowratioof0.65anddiameterofrunnerhubequalto0.3
timestheexternaldiameterofrunner.Alsofindspecificspeedofthe
turbine.

DEPARTMENT OF MECHANICAL ENGINEERING
•Aturbinedevelops1000kWunderaheadof16mat200rpm,
whiledischarging9cubicmetresofwater/sec.Findtheunitpower
andunitdischargeofthewheel.
•Areactionturbineworksat500rpmunderaheadof100m.The
diameterofturbineatinletis100cmandflowareais0.365m2.
Theanglesmadebyabsoluteandrelativevelocitiesatinletare15
0and60degreesrespectivelywiththetangentialvelocity.
Determine:(i)Thevolumeflowrate(ii)thepowerdeveloped(iii)
efficiency.Assumethewhirlvelocityaszeroandunitdischargeof
thewheel.

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DEPARTMENT OF MECHANICAL ENGINEERING
UNIT5

DEPARTMENT OF MECHANICAL ENGINEERING
UNIT 5
CENTRIFUGAL &
RECIPROCATING
PUMPS
CO5:Tobeabletounderstandtheworkingofpowerabsorbingdevices
likepumps&abletoanalyzetheirperformancecharacteristics.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
UNIT –V
Centrifugal Pumps:
•Classification, working-work done.
•Manometric head and efficiencies.
•Specific speed.
•Performance characteristic curves, NPSH.
Reciprocating Pumps:
•Working, discharge.
•Slip and Indicator diagram.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
COURSEOUTLINE
LECTURE LECTURETOPIC KEYELEMENTS Learningobjectives
1 Introduction to Pumps Classification of
centrifugal pumps
Understand types
of pumps (B2)
2 Centrifugal Pump-work done
Manometric Head
& η
Evaluate ηof pump
(B5)
3 Specific Speed of Centrifugal Pump
Performance Understand the
performance of pump
(B2)
4 Characteristic curves of pump
NPSH
Performance Understand the
performance of pump
(B2)
Evaluate NPSH (B5)
5 Problems on Centrifugal pumps
6 Reciprocating pumps Working, Q & η Evaluate η of pump
(B5)
7 Slip& Indicator diagram Positive
displacement
Evaluate slip of pump
(B5)
Understand Indicator
diagram (B2)
8 Problems on Reciprocating pumps

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 1
•Introduction
•Main parts of centrifugal pump
•Heads on Centrifugal pump
•Efficiencies
Introduction to pumps
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
INTRODUCTION TO CENTRIFUGAL PUMPS
•Thehydraulicmachineswhichconvertthemechanicalenergyinto
hydraulicenergyarecalledpumps.
•Thehydraulicenergyisintheformofpressureenergy.Ifthe
mechanicalenergyisconvertedintopressureenergybymeansof
centrifugalforceactingonthefluid,thehydraulicmachineiscalled
centrifugalpump.
•Thecentrifugalpumpactsasareversedofaninwardradialflow
reactionturbine.Thismeansthattheflowincentrifugalpumpsisin
theradialoutwarddirections.
•Thecentrifugalpumpworksontheprincipleofforcedvertexflow
whichmeansthatwhenacertainmassofliquidisrotatedbyan
externaltorque,theriseinpressureheadoftherotatingliquid
takesplace.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
MAIN PARTS OF A CENTRIFUGAL PUMP
•Impeller:Therotatingpartofacentrifugalpumpiscalledimpeller.
Itconsistsofaseriesofbackwardcurvedvanes.
Theimpellerismountedonashaftwhichisconnectedtotheshaft
ofanelectricmotor
•Casing:Itissimilartothecasingofareactionturbine.Itisanair
tightpassagesurroundingtheimpellerandisdesignedinsucha
waythatthekineticenergyofthewaterdischargedattheoutletof
theimpellerisconvertedintopressureenergybeforethewater
leavesthecasingandentersthedeliverypipe.
•Thefollowingthreetypesofthecasingarecommonlyadopted.
Volute
Vortex
Casingwithguideblades

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Suctionpipewithfootvalveandastrainer:Apipewhoseone
endisconnectedtotheinletofthepumpandotherenddipsinto
waterinasumpisknownassuctionpipe.
Afootvalvewhichisanon-returnvalveorone-waytypeofvalveis
fittedatthelowerendofthesuctionpipe.
Thefootvalveopensonlyintheupwarddirection.
Astrainerisalsofittedatthelowerendofthesuctionpipe.
•Deliverypipe:Apipewhoseoneendisconnectedtotheoutletof
thepumpandtheotherenddeliversthewaterattherequired
heightisknownasdeliverypipe.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
CENTRIFUGAL PUMP PARTS

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
HEADS OF A CENTRIFUGAL PUMPS
•SuctionHead&#3627408521;
??????:Itistheverticalheightofthecentrelineof
centrifugalpump,abovethewatersurfaceinthetankorsumpfrom
whichwateristobelifted.Thisheightisalsocalledsuctionlift′ℎ
&#3627408480;′.
•DeliveryHead&#3627408521;
??????:Theverticaldistancebetweenthecentreline
ofthepumpandthewatersurfaceinthetanktowhichwateris
deliveredisknownasdeliveryhead.Thisisdenotedby′ℎ
&#3627408465;′.
•StaticHead??????
??????:Thesumofsuctionheadanddeliveryheadis
knownasstaticshead′??????
&#3627408480;′.
• ??????
&#3627408480;=ℎ
&#3627408480;+ℎ
&#3627408465;
•ManometricHead??????
&#3627408526;:Manometricheadisdefinedasthehead
againstwhichacentrifugalpumphastowork.Itisdenotedby??????
&#3627408474;.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
a)??????
&#3627408474;=Head imparted by the impeller to the water –Loss of head in
the pump=
&#3627408457;
&#3627408484;2
&#3627408482;
2
&#3627408468;
−Loss of head in impeller and casing
=
&#3627408457;
&#3627408484;2
&#3627408482;
2
&#3627408468;
………… If loss of head in pump is zero.
b)??????
&#3627408474;=Total head at outlet of pump –Total head at the inlet of the
pump=
&#3627408451;
0
&#3627409164;&#3627408468;
+
&#3627408457;
0
2
2&#3627408468;
+&#3627408461;
0−
&#3627408451;
??????
&#3627409164;&#3627408468;
+
&#3627408457;
??????
2
2&#3627408468;
+&#3627408461;
??????
Where
&#3627408451;
0
&#3627409164;&#3627408468;
=Pressure head at outlet of the pump = ℎ
&#3627408465;
&#3627408457;0
2
2&#3627408468;
=Velocity head at outlet of the pump
= Velocity head in delivery pipe =
&#3627408457;
&#3627408465;
2
2&#3627408468;
&#3627408461;
0=Vertical height of the outlet of the pump from datum line, and

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
&#3627408451;
??????
&#3627409164;&#3627408468;
,
&#3627408457;
??????
2
2&#3627408468;
,&#3627408461;
??????=Corresponding values of pressure head, velocity
head and datum head at theInlet of the pump,i.e. ℎ
&#3627408480;,
&#3627408457;
&#3627408480;
2
2&#3627408468;
&#3627408462;&#3627408475;&#3627408465;&#3627408461;
&#3627408480;
respectively.
c)??????
&#3627408474;=ℎ
&#3627408480;+ℎ
&#3627408465;+ℎ
&#3627408467;
&#3627408480;
+ℎ
&#3627408467;
&#3627408465;
+
&#3627408457;&#3627408465;
2
2&#3627408468;
Where ℎ
&#3627408480;=Suction head,
ℎ
&#3627408465;=Delivery head,
ℎ
&#3627408467;&#3627408480;
=Frictional head loss in suction pipe,
ℎ
&#3627408467;
&#3627408465;
=Frictional head loss in delivery pipe
&#3627408457;
&#3627408465;=Velocity of water in delivery pipe.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
EFFICIENCIES OF CENTRIFUGAL PUMP
•ManometricEfficiency??????
&#3627408526;??????&#3627408527;:TheratiooftheManometricheadto
theheadimpartedbytheimpellertothewaterisknownas
&#3627408448;&#3627408462;&#3627408475;&#3627408476;&#3627408474;&#3627408466;&#3627408481;&#3627408479;??????&#3627408464;&#3627408440;&#3627408467;&#3627408467;??????&#3627408464;??????&#3627408466;&#3627408475;&#3627408464;??????&#3627409154;
&#3627408474;&#3627408462;&#3627408475;=
&#3627408448;&#3627408462;&#3627408475;&#3627408476;&#3627408474;&#3627408466;&#3627408481;&#3627408479;??????&#3627408464;ℎ&#3627408466;&#3627408462;&#3627408465;
??????&#3627408466;&#3627408462;&#3627408465;??????&#3627408474;&#3627408477;&#3627408462;&#3627408479;&#3627408481;&#3627408466;&#3627408465;&#3627408463;????????????&#3627408474;&#3627408477;&#3627408466;&#3627408473;&#3627408473;&#3627408466;&#3627408479;&#3627408481;&#3627408476;&#3627408484;&#3627408462;&#3627408481;&#3627408466;&#3627408479;
=
????????????
??????&#3627408484;2
&#3627408482;2
??????
=
&#3627408468;????????????
&#3627408457;&#3627408484;2
&#3627408482;2
•MechanicalEfficiency??????
&#3627408526;:Thepowerattheshaftofthecentrifugal
pumpismorethepoweravailableattheimpellerofthepump.The
ratioofthepoweravailableattheimpellertothepowerattheshaftof
thecentrifugalpumpisknownasmechanicalefficiency.
&#3627409154;
&#3627408474;=
&#3627408451;&#3627408476;&#3627408484;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408481;ℎ&#3627408466;??????&#3627408474;&#3627408477;&#3627408466;&#3627408473;&#3627408473;&#3627408466;&#3627408479;
&#3627408451;&#3627408476;&#3627408484;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408481;ℎ&#3627408466;&#3627408480;ℎ&#3627408462;&#3627408467;&#3627408481;
•ThepowerattheimpellerinkW=
&#3627408458;&#3627408476;&#3627408479;&#3627408472;&#3627408465;&#3627408476;&#3627408475;&#3627408466;&#3627408463;????????????&#3627408474;&#3627408477;&#3627408466;&#3627408473;&#3627408473;&#3627408466;&#3627408479;&#3627408477;&#3627408466;&#3627408479;&#3627408480;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;
1000
=
&#3627408458;
&#3627408468;
×
&#3627408457;&#3627408484;2
×&#3627408482;2
1000

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
??????
&#3627408526;=
&#3627408510;
&#3627408520;
&#3627408509;??????
&#3627409360;
×??????
&#3627409360;
&#3627409359;&#3627409358;&#3627409358;&#3627409358;
??????.&#3627408503;
Where S.P. = Shaft power.
•Overall Efficiency??????
&#3627409358;:It is defined as the ratio of power output of
the pump to the power input to the pump.
The power output of the pump in kW=
&#3627408458;&#3627408466;??????&#3627408468;ℎ&#3627408481;&#3627408476;&#3627408467;&#3627408484;&#3627408462;&#3627408481;&#3627408466;&#3627408479;&#3627408473;??????&#3627408467;&#3627408481;&#3627408466;&#3627408465;×??????
??????
1000
=
&#3627408458;??????
??????
1000
The power input to the pump = Power supplied by the electric motor
= S.P. Of the pump
∴ ??????
&#3627409358;=
&#3627408510;??????&#3627408526;
&#3627409359;&#3627409358;&#3627409358;&#3627409358;
??????.&#3627408503;.
??????
&#3627409358;=??????
&#3627408526;??????&#3627408527;×??????
&#3627408526;

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 2
•Work done by Centrifugal pump
•Velocity Triangle diagram
•Derivation
Work done by Centrifugal
pump
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
WORK DONE BY CENTRIFUGAL PUMP
DERIVATION
•Inthecentrifugalpump,workis
donebytheimpelleronthe
water.
•Theexpressionforthework
donebytheimpelleronthe
waterisobtainedbydrawing
velocitytrianglesatinletand
outletoftheimpellerinthe
samewayasforaturbine.
VELOCITY TRIANGLE
DIAGRAM

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Thewaterenterstheimpellerradiallyatinletforbestefficiencyof
thepump,whichmeanstheabsolutevelocityofwateratinlet
makesanangleof90
0
withthedirectionofmotionoftheimpeller
atinlet.
•Henceangle&#3627409148;=90
0
and&#3627408457;
&#3627408484;
1
=0fordrawingthevelocitytriangles
thesamenotationsareusedasthatforturbines.
•LetN=Speedoftheimpellerinr.p.m.
&#3627408439;
1=Diameterofimpelleratinlet
&#3627408482;
1=Tangentialvelocityofimpelleratinlet=
&#3627409163;??????
1&#3627408449;
60
&#3627408439;
2=Diameterofimpelleratoutlet
&#3627408482;
2=Tangentialvelocityofimpelleratoutlet=
&#3627409163;??????
2&#3627408449;
60
&#3627408457;
1= Absolute velocity of water at inlet.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
&#3627408457;
&#3627408479;1
=Relativevelocityofwateratinlet
&#3627409148;=Anglemadebyabsolutevelocity&#3627408457;
1atinletwiththedirection
ofmotionofvane
&#3627409155;=Anglemadebyrelativevelocity(&#3627408457;
&#3627408479;1
)atinletwiththedirection
ofmotionofvane
And&#3627408457;
2,&#3627408457;
&#3627408479;2
,&#3627409149;&#3627408462;&#3627408475;&#3627408465;∅arethecorrespondingvalvesatoutlet.
•Asthewaterenterstheimpellerradiallywhichmeanstheabsolute
velocityofwateratinletisintheradialdirectionandhenceangle&#3627409148;=
90
0
and&#3627408457;
&#3627408484;1
=0.
•Acentrifugalpumpisthereverseofaradiallyinwardflowreaction
turbine.Butincaseofaradiallyinwardflowreactionturbine,thework
donebythewaterontherunnerpersecondperunitweightofthe
waterstrikingpersecondisgivenbyequation.
=
1
&#3627408468;
&#3627408457;
&#3627408484;1
&#3627408482;
1−&#3627408457;
&#3627408484;2
&#3627408482;
2

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
∴Work done by the impeller on the water per second per unit weight of
water striking/second
=−&#3627408484;&#3627408476;&#3627408479;&#3627408472;&#3627408465;&#3627408476;&#3627408475;&#3627408466;??????&#3627408475;&#3627408464;&#3627408462;&#3627408480;&#3627408466;&#3627408476;&#3627408467;&#3627408462;&#3627408481;&#3627408482;&#3627408479;&#3627408463;??????&#3627408475;&#3627408466;
=−
1
&#3627408468;
&#3627408457;
&#3627408484;1
&#3627408482;
1−&#3627408457;
&#3627408484;2
&#3627408482;
2
=
1
&#3627408468;
&#3627408457;
&#3627408484;2
&#3627408482;
2−&#3627408457;
&#3627408484;1
&#3627408482;
1
=
1
&#3627408468;
&#3627408457;
&#3627408484;2
&#3627408482;
2 (1) ∵&#3627408457;
&#3627408484;1
=0
•Work done by the impeller on water per second=
&#3627408458;
&#3627408468;
×&#3627408457;
&#3627408484;2
&#3627408482;
2
Where W = Weight of water =&#3627409164;×&#3627408468;×&#3627408452;, Q = Volume of water

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
Q = Area ×Velocity of flow
=&#3627409163;&#3627408439;
1&#3627408437;
1×&#3627408457;
&#3627408467;
1
=&#3627409163;&#3627408439;
2&#3627408437;
2×&#3627408457;
&#3627408467;
2
•Where &#3627408437;
1and &#3627408437;
2are width of impeller at inlet and outlet and
&#3627408457;
&#3627408467;
1
And &#3627408457;
&#3627408467;
2
are velocities of flow at inlet and outlet
•Head imparted to the water by the impeller or energy given by
impeller to water per unit weight per second
H =
&#3627409359;
&#3627408520;
&#3627408509;
??????&#3627409360;
??????
&#3627409360;

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 3
•Specific speed of centrifugal pump
•Multi stage centrifugal pump
To produce high heads
To produce high discharge
Specific speed of
Centrifugal pump
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
SPECIFIC SPEED OF CENTRIFUGAL PUMP
•Thespecificspeedofacentrifugalpumpisdefinedasthespeedof
ageometricallysimilarpump,whichwoulddeliveronecubicmeter
ofliquidpersecondagainstaheadofonemeter.
•Itisdenotedby′&#3627408449;
&#3627408480;′.
•ThedischargeQforacentrifugalpumpisgivenbytherelation
&#3627408452;=&#3627408436;&#3627408479;&#3627408466;&#3627408462;×&#3627408457;&#3627408466;&#3627408473;&#3627408476;&#3627408464;??????&#3627408481;??????&#3627408476;&#3627408467;&#3627408467;&#3627408473;&#3627408476;&#3627408484;
=&#3627409163;&#3627408439;×&#3627408437;×&#3627408457;
&#3627408467;Or &#3627408452;∝&#3627408439;×&#3627408437;×&#3627408457;
&#3627408467;(1)
WhereD=Diameteroftheimpellerofthepumpand
B=Widthoftheimpeller
•Weknowthat&#3627408437;∝&#3627408439;
•Fromequation(1)wehave&#3627408452;∝&#3627408439;
2
×&#3627408457;
&#3627408467; (2)

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Wealsoknowthatthetangentialvelocityisgivenby
&#3627408482;=
&#3627409163;??????&#3627408449;
60
∝&#3627408439;&#3627408449;(3)
•Nowthetangentialvelocity(u)andvelocityofflow&#3627408457;
&#3627408467;arerelatedto
Manometrichead??????
&#3627408474;as&#3627408482;∝&#3627408457;
&#3627408467;∝??????
&#3627408474; (4)
•Substitutingthevalueof(u)inequation(3),weget
??????
&#3627408474;∝&#3627408439;&#3627408449;Or &#3627408439;∝
????????????
&#3627408449;
•SubstitutingthevaluesofDinequation(2)
&#3627408452;∝
????????????
&#3627408449;
2
×&#3627408457;
&#3627408467;
∝
????????????
&#3627408449;
2
×??????
&#3627408474;∵&#3627408441;&#3627408479;&#3627408476;&#3627408474;&#3627408466;&#3627408478;4&#3627408457;
&#3627408467;∝??????
&#3627408474;
∝
????????????
3/2
&#3627408449;
2
&#3627408452;=&#3627408446;
????????????
3/2
&#3627408449;
2
(5)WhereKisaconstantof
proportionality

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•If ??????
&#3627408474;=1&#3627408474;and &#3627408452;=1&#3627408474;
3
/&#3627408480;&#3627408466;&#3627408464;N becomes &#3627408449;
&#3627408480;
•Substituting these values in equation (5), we get
1=&#3627408446;
1
3/2
??????
&#3627408474;
=
&#3627408446;
&#3627408449;
&#3627408480;
2
∴ &#3627408446;=&#3627408449;
&#3627408480;
2
•Substituting the value of K in equation (5), we get
&#3627408452;=&#3627408449;
&#3627408480;
2
??????
&#3627408474;
3/2
&#3627408449;
2
&#3627408476;&#3627408479;&#3627408449;
&#3627408480;
2
=
&#3627408449;
2
&#3627408452;
??????
&#3627408474;
3/2
??????
??????=
??????&#3627408504;
??????
&#3627408526;
&#3627409361;/&#3627409362;

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
MULTISTAGE CENTRIFUGAL PUMPS
1. TO PRODUCE HIGH HEADS
•Ifahighheadistobe
developed,theimpellersare
connectedinseries(oronthe
same shaft)whilefor
discharginglargequantityof
liquid,theimpellers(orpumps)
areconnectedinparallel.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Fordevelopingahighhead,anumberofimpellersaremountedin
seriesonthesameshaft.
•Thewaterfromsuctionpipeentersthe1
st
impelleratinletandis
dischargedatoutletwithincreasedpressure.
•Thewaterwithincreasedpressurefromtheoutletofthe1
st
impelleris
takentotheinletofthe2
nd
impellerwiththehelpofaconnectingpipe.
•Attheoutletofthe2
nd
impellerthepressureofthewaterwillbemore
thanthewaterattheoutletofthe1
st
impeller.
•Thusifmoreimpellersaremountedonthesameshaft,thepressureat
theoutletwillbeincreasedfurther.
•Letn=Numberofidenticalimpellersmountedonthesameshaft,
??????
&#3627408474;=Headdevelopedbyeachimpeller.
•ThentotalHeaddeveloped=&#3627408475;×??????
&#3627408474;
•Thedischargepassingthrougheachimpellerissame.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
MULTISTAGE CENTRIFUGAL PUMPS
2. TO PRODUCE HIGH
DISCHARGE
•Forobtaininghighdischarge,
thepumps shouldbe
connectedinparallel.
•Eachofthepumpsliftsthe
waterfromacommonsump
anddischargeswatertoa
commonpipetowhichthe
deliverypipesofeachpumpis
connected.
•Eachofthepumpsisworking
againstthesamehead.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Letn=Numberofidenticalpumpsarrangedinparallel.
Q=Dischargefromonepump.
∴TotalDischarge=&#3627408527;×&#3627408504;

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 4
•Performance characteristic curves
•Main characteristic curves
•Operating characteristic curves
•Constant efficiency or Muschel
curves
•NPSH
Performance characteristic
curves
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PERFORMANCE CHARACTERISTIC CURVES
•Thecharacteristiccurvesofacentrifugalpumparedefinedas
thosecurveswhichareplottedfromtheresultsofanumberof
testsonthecentrifugalpump.
•Thesecurvesarenecessarytopredictthebehaviorand
performanceofthepump,whenthepumpisworkingunder
differentflowrate,headandspeed.
•Thefollowingaretheimportantcharacteristiccurvesforthe
pumps:
Maincharacteristiccurves.
Operatingcharacteristiccurvesand
ConstantefficiencyorMuschelcurvesatdifferentflowrate,head
andspeed.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
MAIN CHARACTERISTIC CURVES
•Themaincharacteristiccurves
ofacentrifugalpumpconsistsof
ahead(Manometrichead??????
&#3627408474;)
poweranddischargewith
respecttospeed.
•Forplottingcurvesof
Manometricheadversusspeed,
dischargeiskeptconstant.For
plottingcurvesofdischarge
versusspeed,Manometrichead
(??????
&#3627408474;)iskeptconstant.
•Forplottingcurvespower
versusspeed,Manometrichead
anddischargearekept
constant.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
OPERATING CHARACTERISTIC CURVES
•Ifthespeediskeptconstant,the
variationofManometrichead,
powerandefficiencywithrespect
todischargegivestheoperating
characteristicsofthepump.
•Theinputpowercurveforpumps
shallnotpassthroughtheorigin.
Itwillbeslightlyawayfromthe
originonthey-axis,asevenat
zerodischargesomepoweris
neededtoovercomemechanical
losses.
•Theheadcurvewillhave
maximumvalueofheadwhenthe
dischargeiszero.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
CONSTANT EFFICIENCY OR MUSCHEL CURVES
•Forobtainingconstantefficiencycurvesfora
pump,theheadversusdischargecurvesand
efficiencyv/sdischargecurvesfordifferent
speedsareused.
•Bycombiningthesecurves(??????∼&#3627408452;&#3627408464;&#3627408482;&#3627408479;&#3627408483;&#3627408466;&#3627408480;&#3627409154;∼

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
NET POSITIVE SUCTION HEAD (NPSH )
•ThetermNPSHisverycommonlyusedselectionofapump.The
minimumsuctionconditionsarespecifiedintermsNPSH.
•Itisdefinedastheabsolutepressureheadattheinlettothepump
minusthevapourpressureheadplusvelocityhead.
∴NPSH=Absolutepressureheadatinletofpump–vapour
pressurehead(absoluteunits)+Velocityhead
=
&#3627408477;
1
&#3627409164;&#3627408468;
−
&#3627408477;
&#3627408483;
&#3627409164;&#3627408468;
+
&#3627408483;
&#3627408480;
2
2&#3627408468;
(1) (∵

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•&#3627408449;&#3627408451;????????????=
&#3627408477;??????
&#3627409164;&#3627408468;
−
&#3627408483;&#3627408480;
2
2&#3627408468;
−ℎ
&#3627408480;−ℎ
&#3627408467;
&#3627408480;
−
&#3627408477;&#3627408483;
&#3627409164;&#3627408468;
+
&#3627408483;&#3627408480;
2
2&#3627408468;
=
&#3627408477;
??????
&#3627409164;&#3627408468;
−
&#3627408477;
&#3627408483;
&#3627409164;&#3627408468;
−ℎ
&#3627408480;−ℎ
&#3627408467;
&#3627408480;
=??????
&#3627408462;−??????
&#3627408483;−ℎ
&#3627408480;−ℎ
&#3627408467;
&#3627408480;
∵
&#3627408477;
??????
&#3627409164;&#3627408468;
=??????
&#3627408462;=&#3627408436;&#3627408481;&#3627408474;&#3627408476;&#3627408480;&#3627408477;ℎ&#3627408466;&#3627408479;??????&#3627408464;&#3627408477;&#3627408479;&#3627408466;&#3627408480;&#3627408480;&#3627408482;&#3627408479;&#3627408466;ℎ&#3627408466;&#3627408462;&#3627408465;,
=??????
&#3627408462;−ℎ
&#3627408480;−ℎ
&#3627408467;
&#3627408480;
−??????
&#3627408483;(2) ቀ∵
&#3627408477;
&#3627408483;
&#3627409164;&#3627408468;
=??????
&#3627408483;=

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 5
•Example problems on centrifugal
pump
Problems on Centrifugal
Pump
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM 1

D E PAR TME NT O F M E C HANICAL E NG INE E R ING

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM 2

D E PAR TME NT O F M E C HANICAL E NG INE E R ING

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM 3

D E PAR TME NT O F M E C HANICAL E NG INE E R ING

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM 4

D E PAR TME NT O F M E C HANICAL E NG INE E R ING

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM 5

D E PAR TME NT O F M E C HANICAL E NG INE E R ING

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM 6

D E PAR TME NT O F M E C HANICAL E NG INE E R ING

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM 7

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 6
•Introduction
•Discharge through a reciprocating
pump
•Work done by a reciprocating pump
Reciprocating pump
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
RECIPROCATING PUMPS
•Themechanicalenergyis
convertedintohydraulic
energy(pressureenergy)by
suckingtheliquidintoa
cylinderinwhichapistonis
reciprocating,whichexertsthe
thrustontheliquidand
increasesitshydraulicenergy
(pressureenergy)thepumpis
knownasreciprocatingpump.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Asingleactingreciprocatingpumpconsistsofapiston,which
movesforwardsandbackwardsinaclosefittingcylinder.
•Themovementofthepistonisobtainedbyconnectingthepiston
rodtocrankbymeansofaconnectingrod.
•Thecrankisrotatedbymeansofanelectricmotor.
•Suctionanddeliverypipeswithsuctionvalveanddeliveryvalve
areconnectedtothecylinder.
•Thesuctionanddeliveryvalvesareonewayvalvesornon-return
valves,whichallowthewatertoflowinonedirectiononly.
•Suctionvalveallowswaterfromsuctionpipetothecylinderwhich
deliveryvalveallowswaterfromcylindertodeliverypipeonly.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Whenthecrankstartsrotating,thepistonmovestoandfrointhe
cylinder.WhenthecrankisatAthepistonisattheextremeleft
positioninthecylinder.
•AsthecrankisrotatingfromAtoC(i.e.from&#3627409155;=0&#3627408481;&#3627408476;180
0
)the
pistonismovingtowardsrightinthecylinder.Themovementofthe
pistontowardsrightcreatesapartialvacuuminthecylinder.
•Butonthesurfaceoftheliquidinthesumpatmosphericpressure
inacting,whichismorethanthepressureinsidethecylinder.
•Thustheliquidisforcedinthesuctionpipefromthesump.
•Thisliquidopensthesuctionvalveandentersthecylinder.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•WhencrankisrotatingfromCtoA(i.e.from&#3627409155;=180
0
&#3627408481;&#3627408476;360
0
),the
pistonfromitsextremerightpositionstartsmovingtowardsleftin
thecylinder.
•Themovementofthepistontowardstheleftincreasesthe
pressureontheliquidinsidethecylindermorethanatmospheric
pressure.
•Hencethesuctionvalveclosesanddeliveryvalveopens.
•Theliquidisforcedintothedeliverypipeandisraisedtothe
requiredheight.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
DISCHARGE THROUGH A RECIPROCATING
PUMP
•Considerasingleactingreciprocatingpump.
LetD=Diameterofcylinder
A=Cross-sectionalareaofpistonorcylinder=
&#3627409163;
4
&#3627408439;
2
r=Radiusofcrank
N=r.p.m.ofthecrank
L=Lengthofthestroke=2×&#3627408479;
ℎ
&#3627408480;=Heightoftheaxisofthecylinderfromwatersurfacein
sump
ℎ
&#3627408465;=Heightofdeliveryoutletabovethecylinderaxis(also
calleddeliveryhead)
•VolumeofwaterdeliveredinonerevolutionorDischargeofwater
inonerevolution=Area×Lengthofstroke=&#3627408436;×&#3627408447;

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Number of revolutions per second =
&#3627408449;
60
•Discharge of pump per second Q = Discharge in one revolution ×
No. of revolutions per sec=&#3627408436;×&#3627408447;×
&#3627408449;
60
=
??????&#3627408447;&#3627408449;
60
•Weight of water delivered per second &#3627408458;=&#3627409164;×&#3627408468;×&#3627408452;
=
&#3627409164;&#3627408468;??????&#3627408447;&#3627408449;
60
(1)

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
WORK DONE BY RECIPROCATING PUMP
•Workdonepersecond=Weightofwaterliftedpersecond×Total
heightthroughwhichwaterislifted=&#3627408458;×ℎ
&#3627408480;+ℎ
&#3627408465;_____(2)
Whereℎ
&#3627408480;+ℎ
&#3627408465;=Totalheightthroughwhichwaterislifted
•Fromequation(1)weightofwaterisgivenby
&#3627408458;=
&#3627409164;&#3627408468;??????&#3627408447;&#3627408449;
60
•SubstitutingthevalueofWinequation(2),weget
•Workdonepersecond=
&#3627409164;&#3627408468;??????&#3627408447;&#3627408449;
60
×ℎ
&#3627408480;+ℎ
&#3627408465;
•PowerrequiredtodrivethepumpinkW
&#3627408451;=
&#3627408458;&#3627408476;&#3627408479;&#3627408472;&#3627408465;&#3627408476;&#3627408475;&#3627408466;&#3627408477;&#3627408466;&#3627408479;&#3627408480;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;
1000
=
&#3627409164;&#3627408468;×&#3627408436;&#3627408447;&#3627408449;×ℎ
&#3627408480;+ℎ
&#3627408465;
60×1000
&#3627408451;=
&#3627409164;&#3627408468;×&#3627408436;&#3627408447;&#3627408449;×ℎ
&#3627408480;+ℎ
&#3627408465;
60,000
&#3627408472;&#3627408458;

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 7
oSlip of Reciprocating pump
oNegative Slip of Reciprocating
pump
oIndicator diagram
oIdeal Indicator diagram
Slip & Indicator diagram of
Reciprocating pump
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
SLIP OF RECIPROCATING PUMP
•Slipofapumpisdefinedasthedifferencebetweenthetheoretical
dischargeandactualdischargeofapump.
•Theactualdischargeofpumpislessthanthetheoreticaldischarge
duetoleakage.
•Thedifferenceofthetheoreticaldischargeandactualdischargeis
knownasslipofthepump.
•Hence&#3627408480;&#3627408473;??????&#3627408477;=&#3627408452;
&#3627408481;ℎ−&#3627408452;
&#3627408462;&#3627408464;&#3627408481;
•Butslipismostlyexpressedaspercentageslip
&#3627408451;&#3627408466;&#3627408479;&#3627408464;&#3627408466;&#3627408475;&#3627408481;&#3627408462;&#3627408468;&#3627408466;&#3627408480;&#3627408473;??????&#3627408477;=
&#3627408452;
&#3627408481;ℎ−&#3627408452;
&#3627408462;&#3627408464;&#3627408481;
&#3627408452;
&#3627408481;ℎ
×100=1−
&#3627408452;
&#3627408462;&#3627408464;&#3627408481;
&#3627408452;
&#3627408481;ℎ
×100
=1−&#3627408438;
&#3627408465;×100∵
&#3627408452;
??????&#3627408464;&#3627408481;
&#3627408452;
&#3627408481;ℎ
=&#3627408438;
&#3627408465;
Where&#3627408438;
&#3627408465;=Co-efficientofdischarge.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
NEGATIVE SLIP OF THE RECIPROCATING
PUMP
•Slipisequaltothedifferenceoftheoreticaldischargeandactual
discharge.
•Ifactualdischargeismorethanthetheoreticaldischarge,theslip
ofthepumpwillbecome–ve.
•Inthatcasetheslipofthepumpisknownasnegativeslip.
•Negativeslipoccurswhenthedeliverypipeisshort,suctionpipeis
longandpumpisrunningathighspeed.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
INDICATOR DIAGRAM
•Theindicatordiagramfora
reciprocatingpumpisdefined
thegraphbetweenthepressure
headinthecylinderandthe
distancetravelledbypiston
frominnerdeadcentreforone
completerevolutionofthe
crank.
•Asthemaximum distance
travelledbythepistonisequal
tothestrokelengthandhence
theindicatordiagramisagraph
betweenpressureheadand
strokelengthofthepistonfor
onecompleterevolution.
•Thepressureheadistakenas
ordinateandstrokelengthas
abscissa.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
IDEAL INDICATOR DIAGRAM
•Thegraphbetweenpressureheadinthecylinderandthestroke
lengthofpistonforonecompleterevolutionofthecrankunder
idealconditionsisknownasidealindicatordiagram.
•LineEFrepresentstheatmosphericpressureheadequalto10.3
metersofwater.
•Let??????
&#3627408474;=Atmosphericpressurehead=10.3mofwater
L=Lengthofthestroke
ℎ
&#3627408480;=Suctionheadand
ℎ
&#3627408465;=Deliveryhead
•Duringsuctionstroke,thepressureheadinthecylinderisconstant
andequaltosuctionheadℎ
&#3627408480;,whichisbelowtheatmospheric
pressurehead??????
&#3627408462;&#3627408481;&#3627408474;byaheightofℎ
&#3627408480;.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Thepressureheadduringsuctionstrokeisrepresentedbya
horizontallineABwhichisbelowthelineEFbyaheightof′ℎ
&#3627408480;′
•Duringdeliverystroke,thepressureheadinthecylinderis
constantandequaltodeliveryhead(ℎ
&#3627408465;),whichisabovethe
atmosphericheadbyaheightof′ℎ
&#3627408465;′.
•Thusthepressureheadduringthedeliverystrokeisrepresented
byahorizontallineCD,whichisabovethelineEFbyaheightof
ℎ
&#3627408465;.
•Thusforonecompleterevolutionofcrank,thepressureheadinthe
cylinderisrepresentedbythediagramABCD.
•Thisdiagramisknownasidealindicatordiagram.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•The work done by the pump per second =
&#3627409164;&#3627408468;??????&#3627408447;&#3627408449;
60
×ℎ
&#3627408480;+ℎ
&#3627408465;
=&#3627408446;×&#3627408447;ℎ
&#3627408480;+ℎ
&#3627408465;
∝&#3627408447;×ℎ
&#3627408480;+ℎ
&#3627408465;__(1)
Where &#3627408446;=
&#3627409164;&#3627408468;??????&#3627408449;
60
=&#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;
•Area of Indicator diagram =&#3627408436;&#3627408437;×&#3627408437;&#3627408438;=&#3627408436;&#3627408437;×&#3627408437;&#3627408441;+&#3627408441;&#3627408438;=&#3627408447;×
ℎ
&#3627408480;+ℎ
&#3627408465;
•Substituting this value in equation (1), we get
Work done by pump ∝Area of Indicator diagram

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

DEPARTMENT OF MECHANICAL ENGINEERING
LECTURE 8
•Example problems on Reciprocating
pump
•Applications
•Assignment questions
Problems on Reciprocating
Pump
TOPICS TO BE COVERED

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM 1
•Asingleactingreciprocatingpumprunningat50r.p.m.delivers
0.01&#3627408474;
3
/&#3627408480;ofwater.Thediameterofpistonis200mmandstroke
lengthin400mm.Determinei)Thetheoreticaldischargeofpump
ii)Co-efficientofdischargeiii)Slipandthepercentageslipof
pump.
Sol:Givendata
Thespeedofthepump,N=50rpm
Actualdischarge,&#3627408452;
&#3627408462;&#3627408464;&#3627408481;=0.01&#3627408474;
3
/&#3627408480;
Dia.Ofpiston,D=200mm=0.2m
Area,&#3627408436;=
&#3627409163;??????
2
4
=
&#3627409163;
4
0.2
2
=0.031416&#3627408474;
2
i)Thetheoreticaldischarge,&#3627408452;
&#3627408481;ℎ=
??????&#3627408447;&#3627408449;
60
=
0.031416×0.4×50
60
=&#3627409358;.&#3627409358;&#3627409359;&#3627409358;&#3627409362;&#3627409365;&#3627408526;
&#3627409361;
/??????

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
ii) The Co-efficient of discharge, &#3627408438;
&#3627408465;=
&#3627408452;
??????&#3627408464;&#3627408481;
&#3627408452;
&#3627408481;ℎ
=
0.01
0.01047
=&#3627409358;.&#3627409367;&#3627409363;&#3627409363;
iii)Slip, &#3627408452;
&#3627408481;ℎ−&#3627408452;
&#3627408462;&#3627408464;&#3627408481;=0.01047−0.01
=&#3627409358;.&#3627409358;&#3627409358;&#3627409358;&#3627409362;&#3627409365;&#3627408526;
&#3627409361;
/??????
Percentage Slip =
&#3627408452;
&#3627408481;ℎ−&#3627408452;
??????&#3627408464;&#3627408481;
&#3627408452;
&#3627408481;ℎ
×100=
0.01047−0.01
0.01047
×100
=&#3627409362;.&#3627409362;&#3627409366;&#3627409367;%

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
PROBLEM 2
•Adoubleactingreciprocatingpump,runningat40r.p.m.isdischarging
1.0m3ofwaterperminute.Thepumphasastrokeof400mm.the
diameterofpistonis200mm.thedeliveryandsuctionheadare20m
and5mrespectively.Findtheslipofthepumpandpowerrequiredto
drivethepump.
Sol:Givendata,
Speedofthepump,N=40r.p.m.
Actualdischarge,&#3627408452;
&#3627408462;&#3627408464;&#3627408481;=1&#3627408474;
3
/&#3627408480;=
1
60
=0.01666&#3627408474;
3
/&#3627408480;
Stroke,L=400mm=0.4m
Diameterofpiston,D=200mm=0.2m
∴ &#3627408436;&#3627408479;&#3627408466;&#3627408462;,&#3627408436;=
&#3627409163;
4
&#3627408439;
2
=
&#3627409163;
4
0.2
2
=0.031416&#3627408474;
2
SuctionHead,ℎ
&#3627408480;=5&#3627408474;
Deliveryhead,ℎ
&#3627408465;=20&#3627408474;

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Theoreticaldischargefordoubleactingpump
&#3627408452;
&#3627408481;ℎ=
2&#3627408436;&#3627408447;&#3627408449;
60
=
2×0.31416×0.4×40
60
=&#3627409358;.&#3627409358;&#3627409359;&#3627409364;&#3627409365;&#3627409363;&#3627408526;
&#3627409361;
/??????
•Slip=&#3627408452;
&#3627408481;ℎ−&#3627408452;
&#3627408462;&#3627408464;&#3627408481;=0.01675−0.1666
=&#3627409358;.&#3627409358;&#3627409358;&#3627409358;&#3627409358;&#3627409367;&#3627408526;
&#3627409361;
/??????
•Powerrequiredtodrivethedoubleactingpump
&#3627408451;=
2×&#3627409164;&#3627408468;×??????&#3627408447;&#3627408449;×ℎ
&#3627408480;+ℎ
&#3627408465;
60,000
=
2×1000×9.81×0.031416×0.4×40×5+20
60,000
=&#3627409362;.&#3627409359;&#3627409358;&#3627409367;??????&#3627408510;

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
APPLICATIONS
•Centrifugalpumpsareusedinvarietyofapplications.Almost70-
80%centrifugalpumpsareusedinindustryorfordomestic
purpose.
•Watersupplyandirrigation
•Chemical,food,Petrochemicalindustries
•Mining,domesticappliances
•Reciprocatingpumpsaremainlyusedinoilandgasindustry.
•Theycanalsobeusedsugarindustries,soapanddetergent
industries,watertreatmentplantsandFood&beverages.
•Havehugedemandincryogenicapplications.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
ASSIGNMENT QUESTIONS
•Acentrifugalpumpdeliverswatertoaheightof22mataspeedof
800rpm.Thevelocityofflowisconstantataspeedof2m/sand
theoutletvaneangleis450.Ifthepumpdischarges225litresof
water/second,findthediameteroftheimpellerandwidthofthe
impeller.
•Acentrifugalpumphavingouterdiameterequaltotwotimesthe
innerdiameterandrunningat1200rpmworksagainstatotalhead
of75m.Thevelocityofflowthroughtheimpellerisconstantandis
equalto3m/s.Thevanesaresetbackatanangleof300atoutlet.
Iftheouterdiameteroftheimpelleris600mmandwidthatoutlet
is50mm,determine(i)vaneangleatinlet(ii)workdoneper
secondbyimpeller(iii)manometricefficiency.
•Whatistheworkingprincipleofareciprocatingpump?Explainits
workingwiththehelpofanindicatordiagram.

D E PAR TME NT O F M E C HANICAL E NG INE E R ING
•Asingleactingreciprocatingpumphavingcylinderdiameterof150
mmandstroke300mmisusedtoraisewaterthroughatotal
heightof30m.Findthepowerrequiredtodrivethepump,ifthe
crankrotatesat60rpm.
•Afluidistobeliftedagainstaheadof120m.Thepumpsthatrunat
aspeedof1200rpmwithratedcapacityof300litres/secare
available.Howmanypumpsarerequiredtopumpthewaterif
specificspeedis700.

DEPARTMENT OF MECHANICAL ENGINEERING
THANK YOU

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