Applied Energy Systems — Hydrostatic Fluid Power ...

AppliedEnergySystems—HydrostaticFluidPowerTransmissionandFlowinPipes2012EditionJimMcGovern,DublinInstituteofTechnology

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J.McGovern AES12012r03

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TableofContentsNomenclature.......................................................................................................iii Chapter1. Introduction..................................................................................1 WhatisaFluid?.................................................................................................3 Chapter2. UnitsandDimensions...............................................................5 Chapter3. PressureandCompressibility...............................................9 Pressure................................................................................................................9 CompressibilityandBulkModulus........................................................11 Chapter4. HydrostaticWorkandPowerTransmission...............13 Pascal’sLaw.....................................................................................................13 Continuity.........................................................................................................13 MassandVolumeFlowRates...................................................................15 Work,FlowWorkandPower...................................................................15 DesignandOperatingPrinciplesofaHydraulicJack.....................17 Chapter5. PositiveDisplacementPumps............................................18 SimpleReciprocatingPump......................................................................18 SwashPlatePump.........................................................................................19 RotaryPositiveDisplacementPumps...................................................20 IdealFlowRateofaPositiveDisplacementPump..........................22 EfficiencyofHydraulicPumps.................................................................23 Chapter6. HydraulicSystemComponents.........................................26 CheckValveandPressureReliefValve................................................26 HydraulicCylindersorRams....................................................................27 CylinderForceCalculations......................................................................28 DirectionalControlValves.........................................................................30 TheHydraulicCircuitDiagram................................................................32 Chapter7. DesignofHydraulicSystems..............................................33 FlowControl....................................................................................................33 HydraulicSystemforaLift........................................................................35 HydraulicPress...............................................................................................37 Chapter8. Viscosity......................................................................................40 Newton’sViscosityEquation....................................................................41 KinematicViscosity.......................................................................................44 MoreaboutViscosityUnitsandValues................................................44 Chapter9. PressureLossesinPipesandFittings............................47 LaminarandTurbulentFlow....................................................................47


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ReynoldsNumber..........................................................................................49 HeadLossDuetoFriction..........................................................................50 FlowthroughValvesandFittings...........................................................54 FlowAroundBendsandElbows.............................................................54 EquivalentPipeLength...............................................................................54 PressureLossandPumpingPower.......................................................55 SummaryofKeyFormulaeforPipeFlow...........................................56 Chapter10. ValvesandHeatExchangerFlowPaths....................61 Valves..................................................................................................................61 HeadLossCoefficientsforopenValves...............................................63 HeatExchangers.............................................................................................64 PressureDropsorHeadLossesinHeatExchangers......................68 ImportantFormulae.........................................................................................70 Newton’sViscosityEquation....................................................................70 Kinematicviscosity.......................................................................................70 Newton’sSecondLawAppliedtoS.I.Units........................................71 Pressure.............................................................................................................71 PressureandHead........................................................................................71 CompressibilityandBulkModulus........................................................72 ForceProducedbyaHydraulicCylinder.............................................72 DisplacementofaHydraulicPistonPump.........................................72 IdealFlowRateofaPositiveDisplacementPump..........................73 EfficiencyofaHydraulicPump................................................................73 AreaofaCircle................................................................................................74 MassFlowRate...............................................................................................74 VolumeFlowRate..........................................................................................74 ReynoldsNumber..........................................................................................74 TheDarcy-WeisbachEquation................................................................75 FrictionFactorforLaminarFlow...........................................................76 HeadLossCoefficientforFittingsandFeatures...............................76 Glossary..................................................................................................................77


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Nomenclature Symbol Description SIunits area m vc areaatvenacontracta m o areaoforifice m acceleration m s constant constant distanceordiameter m eq equivalentdiameter m force N frictionfactor accelerationduetogravity m s ℎ head,heightordepth m ℎ headlossduetofriction m bulkmodulus Pa headlosscoefficient surfaceroughness m L lengthdimension length m M massdimension mass kg massflowrate kg s cyl numberofcylinders


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Symbol Description SIunits Power W pressure Nm orPa volumeflowrate m s Re Reynoldsnumber SI SystèmeInternational,InternationalSystem distance m T timedimension time s volume m velocity m s velocityatposition m s work J distance(usuallyhorizontal) m distance(usuallyvertical) m Greekletters compressibility Pa−1 (beta)Δ indicates an increment ofthe following variable e.g.Δ is a difference betweentwovaluesof

(uppercasedelta)Δ hx pressuredropacrossaheatexchanger Pa


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Symbol Description SIunits Δ displacement(changeofposition) m θ temperaturedimension (theta) absoluteviscosity PasorNm sorkgm s

(mu) kinematicviscosity m s (nu) density kg/m (rho) shearstress Nm (tau)


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Chapter 1. Introduction AppliedEnergySystemsmoduleshavedevelopedandevolvedattheDublinInstituteofTechnology(DIT)withintheframeworkofthe Bachelor of Engineering Technology degree and theprogrammes that preceded it. The present work is a basictextbookorsetofnotesforsuchamoduleaddressingbasicFluidMechanicsforincompressiblefluids.Currently ‘energy’ is very topical. Sustainable energy andrenewable energy are buzz words. There is a widespreadappreciation that energymust be used efficiently and a generalwillingnesstomakeuseofallformsofenergythatarenaturalandfree. Likewise there is a generaldesire to avoid,wherepossible,makinguseofformsofenergythatarefinite(i.e.alreadystoredinfixed quantities on the planet Earth) or that damage theenvironmentwhentheyareused.Thispresentworkfitsintothatgeneral picture. The underlying principles and scientificknowledgethatrelatetoenergyusearecommon,whetherenergyisbeingusedwellorineffectivelyandirrespectiveofthesourceoftheenergy.Theuseof fluids, e.g. gases likeairornatural gasorliquids like water or gasoline, forms a major sub-category ofenergysystemsgenerally.An interesting approachwas used by thosewho developed anddeliveredtheAppliedEnergySystemsmoduleforfluidsattheDITin the past. Rather than presenting a course that attempted tocover Fluid Mechanics in a very broad and general way, theyidentified and selected two practical areas in the use of fluids:hydrostatic power transmission systems and the flow of fluidsthroughpipesandfittings.Thisselectionallowedthetwoareastobepresentedinsomedetail,butitalsoallowedthemtobesampleapplication areas that would enable the learners to put theprinciplesof fluidmechanics intocontext,whiledevelopingtheirknowledgeandunderstandinginawaythatwouldpreparethemfor applying the same and similar principles to amuch broaderrangeofpracticalapplicationsinthefuture.Hydrostaticpowertransmissionsystemsareemployed inahugerange of heavy and light machinery used in the constructionindustry.Inmanufacturingindustry,hydraulicpowersystemsarecommon, e.g. for the operation of presses that can apply hugeforcestoformworkpiecesortoallowthepartsofmouldingdiesto be held firmly together and subsequently separated quickly.


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The principles of hydraulic power transmission are applied invehicles such as cars, trains, ships and aircraft. Furthermore,manyoftheprinciplesofhydraulicpowertransmissionalsoapplyin situations where pressures act but there is not necessarilytransmissionofpower.For instance,asubmarine is subjected tohydrostaticpressure.Theflowoffluidinpipesissuchacommonphenomenonthatwetakeitcompletelyforgranted.Airflowsintoandoutofourlungsthrough our bronchial tubes and our blood flows through pipesthat are usually called veins and arteries. Natural gas, oil andwaterarejustthreefluidsthatcommonlyflowthroughpipes.Itisa simple fact that energy is dissipated, or transformed to adifferent form, whenever a fluid flows. Mankind needs to useenergytocausefluidstoflow.Whilethepresentworkislimitedtotheapplicationareaofflowoffluidsinpipesandfittings,theverysameprinciplesarealsocommontootherapplicationareassuchas wind turbines, aerodynamics and biomedical devices such asartificialhearts.Learnerswhohaveambition toworkwith themostexcitingandinnovative ways of harnessing energy will find that what theylearn through taking an Applied Energy Systems module inHydrostatic Fluid Power Transmission and Flow in Pipes willserve them well. Not all applied technologists work at theforefrontoftechnicaldevelopmentandinnovation.Theeverydayworkofsocietymustgoon.Ifthatworkinvolvesfluidsinanyway,the learner will find that their learning within the scope of thepresentworkwillhavebeen relevantandworthwhile.However,the relevanceof thepresentwork is significantlybroaderyet. Incommonwithother subject areas in anengineeringprogramme,this Applied Energy Systems topic involves engineering units(specifically the SI system) and involves important and genericengineering approaches such as experimentation, measurement,representation of practical problems in mathematical form,teamworkandcommunication.Thepresentworkis intentionallycompact and concise. It requires further explanation orelaboration.LearnersareencouragedtomakeuseoftheInternetand library services for additional and supplementaryinformation.


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What is a Fluid? Afluidisasubstancethatcanflow.Generallyitwouldbealiquidoragas.Asolidisnotafluid.So,whatdistinguishesafluidfromanon-fluid?Thefollowingisasuitabledefinition:Afluidisasubstancethatcannotsustainashearstresswithoutundergoingrelativemovement.The definition of a fluid involves the term ‘shear stress,’ whichitselfrequiressomeexplanation.Astress isa forceperunitarea(S.I. units newton/m2). Two main categories of stresses arenormalstressesandshearstresses.Ifaforceisnormaltotheareaoverwhichitactsitissaidtobeanormalforceandthestressisthereforeanormalstress.Atensileforceappliedtoabarwouldbeanexampleofanormalforce.Ifaforceisinthesameplaneasthearea over which it acts it is called a shear force and thecorresponding stress is therefore a shear stress. Figure 1illustrates normal and shear forces and stresses. The normalstressexistsbecauseabarisbeingpulledfrombothends;asthebar is in tension thisnormalstresscanbedescribedasa tensilestress.Theshearstressexistsbecauseabarissubjectedtoforcesbybladesofaguillotineorshearsthataretendingtoshearthebarintheplaneinwhichtheforcesact,theshearplane.

Figure 1 Normalandshearforcesandstresses


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Solid materials can withstand high shear forces without anyrelativemovementoccurringwithinthematerial.However,whena fluid is subjected to a shear stress, relativemovement alwaysoccurs within the fluid, no matter how small the shear stress.Becauseofthischaracteristic,aliquidwillalwaysflowundertheinfluenceofgravityunlessitispreventedfromflowingfurtherbya container wall below the liquid. The surface of a liquid atequilibrium within a container will be horizontal, because if itwere not horizontal, shear forces capable of causing relativemovementwouldexistwithintheliquid.


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Chapter 2. Units and Dimensions A physical quantity such as temperature is quantified as a purenumber such as 301.5 multiplied by a unit such as the kelvin.Hence,atemperaturemightbewrittenas301.5K.Ifwedividethetemperaturebyitsunitweareleftwithjustthepurenumber:301.5[K][K] = 301.5.Table1listsSIandsomenon-SIunitsthatareusedinthepresentwork.Ofallthephysicalquantitiesthatareencountered,theunitsfor these quantities are made up of four fundamental types ofunits, or dimensions. Symbols can be used to represent thesedimensionsasfollows:massM,lengthL,timeTandtemperatureθ.AccelerationwouldhavethefundamentalunitsLT .

Table 1 UnitsPhysicalquantity SIbaseunit Otherunitsmass kg(kilogram) 1 g = 10 kg1 tonne = 10 kglength m(metre) 1 mm = 10 m1km = 10 m1 cm = 10 mtime s(second) 1 min = 60s1 h = 3600sforce N(newton) 1 kN = 10 N1 MN = 10 Narea m 1 cm = 10 m 1 mm = 10 m


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Physicalquantity SIbaseunit Otherunitsvolume m 1 cm = 10 m 1mm = 10 m liter:1L = 10 m milliliter:1 mL = 10 m pressure Pa(pascal) 1 Pa = 1Nm 1kPa = 10 Pa1MPa = 10 Pa1GPa = 10 Pa1hPa = 10 Pa1bar = 10 Paatmosphere:1 atm = 101325Pastress Pa(pascal) 1 Pa = 1Nm absoluteordynamicviscosity PasorNm s,orkgm s 1 poise = 10 Pas1centipoise= 10 poise= 10 Paskinematicviscosity m s 1 stoke = 10 m s 1centistoke= 10 stoke= 10 m s velocity ms acceleration ms


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Physicalquantity SIbaseunit Otherunitsdensity kgm energy,work,heat J 1 J = 1Nm1kJ = 10 J1 MJ = 10 Jpower W 1 W = 1Js 1 kW = 10 Wbulkmodulusofelasticity Pa 1 Pa = 1Nm compressibility Pa conventionaltemperature ℃(degreeCelsius) thermodynamicorabsolutetemperature K(kelvin) torque(moment) Nm angle rad 2 rad = 360°angularvelocity rad s revolutionspersecond:1 r.p.s. = 2 rads revolutionsperminute:1 r.p.m. = 260 rads angularacceleration rad s volumeflowrate m s 1 Ls = 10 m s 1Lmin= 1060 m s


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Newton’s 2nd Law and the SI Units for Force Newton’s second law has a central role in mechanics, includingfluid mechanics. It is mentioned here in order to explain anddefinetherelationshipbetweentheunitsforforce,mass,distanceandtime.Newton’ssecondlawdescribestheforcethatmustbeappliedtoamasstocauseittoaccelerate.Itcanbesummarizedinwordsas:Force equals mass times acceleration. Thiscanbeexpressedmathematicallyas= (1)where = force N = mass kg = acceleration ms Aswithallequationsthatrepresentphysicalquantities,theunitsof the left hand side of the equation must be equivalent to theunitsoftherighthandsideoftheequation.Newton’ssecondlaw,Equation1, therefore defines the relationship between thenewton, the kilogram, the metre and the second. Of these fourunits, anyone canbe expressed in termsof theother three. Forexample 1N = 1kgms 1kg = 1Nm s .Hence,wherevertheunitthenewtonisuseditcanbereplacedbythe combination of units kgms . Also, it can be deduced thatforcehas thedimensionsMLT .Pressurewould thereforehavethedimensionsMLT L ,whichisequivalenttoML T .


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Chapter 3. Pressure and Compressibility

Pressure Pressureisforceexertedperunitareainadirectionthatisnormaltoandtowardsthearea.Itisexpressedas= (2)wherep = pressure PaorNm−2F = force NA = area m2Onepascal(Pa)isaverylowpressure;forinstance,atmosphericpressureisabout105Pa.Absolute Pressure Absolute pressure is true pressure, which is the force per unitareaexertedona surface.Theentire forcenormal to thearea isincluded. The lowest possible value of absolute pressure is zero(inanypressureunits).Atmospheric Pressure Thepressureoftheatmosphereiscausedbytheweightoftheairthat constitutes it. At sea level the atmospheric pressure equalsthe weight of air that is supported by a unit area of horizontalsurface.Athigher levels lessair issupportedso theatmosphericpressureiscorrespondinglylower.1standardatmosphere=1.01325105PaAtmosphericpressurecanbemeasuredbymeansofatraditionalmercury barometer, Figure 2, or, for instance, by an aneroidbarometer,containingasealedcell intheformofabellowsasinFigure3.


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Figure 2 Mercurybarometer

Figure 3 Sealedbellowscellofananeroidbarometer Gauge Pressure Gaugepressure is a relativepressure: it is the amountbywhichthe pressure exceeds the pressure of the atmosphere. Gaugepressure is commonly used in engineering and many commonpressure-measuring devices sense gauge pressure rather thanabsolute pressure. A ‘pressure gauge’ usually measures gaugepressure.Vacuum and Vacuum Pressure Vacuumpressure is theamountbywhichapressure is less thanatmosphericpressure.Inaperfectvacuumtheabsolutepressurewouldbezero.Pressure Head Afluidexertsapressurebyvirtueofitsweight.Atadepthhbelowthefreesurfaceofaliquidatrestthe(gauge)pressureisgivenby:


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= ℎ (3)where = density kg/m3 = accelerationduetogravity ms−2ℎ = depthbelowfreesurface mItiscommontodescribeapressurebyquotingthecorrespondingheadofwateror,sometimes,ofotherfluids,e.g.mercury(density13,546kgm−3).Compressibility and Bulk Modulus Thecompressibilityofafluidisameasureofhowmuchitsvolumechanges with pressure. The volumetric change is described asvolumetricstrain,i.e.thechangeinvolumedividedbytheoriginalvolume.Supposeafluidiscompressed,asshownintheFigure4,duetotheapplicationofasmalladditionalforce,F.

Figure 4 Changeinvolumeowingtoaninincreaseinpressure

Compressibility, = −∆∆ = − 1 ∆∆ (4)The term ‘bulk modulus of elasticity’ is used for the inverse ofcompressibility:


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Bulkmodulus, = ∆−∆ = − ∆∆ (5)Bulk modulus has the same units as pressure, i.e. Pa. The bulkmodulusofliquidwaterisabout2.2GPa(1GPa=109Pa),whichindicates howmuch pressure is required to produce volumetricstrain.Allliquidshavehighvaluesofbulkmodulus:forhydraulicfluidsitranges from about 1.4 to 1.8GPa. Gases have very low bulkmodulus—for isothermal compression the bulk modulus isroughlyequaltothepressure.Inanoil-hydraulicpowersystemthecompressibilitycanusuallybe ignored when sizing components. The fluid is treated as‘incompressible.’Example 1 Bulk Modulus What increase in pressure would be required to cause avolumetric strain of 1% in water (i.e. to reduce the volume by1%)?Solution ∆ = − ∆ = 2.2 × 10 [Pa] × 0.01 = 2.2 × 10 [Pa]Thus,apressureofabout220atmosphereswouldbenecessarytoreducethevolumeofliquidwaterby1%.


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Chapter 4. Hydrostatic Work and Power Transmission

Pascal’s Law It is possible to talk about the pressure at a point in a fluid byconsidering the force on a very small area centred about thatpoint.Pascal’slawcanbestatedasfollows:Pressure within a fluid at rest acts equally in all directions and at right angles to any surface on which it acts. Hydraulicpowertransmissionsystemstakeadvantageofthislawin transmitting pressure from one part of a system to another.While differences in head may exist within a hydraulic powersystem, the overall height of the system is usually such thatpressuredifferencesbetweendifferent levelswithinthefluidarenegligibleincomparisontotheoperatingpressuresthatareused.The term ‘hydrostatic machine’ can be used for a machine inwhichPascal’slawappliestoagoodapproximationwithinbodiesofconnectedfluid.Figure5illustrateshowforcesdependonareasinahydrostaticsystemormachine.

Figure 5 Schematicdiagramofageneralhydrostaticmachine

Continuity In the system shown in Figure 6 an incompressible fluid iscontainedbetween twopistons. The piston on the leftmoves totherightapplyingaforcetothefluid.Thefluidappliesaforcetothepistonontherightcausingittomovetotherightandupwardsinthedirectionofthepipe.


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Figure 6 Anincompressiblefluidbetweenpistons Eachpistonfacemovesthroughthesamevolume,∆ ,suchthat∆ = Δ = Δ . (6)LetΔtbethetimefordisplacementsΔ andΔ ΔΔ = ΔΔ .But = ,thevelocityofpiston1and = ,thevelocityofpiston2,so = = (7)where Q is the volume flow rate just ahead of piston 1 or justbehindpiston2.Equation7isknownasthecontinuityequation.Itappliesequallywelliftherearenopistonsatpositions1and2,asshowninFigure7.ThefundamentalSIunitsforvolumeflowratearem /s.

Figure 7 Flowofanincompressiblefluid

A1

F1

An

A2

Δs1

Δs2

p F2

Position

12

n

A1

v1

An

A2

Δs1

in time Δtin time Δt

Δs2

v2

Position

12

n

vn


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Mass and Volume Flow Rates FromEquation7thevolumeflowrateatagivencrosssectionofapipeisgivenbytheproductoftheareaandtheaveragevelocity,= . (8)Ifthedensityofthefluid isknownthenthemassflowrateatthecrosssectionisgivenby = . (9)ThefundamentalSIunitsformassflowratearekg/s.Work, Flow Work and Power Figure6 andFigure7 canhelpus to grasp themeaningof ‘flowwork.’ In Figure 6 work is done by one piston and on anotherpiston, while in Figure 7 equivalent ‘flowwork’ or ‘flow power’occurswithoutthepresenceofpistonsattheflowcrosssectionsof interest. In a sense, the flowwork for a volume of fluid thatpasses through a cross section is the energy being transferredthrough that cross section by the hydraulic fluid and the flowwork per unit time is the power being transmitted by thehydraulicfluid.Theworkdoneonthefluidbypiston1inFigure6isgivenby∆ = Δ = Δ .Similarly,theworkdoneonpiston2bythefluidisgivenby∆ = Δ = Δ .From Pascal’s law and neglecting friction, the same pressure, p,acts at positions 1 and 2 and throughout the connected fluid.Hence ∆ = ∆ = ∆ = ∆ (10)where ∆ is the volume through which each piston sweeps.Equation10describes theworkdoneby the left-handpistononthe fluid in Figure6 and also theworkdoneby the fluidon the


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right-hand piston. In each case the volume displacement at thepistonisΔ .IftheworkatthepistonsoccursoverasmalltimeintervalΔ thenthe rate at which work is done is given by the work amountdividedbythetimeinterval:= ∆∆ = ∆∆ = (11)where = power W = pressure Pa = volumeflowrate m3/s = time s = volume m3 = work JExample 2 Piston Work a. What work is done by a piston of diameter 0.05m when itexertsapressureof200kPaandmovesthroughadistanceof0.04m?b. If the movement occurs in 2s, what is the average rate ofwork,orpower?Solution W = AΔs = 200 × 10 [Pa] × 0.054 [m ] × 0.04[m]= 15.7Janswera.

= Δ = 15.7[J]2[s] = 7.85Wanswerb.


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Design and Operating Principles of a Hydraulic Jack

Figure 8 Schematicrepresentationofahydraulicjack ThehydraulicjackshowninFigure8isinastationarystate,withthe throttle release valve closed. The oil in the space below thepistonontheright isunderpressure,owingtotheweightof thepistonandtheload.Ifanoperatorappliesdownwardpressureonthe lever, the plunger on the left will move downwards in thecylinder.Oilwillflowtowardstherightthroughtheone-wayvalveinthebottompipeandwillcausetheloadtoberaised.Thepressurereliefvalve issimilar toaone-wayvalve,buthasaveryhighspring forceappliedsothat itwillonlyopenwhenthemaximumdesiredsystempressureisreached.If the throttle release valve is opened, the load on the rightwillcausetheoiltoflowthroughtherestrictionofthethrottlereleasevalve into the hydraulic oil tank. In this way the load can beloweredsmoothly.If the lefthandplunger is in its lowestpositionand is caused torisebyanoperatorliftingthelever,oilwillflowintothelefthandcylinderfromthehydraulicoiltank(orreservoir).Therefore the machine (a hand-operated hydraulic jack) allowsthe load to be raised or lowered in a controlled manner. Thepressurereliefvalveprovidesprotectionagainstoverpressure.

Plunger

Lever

Air vent

Hydraulic

oil

Load

Throttle release valve

Pressure relief valve

(strong spring)

One-way valve

(weak spring)


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Chapter 5. Positive Displacement Pumps Apositivedisplacement pump is a pump that discharges a fixedvolumeofpressurizedfluidforeachrotationoftheinputshaftorforeachcycle in the caseof a reciprocatingpump.The standardhydraulicsymbolforapumpisshowninFigure9.

Figure 9 Hydraulicpumpsymbol Simple Reciprocating Pump

Figure 10Schematicrepresentationofahydraulichandpump A reciprocatingpositive displacement pump, as shown in Figure10,dischargesafixedamountofpressurizedfluidforeachcycleofthe piston. The cycle is made up of an intake stroke and adischargestroke.Thedisplacementofthepumppercycleisequalto the product of the stroke and the cross-sectional area of thecylinder,Equation12.

displ = cyl stroke (12)wheredispl = pumpdisplacementpercycle


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cyl = cross-sectionalareaofcylinderstroke= pumpstroke

Example 3 Hand Pump Displacement If thepistonof ahand-operatedpumphasadiameterof30mmandastrokeof20mm,determinethedisplacementpercycle.= 4 = 0.03 [m ]4 = 0.7068 × 10 m Hence,

displ = 0.7068 × 10 [m ]0.02[m] = 14.14 × 10 m Swash Plate Pump Theswashplatepump,Figure11,containsarotatingpistonblockthathousesanumberofpistons.Thepistonendspressagainstaswash plate that causes the pistons to undergo a reciprocatingmotion.Notethefollowingfeatures:Aswashplate,whichisaflatthrustplateatanangletotheaxisofrotationofthepumpCheckvalvesforintakeanddischargeA fixed valve plate that contains two curved slots: one todistribute the intake fluid and the other to collect the dischargefluidApistonblock,pistons,ashoeplateandpistonshoes,allofwhichrotatewiththeshaftA spring that pushes the shoeplate against the swashplate andthepistonblockagainstthevalveplateAdrainportsothatanyleakagepastthepistonscanbereturnedtothetankofthehydraulicsystem


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Figure 11 Schematicdiagramofaswashplatepistonpump Thedisplacementof theswashplatepumpper revolutionof thepumpdriveshaftisgivenby

displ = cyl cyl stroke (13)wherecyl = numberofcylinders

Rotary Positive Displacement Pumps Rotary positive displacement pumps, e.g. gear pumps or vanepumps, discharge a fixed amount of pressurized fluid for eachrotationoftheinputshaft.Gear Pumps Inagearpump,Figure12,thefluidistransportedfromtheintakeporttothedischargeport.Notecarefullythedirectionsofrotationofthegearsandthepathfollowedbythefluid.Thedisplacementof the gear pump per revolution of the drive shaft can becalculatedfromthedimensionsandgeometryofthegearsandthehousing.

Discharge valve

Intake valve

Swash plate

Shoe plate

Piston shoe

Shaft

Dischargeslot

Intakeslot

Piston block(rotating)

Piston Valve plate(fixed)

Spring

Drainto tank


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Figure 12 SchematicrepresentationofagearpumpVane Pump Abasic type of vane pump is shown schematically in Figure 13.Volumes of fluid are enclosed on the inlet side of thepumpandtransportedtothedischargeside.As with other rotary positive displacement pumps, thedisplacement per rotation of the drive shaft can be calculatedfrom the dimensions and geometry of the housing, rotor andvanes.NotethatthecasingrecessesinFigure13arenecessaryasthe fluid is incompressible and the enclosed volume should notchange as it is brought from the inlet to the outlet side of thepump.

Figure 13 Schematicdiagramofavanepump

VanesHousing

Rotor

Inlet Discharge

Recess

Recess


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Ideal Flow Rate of a Positive Displacement Pump The ideal flow rate of a positive displacement pump can becalculated as the product of the displacement of the pump perrotation(orcycle)andthespeedinrotations(orcycles)perunittime,ideal = displ (14)where

displ = pumpdisplacementperrevolution(orcycle) = speed(rotationsorcyclesperunittime)Example 4 Ideal Flow Rate of a Vane Pump Avanepumphavingadisplacementof25mLperrevolutionrunsataspeedof1750r.p.m.Whatflowratewoulditideallyproduce?Expresstheanswerinm /sandalsoinlitresperhour.Solution

displ = 25mLrev = 25 × 10 mrev= 1750 revsmin = 175060 revs = 29.17 revs ideal = displ = 25 × 10 mrev × 29.17 revs

= 729.3 × 10 ms orideal = 729.3 × 10 ms × 1000 Lm × 3600 sh

= 2625 Lh


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Efficiency of Hydraulic Pumps

Mechanical Efficiency of a Pump Themechanicalefficiencyofapumpisdefinedaspump,mech = ideal ∆actual (15)where

pump,mech = pumpmechanicalefficiencyideal = idealvolumeflowrate m /s∆ = pressuredifferenceacrosspump Paactual = actualpumpinputpower WItisimportanttonotethattheflowrateusedinthedefinitionofthe mechanical efficiency, Equation 15, is the ideal flow rate,basedonthetheoreticaldisplacementofthepump.

Volumetric Efficiency of a Pump An ideal positive displacement pump would have a fixeddisplacementandwouldalwaysdischargethatamountoffluidforeach rotation (or cycle, in the case of a reciprocating pump). Inpractice, there is usually some leakage backwards through thepumpsothattheamountoffluiddischargedislessthantheidealamount (corresponding to the theoretical displacement of thepump). The leakage within a pump depends on the pressuredifference:higherleakageratesoccurwhenthepumpisoperatingagainsthigherpressuredifferences.Thevolumetricefficiencyofapumpisdefinedaspump,vol = actualideal (16)where

pump,vol = pumpvolumetricefficiency


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actual = actualvolumeflowrate m /sOverall Efficiency of a Pump Theoverallefficiencyofapump,Equation17,istheproductofthemechanical and volumetric efficiencies. Some typical values ofhydraulicpumpefficienciesareshowninTable2.

pump,overall = pump,mech × pump,vol= ideal∆actual actualideal = actual∆actual (17)where

pump,overall = pumpoverallefficiencyTable 2Typicalhydraulicpumpefficiencies PumpType MechanicalEffic./[%] VolumetricEffic./[%] OverallEffic./[%]Gear 90–95 80–90 72–86Vane 90–95 82–92 74–87Piston 90–95 90–98 81–93 Example 5 Gear Pump Efficiency Agearpumpoperatesat1,500r.p.m.withapressuredifferenceof15MPa and has a displacement of 12mL per revolution. Themeasuredflowrateis0.948m /h.Themechanicalpowerinputtodrivethepumpis4.77kW.Calculatethemechanicalefficiency,thevolumetricefficiencyandtheoverallefficiencyofthepump.Solution Theidealflowrateofthepumpisgivenby

ideal = 12 × 10 [m ] × 150060 [s ]= 300 × 10 ms


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pump,mech = ideal∆actual = 300 × 10 ms × 15 × 10 Nm4.77 × 10 [W]

= 0.943 = 94.3%actual = 0.9843600 ms = 273.3 × 10 ms

pump,vol = actualideal = 273.3300 = 0.911 = 91.1%pump,overall = pump,mech × pump,vol= 0.943 × 0.911 = 0.859 = 85.9%


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Chapter 6. Hydraulic System Components

Check Valve and Pressure Relief Valve Figure14illustratestheconstructionofaspringloadedone-wayvalve. Figures 15 and 16 are the schematic representations ofcheckvalveswithandwithoutspringsrespectively.

Figure 14Simpleone-wayorcheckvalve

Figure 15 Checkvalvesymbol(withspring)

Figure 16Checkvalvesymbol(withoutspring)

Figure 17Operatingprincipleofapressurereliefvalve Figure 17 illustrates the construction principles of a pressurereliefvalve.Thepressurereliefvalvecanbeadjustingbyturninga

Adjustment screw

Out

In

Spring


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knob. The resulting force, exerted by the compressed spring,determines the pressure at which the valve will open. ThestandardsymbolforapressurereliefvalveisshowninFigure18.

Figure 18Pressurereliefvalvesymbol Hydraulic Cylinders or Rams Hydrauliccylindersorramsarelinearactuatorsthatcanbeusedtoapplyhighforcesandprovidepushingorpullingmovements.AtypicalconstructionforahydraulicramisillustratedinFigure19.

Figure 19 Typicalconstructionofacommercialhydrauliccylinder A ram is double acting if it can apply forcewhile extending andwhileretracting.Iftheramcanonlyapplyforceinonedirectionitissaidtobesingleacting—someexternalmeanswillberequiredtoreturnthepistonto itsoriginalposition.Ahydraulicrammayhavearodthatextendsfromoneendonlyormayhavearodthatextends from both ends. These configurations are described as‘singlerod’and‘doublerod’respectively.Figures20to22aretheschematicrepresentationsofthreepossibleconfigurations.

Figure 20 Doubleactingsinglerodcylinder

Cylinder cap

Piston

seal

Rod seal

Cylinder seal

Rod

Piston

Port

Port

Cylinder

head

Cap eye

or bearingRod eye

or bearing

Cylinder barrel


28

Figure 21 Singleactingcylinder

Figure 22 Doubleactingdoublerodcylinder Cylinder Force Calculations The main parameters of a hydraulic cylinder are the bore, thecylinder rod diameter and the stroke of the piston. Theseparametersdeterminethevolumesofhydraulic fluidrequiredtofill the cylinder. The cylinder bore and the piston rod diameterdetermine the thrust or tension generated at a given operatingpressure.

Figure 23ForcecalculationforthecapendofacylinderFigure23showstheforceproducedbythepressureactingonthepistonatthecapendofasingle-actingcylinder.Itiscalculatedas= cyl (18)where = forceproducedinthepistonrod = pressurecyl = cylindercross-sectionalarea


29

Figure 24ForcecalculationfortherodendofacylinderFigure24showstheforceproducedbythepressureactingonthepistonattherodendofasingle-actingcylinder.Itisatensileforcewithintherodandiscalculatedas= cyl − (19)whererod = rodcross-sectionalarea

Example 6 Cylinder Forces a. Asingle-roddouble-actingcylinderwithaboreof160mm isrequired to exert a force of 200kN on the outstroke.Determine the required oil pressure in the cap end of thecylinder.b. If thediameterof the cylinder rodof the samedouble-actingcylinder is 90mm, determine the pulling force of the rod ontheretractstrokewiththesameoilpressure, thistimeintherodendofthecylinder.Solution

cyl = 4 = 0.164 [m ]= 0.02011m

= cyl = 200 × 100.02011 Nm = 9.945 × 10 Pa= 9.95MPa a.


30

rod = 4 = 0.094 [m ]= 0.006362m eff = cyl − rod= 0.02011[m ] − 0.00636[m ] = 0.01375m = eff = 9.945 × 10 [Pa] × 0.01375[m ]= 136,743N = 136.7kN b.

Directional Control Valves In order to be able to switch between extending or retracting ahydraulic cylinder, a directional control valve is required. In thevalvesymbolsshowninFigure25andFigure26eachsquareboxrepresentsapossibleconfigurationofflowpathsbetweenthefourportsofthevalve.

Figure 25Twopositiondirectionalcontrolvalvesymbol

Figure 26Threepositiondirectioncontrolvalve Manydifferentmethodsofactuatingavalvearepossible,someofwhichareillustratedbythestandardsymbolsshowninFigure27.


31

Figure 27 Symbolic representations of different valve actuationtypes Directional Spool Valve Variousdesigns fordirectional control valves arepossible, but acommonarrangementmakesuseofaspool,whichisasolidobjectin the formofaseriesofcylindersofdifferentdiameters.Figure21includesaschematicillustrationoftheconstructionofathreeposition,springreturn,solenoidactuateddirectionalcontrolspoolvalve aswell as the standard symbolic representation of such adevice.LettersAandB label theconnections to thedevicebeingcontrolled,suchasahydraulicramorahydraulicmotor.PlabelstheconnectionfromthepumpandTlabelstheconnectiontothetank.

Figure 28Three-positiondirectionalspoolvalve

Solenoid

Lever

Manuallyoperated

Spring

ButtonMechanicallyoperated

BA P

T

Solenoid

Spool

BA

TP


32

The Hydraulic Circuit Diagram Figure29isanexampleofahydrauliccircuit foradoubleactingcylinder and illustrates how a directional control valve isconnected. This circuit ismade up entirely of standard symbolsfor hydraulic components such as the cylinder, the pump, thepressurereliefvalve,afilterandthehydraulictankorreservoir.Itshouldbenotedthatthetankisshownthreetimes inthecircuitdiagram, although there is only one hydraulic fluid tank. Thisavoids the clutter that would be caused if all return lines wereshownconnectedtojustonesymbolforthetank.

Figure 29 Basic hydraulic circuit for a double acting single rodcylinder

Filter

Tank


33

Chapter 7. Design of Hydraulic Systems Control of hydraulic systems is introduced through the topic offlow control. The principles of hydraulic system design areintroduced by means of two worked examples, for a lift and apressrespectively. It is important tokeep inmindthat indesigntherearealwayschoicesandjudgementstobemade.

Flow Control

Flow Restriction and Flow Control Theflowoffluidinahydrauliccircuitcanbereducedbymeansofa flow restrictor, Figure 30. This causes a pressure drop in thedirectionofflow.Bymeansofavariableareavalve,theflowratecanbecontrolled.Generally, the flowrate througharestrictor isproportional to the effective flowarea and to the square root ofthepressuredropacrossit.

Figure 30Symbolsforafixedareaflowrestrictorandavariableareaflowrestrictor Powerisalwaysdissipatedinaflowrestrictor.Thishastheeffectofincreasingthetemperatureofthehydraulicfluid.Hydraulic Ram Speed Control The speed of a hydraulic ram can be controlled by the use of aflowcontrolvalve in conjunctionwithapressure relief valve, asillustrated inFigures31 to33.Threepossible arrangements areillustrated, named ‘meter-in,’ ‘meter out’ and ‘bleed-off’ flowcontrol.


34

Figure 31Meter-inspeedcontrolmeteringcircuit

Figure 32Meter-outspeedcontrolmeteringcircuit

Figure 33Bleed-offspeedcontrolmeteringcircuitWith meter-in and meter-out flow control, a proportion of theflow is forced through the pressure relief valve. With meter-incontrol the forceavailable todrivethe load isgreater.Meter-outcontrolcreatesabackpressureontherodsideofthepistonwhichcanhelppreventlungingiftheloaddecreasesrapidlyorreverses.Withbleed-offcontrol,fluidisbled-offthroughtherestrictoratalower pressure than the pressure relief valve setting so there islesspowerwastage.Bleed-offcontrolcanonlybeusedwheretheloadopposesthemotion.


35

Powerwastage is inevitablewhenmetering flowcontrol isused.An alternative would be to use a hydraulic pump whosedisplacementcanbevaried.Figure34isthestandardsymbolforsuchapump.

Figure 34StandardhydraulicsymbolforavariabledisplacementpumpHydraulic System for a Lift

Example 7 Component Sizing for a Hydraulic Lift Ahydraulicliftisrequiredtoraiseloadsofupto20tonnethroughadistanceof4m in30s. Select a reasonableoperatingpressureandusethistosizeasuitablecylinderandpump.Donotexceedarunning pressure of 120bar. Determine the exact runningpressure and specify a relief valve setting. Assume an overallpumpefficiencyof60%andestimatetherequiredelectricmotorpoweroutput.Thefollowingstandardboresshouldbeused: 25,32,40,50,63,80,100,125,160,200,250,320mm.Solution First, to findtherequiredcylinderbore,use = ⁄ ,anda trialpressureof120bar. = 20 × 1,000[kg] × 9.81[ms ] = 196,200N = 120 × 10 Nm

= 1.962 × 10 [N]120 × 10 [Nm ] = 0.0164m and = 4 / = 4 × 0.0164/ m = 0.1445m = 145mm.Wemustchooseastandardsize,thenearestonesbeing125and160mm.


36

Inordertokeeptherunningpressurebelow120bar,wechoosethelarger(160mm)bore.Runningpressurerun == = 1.962 × 10 [N]0.1604 [m ] = 9.758 × 10 Nm = 97.6bar

It would be reasonable to set the pressure relief valve at apressure that is 10% higher than the pressure required by themaximumload.Hence,thesettingwouldberv = 97.6[bar] × 110% = 107bar.Pumpsize:

= × = 0.1604 [m ] 430 [ms ]= 0.00268ms = 2.68 Iitres = 161 litreminElectricmotorsize:Hydraulicpower = × = 107 × 10 × 0.00268[W] = 28,676W = 28.7kWEstimatedelectricmotorpower

= HydraulicoutputpowerOverallpumpefficiency = 28.7[kW]0.6 = 47.8kWThe next standard motor power rating above this would beselected,e.g.50kW.


37

Hydraulic Press Figure 35 is a possible circuit for a hydraulic press. With thevariableareaflowrestrictorfullyopenonlyaslightpressurewillbe applied to the piston. The pressure can be increasedprogressivelybygraduallyclosingtherestrictor.

Figure 35Circuitforahydraulicpress Figure36showshowahydrauliccylindermightbeincorporatedintoaloadframeofapress.

Figure 36Loadframeofahydraulicpress Example 8 Design of a Simple Hydraulic Press A25tonnehydraulicpressbasedonasingle-actingspring-returncylinderisrequiredtostrokethrough300mmin6seconds.Ifthemaximumallowablepressureratingis150bar(a)Specifyasuitablestandardcylinderboreand(b)Sizeasuitablepump.

Vent


38

Thefollowingstandardboresshouldbeconsidered:25,32,40,50,63,80,100,125,160,200,250&320mm.Solution Asuitabledesignforthepressmightbeasfollows:(a) Specify a suitable standard bore for the cylinder The force exerted by the press is 25 tonneweight (strictly, theweightofamassof25tonne).Thismustbeconvertedtonewtons.1 tonne weight = 1 × 1,000 × 9.81N, where 9.81 m/s is theaccelerationduetogravity.Hence= 25 × 1,000 × 9.81 = 245,250N.Themaximum pressure of 150 barmust be converted to N/m2.Hence = 150 × 10 Nm .Theforceisalsoequaltotheproductofthepressurebytheareaof the piston. Hence in terms of the piston diameter or cylinderbore,d, 4 =

or = 4 .Hence= 4 × 245,250150 × 10 [N][Nm ] = 0.1443m.

Thereforeastandardboreof0.160mcouldbeused.Therequiredoperatingpressurewouldthereforebeop = 150[bar]× 0.14430.16 = 122.01barThepressurereliefvalvewouldbesetatperhaps10%or10barhigher,whichever is the smaller.Hence thepressure relief valve


39

wouldbesetatabout132bar.Notethat132bar is lessthanthestatedmaximumallowablepressurerating.(b) Size a suitable pump The volume flow rate into the cylinder is the product of thevelocityandthecross-sectionalarea,i.e. × /4.Thevelocity,,is

= 0.36 ms = 0.05ms .Thevolumeflowrateisgivenby= π 4 .Thereforethevolumerateinlitresperminuteis

= 0.05 0.164 ms × 60 smin × 1,000 litrem = 60.32litremin .Asuitablepumpwouldbecapableofoperatingatup to132barandwithaflowrateofabout61litresperminute.


40

Chapter 8. Viscosity Viscosity is a property of a fluid,which quantifies how stronglythatfluidresistsrelativemotion.Afluidthathasahighresistancetorelativemotion,suchascrudeoil,wouldhaveahighviscosityandwouldbedescribedasbeingviscous.Afluidthatpresentednoresistance to relativemotionwouldbedescribedas inviscid,butunder normal circumstances all fluids have a finite viscosity,howeversmall.Figure 37 illustrates a block sliding horizontally over a layer ofliquid that covers ahorizontal surface. Figure38 illustrateshowthevelocityvarieswithinthefluidlayer:thefluidvelocityiszeroatthestationarysurfaceandincreaseslinearlywithdistancefromthesurfacetothevelocity, ,ofthemovingsurface.

Figure 37 Blockslidingonafilmofliquid

Figure 38 VelocityprofilewithinafluidundergoingshearmotionIt is foundthattheforce requiredtomaintainthevelocity isproportionaltothesurfacearea ofthemovingblockandtothevelocity , while it is inversely proportional to the distance betweenthestationarysurfaceandthemovingblock.Thiscanbewrittenas ∝ A .

Area AVelocity v

Thickness d

Area A

Distance d

Stationary

Force F

Velocity profile

Velocity v

y

x


41

Thisrelationshipofproportionalitycanbemadeintoanequationbyincludingaconstantofproportionality, ,asfollows:= A . (20)Equation 20 is known as Newton’s viscosity equation. Theconstant (the Greek letter mu) is a fluid property, which isknownastheabsoluteordynamicviscosity.

Newton’s Viscosity Equation Whereas in Figure 38 the velocity profile is a straight line, ingeneralwhena fluid flowsovera stationary surface thevelocityprofile is likely toberepresentedbyacurve,asshowninFigure39. It is found thatNewton’s viscosity equation applies for eachthinlayeroffluid,e.g.forthelayershowninFigure39thathasathicknessofΔ .Overthissmallthickness,thevelocityofthefluidincreasesbyanamountthatisrepresentedbyΔ .

Figure 39 Generalvelocityprofilewithinafluid Newton’sviscosityequationcanbewritteninamoregeneralformthanEquation20as

= ΔΔ . (21)Atthispointitisworthpausingtomakesurethatweunderstandwhat the force in Equation 21 represents. It represents theshear force, owing to viscosity, between adjacent layers of themoving fluid. The layers can be regarded as sliding, one overanother.

Moving fluid

Stationary

surface

Velocity profile

y

x

�y

�v


42

ForthevelocityprofileshowninFigure39,wecanseethataswemove away from the stationary surface, Δ becomes smaller forsuccessive equal steps of , the distance from the surface.Therefore Δ /Δ becomes smaller as we move away from thesurface. Hence, the shear force F between layers of the movingliquiddecreasesaswemovefurtherfromthefixedsurfacewhere= 0.Newton’sviscosityequation,Equation21,givestheforcethatactsoveraplaneofarea withinthefluid.Wecaneasilymodifyittogive the force per unit area over the same plane, which is theshearstressover theplane.Wedo thisbydividingbothsidesofthe equationby the area .Hence, the shear stress (theGreeklettertau)isgivenby= ΔΔ . (22)

The SI Units Involved in Newton’s Viscosity Equation The SI units for distance, velocity, area and force are relativelystraightforward,assummarizedhere: = distanceindirection normaltodisplacement m = distancebetweensurfaces m = fluidvelocity ms = force NWecannotethatNstandsfortheSIunitofforce,thenewton.Wecan also note that ms−1 is another way of writing m/s, whichstands for ‘metres per second.’ As stress is defined as forcedividedbyarea (or forceperunitarea) itsunitsare theunitsofforcedividedbytheunitsofarea.Hence = shearstress Nm Even ifwedonotknoworrememberwhat theunitsofabsoluteviscosity are, we can determine the units easily from Newton’sviscosity equation, Equation 21 or Equation 22. RearrangingEquation21,


43

= ΔΔ . (23)Theabsoluteviscosity mustthereforehavethesameunitsastherighthandsideofEquation23.TheseunitsareN mm ms = Nm s.Therefore the correct SI unit for absolute viscosity is Nm s orPas,whichcouldalsobewrittenasNs/m2.Tosummarize,basedon what has been covered so far, the symbol and units forabsoluteviscosityare: = absoluteviscosity Pas(orNm s)Newton’s 2nd Law and the SI Units for Absolute Viscosity TheSIunitsforabsoluteviscositycanbeexpressedinaformthatincludes the unit for mass, the kilogram. Newton’s second law,Equation1,Page8, is thekey to the equivalencebetween formsthatincludemassandthosethatincludeforce.= (1)(repeated)Of the four units the newton, the kilogram, the metre and thesecond,anyonecanbeexpressedintermsoftheotherthree.Forexample 1N = 1kgms .This allows a version of the SI units for absolute viscosity to beused that does not include the newton, but does include thekilogram. 1Nm s = 1kgms m s = 1kgm s .Summarizing again, the symbol and units for absolute viscosityare: = absoluteviscosity PasorNm sorkgm s ItisimportanttobeawareofthesedifferentwaysofwritingtheSIunits forabsoluteviscosity.All threeversionsgivenaboveare inuse.


44

Kinematic Viscosity Kinematicviscosity (theGreekletternu)isdefinedas= where = kinematicviscosity m2s−1 = density(= massvolume) kg/m3We can note that the units for kinematic viscosity are got bydividingtheunitsofabsoluteviscositybytheunitsofdensity.Asthedensityunitsincludekg,itisconvenienttousetheformoftheabsolute viscosity units that includes kg. Thus, the units ofkinematicviscosityarerepresentedaskgm skgm = m s .More about Viscosity Units and Values Absolute(ordynamic)viscosityvaluesforrealfluids,expressedinthefundamentalSIunits,areusuallyverysmall,e.g.1×10−3Pasfor water. Kinematic viscosity values are even smaller, e.g.1 × 10 m s for water. Other viscosity units are also in usesuchas1poise=10−1Pas(absoluteviscosity)1centipoise=10−2poise=10−3Pas(absolute viscosity)1stoke=10−4m2s−1(kinematicviscosity)1centistoke=10−2stoke=10−6m2s−1(kinematic viscosity)


45

Figure 40 SchematicdiagramofaRedwoodviscometer Another scale for viscosity is the Redwood scale in ‘Redwoodseconds.’TheRedwoodviscosityisdeterminedbymeasuringhowlong it takes for a fixed amount of liquid to flow through acalibrated orifice, Figure 40. The Redwood scale is a practicalscale for comparing viscosities of different liquids, which isessentiallyakinematicviscosity scale.However, the relationshipbetweentheRedwoodscaleandatruekinematicviscosityscaleisnotquitelinear.Therefore,valuesfromtheRedwoodscalearenotused directly for engineering calculations, but a calibrationformula can be used to convert test readings from a Redwoodviscometertokinematicviscosity.For liquids generally the viscosity decreases as the temperaturerises. For gases, the opposite is generally the case: viscosityincreaseswithincreasingtemperature. Example 9 Newton’s Viscosity Equation Calculate the force required tocause relativemotionata rateof1.7m/sofaflatplateofarea0.36m2withrespecttoaparallelflatsurfacewheretheplateisseparatedbyadistanceof0.6mmandwherethespacebetweentheplateandthesurfaceisfilledwithoilhavinganabsoluteviscosityof64centipoise.(1centipoise=10−2poise=10−3Pas).Includeasketchinyoursolution.

Thermometers

Water

Plug

Stirrer

handleLevel

pointer

Oil

Stirrer vane

Insulation

Graduated

cylinder

Electric

heater

Calibrated

orifice


46

Solution

Theshearforceisgivenby= = ∆∆ =

= 64 × 10 [Nm s] 1.7[ms ]0.0006[m] 0.36[m ]= 65.3N

Area AVelocity v

Thickness d


47

Chapter 9. Pressure Losses in Pipes and Fittings

Laminar and Turbulent Flow Theeasiestwaytovisualisetheflowofafluidwithinapipeistoimagine that thevelocityof the fluid is the sameover the entirecrosssectionofflow,asillustratedinFigure41.Thisisknownasuniformflow.

Figure 41 UniformflowFor uniform flow, the velocity can be calculated by dividing thevolumeflowratebythecrosssectionalarea. = (24)Inthecaseofarealfluid,withfiniteviscosity,thevelocityisneverperfectlyuniform:thefluidincontactwiththepipewallwillhavezerovelocityand,generally,thefluidinthemiddleofthepipewillhave the highest velocity. For a fluid that has a relatively lowvelocityandarelativelyhighviscositythevariationofthevelocityacrossthecrosssection,whichiscalledthevelocityprofile,wouldbeasshowninFigure42.Inthistypeoffloweveryparticleoffluidmoveswithavelocitythatisparalleltothecentrelineofthepipe.A lamina is a thin layer. Thin cylindrical laminae of fluid can beimagined to slide over one another, the laminaenear the centremovingfasterthanthoseneartheedges.

Figure 42Laminarvelocityprofile Where the velocity is relatively high and the fluid viscosity isrelatively low, a different type of velocity profile, as shown inFigure 43, is produced. In addition to an average velocity in the


48

direction of flow, each particle of fluid moves with a chaoticvelocity that causes mixing over the cross section of the pipe.Local eddies occur and continue to change with time. Theturbulentvelocityflowprofilecomesclosertomatchingauniformflowprofilethanthelaminarprofiledoes.

Figure 43 TurbulentvelocityprofileOsborneReynoldscarriedoutanexperimentinwhichhewasableto visually compare laminar and turbulent flow through a glasspipe. This experiment is shown schematically in Figures 44 and45.Whendye is injected fromaneedle into laminar flow, a thindyefilamentcanbeseentocontinueoveralonglengthwithintheflow,Figure44.

Figure 44Reynolds’experimentwithlaminarflow

Figure 45Reynolds’experimentwithturbulentflow Whendyeis injectedintoahighflowrateoffluid,Figure45,thedyefilamentismaintainedforacertainlengthfromthepipeinlet(indicatinglaminarflow),but,furtheron,turbulentflowdevelopsandcausesmixingofthedyeoverthefullboreofthepipe.In laminar flow, Figure 44, viscous forces dominate, whereas inturbulent flow inertia forces (i.e. acceleration or decelerationforces) dominate. In laminar flow there is no cross mixingbetweenthe layers. In turbulent flowathoroughcross-mixingof

Laminar flow at very low velocity

Velocity

profile in pipeDye injection

Dye injection

Turbulent flow

at high velocity

after entry length

Velocity

profile in pipe


49

thefluidoccursandindividualsmallvolumesoffluidhavechaoticmotionsuperimposedontheiraverageforwardvelocity.Itisworthnotingatthispointthat,irrespectiveofthetypeofflow,theaveragefluidvelocityisgivenbyEquation24.Reynolds Number Whethertheflowinapipeislaminarorturbulentdependsonthedensityof the fluid, thevelocityof flow, thediameterof thepipeand the absolute viscosity of the fluid. These parameters can becombined in a dimensionless number, called the Reynoldsnumber.Thevalueofthisnumbercanbeusedtopredictthetypeofflowthatwillexist.TheReynoldsnumberisdefinedas

Re = = (25)whereRe = Reynoldsnumber = density kg/m = averagefluidvelocity ms = diameter m = absoluteviscosity Pas(orNm s orkgm s ) = kinematicviscosity m s WhethertheflowislaminarorturbulentdependsontheReynoldsnumber,asfollows:LaminarflowexistsforRebelow2,000.TurbulentflowexistsforRegreaterthan4,000.Betweenthesevalueslaminarorturbulentflowispossible.Example 10 Laminar or Turbulent Flow? Ittakes30secondstofillakettlewithtwolitresofwaterfromatap.Thetapissuppliedbyapipethathasaninternaldiameterof


50

12.7mm.Determinewhether the flow is laminaror turbulent inthesupplypipewhilethekettleisbeingfilled.Takethedensityofwater as 1,000kg/m and take its absolute viscosity as1.002 × 10 kg/m s.Solution Calculatethevolumeflowrateas

= = 2 × 10 [m ]30[s] = 66.67 × 10 m /s.Calculatethecrosssectionalareaas

= 4 = 0.01274 [m ] = 126.7 × 10 m .Calculatethemeanvelocityas= = 66.67 × 10 [m ⁄ ]126.7 × 10 [m ] = 0.5262ms .CalculatetheReynoldsnumberas

Re = = 1000 kgm 0.5262 ms 0.0127[m]1.002 × 10 kgms = 6,669 ≅ 6.7 × 10 AsRe>4,000,flowisturbulent.

Head Loss Due to Friction Different lawsoffluidresistanceapplyfor laminarandturbulentflows, although one common equation (the Darcy-Weisbachequation,whichissometimesjustcalledtheDarcyequation)canbeused for fluid flow inpipes—it incorporatesa ‘friction factor’thatdependsonthenatureoftheflow.


51

TheDarcy-Weisbachequationhastheformℎf = 4 2 (26)wherethefrictionfactorf isobtainedasdetailedbelowandwhereℎf = headlossduetofriction m = frictionfactor = lengthofpipe m = diameter m = meanfluidvelocity ms−1 = accelerationduetogravity ms−2

Note: The friction factor used here is known as the Fanningfriction factor. In some textbooks and references the Darcy-Weisbach equation is written without the integer 4 in thenumerator, but the friction factor used is four times the frictionfactorusedhere.ThatfrictionfactorisknownastheDarcyfrictionfactor.Asmightbeexpected,theinternalroughnessofapipe,knownasitssurfaceroughness,canhaveaninfluenceonthepressuredropthat occurs in the pipe. In fact, it is the ratio of the surfaceroughnesstothediameterthatisofimportance.Thisratio,knownas the relative roughness, influences the pressure drop forturbulent flow.However, for laminar flowtherelativeroughnessdoesnotinfluencethepressuredrop.

Figure 46Diameterandsurfaceroughnessofapipe

dk


52

Forturbulent flowthefrictionfactorfiscommonlyobtainedfroma friction factor diagram,which is often referred to as aMoodydiagram or chart, Figure 47. The friction factor depends on theReynolds number and on the relative roughness of the pipek/dwhere = Pipesurfaceroughness m = Pipeinternaldiameter m The friction factor can also be found by using appropriateempiricalformulae,e.g.withinaspreadsheet.Theseformulaearetoo cumbersome for hand calculations; hence the friction factordiagramisuseful.Forfullyturbulentflowtheheadlossperunitlengthofpipe,ℎf⁄ ,isproportionaltothemeanvelocitysquaredanddependsontherelativeroughnessofthepipe.Forflowthatisnotfullyturbulentthere isnotadirectproportionality tovelocitysquared(roughlyspeaking, the head loss is proportional to velocity raised to apowerthatisclosetoandalittlelowerthan2).Forlaminar flowthefrictionfactorisgivenby= 16Re . (27)

Note 1: Equation27 iswrittenon the friction factordiagram,sothereisnoneedtorememberit.Note 2: In referenceswhere the friction factor is four times thevalueusedhereEquation27iswrittenwith64inthenumeratorinsteadof16.Itcanbeshownthatforlaminarflowtheheadlossperunitlengthof pipe, ℎf⁄ , is proportional to themean velocity (or themeanflow rate) and independent of surface roughness. The frictionfactorforthelaminarflowregioncanalsobereadfromafrictionfactordiagram.


53Figure47Frictionfactordiagram(Moodydiagram).

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Flow through Valves and Fittings The presence of a valve in a pipe, even when it is fully open,almostalwayscausesaheadlossowingtodisruptionoftheflowand the formation of a vena contracta (reduced flow area). Thesameistrueformanyothertypesoffittings,e.g.afittingthatjoinspipesofdifferentdiameters.Inthesecasesaheadlosscoefficientisused,Equation28.ℎ = 2 . (28)Intheaboveexpressionthevelocity isthemeanvelocityinthefullborepipeupstream.

K is a dimensionless constant for a particular configuration andwilldependonthenatureofthevalveorfitting.Hence,iftheheadlossatavalveorfittingismeasuredforagivenmeanvelocityofflowtheheadlosscoefficientcanbedeterminedas= 2 ℎ .

Flow Around Bends and Elbows Whenever thedirectionof flow isabruptlychangedatabendorelbow, a loss of head occurs due to the formation of a venacontracta on passing the bend or elbow and the turbulenceassociated with the subsequent enlargement. Surface roughnesswithinthebendalsohasaninfluenceontheheadloss.Theheadlosscanbecharacterizedbyaheadlosscoefficient,asinEquation28.Equivalent Pipe Length Commercially, the head loss caused by flow through fittings issometimespresentedasan ‘equivalent lengthofstraightpipe’tocausethesameheadloss.Inthiscasetheheadlossforapiperunthat includes fittings (elbows, bends, reducers etc.) can becalculatedusingtheDarcy-Weisbachequation(Equation26),butthe pipe length used is the actual length of the pipe plus theequivalent lengths of any fittings used. Head losses associatedwithvalves, fittingsandbendsareoftendescribedasminorpipelosses.Major losses are those associatedwith fluid friction overthefulllengthofapipe.


55

Pressure Loss and Pumping Power TheDarcy-Weisbachequation,Equation26,allowsustocalculatethehead lossdueto friction inapipe.Head losshastheunitsoflength, the metre. If we require the pressure loss, or pressuredrop,overthe lengthofpipewecanuseEquation3, = ℎ, tofindit:loss = ℎf (29)where

loss = pressurelossduetofriction Pa = density kg/m = accelerationduetogravity m/s ℎf = headlossduetofriction mCombiningEquations26and29loss = 4 2 orloss = 4 2 (30)Equation30 is an alternative form of the Darcy-Weisbachequation,whichgivesthepressurelossratherthantheheadlossduetofluidfriction.In some cases flow through a pipe is caused directly by adifference in liquid levels in two tanksat either endof thepipe.The flowmight alsobe causeddirectlybyapump,whichwouldhavetoprovideapressuredifferencetomatchthepressure lossin the pipe. The pressure difference across the pump is oftenrepresented by Δ , Figure 48. If a pump is required to exactlymeetapressuredropduetofrictioninapipethenΔ = loss.


56

Figure 48PressuredifferenceacrossapumpThe ideal pumping power equals the product of the pressuredifferenceacrossthepumpandthevolumeflowrate,Equation31. = Δ (31)where = idealpumpingpower WΔ = pressuredifference Pa = volumeflowrate m /sSummary of Key Formulae for Pipe Flow ThekeyformulaeforpipeflowaresummarizedinTable3.

Table 3Summaryofkeyformulaeforpipeflow Meanvelocityofflowinapipe: =ReynoldsNumber: Re = =Re<2,000 LaminarFlowRe>4,000 TurbulentFlowHeadlossduetofrictioninapipe(Darcy-Weisbachequation):ℎf = 4 2Forlaminarflow canbecalculatedfrom= 16Re .For turbulent flow, f depends on Re and on the relativeroughnessk/d.


57

Headlossesatbendsorfittingsℎ = 2 .Pumpingpower = Δ .Example 11 Head Losses due to Friction Apipeofabsoluteroughness0.0030mmandbore50mmconveyswater at a rate of 210 litres per minute. If the viscosity of thewateris0.0013kg/msdetermine(a) thefrictionfactor(b) theheadlosspermetrerunofpipe(c) the maximum allowable flow rate if the flow were to be laminar(d) the theoretical pumping power per metre length for theturbulentflowinthepipe.Solution Thevolumeflowrateisgivenby

= 21060 × 1,000m /s=0.0035m /s.Thecross-sectionalareais= 0.054 m = 1.963 × 10 m .Themeanvelocityofflowinthepipeis

= = 0.0035[m /s]1.963 × 10 [m ] = 1.783m/s.TheReynoldsnumberis

Re = = 1,000 × 1.783 × 0.050.0013 kgm ms mkgm s = 68,577 ≅ 6.9 × 10 AsRe>4,000,theflowisturbulent.


58

Therelativeroughnessis= 0.003mm50mm = 0.00006For Re = 68,600 and k/d = 0.00006 the friction factor is foundfrom the friction factor diagram to be 0.0049 (see diagram onpage59). (a)Theheadlossduetofrictioninapipe(Darcy-Weisbachequation)is:

ℎf = 4 2 Hence ℎf = 4 2 = 4 × 0.00490.050 1.7832 × 9.81m s mms

= 0.0635m/mThatis,theheadlossis63.5mmpermetreofpipelength. (b)Themaximumallowableflowrate if theflowweretobelaminarwould require Re to equal 2,000. From the definition of theReynoldsnumber(Re = )themaximumlaminarflowvelocityis= Re = 2,000 × 0.00131,000 × 0.05 kgm skgm 3m

= 0.0520m/sQ = vA = 0.0520[m/s] × 1.963 × 10 [m ]= 0.1021 × 10 m /s= 0.1021 × 60L/min = 6.13L/min (c)Thepumpingpowerfortheturbulentflowcaseisgivenby= ∆ = ℎf


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61

Chapter 10. Valves and Heat Exchanger Flow Paths This chapter describes valves and the flow paths of some heatexchangers.Thebasis for calculatingpressure losses for these isexplained.

Valves Valvesareused for isolation (i.e. shuttingoff flow inapipe), forthrottling (i.e. causing a pressure drop in the direction of flow)andforflowregulation(i.e.allowingtheflowratetobevaried).Mostvalvesrequireastemseal(orpacking)topreventfluidfromleaking out. Valves that are to achieve a tight shut-off require asuitableseat(theareawithwhichtheplugelementmakescontactwhen the valve is closed) to maintain leak-tightness. This mayinvolve precision-machined matching surfaces or one of thesurfacesmaybecompliant,e.g.madefromrubber.Gate Valve Thisvalve,Figure49,containsaslidinggatethatcanbeopenedorclosedby turning a handwheel attached to a threaded stem formanual operation. For automation purposes, a rotary actuatormay be used or a version without threads on the stemmay beconnectedtoalinearactuator.Thegateslidesintoarecessinthehousing,allowingthefluidtoflow.Openingorclosingisrelativelyslow.Afullboreopeningcanbeprovidedbutthehousingmustbeveryhightoaccommodatetheslidinggateandthescrewthread.

Figure 49Gatevalvediagram

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62

Applications: mainlyshut-offbutcanbeusedforflowregulationorthrottling.Advantages: Canprovidetightshut-off(withsuitableseat)MinimalpressuredropwhenfullyopenDisadvantages: Slowoperation(manyturnsofscrewthreadtoopenorclose)LargesizeGlobe Valve In this valve, Figure 50, there is an orifice in the valve bodythroughwhich the fluid flows,which canbe shutby aplug.Theplug is connected to a stem or spindle which has a screwmechanismformanualorrotary-actuatoradjustmentorshut-off.Forautomationpurposes,aversionwithout threadsonthestemmaybeconnectedtoalinearactuator.

Figure 50Globevalvesketch Applications:shut-off,throttling,flowcontrol.Advantages: Canprovidetightshut-off(withsuitableseat)Canprovidefinecontroloffloworthrottling(withsuitabledesignofplugandseat)Moderatelycompact.Disadvantages: Significantpressuredropwhenfullyopen.

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63

Ball Valve Withinthisvalve,Figure51,thereisaballwithaportthroughit.Theballiskeptinplacebyaclose-fittingsphericalseating.Afull-boreportthroughtheball ispossibleandshut-off isachievedbyrotating the ball through a quarter turn. A handle or lever isattachedbya short stem to theball foractuation.Thestemalsolocatedtheballandallowsittopivot.Ifthereisonlyastemtothehandletheballissaidtobefloating.Otherwisetheremaybetwocollinearstemsandtheballissaidtobetrunnionmounted.

Figure 51Ballvalvesketch

Applications:mainlyshut-off.Advantages: Canprovidetightshut-off(withsuitableseat).Quickoperation(quarterturntoopenorclose).Minimalpressuredropwhenfullyopen.Verycompact.Disadvantages: Notgoodforflowregulationorthrottling.Precisionsphericalsurfacesarerequiredwithinthevalve.Head Loss Coefficients for open Valves Some indicative head loss coefficients for valves for use inEquation28arepresentedinTable4.

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64

Table 4Indicativeheadlosscoefficientvaluesforfullyopenvalves HeadLossCoefficient,K Globevalve 4to8dependingonpipediameterBallvalve verylowGatevalve 0.3(Figures for globe and gate valves fromCIBSEGuide C,2007)Example 12 Head or Pressure Loss Calculation Ifaglobevalvefora40mmpipehasaheadlosscoefficientof4.9,calculate the head loss inmetreswhen themean velocity of thewater flowing in thepipe is1.3m/s.Also calculate thepressuredropinpascal.Solution ℎ = 2 = 4.9 1.32 × 9.81m sms = 0.422mThecorrespondingpressurelossiscalculatedasΔ = ℎ = 1,000 kgm × 9.81 ms × 0.422[m] = 4,140Pa.

Heat Exchangers Aheat exchanger is a device inwhichheat transfer occurs fromone fluid to another through a separatingwall or partition. Onefluid is heated as it flows through the heat exchangerwhile theotherfluidiscooled.Heat transfer is enhancedwhen the flow ishighly turbulent andwhen the velocity is relatively high. However, there is acompromisebetweenthepressurelossduetofrictionintheheatexchangerandtheheattransferperformance.Tube in Tube Heat Exchanger A tube-in-tube heat exchanger, Figure 52, is made up of twoconcentric tubes. One fluid flows through the inner tube. The


65

secondfluidflowsintheoppositedirectioninthespacebetweentheinnerandoutertubes.Extendedsurfaces,e.g.spiralfins,maybeusedwithinbothfluidspacestoenhanceheattransfer.

Figure 52Tubeintubeheatexchanger(straight)Applications:Variousfluids,moderatepressures.Advantages: Variousconfigurationsareavailable,savingspaceSingleunitwithnoremovablepartsLightweightDisadvantages: CannotbedismantledforcleaningorrepairMaytakeupmorespacethanothertypesShell and Tube Heat Exchanger Ina shell and tubeheatexchanger,Figure53,a largenumberoftubesrunbetweenheaderplatesateitherendofacylinder.Onefluidpassesthroughthesetubeswhiletheotherfluidpassesoverthetubes,betweenthetubesandtheshell.Theremaybemultiplepasses on the tube side, depending on the arrangement of theheaderboxesatthetwoends.Usuallythereisjustonepassontheshellside.

Figure 53Shellandtubeheatexchangerschematic


66

Applications: Various fluids, wide range of pressures includinghighpressures.Advantages: CompactCanwithstandhighpressuresCanoftenbepartiallydismantledforinspection,cleaningorrepairLeakingtubescanbepluggedorreplaced(dependingondesign).CanbecustombuiltinverylargesizesDisadvantages: DifficulttoinspecttheshellsideoftheheattransfertubesHeavyPlate Heat Exchanger

Figure 54Plateheatexchangerassemblyexplodedview Aflatplateheatexchangerisconstructedfrommultipleplates,asshown in Figure 54 and Figure 55. Between each pair of platesthereisaflatregioninwhichoneofthetwofluidsflows.Theinletport for the each fluid admits the fluid to every second spacebetweentheplatessothefluidsareseparatedfromeachotherbutexchangeheatthroughtheplates.Specialsurfacepatterns,suchasa chevron pattern, can be used to enhance heat transfer. In aframe andplate heat exchanger theplates are clamped together

Hot streamHeat exchanger plates

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67

by tie rods andnuts,while in abrazedplateheat exchanger theplatesarejoinedbybrazing.Frameplateheatexchangerscanbedismantled for cleaning or inspection, whereas brazed heatexchangers cannot. It shouldbenoted that the endplates of theheatexchangerarenotshowninFigure54.

Figure 55PlateheatexchangerflowpathsApplications:Mediumtolowpressureliquids.Advantages: Compactblockshapeandhighheattransferrates.Platescanbeaddedorremovedtochangecapacity(frametype).Disadvantages: Cannotbedismantledforcleaningorrepair(brazedtypes)Shell and Coil Heat Exchanger In a shell and coil heat exchanger, Figure 56, a bundle of longtubes is formed into a helical shape to fit inside a tubular shellwith hemispherical end caps. One fluid flows through the coilbundlewhiletheotherfluidflowsoverthecoilbundle.Thetubesmay be corrugated to promote turbulence and enhance heattransfer.

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68

Figure 56 Shellandcoilheatexchanger.Inrealitythecoilswouldbemuchmoretightlypackedthanshowninthisschematicrepresentation.Applications:Variousfluids,widerangeofpressures.Advantages: CompactlongcylindricalshapeCanbeinstalledvertically,savingspace.SingleunitwithnoremovablepartsLightweightDisadvantages: CannotbedismantledforcleaningorrepairPressure Drops or Head Losses in Heat Exchangers In general, separate head loss calculations are necessary for thetwofluidsthatpassthroughaheatexchanger. InsomecasestheDarcy-Weisbach equation can be used directly, e.g. for flowthroughstraighttubesinashellandtubeheatexchanger.Insomeothercaseswheretheflowpassagesarenotroundlikeapipe,anequivalentdiametercanbeusedintheDarcy-Weisbachequation.If fins or roughened surfaces are used to increase heat transfer,therewillinevitablybeincreasedpressureloss.

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69

Manufacturersofheatexchangerswillnormallyprovidedetailsofthepressuredrops. Ingeneral, theamountof turbulence inheatexchangers is likely to be high and so, as can be seen from thefriction factor diagram for pipe flow, the friction factor will beinvariantwithReynoldsnumberor flow rateonce the flow is inthe fully turbulent region. Hence, from the Darcy-Weisbachequation,thepressuredropwillbeproportionaltothesquareofthevelocityorflowrate. Δ hx ∝ (32)whereΔ hx = pressuredrop Pa = volumeflowrate m s Example 13 Heat Exchanger Pressure Drop Waterflowsthroughaheatexchangerattherateof126litresperminute and the pressure drop is 28kPa. What would be thepressure drop through the heat exchanger if the flow ratewerereducedto86litresperminute?Solution LetΔ 1betheinitialpressuredropandΔ 2bethenewpressuredrop.Hence Δ 2Δ 1 = 21 Hence,

Δ 2 = Δ 1 21 = 28 86126 kPa = 13.0kPa.


70

Important Formulae

Newton’s Viscosity Equation

∝ A = ∆∆

where = distance m = force N = area m2 = shearstress Nm−2 = absoluteviscosity Pas(orNm−2sorkgm−1s−1) = fluidvelocity ms−1 = distanceindirection normaltoflow mKinematic viscosity Kinematicviscosity, ,isdefinedas=


71

where = density kg/m3 = kinematicviscosity m2s−1Newton’s Second Law Applied to S.I. Units = where = force N = mass kg = acceleration ms−2Hence, 1N = 1kgms Pressure = Awherep = pressure Nm−2orPaF = Force NA = area m2Pressure and Head = ℎwhere = density kg/m3g = accelerationduetogravity ms−2h = heightorhead m


72

Compressibility and Bulk Modulus

Compressibility, = −∆∆ = − 1 ∆∆ Bulkmodulus, = ∆−∆ = − ∆∆

where = compressibility Pa−1 = bulkmodulusofelasticity Pa = volume m3p = pressure Nm−2orPaForce Produced by a Hydraulic Cylinder Force produced by the pressure acting on the piston at the capendofasingle-actingcylinder:= cylwhere = forceproducedinthepistonrod = pressure

cyl = cylindercross-sectionalareaForce produced by the pressure acting on the piston at the rodendofasingle-actingcylinder:= cyl − whererod = rodcross-sectionalarea

Displacement of a Hydraulic Piston Pump Withonecylinder:displ = cyl stroke


73

wheredispl = pumpdisplacementpercyclecyl = cross-sectionalareaofcylinderstroke= pumpstrokeWith cylinders:

displ = cyl cyl strokewherecyl = numberofcylinders

Ideal Flow Rate of a Positive Displacement Pump ideal = displ wheredispl = pumpdisplacementperrevolution(orcycle) = speed(rotationsorcyclesperunittime)

Efficiency of a Hydraulic Pump Hydraulicpower(forincompressibleflow)= ∆ where = power Js−1orW = volumeflowrate m3/s∆ = pressuredifference Papump,mech = ideal∆actual pump,vol = actualideal

pump,overall = pump,mech × pump,vol


74

= ideal∆actual actualideal = actual∆actual Area of a Circle = π = π4 where = area m2 = radius m = diameter mMass Flow Rate = where = velocity ms−1 = massflowrate kg/s = area m2 = density kg/m Volume Flow Rate =

= where = velocity ms−1 = Volumeflowrate m3/s = area m2Reynolds Number Re = =


75

Forlaminarflowinapipe: Re<2,000Forturbulentflowinapipe:Re>4,000where = density kg/m3 = fluidvelocity ms−1 = diameter m = absoluteviscosity Pas(orNm−2s orkgm−1s−1) = kinematicviscosity m2s−1The Darcy-Weisbach Equation Theheadlossduetofrictionisgivenby

ℎf = 4 2 dependson / andRewhereℎf = headlossduetofriction m = frictionfactor = lengthofpipe m = diameter m = meanfluidvelocity ms−1 = accelerationduetogravity ms−2 = equivalentsurfaceroughness mRe = ReynoldsnumberThepressuredropduetofrictionisgivenby

Δ = ℎf = 4 2


76

whereΔ = pressuredropduetofriction PaFriction Factor for Laminar Flow = 16Re = 16 Head Loss Coefficient for Fittings and Features (forreducers,bends,etc.) ℎ = 2 whereℎ = headlossforfittingorfeature m = headlosscoefficient = meanvelocityinpipe ms−1g = accelerationduetogravity ms−2


77

Glossary Absolute Pressure Absolutepressureisthetruepressure.Thistermisusedwhenitis necessary to emphasise that the pressure being referred to isnot a relative pressure, such as gauge pressure, or a differentialpressure, such as the pressure drop across a valve. The lowestpossiblevalueofabsolutepressureiszero(inanypressureunits).Absolute Viscosity Fluidpropertywith SIunitsPas.This is also knownasdynamicviscosity.EquivalentSIunitsareNm−2sandkgm−1s−1.Atmospheric Pressure The pressure of the atmosphere, which can be measured by abarometer.1standardatmosphere=1.01325105PaBulk Modulus of Elasticity Thebulkmodulusofelasticityofafluidistheratioofachangeinpressuretothefractionalvolumechange(orvolumetricstrain)itcauses.Itistheinverseofthecompressibility.TheunitsarePa.

= ∆−∆ = − ∆∆ = bulkmodulus Pa = volume m Δ = pressurechange PaΔ = volumechange m

Compressibility The compressibility of a fluid is the ratio of a fractional volumechange(orvolumetricstrain) tothepressurechangethatcausesit.Itistheinverseofthebulkmodulusofelasticity.TheunitsarePa−1.= −∆∆ = − 1 ∆∆


78

= compressibility Pa Continuity Equation Thevolumeflowrateisthesameateverytransversecrosssectionof a pipe. If the velocity can be considered uniform over eachtransversecrosssectionthen= = = . = volumeflowrate m3/s = meanvelocityovercross section m/s = areaofcrosssection m2Darcy-Weisbach Equation This equation gives the head loss due to friction in a pipe. Thefrictionfactordependsonthenatureoftheflowandcanbefoundfromafrictionfactordiagram.Theheadlossalsodependsonthelengthofthepipe,itsdiameter,thesquareofthevelocityandtheaccelerationduetogravity.

ℎf = 4 2 ℎf = headlossduetofriction m = frictionfactor = lengthofpipe m = diameter m = meanfluidvelocity ms−1 = accelerationduetogravity ms−2Dynamic Viscosity Fluidpropertywith SI unitsPas.This is also knownas absoluteviscosity.EquivalentSIunitsareNm−2sandkgm−1s−1.Fixed Displacement Pump Afixeddisplacementpumpisapositivedisplacementpumpthatdischargesfluidatafixedflowrateandwherethefixedflowratecannotbevariedbychangingthesettingofthepump.


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Fluid A fluid is a substance that cannot sustain a shear stresswithoutundergoingrelativemovement.Liquidsandgasesarefluids.Friction Factor ThisisadimensionlessfactorintheDarcy-Weisbachequationforthe head loss associated with pipe flow. The form used in thisworkisknownastheFanningfrictionfactor.Somereferencesusethe Darcy friction factor which equals four times the FanningFriction factor—in these cases the constant ‘4’ is replaced withunityintheDarcy-Weisbachequation.Friction Factor Diagram (Moody Diagram) This isadiagramfor lookingupthe friction factor foruse in thecalculation of pressure or head losses in pipes. Lines on thediagram represent the friction factor as a function of Reynoldsnumberandrelativeroughness.Gauge Pressure Gaugepressure is a relativepressure: it is the amountbywhichthepressureexceedsthepressureoftheatmosphere.Head Theterm‘head’referstotheheightfromapositionwithinaliquidat rest to a connected free surface of the liquid. The concept ofheaddifferencecanalsobeusedtodescribeapressuredifferenceintermsofanequivalentheadofliquid.Incompressible An incompressible fluid is one whose volume does not changewithchangesinpressure.Ingeneral,gasesarenotincompressibleasamoderatechangeinpressurecancauseasignificantchangeinvolume.However,liquidshaveverylowcompressibilityand,asasimplification,canbeconsideredincompressible.Kinematic Viscosity FluidpropertywithSIunitsm2s−1.Major and Minor Head or Pressure Losses in Pipes Major lossesare thoseassociatedwith fluid frictionover the fulllengthofthepipe.Minorlossesarethoseassociatedwithfittingssuchasvalvesandconnectorsandflowchangingfeaturessuchasbends,elbows,suddenexpansionsandsuddencontractions.


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Mass Flow Rate Thisistheamountofmassthatflowsthroughaspecifiedsystem(suchasaheatexchanger)orcrosssection(ofapipeorduct,forinstance)perunittime.Moody Diagram or Chart (Friction Factor Diagram) This isadiagramfor lookingupthe friction factor foruse in thecalculation of pressure or head losses in pipes. Lines on thediagram represent the friction factor as a function of Reynoldsnumberandrelativeroughness.Newton’s Viscosity Equation = ∆∆ = shearstress Nm−2 = absoluteviscosity Pas(orNm−2sorkgm−1s−1) = fluidvelocity ms−1 = distanceindirection normaltoflow m

Normal Stress Anormalstressisastresswheretheforce involvedisnormaltotheareaoverwhichitacts.Pascal’s Law Pressurewithinafluidatrestactsequallyinalldirectionsandatrightanglestoanysurfaceonwhichitacts.Positive Displacement Pump Apositivedisplacement pump is a pump that discharges a fixedvolumeofpressurizedfluidforeachrotationoftheinputshaftorforeachcycleinthecaseofareciprocatingpump.Pressure Pressureisforceexertedperunitareainadirectionthatisnormaltoandtowardsthearea.Itisexpressedas=


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Pressure and Head Thepressureowing to theweightofacolumnof incompressiblefluid(suchasaliquid)ofheighthisgivenby = ℎHead is a way of expressing the pressure of a fluid that has aconstantdensityintermsofaverticaldistance,equivalenttohinthe equation above. The relationship between head h andpressure is ℎ = /( )or Head=Pressure/(DensityAccelerationduetogravity)In the context of this term, the pressure is normally the gaugepressure. For a liquid at rest at a point within a pressurisedsystemorpipetheheadistheheightabovethepointtowhichtheliquidwouldriseinaverticaltubeofsufficientlengthandopentothe atmosphere that might be attached to the system at thatposition.Pressure Relief Valve A pressure relief valve is a spring-loaded valve that only allowsflowwhenacertainpressurelevelisexceeded.Theopeningofthevalverelievesthepressure,preventingitfromgoingsignificantlyhigher.Relative Roughness Therelativeroughnessofapipe,k/d, is theratioof theabsoluteroughnessof itssurfacetoitsdiameter(orbore).Theconceptofabsolute roughnesswas originally based on roughness thatwascreated artificially with grains of sand: the absolute roughnesswasequivalenttotheheightofthesandgrainsusedtocreatetheroughness.Asboththenumeratorandthedenominatorhavetheunitsoflength,therelativeroughnessisdimensionless.Reynolds Number Adimensionlessnumberusedtocharacterisethetypeofflow.

Re = = = density kg/m3


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= fluidvelocity ms−1 = diameter m = absoluteviscosity Pas(orNm−2s orkgm−1s−1) = kinematicviscosity m2s−1Shear Stress Ashearstressisastresswheretheforceinvolvedisinthesameplaneastheareaoverwhichitacts.Steady Flow Flow is said to be steady if the velocities and other parameters,e.g.densityandpressure,tonotvarywithtime.Thevelocityandotherparametersmayvaryfrompointtopointwithintheflow.Stress Astressisaforceperunitarea(S.I.unitsN/m ).Throttling Throttling is said to occur wherever a flow restriction causes apressuredrop.Uniform Flow Flowissaidtobeuniformifthevelocitiesandotherparameters,e.g. density and pressure, do not vary over a region, usually atransversecross-section,ofaflowstream.Vacuum and Vacuum Pressure Vacuumpressure is theamountbywhichapressure is less thanatmosphericpressure.Inaperfectvacuumtheabsolutepressurewouldbezero.Viscosity Viscosity is ameasure of the resistance that a fluid provides toshear, i.e. the sliding of one fluid layer over another. AbsoluteviscosityhastheunitsofPasorkgm−1s−1.Kinematicviscosityhastheunitsofm2s−1.Volumetric Strain Volumetric strain is defined as a change in the volume of asubstance divided by the original volume. Volumetric strain can


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be caused in a gas or a liquid by increasing or decreasing thepressuretowhichitissubjected.

Applied Energy Systems — Hydrostatic Fluid Power ...

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