工程流体力学(英文版)第一章.pdf

contents1 The definition of fluids andcontinuum Hypothesis (2 Physical Properties3 Forces acted on a fluid4 Flow of a fluidof a fluid **5 Viscosity (6 Compressibility and incompressible flow7 Surface Tension1. Definition of a fluid:In the earth there are three main substance states1 The definition of fluids and continuumHypothesisSolid liquidgas.Fluidfluidsoliddeformation, no shape•Distinction Between a Solid and a FluidSolid can not only bear pressure and pulling force , but also resisting tangential stress.Fluid can bear pressure, but cannot bear pulling deformation force and resist tangential stress .Microcosmic A fluid is composed of a large amount of molecules in irregular motion with gap between molecules.Macroscopic much larger than inter molecular distance mean free pathδδ If δ δ ρ=δ δIf δ δ δ δ change so, ρis defined asAssumption of Continuum ModelAssumption of Continuum ModelContinuum /Continuous substance Fluids or solid particles occupy space continuously with inter molecular distance and molecular motion ignored 2 Definition of Continuum Modelδδ 1 Definition fluid particleA fliud element with the volume δ is defined as a fluid particle ,δ is a volume that is small enough with enoughmoles to make sure that the macroscopic meandensity has definite value.1cm 3liquid : 3.3x1022molecules1 cm 3gas : 2.7x1019 molecules10-19cm 3gas : 2.7x1010moleculesContinuum Model An assumption model that assumes a fluid as a continuum. Under this assumption, all physical properties are the continuous function of space coordinates and timeExerciseAccording to the concept of continuum, fluid particle isA fluid moleculeB solid particle in fluid Cgeometric point D differential element whose geometric scale is infinitesimal , yet includes a large amount of molecules.3 Advantages1 Excluding complexibility of molecules in motion.2 Physical properties are the continuous function of space and time,so we can use mathematical tool of continuous function to solve problems.Density is mass of the substance per unit volume Unit kg/m3 .1. Density2 Physical Properties( Density, Relative density, Specific Volume and Density of mixture gas)For ideal gasδδ'limm→=ΩΩΩρ2. Relative densitywhere ρw ——density of 4 water kg/m 31.2 Physical Properties3. Specific Volume of Gas (Specific Volume is the volume occupied by a mass of gas, specific volume is the reciprocal of density, m 3/kgRelative density of a fluid is the ratio of its density to that of pure water at 4 It has no unit and dimensionless.4. Density of mixture gas Density of mixture gas is calculated as volume fraction:where ρi ——density of gas i——volume fraction of gas i1122331ni i i ρραραραρα==++⋅⋅⋅⋅⋅⋅= i α 1.2 Physical PropertiesSpecific Weight is weight per unit volume. Unit 30lim G∆Ω∆γ∆Ω→=For homogenous fluid, each point has the same specific weightγ=G/V =ρgFor water, the nominal value of specific weight isγ=9800 N/m 35. Specific Weight ***1.2 Physical Properties1 Classification1 According to physical properties of fluids:gravity, friction force, inertia force,elastic force and surface tension.2 According to acting method:Mass force and surface force .2 Mass force 2 Unit mass force is the mass force acting on per unit mass fluid.Unit of unit mass force [m/s 2] It’s the same as the unit of acceleration unit.dF f d ρ=Ω x y x f f i f j f k=++3 Surface Force2/m N orPaNormal Compressive Force : perpendicular to acting surfaceShear Force : parallel to acting surfaceAccording to the acting directions, surface forces can be classified into:2 Stress is the surface force on per unit area,unit:Shearing stress Pressure n F F F τ=+1. Mechanical characteristics:Solid can not only bear pressure and pulling force , but also resisting tangential stress.Fluid can bear pressure, but cannot bear pulling deformation force and resist tangential stress .4 Flow of a fluidpressure pulling force tangential force.Solid:Fluid: √√√√fluidsolid2. Relationship between deformation and force:F x ∝∆x∆F2. Relationship between deformation and force:≠∆→∞0,F xExerciseEnglishe Text:P2: 1.1.1,P17:1.7.1, 1.7.2, 1.7.3Chinese Text:P16: 1-1, 1-2, 1-4, 1-7,1-85 Viscosity of a FluidThe definition of viscosity2. Newton’s Viscosity Law *3. Newtonian and non-Newtonian Fluids4. Coefficient of Viscosity5. Affecting Factors of Viscosity5.1 Definition of viscosity Viscosity is that property of a fluid by virtue of which it offers resistance to shear. (the fluid sticking action)The most important property of a fluid, any real fluid is viscous fluid.Kinescope2Kinescope1According to whether fluids have viscosity, fluids can be classified intoI deal fluid 1. µ 0.2. du/dy=0.Real fluid Viscous fluid, µ≠0.1. Viscosity:Question : What is ideal gas ?2. Cause of viscosity:(a) attractive, cohesive forces between the molecules(b) molecular interchangereturn5.2 Newton’s Viscosity Law1686, Newton suggested: Fluid sticking action:1. Velocity gradient 2. Surface area 3. Fluid propertyduA dyFµ=±du F A dyτµ==±Unit of ?returnNewtonian Fluids The shear stress of a fluid is directly proportiond to therate at each point, that is , fluids conform Newtons Law of Viscosity.Non-Newtonian Fluid:fluids that don ′t meet the above conditions.dydu τsolidfluid fluid5.3 Newtonian Fluid and Non -Newtonian Fluiddu dyF∝()0ndu=+ττµFluid Classificationreturn5.4 Coefficient of ViscosityDynamic Viscosity µdynamic viscosity or viscosity, is the measurement of fluid viscosity. Unit: N•s/m 2 .ρµ=v (m 2/s)Kinematic Viscosity νdyµτ=(),Pa s kg m s ⋅⋅Reletive Viscosity EE t t=t’: fluid, 200cm 3, =2.8mm t : 20 C water , 200cm 3, =2.8mmE- 0.07310.0631Eν=return5.5 Affecting Factors of ViscosityCause of Viscosity(a) attractive, cohesive forces between the molecules(b) molecular interchangeAffecting Factors of ViscosityThe value of viscosity of a fluid µvaries due to the change in pressure or temperature. Different fluids has different viscosity.1 Fluid type. The viscosity of liquids (ordinarily ) > gases2 Pressure. µvaries little , can be negligible3 Temperature. the main factor to affect viscosity.a. LiquidFor liquid,Cohesive Force is the main factor to produce viscosity .When temperature increases, molecular distances increase cohesive forces diminishes, so shear force produced by shear deformation rate diminishes and µdecreases.When temperature increases , viscosity of liquids decreasewhile that of gases increases .2000221.00337.0101775.0tt v ++=(cm 2/s)Kinematic viscosity of water νis usually calculated:Correct: P10, vb. GasesGas molecular distance is big ,therefore cohesive force is small. So for gases the predominating factor is the interchange of molecular momentum. When temperature increases, molecular activities increase, momentum interchanges also increase, so µincreases.returnFor most gas, dynamic viscosity can be calculated as Sutherland’formula:1.512T C T C µ=+For air :1.561.45810110.4TT µ−=×+ExerciseThe wrong statement about the viscosity of a fluid isA Viscosity is a property of fluidB Viscosity is a measure of its resistance to sheardeformation during motion.C Viscosity of a fluid can both speed up and slowdown movementD Viscosity of a fluid increases as thetemperature increases.returnExample 2 As shown in figure, the mass of a 1-cm-height and 40 45cm2 bottom area, wood board is 5kg. It moves at a fixed velocity along a slope with lubricating oil. Wood velocity u=1m/s, the oil thickness δ=1mm, oil velocity gradient caused by wood is a straight line.What is viscosity of oil?Solution constant velocity is fixed as=0Based on Newton Second LawF s =ma s =0mgsin θ ·A=0δµµτudydu == Velocity gradient is a line2/105.0sin m s N Au mg ⋅=⋅=∴δθµ θ=tg -1(5/12)=22.62°66 Elasticity (Compressibility and incompressible flow) 1. Elasticity ( Compressibility)1.The elasticity of a fluid is related to the amount of deformation(expansion or contraction) for a given pressure change.The degree of elasticity is given by E(Bulk Modulus of Elasticity)returnreturnFor water:92.010E Pa=×5491.013100.5102.010P E −∆Ω∆×===×Ω×51.01310P Pa∆=×if:return2. incompressible flowLiquids: E ~109, incompressible flowM<0.3: 100m/s, <3%0.3 M<0.75 :100-250m/sM 0.75:250m/sGas: V---dPExample 1 What pressure must be applied to waterrespectively in order to reduce its volume by 0.1% or 1% K =2000MPap E =−ΩΩ p E =−⋅ΩΩ()962.0100.1% 2.010Pa=2.0MPa p =−××−=×()97 2.0101% 2.010Pa=20MPa p =−××−=×reduce 0.1%reduce 1%。