FLUID MECHANICS NOTESPeter E. ClarkDepartment of Chemical EngineeringAugust 3, 20071Contents1 Introduction 31.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.1 Density . . . . . . . . . . . . . . . . . . . . . 41.1.2 Specic Gravity. . . . . . . . . . . . . . . . . 41.1.3 Specic Weight . . . . . . . . . . . . . . . . . 41.2 Newtonian Fluids . . . . . . . . . . . . . . . . . . . . 51.3 Non-Newtonian Fluids . . . . . . . . . . . . . . . . . 61.3.1 Power Law Fluids . . . . . . . . . . . . . . . . 61.3.2 Bingham Plastic . . . . . . . . . . . . . . . . . 71.3.3 Herschel-Bulkley Fluids . . . . . . . . . . . . 81.3.4 Dilatant Fluids . . . . . . . . . . . . . . . . . 91.3.5 Time Dependent Fluids . . . . . . . . . . . . . 91.4 Kinematic Viscosity. . . . . . . . . . . . . . . . . . . 91.5 Surface Tension . . . . . . . . . . . . . . . . . . . . . 101.6 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . 102 Fluid Statics 112.1 Basic Equation of Fluid Statics . . . . . . . . . . . . . 122.2 Pressure - Depth Relationships . . . . . . . . . . . . . 122.2.1 Constant Density Fluids . . . . . . . . . . . . 122.2.2 Variable Density Fluids. . . . . . . . . . . . . 132.3 Pressure Forces . . . . . . . . . . . . . . . . . . . . . 142.4 Buoyancy . . . . . . . . . . . . . . . . . . . . . . . . 152.5 Pressure Measurement . . . . . . . . . . . . . . . . . 162.5.1 Manometers. . . . . . . . . . . . . . . . . . . 162.5.2 Bourdon Tube. . . . . . . . . . . . . . . . . . 192.5.3 Pressure Transducers . . . . . . . . . . . . . . 202.6 Accelerated Rigid-Body Motion . . . . . . . . . . . . 2023 Balance Equations 233.0.1 Equation of Continuity . . . . . . . . . . . . . 243.1 Control Volume . . . . . . . . . . . . . . . . . . . . . 253.2 Fluid Velocity in a Conned Region . . . . . . . . . . 263.2.1 Flow Regimes. . . . . . . . . . . . . . . . . . 273.2.2 Plug or Creeping Flow . . . . . . . . . . . . . 283.2.3 Laminar Flow. . . . . . . . . . . . . . . . . . 283.2.4 Turbulent Flow . . . . . . . . . . . . . . . . . 293.3 Unsteady-State Mass Balances . . . . . . . . . . . . . 293.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . 314 The First Law of Thermodynamics 334.1 Energy Transfer . . . . . . . . . . . . . . . . . . . . . 334.2 Energy Balance . . . . . . . . . . . . . . . . . . . . . 344.2.1 Sign Conventions . . . . . . . . . . . . . . . . 354.2.2 Potential Energy . . . . . . . . . . . . . . . . 354.2.3 Kinetic Energy . . . . . . . . . . . . . . . . . 364.3 Internal Energy . . . . . . . . . . . . . . . . . . . . . 364.4 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . 375 Bernoulli Equation 395.1 Applying the Bernoulli Equation . . . . . . . . . . . . 405.2 Bernoulli Equation With Friction. . . . . . . . . . . . 435.3 Gas Flows . . . . . . . . . . . . . . . . . . . . . . . . 445.4 Non-Flow Work. . . . . . . . . . . . . . . . . . . . . 465.5 Flow Measurement . . . . . . . . . . . . . . . . . . . 465.5.1 Pitot Tube. . . . . . . . . . . . . . . . . . . . 475.5.2 Static Pitot Tube . . . . . . . . . . . . . . . . 485.5.3 Venturi Meter . . . . . . . . . . . . . . . . . . 495.5.4 Orice Meter . . . . . . . . . . . . . . . . . . 505.5.5 Rotameters . . . . . . . . . . . . . . . . . . . 515.6 Unsteady Flows . . . . . . . . . . . . . . . . . . . . . 525.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . 566 Fluid Friction in Steady One-Dimensional Flow 577 Momentum Balance 593Contents7.1 Newtons Laws . . . . . . . . . . . . . . . . . . . . . 597.2 Control Volumes . . . . . . . . . . . . . . . . . . . . 597.3 Forces on a Control Volume . . . . . . . . . . . . . . . 617.4 Steady Flow. . . . . . . . . . . . . . . . . . . . . . . 627.5 Rotational Motion and Angular Momentum . . . . . . 697.5.1 Review . . . . . . . . . . . . . . . . . . . . . 6941 IntroductionOutline1. Reviewa) Densityb) Specic gravityc) Specic weight2. Newtonian Fluidsa) Stressb) Strainc) Strain rate or shear rated) Viscosity3. Non-Newtonian Fluidsa) Pseudoplasticb) Bingham Plasticc) Yield pseudoplastic or Herschel-Bulkly Fluidd) Dilatante) Time dependenti. Rheopeticii. Thixotropic4. Kinematic Viscosity5. Surface Tension51. Introduction6. Pressure7. Computer Problems1.1 ReviewThere are a few concepts that need to be reviewed to aid inunderstanding the text.1.1.1 Density =massvolumeUnits :kgm3;gcm3 :lbmf t3(1.1)Density is an extremely important property of matter. The density of amaterial can be considered continuous except at the molecular level.Density can also be thought of as the constant that relates mass tovolume. This makes it easy to convert between the two.1.1.2 Specic GravitySG =re fWhere re f = 1000kgm3; 1gcm3; 62.4lbmf t3; 8.33lbmgal(1.2)Specic gravity is used instead of density to tabulate data for differentmaterials. Using the specic gravity, the density in any set of unitsmay be found by picking the reference density in the desired units.Note: When the reference density is expressed ingcm3, the density andspecic gravity have the same numerical value.1.1.3 Specic Weight = ggcUnits :Nm3;dynecm3 ;lbff t3(1.3)The specic weight (1.3) is a quantity that is used frequently in uidmechanics. In the American Engineering Series (AES) of units, it isnumerically equal to the density. The units are lbfrather than lbm.The choice of (gamma) for the symbol for specic weight was6Newtonian Fluidssomewhat unfortunate since another important variable in uidmechanics (strain) also uses . To avoid this problem, your book uses for strain. This is non-standard and can cause confusion since israrely used for this purpose in the literature.1.2 Newtonian FluidsA uid is dened as a material that can not support a stress or as amaterial that is continuously deformed by the application of a stress.Figure 1.1: A uid element before deformation.Figure 1.2: Fluid element after the application of a force acting tan-gentially on the top of the element.shear stress( or ) =f orceareaUnits :Nm2;dynecm2 ;lbff t2;lbfin2(1.4)strain () =displacementelement height= xy(1.5)Since a uid is continuously deformed by a force acting on it, straindoesnt mean much, but the rate of strain or shear rate ( ) is important.strain rate or shear rate ( ) = ddt= dxydtUnits :1time(1.6)Newton rst proposed that the shear stress could be related to theshear rate by =Constant 71. IntroductionThe constant is termed the viscosity (). It is a constant ofproportionality between shear stress and shear rate. Viscosity isanalogous to a modulus. = = = Units :F tarea;dyne scm2;N sm2or Pa s (1.7)The unit (dyne scm2) is called a Poise. It is more common to usecentipoise (cp) or 0.01Poise. Water has a viscosity of 1 cp whilehoney has a viscosity of about 400 cp. It is easy to confuse Poise andcentipoise when making calculations. Remember, a Poise is equal to1, dyne s1cm2. A Pa s is equal to 0.1 Poise. Viscosity can also beexpressed inlbf sf t2. The viscosity of water in the AES system of unitsis 2.1x105lbf sf t2. Converting from Poise tolbf sf t2is accomplished bymultiplying Poise by 1lbf sf t2/478.8 Poise. The equivalents forviscosity are1 lbf sf t2= 47.88 N sm2= 478.8 Poise = 47880 centipoise1.3 Non-Newtonian FluidsFluids that exhibit a nonlinear relationship between stress and strainrate are termed non-Newtonian uids. Many common uids that wesee everyday are non-Newtonian. Paint, peanut butter, and toothpasteare good examples. High viscosity does not always implynon-Newton behavior. Honey is viscous and Newtonian while a5W30 motor oil is not very viscous, but it is non-Newtonian. Thereare several types of non-Newtonian uids. Figure 1.3 shows severalof the more common types.1.3.1 Power Law FluidsPower law uids are uids that follow the power law (Equation 1.8)over part or all of the shear rate range. These uids are also known aspseudoplastic uids. Where m is the consistency index (K is also usedin the literature) with units ofF snareaand n is the power law or ow8Non-Newtonian FluidsFigure 1.3: Types of non-Newtonian uids.behavior index. For power law uids, n ranges from 0 to 1. While it isgreater than one for dilatant uids. The value of m or K depends uponthe system of units. Viscosity of a power law uid is obviously afunction of shear rate and not constant as it is for Newtonian uids.There are two different ways to describe the viscosity: the slope of thetangent line at any point on the curve and the slope of the line drawnfrom the origin to the shear rate of interest (Figure 1.4). The latter isthe preferred method and is termed the apparent viscosity and giventhe symbol (a). This is short for the apparent Newtonian viscositybecause it is the viscosity that a Newton uid would have if the line isbased on a single point measurement. Because the viscosity is afunction of shear rate it is necessary to specify the shear rate at whichthe viscosity is reported (a( )). = m n(1.8)a( ) = = m n =m 1n(1.9)1.3.2 Bingham PlasticBingham plastic uids are uids that exhibit a yield stress (Figure 1.3.This means that the uid will support a stress up to a point before ow91. IntroductionFigure 1.4: Apparent Newtonian viscosity.begins. Good paints are Bingham plastics. The ow behavior ofBingham plastic uids is described by = 0 +p (1.10)Where 0 is the yield stress (the stress that must be exceeded beforeow begins) and p is the plastic viscosity.1.3.3 Herschel-Bulkley FluidsThe Herschel-Bulkley formulation is a generalization of the Binghamplastic equation. In the Bingham plastic model, the viscosity isconstant after the yield stress is exceeded while the Herschel-Bulkleymodel allows for power law behavior. = 0 +m n(1.11)Looking at the various parameters in the Herschel-Bulkley equation,we can make the following observations for = 0 +m nwhen0 = 0 = = m nPowerLaw0 = 0 and n = 1 = = Newtoniann = 1 = = 0 +p BinghamPlastic10Kinematic ViscosityThe Herschel-Bulkley uid model can be reduced to the three othermodels and is therefore the most general of the simple uid models.1.3.4 Dilatant FluidsDilatant uids are shear thickening uids. This means that theviscosity increases with shear. There are few examples of dilatantuids. Most dilatant uids are concentrated slurries. The power lawcan be used to describe dilatant uid behavior (n>1).1.3.5 Time Dependent FluidsTime dependent uids are uids that increase or decrease in viscosityover time at a constant shear rate. The construction industry uses acement slurry that thins with time at a constant pump rate because it iseasier to pump and lls forms easily. There are two types of timedependent uids: Rheopectic and Thixotropic.Rheopectic FluidsRheopectic uids increase in viscosity with time at constant shearrate. There are few examples of rheopectic behavior. In the Britishliterature, rheopectic behavior is called anti-thixotropic behavior.Thixotropic FluidsThese are uids that lose viscosity over time at a constant shear rate.Dispersing agents for cements tend to make them thixotropic.Viscosity is recovered after the cessation of shear. In some systems,the time it takes to recover is so short that a sheared sample can not bepoured out of a container before it gets too thick to ow.1.4 Kinematic ViscosityThe kinematic viscosity, (), is derived from gravity driven owmeasurements. Usually a glass capillary viscomer is used. Kinematicviscosity can be derived from the shear viscosity () by dividing by111. Introductionthe uid density. = Units areareatime( f t2s,cm2s, etc).1.5 Surface TensionSurface tension, (), is a property of a liquid surface. It describes thestrength of the surface interactions. The units on surface tension areForcelength. Surface tension is the driving force for water beading on awaxy surface and free droplets of liquid assuming a spherical shape.Lowering of surface tension can be accomplished by adding speciesthat tend collect at the surface. This breaks up the interactionsbetween the molecules of the liquid and reduces the strength of thesurface. Pure water has a surface tension that approaches 72dynecm.Surface tension can be reduced by adding surface active agents(surfactants) to the liquid.1.6 PressurePressure is dened as force divided by the area that the force acts overand therefore has units ofFA. It can be a result of an applied force (forexample pumping) or hydrostatic (weight of a column of uid). Thetotal pressure is the sum of the applied and hydrostatic pressure.P = Papplied +Phydrostatic122 Fluid StaticsOutline1. Basic Equation of Fluid Statics2. Pressure - Depth Relationshipsa) Constant density uidsb) Ideal gases3. Pressure4. Pressure Vessels and Piping5. Buoyancy6. Pressure Measurementa) Manometersb) Pressure gagesc) Pressure transducers7. Static Pressure Head8. Acceleration132. Fluid Statics2.1 Basic Equation of Fluid StaticsThe pressure at a point in a static uid is the same in all directions.What this means that the pressure on the small cube in Figure 2.1 isthe same on each face. Since depth increases in the downwardz-direction, the zero point will be at the top of the uid.Figure 2.1: Hydrostatic pressureA simple force balance and some algebra yieldsdPdz= g = (2.1)For angles, (), that deviate from vertical the equation becomesdPdz= g cos = cos (2.2)2.2 Pressure - Depth Relationships2.2.1 Constant Density FluidsTo establish the relationship between pressure and depth for aconstant density uid, the pressure - position relationship must beseparated and integrateddPdz= g =_dP =_ gdz14Pressure - Depth RelationshipsP = gz = z (2.3)Equation 2.3 is one of the most important and useful equations in uidstatics.Example 2.1. A 5000f t well is lled with a drilling mud thathas a specic weight of 11.2 lbff t3. What is the pressure in psig atthe bottom of the well?h =z = 0 f t 5000 f t =5000 f tP =hP =h12inf t|11.2lbfgal|7.48galt3|f t2144in2P = (5000 f t)|12inf t|11.2lbfgal|7.48galt3|f t2144in2 = 2909 psig2.2.2 Variable Density FluidsMost uids are relatively incompressible, but gases are not. Thismeans that the density increases with depth so we cannot use thespecic weight as a constant in determining the pressure. If weassume an ideal gas, then the density is given by = PMRTwhere M is the molecular weight, T is the absolute temperature, and Ris the gas constant in appropriate units.Replacing in the differental equation we getdPdz= g =dPdz=PMRTg (2.4)152. Fluid StaticsdpP=gMRTdz =_P2P11P dP =gMRTdz (2.5)Which yieldsln(P2P1) =gMRT(z2z1) = P2 = P1egMRT z(2.6)Because the it is possible for the temperature to be non-uniform andgases are poor conductors of heat, the correct equation to use isP2 = P1_1k 1k gMzRT1_kk1(2.7)T2 = T1_1k 1k gMzRT_(2.8)Where k is the ratio of the heat capacities.2.3 Pressure ForcesPressure acting on a surface exerts a force. This force can becalculated fromdF = PdA =_dF =_PdA = F =_PdAFor constant pressureF = PA (2.9)Example 2.2. Using the result of the previous example(P = 2900 psi), what is the force acting on the bottom of thehole if the hole is eight inches in diameter?F = PAF = 2909 lbfin24(8in)2= 877334lbf16BuoyancyExample 2.3. You are the proud possessor of a 200 f t tall watertower. At the base of the tower (1 f oot above the ground) is aseven foot tall by two foot wide access panel that is held in placeby bolts. What is the force on the plate?F =(7 f t)(2 f t)_192 f t199 f twdhF = 170789lbfSince the panel is rectangular, the average pressure can be used.P =192 f t +199 f t2w = 12199 lbff t2F = 12199lbff t2|7 f t|2 f t= 170789lbf2.4 BuoyancyRemembering that in a liquid at rest the pressure at a point is the samein all directions, a body submerged in the liquid will experience aforce on the top and bottom of the body. Because the body is nite inlength, the force on the bottom will be greater than the force on thetop. The easy way to think about the magnitude of the buoyant forceis that it is equal to the weight of the liquid displaced. It is probablybetter to consider it as a force balance. To see how this works we willuse a length of pipe suspended in a well.172. Fluid StaticsExample 2.4.A 5000 f twell is lled with a drilling mud that has a specicweight of 11.2 lbfgal. A 1000 f t piece of pipe (od =8in, id =7in)is submerged at the surface with only the top of the pipe exposed.What is the buoyant force on the pipe?h =z = 0 f t 1000 f tPb =1000 f t|12inf t|11.2lbfgal|7.48galf t3|f t3(12in)3 = 582lbfin2FB = Pb4_(8in)2(7in)2_= 5940lbfWhat happens if the bottom of the pipe is at 3500 f t.Pb =3500 f t|12inf t|11.2lbfgal|7.48galf t3|f t3(12in)3 = 2036lbfin2Pt =2500 f t|12inf t|11.2lbfgal|7.48galf t3|f t3(12in)3 = 1454lbfin2FB = (PbPt)4_(8in)2(7in)2_= 5940lbf2.5 Pressure Measurement2.5.1 ManometersThere are numerous ways that pressure can be measured. Twoexamples are shown in Figures 2.2 and 2.3.Most books, yours included, present a manometer equation, whichfor simple manometers is easy to use. For a manometer having morethan one uid and a complex topography, the following three rules are18Pressure MeasurementFigure 2.2: Open-end manometer.Figure 2.3: Differential manometer.far easier to use.Starting at one end of the manometer1. Work through the system considering only one uid at a time2. Add pressure differences as you proceed down through a uidfrom the starting point (or subtract when working upwards).3. Move horizontally through a uid without change in thepressureSet the resulting equation equal to the pressure at the other end of themanometer.192. Fluid StaticsApplying these rules is simple. In Figure 2.4, the tank and tube form asimple manometer. To nd the pressure at point A, we start by writingthe pressure at point A as PA. Appling rule 3, we move from point Ato point B. Since both points are at the same level, the pressures areequal.Figure 2.4: A tank with a standpipe (simple manometer).Rule 2 says to subtract going up so h is subtracted from PA and sincethere is no ow PAh must be equal to the pressure at the top of thetube (Patm). This produces the equationPAh = PatmFor more complicated manometers containing more uids and morelegs, just add terms to the equation to describe each one.20Pressure MeasurementExample 2.5. Looking at the manometer in Figure 2.3 we can,by applying the rules, write an equation for the manometer.Begin by designating the pressure at point A as PA.Applying rule 3, move horizontally to point F where the pressureis equal to PA.Applying rule 2, add the contribution to the pressure from thecolumn of uid (F E), (b+h)w.PA +(b+h)wApplying rule 3, move horizontally from E to DApplying rule 2, subtract the contribution from the mercury col-umn, hHgPA +(b+h)whHgApplying rule 2, subtract the contribution fromthe water columnC B, (b+a)wPA +(b+h)whHg(b+a)wSince there is no ow, set the equation = to the pressure at B,PB.PA +(b+h)whHg(b+a)w = PBThis becomes the manometer equation. Some algebra allows usto determine the pressure difference between points A and B, P2.5.2 Bourdon TubeBourdon tube pressure gages (Figure 2.5) consist of a closed-endmetallic tube that is curved in to almost a circle. Application ofpressure causes the tube to straighten a small amount and the motionis converted into a rotation of the needle on a dial. These gages aresimple and rugged, so they are used in a wide range of applications ina plant environment.212. Fluid StaticsFigure 2.5: Bourdon tube pressure gage2.5.3 Pressure TransducersA transducer is a device that turns a mechanical signal into anelectrical signal or an electrical signal into a mechanical response.There are a number of ways to accomplish this.1. Strain gage2. Capacatance3. Variable reluctance4. Optical2.6 Accelerated Rigid-Body MotionChanges in motion bring about acceleration or deceleration. Agasoline truck accelerating or decelerating will cause the uid in thetanks to move to one end of the tank or the other. The major change inthe analysis of pressure is that the force balance becomes equal to themass times accleration rather than zero.dPdz=_g+ d2zdt2_(2.10)This equation can be integrated for constant density uids to yieldP2P1 =_g+ d2zdt2_(z2z1) (2.11)22Accelerated Rigid-Body MotionFor gauge pressureP = h_g+ d2zdt2_=h +_ d2zdt2_(2.12)Equation (2.12) has the static pressure term (h) and anacceleration term,_d2zdt2_. Evaluating the static pressure term isrelatively easy, the acceleration term can be a little more difcult.232. Fluid Statics243 Balance EquationsAlmost all of engineering comes down to balances mass,momentum, and/or energy. We have already seen in developing themanometer equation that it is nothing more than a pressure balance.In the beginning chemical engineering course, the general balanceequation was presented. Evaluating the various components of theequation is the challenge in various aspects of chemical engineering.FlowinFlowout +GenerationConsumption =Rateof AccumulationSimple mass balances without chemical reactions leave out thegeneration and consumption terms. min mout = Rateof AccumulationSo, if the rate of accumulation is zero (steady-state), we have min = moutBased on balance arguments, the following important relationshipscan be derived.Q = vA (3.1)In Equation 3.1, v is the velocity and A is the cross-sectional area. Forconstant density uidsQin = QoutsovinAin = vout Aout(3.2)253. Balance Equations3.0.1 Equation of ContinuityBecause mass must be conserved m1 = m2and m = vA (3.3)the continuity equation becomesinvinAin = out vout Aout(3.4)or vA = 0 (3.5)The equation of continuity can also be written in terms of the specicweight.Example3.1. Water ows intoacylindrical tankthroughpipe1at therateof20ft/sandleavesthroughpipes2and3at therateof 8ft/s and10ft/s, respectively. At 4wehaveanopenairvent. Thefollowingaretheinsidediame-ters of the pipes: D1 =3in, D2 =2in, D3 =2.5in, and D4 =2inFigure 3.1: Example: Flow into and out of a tank.26Control VolumeSolution1. Calculate the mass ow rates in using m = vA m1 m2 m3 = mtank v1A1 v2A2 v3A3 = vtankAtankIt is clear that the value of is the same on both sides of the equationso it can be eliminated. Since area =4d2and the area term is on bothsides of the equation, the4term cancels.20 f ts_3in|f t12in_28 f ts_2in|f t12in_210 f ts_2.5in|f t12in_2=vtank (12 f t)2Solving for vtank yieldsvtank = 0.148 f tsLater it will be necessary to think of the vtank term as the differential_dhtankdt_ . Using Equation 3.2 and assuming that the air in the tank isincompressiblevtankAtank = v4A40.148 f ts|4|_2 f t_2= v4|4|_2in|f t12in_2Solving for v4 givesv4 = 21.38 f ts3.1 Control VolumeThe control volume is an imaginary sub-volume in a ow systemthrough which ows occur. It can be denoted with a dotted or dashedline enclosing the volume (Figure 3.2).This may seem like an unnecessary complication, but with anothersimple addition the control volume concept becomes quite useful. If273. Balance EquationsFigure 3.2: Control volumethe area through which the ow occurs is declared to be a vectorquantity directed outward from the control volume_

A = An_, it canbe combined with the velocity vector using the dot product to producethe proper sign for the ow when Equation 3.5 is used_ v

A = 0_.The dot product produces v

A =vA cos(), where is the anglebetween the area and velocity vector. Since the cosine of zero is oneand the cosine of 180 is (1), the direction of ow is set relative tothe area vector. It is important to draw the control volume such thatthe velocity vector is perpendicular to the control volume surface(Figure 3.3).Figure 3.3: Control volume with area and velocity vectorsThe previous example would become v1A1cos180+ v2A2cos0+ v3v3cos0+ v4A4cos = 0Where the sign of the nal answer would determine the direction ofow. Plugging in numbers we nd that A4cos is positive so the owis outward.3.2 Fluid Velocity in a Conned RegionThere are two ways to consider the velocity of a uid in a owchannel. The average velocity is probably the most straightforward. Itis simply the volumetric ow rate divided by the area of the conduit.vave = QA(3.6)28Fluid Velocity in a Conned RegionIn reality, unless the uid is owing very fast or very slow, there is asignicant velocity distribution that for ow of a Newtonian uid in apipe is parabolic (Figure 3.4). One of the basic assumptions of uidmechanics is that the uid velocity at the wall is zero i.e. no slip.With the velocity equal to zero at the wall, the uid velocity is at amaximum at or near the centerline of the ow (depends upon thegeometry of the ow channel).Figure 3.4: Velocity distribution for a Newtonian uid in laminar ow.3.2.1 Flow RegimesDepending upon the uid velocity, the uid can ow in different owregimes. At low ow rates, the uid particles move in a uniformfashion along the channel with only a small difference in the velocitybetween the wall and the centerline. As the velocity increases, theow remains uniform, but the difference between the velocity at thecenterline and the wall becomes large. This ow regime is termed thelaminar region or simply laminar ow. At some point the velocityincreases enough that the ow becomes chaotic and the uniformity inthe ow disappears (termed turbulent ow). The velocity prolebecomes at and the ow is strongly mixed. Flow regimes can beseparated by the value of the Reynolds number. This is adimensionless group that represents the ratio of the inertial forces tothe viscous forces and is given by293. Balance EquationsNRe = d vwhere d is the characteristic diameter of the ow channel, is theuid density, v is the velocity, and is the uid viscosity.3.2.2 Plug or Creeping FlowAt extremely low ow rates, the uid is normally in plug or creepingow. This ow regime has a at velocity prole with much smallerdifference between the velocity of the uid owing near the wall andat the centerline (Figure 3.5). Experimentally, it is a difcult owregime to maintain. This ow regime is mathematically simpler thanthe other two ow regimes so it has been the subject of a great deal ofstudy. Reynolds numbers are typically less than one for plug orcreeping ow.Figure 3.5: Velocity distribution for a Newtonian uid in plug ow andlaminar ow.3.2.3 Laminar FlowAs velocity increases above the plug ow limit, a velocity prolesimilar to that labeled as laminar in Figure 3.5. The Reynolds numberrange for laminar ow is one to 2,100 (it may be somewhat higher in30Unsteady-State Mass Balancesnon-Newtonian uids). In steady (time derivative equal to zero),uniform ow a uid particle will ow along the same streamline. Thismeans that there is no mixing of uid across streamlines in laminarow. In extreme cases where vibrations have been excluded, thelaminar ow regime can extend to a Reynolds number of 60,000 or so.3.2.4 Turbulent FlowTurbulent ow proles exhibit a at velocity prole with a thinlaminar layer near the wall. The motion of uid particles in turbulentow is chaotic and the ow is strongly mixing. If you have everwatched smoke rise from a cigarette, you should have noticed that itrises in a narrow channel (laminar ow) and then breaks into a plumewhen the velocity becomes high enough to reach the turbulent owregime. Almost all ow in water pipes, air ducts, and pipelines isturbulent.3.3 Unsteady-State Mass BalancesThere are a number of real world problems that are not steady-stateproblems. What this means is that the rate of accumulation term in theequation below is not zero but it is equal todmdt . min mout = Rateof AccumulationSo the equation becomes min mout = dmdtExample 3.7 in de Nevers needs some explanation. In this example, atank that is initially full of air is being pumped down with a vacuumpump. At a constant volumetric ow rate (Q =1 f t3min ), how long does ittake to reduce the pressure to 0.0001 atm? As with any balanceproblem, the best place to start is by writing the balance equation min mout = dmdt313. Balance EquationsIn this system, there is no ow in so the min term is zero and theequation reduces todmdt= moutAt this point, use the denition of density and rearrange the equationto nd the mass of gas in the system using the density and systemvolumemsystem = systemVsystemTaking the time derivative of both sides of the equation yields_dmdt_system= Vsystemdsystemdt+systemdVsystemdtbut the system volume is constant so the last term is equal to zero andthe equation becomes_dmdt_system= VsystemdsystemdtAssuming that the gas is not compressible, m = V , however, we areinterested in the mass ow rate m so Q is used instead of V , yielding mout = QoutoutSince Q is constant out = system which leads to mout = QoutsystemWe can now write the differential equation for the systemVsystemdsystemdt=QoutsystemSeparating the variables yieldsdsystemsystem=QoutVsystemdtIntegrating the equation yields32Summaryln_systemfsystemi_=QoutVsystemtRemembering thatgas = PMRTthis equation can be solved for t.3.4 SummaryMass balances are extremely important in uid ow. Missing massusually means that there is a leak. Using the equation of continuityseems difcult at rst and dealing with the dot product of velocity andarea_v

A_ appears to add an unnecessary complication, learninghow to set up problems using this formulation will help to minimizemistakes in the long run. In Chemical Engineering, unsteady statemass balances are important. It is worth spending some time to try tounderstand them.334 The First Law of ThermodynamicsIn the last chapter, we covered mass balances. As a prelude to theBernoulli equation in the next chapter, we need to discuss energybalances. The energy balance, like the mass balance, does not havecreation or destruction terms and so the general balance equationreduces tof lowin f lowout = AccumulationThere are several forms of energy that we must deal with in uidmechanics.1. Kinetic energy energy due to motion2. Potential energy energy due to position3. Internal energy energy of the atoms and molecules that makeup a system4. Surface energy energy due to surface interactionsThe other forms of energy are generally not useful in uid mechanics,but may nd specialized uses in areas such as magentohydrodynamics.4.1 Energy TransferEnergy may be transfered through heat ow or work. Heat ows fromhot bodies to colder bodies. This form of transfer forms the basis forthermodynamics. Work is dened asWork =_f orce d(distance) =_F dxWork can also be done by rotating shafts.354. The First Law of Thermodynamics4.2 Energy BalanceReturning to the general balance equation and remembering thatenergy cannot be created or destroyed we getf lowin f lowout = AccumulationYour book uses lower case symbols for specic values as opposed tothe carrot or hat above the symbol. For instanceu UIt also used upper case symbols for the non specic values as shownbelowU = muEnergy may enter a system in several ways. It may be carried in acrossthe ow boundary by mater entering or leaving the system. Heattransfer to or from the system is also a conduit for energy transfer.Work may done on or by the system can also serve as a transfermechanism. Figure 4.1 shows an example of each of the mechanisms.Figure 4.1: Energy input from ow, shaft work, and heat transferSo the overall energy balance for the system becomesd[m(u+pe+ke)]sys = [(u+pe+ke)indmin+dQin+dWin[(u+pe+ke)outdmout +dQout +dWout]36Energy BalanceThis equation can be simplied by deningdQ = dQindQoutanddW = dWindWoutyieldingd[m(u+pe+ke)]sys = [(u+pe+ke)indmin[(u+pe+ke)outdmout +dQ+dW(4.1)4.2.1 Sign ConventionsSign conventions have always been a bit confusing. It seems difcultto imagine that one can just arbitrarily pick a sign for ow into or outof the system and get the correct answer. But, it works as long as youpick a sign and stick with it. The sign convention in your book is Work done on the system (owing in) is positive Heat transfered to the system (owing in) is positiveThis results indQ+dW = 04.2.2 Potential EnergyThe specic form of the potential energy equation simplies tope = gzIf one looks at a owing system, the specic form of the potentialenergy equation becomespe = peinpeout = g(zinzout)To get the total potential energy change we need to multiply thespecic quantity by the mass of the system.PE = mpe374. The First Law of Thermodynamics4.2.3 Kinetic EnergyThe kinetic energy term reduces toke = v22orKE = mke = mv22For systems with velocity changeKE = m_v2in2 v2out2_4.3 Internal EnergyFor a closed system of constant mass with no kinetic or potentialenergy, the internal energy can be written asmdu = dQ+dWIntegration yieldsU = mu = Q+W +constantThe constant is based on some reference point. Typically, it is water atthe triple point where u is set equal to zero.4.4 WorkWork on a compressible system is given bydW = Fdx = PAdx =PdVSince the volume decreases, the sign is negative. Injection work is thework required to inject a mass across system boundaries. Rather than38Summarydeal with the injection work and the internal energy, the enthalpy(h = u+PV ) is used. So that the energy balance equation becomesd_m_h+gz + v22__sys=_h+gz + v22_indmin_h+gz + v22_outdmout+ dQ + dW(4.2)4.5 SummaryThe rst law can be used to analyze energy ow into and out of asystem. Equation 4.2 provides the basis for the Bernoulli equationdiscussed in the next chapter.395 Bernoulli EquationOne of the most useful equations in uid mechanics is the Bernoulliequation (BE). Your book presents it in the differential form_P +gz + v22_= 0where P is PoutPin. Looking at the units we ndFl2|l3mass +lt2|l+ l2t2|2In order to straighten this mess out, it is necessary to divide bygc orto express force in terms of mass acceleration. Either way yieldsunits ofl2t2.Fl2|l3mass|masslt2F+lt2|l+ l2t2|2There is a better way to think about the Bernoulli equation. Normally,the BE is written between two points in the ow.P1+gz1 + v212= P2+gz2 + v222(5.1)Neglecting friction and work and dividing both sides by g theBernoulli equation may also be written asP1+z1 + v212g = P2+z2 + v222g(5.2)This form has the advantage of having units of length or head.Otherwise, in Equation 5.1 has to be divided by gc to make the unitswork out.415. Bernoulli Equation5.1 Applying the Bernoulli EquationIn applying the BE, it is best to pick the two points (1 and 2) to zeroout as many terms as possible. For instance picking the two pointsalong the same horizontal line eliminates z1 and z2. As an example,consider a tank full of water that is draining from an opening near thebottom of the tank (Figure 5.1)By selecting points 1 and 2 (point 2 is just at the exit) as shown inFigure 5.1: Bernoulli equation tank draining exampleFigure 5.1, z1 and z2 are equal and cancel each other out. V1 is smallrelative to V2 and can be ignored and P2 is equal to Patm so for gaugecalculations it is equal to zero. We are left with P1 and v2 to evaluate.The position of the points can be selected in a slightly differentmanner to make the problem even easier as shown in Figure 5.2.In this case, P1, z1, V1, and P2 are all equal to zero. This leaves z2 andFigure 5.2: Bernoulli equation tank draining examplev2 to be evaluated. The quantity, z2, is just the distance from thesurface of the liquid to the midpoint of the ow and is equal to (h).Evaluating the velocity, v242Applying the Bernoulli Equation0+0+0 = 0+(h) + v222gSolving for v2v2 =_2gh (5.3)Which is known as Torricellis equation.Example 5.1. Example: In the venturi meter shown below, theD=40cmand d =10cm. What is the water discharge (owrate)in the system, if the pressure at A is 100kPa, Patm =101.325kPa,the water temperature is 10C, and cavitation has just started atpoint 2?Figure 5.3: Venturi tube.Solution: The pressure at point 1 is given byP1 = 100, 000Pa+101, 325Pa The pressure at point 2 is set by the vapor pressure of water at10C. P2 = 1230Pa Since the points are along the same horizontal line, z1 = z2.Writing the Bernoulli equation between points 1 and 2P1+z1 + v212g = P2+z2 + v222gcanceling the elevation terms and rearranging the equation we get435. Bernoulli Equation_P1 P2_=v222g v212gNow we must express v2 in terms of v1.v1A1 = v2A2 =v2 = v1A1A2and solve the BE equation for V1_P1 P2_=_v1A1A2_22g v212gv212g =_P1 P2__A1A2_2V1 =_2g_P1 P2__A1A2_2substituting in the numeric values yieldsv1 = 1.25Since we know Q = vAQ = 0.157m3sIt should be noted that cavitation is just boiling. When the pressureequals the vapor pressure of the liquid at the specied temperature,boiling occurs. So when a problem species cavitation, the pressureat the point of cavitation is xed by the vapor pressure of the liquid atthe specied temperature.44Bernoulli Equation With Friction5.2 Bernoulli Equation With FrictionUp to now we have neglected frictional forces. These forces increasepressure drop and in some cases cause heating. The Bernoulliequation can be modied to include friction by adding a friction term(F) to the right side of the equation. Note: In deNevers the frictionterm has a negative sign because he has rearranged the equation toyield the values to be nal state minus initial state i.e. P2P1 sodont be confused. The Bernoulli equation becomesP1+z1 + v212g = P2+z2 + v222g +Fg(5.4)The friction term Ftakes the form of a constant times_v22g_.Frictional forces arise from simple ow, ow through valves, elbows,orices, etc. Values for the constant can be obtained from tables orfrom the friction factor charts.Example 5.2. Fluid owing through the pipe and elbow shownbelow exhibits frictional losses both in the pipe and the elbow.The pipe friction will be treated in Chapter 6 and will be ne-glected in this example.Figure 5.4: Flow through an elbow.The Bernoulli equation for this ow system becomesP1+z1 + v212g = P2+z2 + v222g +Fg455. Bernoulli EquationIf the elbow is in plane perpendicular to the page, z1 = z2. Since thereis no change in velocity, v1 = v2. So the Bernoulli equation reduces toP1= P2+Fgwhich simplies toP1P2g =FbMeasuring P then allows us to calculate the friction associated withthe elbow. If the ow rate is known then the constant, Kb in theequation below can be evaluated.F = Kv225.3 Gas FlowsGasses at low velocities can be treated as incompressible uids. Thepressure drop must also be low or the change in volumetric ow rateand thus, the uid velocity will cause problems. This means that theBE can be used in many situations that are of interest to ChemicalEngineers.When gases ow through a venturi meter, the velocity increases as thepassage gets smaller and decreases as passage returns to the originalsize. Figure 5.5 is an idealized venturi tube with sharp changes in theshape of the tube. Normally, the changes are more gradual, withoutabrupt changes (like those shown in the gure). This means that thepressure will decrease as the velocity increases on the left side of therestriction and recover on the right side of the restriction. This can beseen in Figure 5.6. Pressure recovery on the downstream side of arestriction is a common occurrence. To understand this, we need tolook at a cross-section of the ow in the venturi tube at several pointsAt the rst increment, x, the pressure, P1, must be greater than P246Gas FlowsFigure 5.5: Idealized venturi tube.Figure 5.6: Pressure prole in a venturi tube.Figure 5.7: Flow in a venturi tube.475. Bernoulli Equationfor ow to occur. The increment near 2 in Figure 5.7 is characterizedby an increase in velocity that necessitates an increase in the pressuredifferential across the increment. In the third increment, the velocityis decreasing and P3 must be greater than P2 to provide the forcenecessary to decelerate the ow.5.4 Non-Flow WorkIn the previous chapter, it was shown that the ow work could beburied in the enthalpy. This left all of the non-ow work to beaccounted for. Non-ow work is often called shaft work, since in uidow it is usually the result of a shaft turning. The Bernoulli equationcan be extended to include the work term by adding it to the left sideof the equation. Again be careful when comparing Equation 5.5 withthe equation in your book.P1+z1 + v212g + dWn fg= P2+z2 + v222g +Fg(5.5)The work term consists of all of the non-ow work. It includes workdone on the system by stirrers, mixers, pumps, etc. or work done bythe system turning turbines, shafts, or other rotating machinery. Wewill return to non-ow work in Chapter 6.5.5 Flow MeasurementThe Bernoulli equation can be exploited to measure uid ow rates.Before we explore the different ways that the BE equation can beused, we need to talk about the nature of ow around bodies. For axed body in a ow, the uid will approach the body and ow aroundit as shown in Figure 5.8. The dark point in the center of the bodyrepresents the stagnation point. At this point, the ow has a velocityof zero i.e. the ow is stagnant. Since the uid velocity at thestagnation point is zero, it gives us a point in the system to anchor theBE equation.48Flow MeasurementFigure 5.8: Stagnation point.5.5.1 Pitot TubeA pitot tube is a simple device used to measure ow rates. It worksfor both liquid and gas ows. The device, in its simplest form,consists of a small diameter hollow tube bent into the shape of a L.Usually the upstream opening is smaller than the diameter of the tube.The height, above the center-line, of the uid in the in the vertical legof the tube is related to the velocity of the uid in the ow. Anexample is shown in Figure 5.9.Figure 5.9: Pitot tube495. Bernoulli EquationWriting the Bernoulli equation between points 1 and 2P1+z1 + v212g = P2+z2 + v222gand zeroing out z1, z2, and v2 we getP1+ v22g = P2Assigning values to the various parametersP1 = Patm +bP2 = Patm +(b+h) Substituting into the Bernoulli equation, neglecting friction, andsolving for v1 we obtainv1 =_2gh (5.6)5.5.2 Static Pitot TubeSome books refer to the rst example of a pitot tube as a stagnationtube and the static pitot tube (pitot static tube) as a pitot tube. We willuse deNevers nomenclature. The difference between the pitot tubeand the static pitot tube is thesmall opening on the side of the submerged part of the tube. Unlikethe stagnation tube (Figure 5.9), a direct measurement of P ispossible. Using the Bernoulli equation and neglecting friction,P1+z1 + v212g = P2+z2 + v222g = P1+ v212g = P2solving for v1v1 =2gP=2P(5.7)50Flow MeasurementFigure 5.10: Pitot static tube.Figure 5.11: Venturi meter5.5.3 Venturi MeterWe have previously discussed the venturi tube, but not as a method ofow measurement. It has the advantage of no moving parts and if it isdesigned properly friction can be ignored.The Bernoulli equation can be written between points 1 and 2. Forhorizontally mounted tubes, the elevation terms can be zeroed out.P1+z1 + v212g = P2+z2 + v222gIf the diameter of the tube at 1 is d1 and the diameter at 2 is d2, thevelocity at point 1 can be expressed in terms of the velocity at point 2by solving v1A1 = v2A2 for v1. Furthermore, the area ratio can beexpressed as the ratio of the squares of the diameters. This leads tothe expression of the Bernoulli equation515. Bernoulli EquationP1+_v2d22d21_22g= P2+ v222gThis equation can be solved for v2v2 =_2 (P1P2)_1d22d21_ (5.8)Application of the Bernoulli equation to the venturi tube in thismanner ignores friction and other important factors. This results in asmall error in the measured ow rates. To correct for this error, themeters are calibrated and a calibration constant provided. This resultsin the corrected venturi tube equation 5.9.v2 =Cv_2 (P1P2)_1d22d21_ (5.9)5.5.4 Orice MeterOrice meters are, in some respect, an extreme example of a venturimeter. The orice meter has an abrupt change in diameter rather thanthe more gradual change in the venturi meter. An orice meter has aplate, usually removable, with a small hole machined in the center.The upstream side of the hole is at, while the downstream side iscone-shaped. One of the common mistakes in installing an oricemeter is reversing the plate. This results in erroneous measurements.Gas measurements made with orice meters are analyzed using acomplex formula (Figure 5.12) that goes far beyond the Bernoulliequation. The full orice meter equation even includes the localgravitational acceleration!Needless to say the full equation is seldom used, but it is easy to seewhy a great deal of accuracy is needed. Consider the case of a naturalgas well owing into a pipeline. If the well is owing 5MMc f d (5million cubic feet/day) and there is a 0.1 percent error in the meterreading, the gain or loss in gas sales is about thirty dollars per day or52Flow MeasurementFigure 5.12: Full orice meter equationeleven thousand dollars per year at the current price of natural gas. Fora eld that produces a billion cubic feet per day (there are elds thatproduce this much) the difference is in excess of two million dollars!5.5.5 RotametersA rotameter is a simple device that is used when a high degree ofaccuracy is not needed. The gas or liquid ow is used to levitate a ballof known diameter and density. Usually the rotameter is calibrated sothat a measured position of the ball corresponds to a ow rate. Caremust be exercised when using rotameters. They are most accuratewhen the ball is located in about the middle two-thirds of the meter.535. Bernoulli EquationMeasurement error increases at the ends.5.6 Unsteady FlowsThere are a number of situations where unsteady ows areencountered. The simplest is the draining of a tank where the owrate varies with the height of the uid in the tank. The Torricelliequation (5.3) really only gives you a snapshot of the ow rate intime. If the tank diameter is large the ow rate changes slowly, but ifit is small the ow rate change can be signicant.For a simple tank (Figure 5.13) Since Torrecellis equation gives theFigure 5.13: Unsteady ow example.instantaneous ow rate we can write the velocity at 2 asv2 =_2ghand if we solve v1A1 = v2A2 for v2, we now havev1A1A2=_2ghv1 is only the rate at which the height is dropping _dhdt_. Rearrangeand solve for _dhdt_ yields the differential equationdhdt= A2A1_2g, h (5.10)54Unsteady FlowsThis equation can be solved by separating the variables and integratingor by using the ordinary differential equation solver in MatLab._h2h11h dh = A2A1_2g_t2t1dt (5.11)When the tank has parallel sides this problem is relatively easy,however, if the tank is cone shape the area of the surface becomes afunction of height. Finding the functional form of the diameter is rstproblem and integrating it is the second. For the cone-shaped tankshown in Figure 5.14 as the level in the tank drops, the cross-sectionalarea A1 decreases.Figure 5.14: Cone shaped tank.The area of a cone as a function of height can be derived using the twodiameters and the height. Since we are dealing with a truncated cone,we have to break the problem down into two parts (Figure 5.15). Firstthe area between the two horizontal lines is constant and equal to4d22.Next we have to determine angle,. We know the value of hinitial andwe can derive the length of the opposite side of the triangle (o) byo = d1d22Using h and o, tan can be calculatedtan = oh555. Bernoulli EquationFigure 5.15: Cone shaped tank.Combining the numeric value for tan with any value for the heightyields the length of o at the selected height. The diameter, d1 at anyheight between h0 and h can now be expressed byd1(h) = 2(tan)(h) +d2Integrating Equation 5.11 becomes a bit more difcult because A1 isnow a function of height. The A1 terms now looks likeA1 = 4(2(tan)(h) +d2)2and the integral is_h2h1(2(tan)(h) +d2)2hdh = A24_2g_t2t1dt (5.12)This integral is somewhat more difcult to integrate than the one inEquation 5.11, but there are a number of calculators or computerprograms (MatLab, Maple, or Mathematica) that can do the integral.56Unsteady FlowsExample 5.3. Example: In Figure 5.16, the diameter of thetop of the tank is 2 f t and the diameter of the bottom is 0.5 f t.The tank is 15 f t high. Derive an equation to calculate the timeneeded to empty the tank givendhh= A2A1_2gdtFigure 5.16: Conical tank example.Expressing A1 asA1 = 4(2(tan)(h) +d2)2The integral that must be solved is_150A1h dh = A2_2g_t2t1dtAssigning values to the various parametersA2 = 4(0.5 f t)2= 0.196 f t2tan = 0h =d2d12h=2 f t0.5 f t215 f t= 0.05A1 = 4(2(0.05)(h) +0.5 f t)2= (0.1h+0.5)2575. Bernoulli EquationThe integral is now_150(0.1h+0.5)2hdh = 0.196 f t22_32.2 f ts2__t2t1dtt =_150(0.1h+0.5)2hdh0.196 f t2_2_32.2 f ts2_Analytical solution of this equation is difcult, but it is easy toevaluate numerically. The answer ist = 4.6s5.7 SummaryThe Bernoulli equation describes the relationship between velocityand pressure. It is useful in a wide variety of ow problems. Caremust be taken in choosing the start and end points. In general, thepoints should be selected to minimize the number of variables thatmust be evaluated. In the next chapter we will see how the Bernoulliequation can be extended to more complex ow systems wherefriction is important and work is done on or by the system.586 Fluid Friction in SteadyOne-Dimensional FlowUnlike the systems that were covered in the last chapter, real uids donot ow without frictional losses. Flow through pipes, valves,expansions, contractions, bends or into or out of tanks are just a fewexamples of sources of friction. One of the results of friction ispressure drop.597 Momentum Balance7.1 Newtons Laws1. A body at rest remains at rest and a body in motion remains inmotion at the same velocity in a straight path when the net forceacting on it is zero.2. The acceleration of a body is proportional to the net force actingon it and is inversely proportional to its mass.3. When a body exerts a force on a second body, the second bodyexerts an equal and opposite force on the rst. This is the socalled reaction force.Expressing the second law in equation formF = ma = mdvdt= d (mv)dt(7.1)The mv term is called the linear momentum or simply the momentum.Looking at the second law equation (7.1), it can also be thought of asthe rate of change of the momentum of a body is equal to the net forceacting on the body. For our purposes, this is probably a better way ofthinking about the second law. It is important to remember that theforce, acceleration, velocity, and momentum are vectors, i.e. theyhave both a magnitude and direction.7.2 Control VolumesMost of the work in working momentum problems is in thebookkeeping. Properly selecting a control volume and control surface617. Momentum Balancemakes the process easier. If the subject of the analysis is moving, thenit is usually better to let the control volume move. For a xed owsystem, the control volume should be xed. For example, the systemshown in 7.1 should have a control volume that encompasses thenozzle and the plate. For a jet exiting a moving airplane, the velocityFigure 7.1:Force on a plate exerted by a jet of uid impinging on theplate at a 90 angle.of interest is the relative velocity. An example is shown in Figure 7.2.In this case, the velocity is given byvx = vjetvCVWhere the jet is moving in the negative x-direction while the controlFigure 7.2: Moving airplanevolume is moving in the positive x-direction. This yields a negativeoverall velocity _v =vjetvCV_. An example of a deformablecontrol volume is the control volume inside a piston chamber in amotor (Figure 7.3). The control volume expands and contracts as thepiston moves up and down.62Forces on a Control VolumeFigure 7.3: Deformable control volume.7.3 Forces on a Control VolumeThere are two types of forces that act on a control volume: Body andSurface. The body forces act throughout the whole body while thesurface forces act on the control surface. Body forces can be a gravity,electric, or magnetic eld. Surface forces of interest in uid ow arepressure and viscous forces. Normally, we can ignore all of the bodyforces except gravity.F =Fbody +Fsur f ace(7.2)Gravitational force on a uid elementdFgravity = gd V (7.3)Total force acting on a control volumeFbody =_CV gd V= mCVg (7.4)Surface force acting on a small surface elementd

Fsur f ace = ndA (7.5)Wheren is an outwardly pointing vector normal to the surface dA and is the stress on the surface, dA (Figure 7.4).The total force on a body is

F =

Fbody +

Fsur f ace =_CV gd V +_CVndA (7.6)637. Momentum BalanceFigure 7.4: Surface forces acting on a small area.7.4 Steady FlowIn steady ow, the equation reduces to

F =out mv in mvWhich reduces toF = m(voutvin) (7.7)Example: Liquid Jet Striking a FlatPlateWater strikes a at plate at a rate of 10kg/s with a velocity of 20m/sand exits in all directions in the plane of the plate. What is the forcenecessary to hold the plate in place?Solution: All of the entering velocity is in the positive x-direction andnone of the uid exits in the x-direction. This means that v2x is equalto zero.F = m(v2xv1x)reduces toFx = 0 mv1Fx = mv1Fx = 10kgs|20ms|N s2kg m. .gc=200 NExample: Liquid Jet Striking aCurved Plate64Steady FlowFigure 7.5: Force on a curved vaneA 3-inch diameter jet with a velocity of 50 ft/s is deected through anangle of 70 when it hits a stationary vane, determine the horizontaland vertical components of the force of the water on the vane.Solution: Neglecting the weight of the uid, height changes in theuid, the weight of vane and assuming no frictional losses, thevelocity of the exiting liquid jet must be the same as the entering jet.This does not preclude changes in direction or cross-sectional shape.Vector analysis must be used to determine the x and y velocities.F =out mv in mvThe x-component of the velocity is given byv2,x = v2cos = 50 f ts| cos_70|2360_= 17.1 f tsThe y-component of the velocity is given byv2,y = v2sin = 50 f ts| sin_70|2360_= 46.98 f tsCalculating the mass ow rate m = q where q = v1Aq = 50f ts|0.0491 f t2= 2.455 f t3s m = 62.4lbmf t3|2.455f t3s= 153.19lbms657. Momentum BalanceThe reaction force in the x-direction is given byFx = m(v2,xv1,x)Fx = 153.19lbms|1gc|_17.1 f ts 50 f ts_=-156.5 lbfThe reaction force in the y-direction is given byFy = m_v2,yv1,y_Fy = 153.19lbms|1gc|_46.98f ts0 f ts_=223.5 lbfIf we want the force of the uids acting on the plate, the signs arereversed.Now we will complicate the problemby having the vane move with a velocity in the x-direction with avelocity of 15 f t/s.Example: Liquid Jet Striking aMoving Curved PlateFigure 7.6: Force on a moving curved vaneUsing the relative velocity in place of v1v2,x = v2cos = 35 f ts| cos_70|2360_= 11.97 f tsv2,y = v2sin = 35 f ts| sin_70|2360_= 32.89 f tsFx = 153.19lbms|1gc|_11.97 f ts 35 f ts_=-76.7 lbf66Steady FlowFy = 153.19lbms|1gc|_32.89f ts0 f ts_=109.5 lbfSince the vane is moving away from the jet, the forces will decrease asthe jet lengthens (the distance between the jet and the vane increases).There is another way to think aboutthe momentum equation. If we write the equation asF = vv A (7.8)This really is not that different. Remember that m = vA and bytaking the dot product between the velocity and the area vector (italways points outward from the control surface) we can keep the signsstraight. Reworking the rst part of the last example we get thefollowing.Example: Force on a Curved VaneF = vv AThe x-component of the velocity is given byv2,x = v2cos = 50 f ts| cos_70|2360_= 17.1 f tsThe y-component of the velocity is given byv2,y = v2sin = 50 f ts| sin_70|2360_= 46.98 f tsThe reactive force in the x-direction is given byFx =gcv1,xv1 A+ gcv2,xv1 AFx = 62.4lbmf t3|lbfs232.2lbmf t|50 f ts|50 f ts|cos(180)|0.0491 f t2+ 62.4lbmf t3|32.2lbfs2lbmf t|17.1 f ts|50 f tscos(0)|0.0491 f t2=-156.5 lbf677. Momentum BalanceThe reactive force in the y-direction is given byFy =gcv1,yv1 A+ gcv2,yv1 AFy = 62.4lbmf t3|lbfs232.2lbmf t|0 f ts|50 f ts|cos(180)|0.0491 f t2+ 62.4lbmf t3|32.2lbfs2lbmf t|46.98 f ts|50 f tscos(0)|0.0491 f t2=223.5 lbfLet us look at a slightly different problem involving ow of a uidaround a bend. The ow is not open to the atmosphere as it was in theprevious examples so pressure forces must be accounted for.Example: Force on a Pipe BendWater ows through a 180 vertical reducing bend. The discharge is0.25m3/s and the pressure at the center of the inlet of the bend is150kPa. If the bend volume is 0.1m3and it is assumed thatBernoullis equation is valid, what is the reaction force required tohold the bend in place. Assume the metal in the bend weighs 500N.Figure 7.7: Force on a bend.Solution: Starting with the basic momentum equation, evaluate all ofthe variables that can be easily evaluated.F = vv AA1 = 4d21 = 0.071m3A2 = 4d22 = 0.0177m368Steady Flowv1 =qA1= 3.54 msv2 =qA2= 14.15 msUsing the Bernoulli equation to evaluate the missing pressure.P1+z1 + v212g = P2+z2 + v222gP1 = 150kPa z1 = 0m z2 =_d12+0.1m+ d22_=0.325mSolving for P2P2 = 59.37kPaReturning to the momentum equation we can writeFx +P1A1 +P2A2 =gcv1v1 A1 + gc(v2) v2 A2Inserting values for the variables and solving for Fx we getFx =10.60kN1.049kN0.884kN3.537kN = 16.07kNThe force in the y-direction is simply the sum of the weight of theuid in the bend and the weight of the bend.Fy =WB +V ggc= 500N+ 0.1m3|1000kgm3|9.81ms2|Ns21kgmFy =1.48kNNote: If the bend in the previous example is turned around, the PAforces have a negative sign in front of them and v1 is negative ratherthan v2. This results in a change in sign, but not magnitude, of Fx.The analysis of a bend of less than 180 is similar, but the angles haveto be taken into account when the velocities are calculated.Example: Flow in a 30 Bend.A pipe that is 1 m in diameter has a 30 horizontal bend in it, as shown,and carries crude oil (SG = 0.94) at a rate of 2 m3s. If the pressure inthe bend is assumed to be essentially uniform at 75kPagage, if thevolume of the bend is 1.2m3, and if the metal in the bend weighs 4kN,697. Momentum BalanceFigure 7.8: Forces on a 30 bend.what forces must be applied to the bend to hold it in place?Solution: Starting with the momentum equationF =v v AA1 = A2 = 4d2= 0.785m2v1 =qA1= 2.55 msThe velocity at the exit, v2,x, is not equal to v1, but is equal tov2,x = v1 cos_302360_= 2.205 msSo the momentum equation becomesFx+P1A1P2A2cos_30 2360_= oilgcv1,xv1A1 cos()+oilgcv2,xv2A2cos(0)Solving for FxFx =59.05kN+51.01kN4.79kN+4.15kN =8.64kNThere is no pressure force acting in the y direction on the entranceside of the bend nor is there a y component of the velocity soFy +P2A2sin_30 2360_= oilgcv2,yv2A2cos(0)Fy = 29.45kN+2.39kN =31.84kN70Rotational Motion and Angular MomentumThe force in the z direction results from the weight of the bend andthe weight of the uid enclosed within it.Fz =WB +Wf = 4kN+ 1.2m|(0.94)1000kgm3|9.81ms2|Ns21kgmFz =15.1kN7.5 Rotational Motion and Angular Momentum7.5.1 ReviewAngular velocity The angular distance traveled per unit time. = ddt= d (l/r)dt= 1rdldt= vr = ddt= d2dt2= 1rdvdt= atrv = rat = rWhere at is the acceleration in the tangential direction and v is thelinear velocityFor constant rotational motion there must be a force actingtangentially to maintain the angular acceleration. The moment ortorque (M) is proportional to the length of the arm that connects thepoint of action and the center of rotation.M = rFi = rmat = mr2In integral formM =_r2dm =__r2dm_. .I = I717. Momentum BalanceFigure 7.9: Angular velocityI is the moment of inertia. Remembering that mv is the momentum ina linear system, the moment of momentum or angular momentum fora rotating point mass is given byH = rmv = r2mFor a rotating rigid body, the total angular momentum is given byH =_r2dm =__r2dm_. .I = IAngular velocity =2 n60yields rad/s.In vector form, the momentum force equation becomes

M =r

FWhere the magnitude of the moment of force is given byM = Fr sinThe moment of momentum in vector form becomes

H =r mvFor a differential mass dm72Rotational Motion and Angular Momentumdm = (r v) dVIntegration yields

Hsys =_sys(r v) dVFor steady ow the net torque can be written as

M =outr mv inr mvExample: A large lawn sprinklerwith four identical arms is to be converted into a turbine to generateelectrical power by attaching a generator. Water enters the sprinklerfrom the base along the axis of rotation at a rate of 20 liters/s andleaves the nozzles in the tangential direction. The sprinkler rotates ata rate of 300 rpm in a horizontal plane. The diameter of each jet is1cm, and the normal distance between the axis of ration and the centerof each nozzle is 0.6m. What is the torque on the shaft available togenerate electrical power?Solution: We want to solve the equationM =outr mv inr mvThe last term is equal to zero since rin = 0. Calculate m m = q = 1000kgm3|0.02m3s= 20kgsCalculate the nozzle velocityvn = 0.02m3s|14|4 (0.1m) = 63.66msCalculate the angular velocity = (2)300revmin|min60s = 31.42radsCalculate the tangential velocity737. Momentum Balancevnozzle = r = (0.6m)_42s_= 18.85msCalculate the relative velocityvrel = vnvr = 63.6618.85 = 44.81msCalculate the torqueTshaf t = r mtotalvr = 0.6m|20kgs|44.81ms|N s2kg m =537.7N mCalculating the powerPo = Tsha f t = 31.42s|537.7N m|kW s1000N m = 16.9kW74