Chapter4 Basics of fluid flow

1、equationicHydrodynamequationaticHydrokinemequationsBasic4.1 Description of Fluid Motion nMethod：Describe the variation of the physical parameters of a given particle (fluid parcel) with time , such as displacement、velocity、acceleration、pressure、density、temperature. 4.1.1 Lagranges Method ( Particle

2、Method ) 4.1.1拉格朗日方法ABCDt1时刻ABCDt2时刻fafbfcfd(t)(t)(t)(t)n具体表示方法:n一、用流体质点在t0时的 位置标识不同的质点qt=t0时流体质点的坐标是 ( a,b,c)qa,b,c可以是直角坐标的(x0,y0,z0), 也可以是曲线坐标（x1,y1,z1）q不同的a,b,c代表不同的质点n二、流体质点的运动规律数学上 可表为下式：( , , , )rr a b c t( , , , )a b c t拉称为格朗日变数4.1.1拉格朗日方法n将 在笛卡尔坐标中展开( , , , )rr a b c t( , , , )( , , , )( , ,

3、 , )xx a b c tyy a b c tzz a b c t( , , , )rr a b c t（位移函数）（位移函数）( , , , )xx a b c tVt( , , , )yy a b c tVt( , , , )zz a b c tVt( , , , )r a b c tVt流体质点的位移函数速度函数22( , , , )xxx a b c taVt22( , , , )yyy a b c taVt22( , , , )zzz a b c taVt22( , , , )r a b c taVt加速度函数4.1.1拉格朗日方法n拉格朗日方法 自然、直观；实现非常困难；无法对成

4、千上万的流体质点进行跟踪。n实际问题：固定空间区域内固体与流体的作用。n实验测量值：空间固定点的参数。),(),(),(321tcbaFztcbaFytcbaFxttcbaFtzvttcbaFtyvttcbaFtxvzyx),(),(),(321232222222221222),(),(),(ttcbaFtztvattcbaFtytvattcbaFtxtvazzyyxxnDistinguishing of different particle：by the coordinates of particle（a，b，c）at the beginning.nSimilarly, for an arb

5、itrary physical parameter nThe method is rarely used because of the difficulty in deriving the function B),(tcbaFB ttcbaFtB),(F4.1.2 Eulers Method ( Field Method ) nMethod：Describe the variation of the physical parameters of fluid in a fixed flow field with time, such as displacement、velocity、accele

6、ration、pressure、density、temperature.nFor an arbitrary physical parameter B),(tzyxFB nFor velocityn For a fixed space, section or point, Eulers method is usually used. For example, a weather map at any instant can be drawn with Eulers method for weather forecast.),(),(),(tzyxFvtzyxFvtzyxFvzzyyxx),(tz

7、yxFvv4.1.3 Particle Derivative nThe physical parameters of particle, whether they are scalars or vectors, can be differentiated with respect to time.nOnly Eulers method is discussed here.0(,)(, )limtDVV M ttV M tDtt ,V M tV M ttMt 设此质点在场内运动,其运动轨迹为L,在t时刻位于M点,速度为过了后 该质点运动到点 速度为根据定义 加速度的表达式为),(tMV), (t

8、tMV0(,)(, )limtV M ttV M tt 0(, )(, )limtV M tV M tt 0(,)(, )limtDVV M ttV M tDtt 按照时间和空间引起速度变化,把极限分为两部分场的非定常性场的不均匀性流体质点的加速度dtVda dtdVaxxdtdzzVdtdyyVdtdxxVtVxxxx),(tzyxVV ),(tzyxVVxx流体质点的加速度zVVyVVxVVtVayzyyyxyyzVVyVVxVVtVazzzyzxzz同理4.1.3随体导数随体导数xxVVxxyVVyxzVVzdtdVaxxdtdzzVdtdyyVdtdxxVtVxxxxtVxzyxxxx

9、VVtVa)(yyyVVtVa)(zzzVVtVa)(tBBtzyxttzzyyxxt),(),(0limtBdd1lim),(),(0tzyxtzyxtBttBzzByyBxxBBt1lim0ttBzzByyBxxBttzBvyBvxBvtBzyx General formulain which -Local derivative that is caused by unsteady flow . -Transfer derivative that is caused by position change of particle because of nonuniform flow fiel

10、d or the motion of particle. BvtBdtBd)(zkyjxitBBv)(nMeaning：A connection is made between Lagranges method and Eulers method.If is Bvvvtvdtvda)(zvvyvvxvvtvzyxzvvyvvxvvtvazvvyvvxvvtvazvvyvvxvvtvazzzyzxzzyzyyyxyyxzxyxxxxnExample: Discuss ax at x axis in the following figure.q If the free surface goes d

11、own in the water tank, xvvtvaxxxx00 xvvtvxxxqIf H is a constantqIf H is a constant and the water pipe is a cylinder.0tvxxvvaxxx00 xvvtvxxx0 xaxvvtvaxxxx例题 1DVDtt求质点的密度变化率.222,.A xyztVxtiytjztkA已知密度场速度场为求流体质点的密度变化率 其中 为常数222222A xyztxtiytjztkA xyztt2222A xyzxtiytjztkA xiyjzk t222212A xyzt4.2 Basic Co

12、ncepts for Description of Fluid Motionn(1) According to the number of independent coordinates ),(Flow3D),(Flow2D),(Flow：1DtzyxFBtyxFBtxFB：4.2.1 Types of Flow nDimension Simplification : Example: the flow in a cylinder),(:Flow2DtrxFB),(:Flow1DtxFBwhen the differences of each physical parameter at dif

13、ferent point on the same section are neglected. qDiscuss: Determine the dimensions of flow and the number of flow direction in velocity field.n(2) According to whether the physical parameters change with time kxyjyiyxv42221),();,();,(FlowUnsteadytzyxFBtyxFBtxFB：),();,();(FlowSteadyzyxFByxFBxFB： Stea

14、dy FlowUnsteady FlownState Simplification Unsteady Flow Steady Flowq A flow through a small hole on a container with great horizontal section.q A system of coordinates is fixed on a container moving at uniform linear acceleration or rotating about a vertical axis at constant angular velocity.n(3) Ac

15、cording to whether the velocity is the same at a given instant at every point in the fluid.n(4) According to whether a flow shows its viscosity Flow)(ViscousFlowRealFlowIdealFlowNonuniformFlowUniformn(5) According to whether a flow is laminarn(6) According to whether a fluid is compressibleFlowTurbu

16、lentFlowLaminarFlowibleIncompressFlowleCompressib Example 2 In a velocity field, Determine (1) the acceleration at the point (2, 4) for t=2s？(2) unsteady or steady? (3) the dimensions of flow and the number of flow direction in velocity field.n Solution (1)jtxyi txyv)96()64(xyttxyttxyttxyxyyvvxvvtva

17、xyxxxx64)661)(64()4()96()6()64()64(22For t=2s，x=2, y=4, then ax= 4 m/s2，ay=6m/s2So, （2）Because the velocity changes with time, the flow is unsteady. (3) two dimensions (x, y) and two directions (i, j).yvvxvvtvayyyxyy222/21. 7smaaayx)6()96()9()64()96(ttxyttxyxyxyttxy96)661)(96(224.2.2 Path Lines and

18、Streamlines n(1) Path LinesThe path traced by moving fluid particle. It can be described with Lagranges method.q Notes: t-independent variable; q x、y、z-dependent variablesdtvdzvdyvdxzyxn(2) StreamlinesA continuous line. It can show the direction of the velocity vector at each point in a flow field.

19、A particle always moves tangent to the streamline. It can be described with Eulers method.qCharacteristic of streamline: Because every particle has only fixed velocity vector, streamlines cant intersect one another.qDifferential Equations：two kinds of methodszyxvzvyvxvsdddEquationProportion0dEquatio

20、nVector：vqvsqExample 3 In a velocity field, Determine（1）streamline equation and streamline chart for t =0，t =1，t =2.（2） path line equation and the path line of particle passed through point (0,0) for t =0.jbtiavnSolution（1）Streamline Equation and Streamline ChartbtdyadxCxabty(a) Streamlines for t=0n

21、The streamlines are some of straight lines. nFor t=0，t=1，t=2, the charts of streamline are a group of straight lines with different slope. (b) Streamlines for t=1 (c) Streamlines for t=2 （2）Pathline Equation and Path Line of Particle and for t =0，x =0，y =0，then C1=0，C2=0. 22tbyatxdtbtdyadxbtdtdyadtd

22、x2212CtbyCatx222xaby thenorExample 4 In a velocity field ,and a is a constant. Determine （1）streamline equation； （2）pathline equationnSolution(1) Streamline Equationaxu ayv0w0y0w0y and aydyaxdxCyxlnlnln alsoThe streamlines are a group of hyperbolas.q（2）Pathline EquationCxy dtaydyaxdxatateCyeCx21then

23、 CeCCxy021then 4.2.3 Streamtube （1）StreamtubeA tube made by all the streamlines passing through a small closed curve.nCharacteristicsq1）There is no flow through the streamtubes walls because the velocity vector has no component normal to the tube surface. q2）In unsteady flow, the shape of streamtube

24、 changes with time. q3）In steady flow, the shape of streamtube doesnt change with time.(2) Infinitesimal element tubenIts section is infinitesimal. The particles on the same section of the tube have the same velocity.Infinitesimal element tubeStreamtubStreamlineThe section gets smaller.fluxVolumetri

25、c fluxMass fluxIncompressible fluid AudAQAmudAQQQmsm /3skg /Average velocity on cross section AQv vAQ dA1dA2u1u2dAuAnCross Section12注意：只有均匀流的过流断面才是平面在流束上作出与流线正交的横断面4.3 Fluid System and Control Volume nFluid SystemAn aggregation of some particles within the boundaries defined by a closed surface. The

26、 form of system changes in motion, while the particles in the system never change. Lagranges method is usually applied to system.nControl VolumeA fixed region in flow field. The surface of control volume is called control surface. The form and position of control volume never change in flow field, w

27、hile the number of particles in it usually changes. Eulers method is usually applied to control volume.SystemControl Volumesystem control volume p a r t i c l e and mass invariablevariablemomentum and energy interacting at its boundary 1）interacting at its boundary； 2）inflow and outflow across its b

28、oundary nTransport EquationIn which,Xthe total amount of some property (e.g., mass, energy, or momentum) within the system at time t. the amount of property per unit volume. CVthe total volume of control volume CSthe total surface area of control volume velocity of fluid AareaCSCVAddVtdtdXdtdXdtdXdt

29、dXdtdXinCVoutCVCVsnIndication: The time rate of increase of X within a system is just equal to the time rate of increase of the property X within the control volume, which exactly coincides with the system at time t, plus the net efflux of X per unit time across the control volume boundary. System D

30、erivative ( ) = Local Derivative+ Transfer Derivative nMeaning：A connection is made between Lagrange method and Euler method.dtdX4.4 Equation of Continuity nContinuity Equation of Mass： 3.4.1 Integral Form The change of fluid mass within a control volume satisfies the law of conservation of mass. Ti

31、meUnitperMassinIncreaseTimeUnitperMasseffluentTimeUnitperMassInfluentCVCSCSVtAAddd21CVCSnnCSVtAvAvddd2211tmtmtmCVoutCVinCVddddddin which -a component of in the inward normal direction of ; - a component of in the outward normal direction of n1）If the fluid is compressible and steady,1nv1vAd2nv2vAd21

32、21CSnCSndAvdAvm21mqqCVCSnnCSVtAvAvddd2211n2）If the fluid is incompressible, =constantn3）Specially, if the flow is one dimensionaldAvdAvCSnCSn212121qq 222111AvAv2m1mqq2211AvAv21qq nExample 5 There is a three-way pipe as shown in the following Fig. d1=d2=200mm，d3=100mm, v1=3m/s, v2=2m/s. Determine v3.

33、nSolution321qqq332211AvAvAvsmddvvv/4221213 then 4.4.2 Differential form nAccording to the integral formCVCSCSdVtAdAd21CVCSdVtAdCVCSndVtdAvin which vn- a component of in the outward normal direction of . vAdnAnd according to Gauss formula, in whichand div is called divergence CSCVndVvdivdAv)(zvyvxvvd

34、ivzyxCVCSndVtdAvHence or Meanwhile, because the expression can be applied to any control volume, then 1) In XYZ rectangular coordinate system0dVvdivdVtCVCV0zvyvxvtzyx0dVvdivtCV0)(vdivtor 2) In cylindrical-coordinate systemnSpecially, in XYZ systemq 1）If a flow is steady, 011zrvzvrvrrrt0t0zvyvxvzyxq

35、2）If a flow is incompressible, nMeaning: Determine whether a flow satisfies the continuity in physics.nExample 6 In a velocity field, Does the flow satisfy its continuity equation?constant0zvyvxvzyxkztjxyixyv)2(1)2(1)(1222tnSolution The Flow is compressible and unsteady.tt2xxvx2xyvy2tzvz20zvyvxvtzyx

36、 Then The flow satisfies its continuity equation. n The other applications (Continuity Equation) and (Two Components of Velocity) (The third Component)n思考题思考题 试证不可压缩流体的运动是可能存在的：试证不可压缩流体的运动是可能存在的：n（1 1）n（2 2）222,2,4xyzvxy vyz vxy zxy ,xyzvyzt vxzt vxyt4.5 Energies of A Flowing FluidgvgVmv221WeightKE2

37、2221MassKE22vmmv2)(2121VKE222vVvVVmvolume(1) Kinetic Energy(4) Internal EnergypgAspAsWeightPHzmgmgzWeightPE(2) Potential Energy(3) Pressure Energy (Pressure Head)EnergyticElectrostaEnergyChemicalEnergyNuclear(Here)EnergyThermalgTTcmgTTcm)()(WeightIE12124.6 Equation for Steady Motion of an Ideal Flui

38、d Along a Streamlinen Take an infinitesimal element tube in steady flow field as shown, and regard the segment ds of the tube as control volume. The segment ds becomes a truncated cone when its curvature is neglected.3.6.1 Differential Motion Equation for One-Dimensional Steady flow of Incompressibl

39、e Ideal Fluid nBecause the ideal fluid is frictionless and steady, the momentum equation in the flow direction is)d(cos)d)(d(d)d21(vvvAvwAAppApppAzgAsgAwdcosdcos)d(d)d)(d(d)d21(vvvAvzgAAAppApppAand tvvmF)(12nBecause the segment is infinitesimal, the higher order infinitesimals may be omitted. Hencen

40、Indication: The differential relations are established among pressure, velocity and position of particle along an arbitrary streamline.0dddpzgvvvvApAzgAApddddd214.6.2 Bernoullis Equation nand for incompressible fluid, =constant，thenCpgzvd22Cpgzv220dddpzgvvSo, Bernoullis Equation nApplication conditi

41、ons：Hgpzgv22flowSteady) 1 (Cpgzv22forcemassisgravityOnly)2(ibleIncompress)3(flowIdeal)4(removedor added isenergy No (5)nPhysical meaning：nin which, pressure potential energy ：The work is done when the pressure brings a unit mass fluid from a position of p pressure to the position of zero pressure. C

42、onstantmassunitperenergyPressuremassunitperenergypotentialPositionmassunitperenergyKineticCpgzv22pConstantweightunitperenergyPressureweightunitperenergypotentialPositionweightunitperenergyKineticHgpzgv22In a word, three kinds of energies can be changed, and the sum is a constant along a streamline i

43、n steady, frictionless and incompressible flow.Hvpgz常数22bc1aa2cbH总水头线静水头线gv2/21gp/11zgv2/22gp/22z速速度度水水头头位位置置水水头头压压强强水水头头总总水水头头不可压缩理想流体在重力场中作定常流不可压缩理想流体在重力场中作定常流动时，沿流线单位重力流体的总水头线动时，沿流线单位重力流体的总水头线为一平行于基准线的水平线。为一平行于基准线的水平线。nGeometrical meaning：ConstantheadwaterTotalpressureofheadWaterpositionofheadWat

44、ervelocityofheadWater4.6.3 Application of Bernoullis Equation nExample 7 In 1773, Pitot used a bent tube to measure the velocity of stream in a river.flowofquantitytheMeasureflowofvelocitytheMeasurenSolutionnFor a stream passing through point A and Point BgpzgvgpzgvAAABBB22220AvBAzzABBppv22)(2BABppv

45、because then So nAnd because 0gHpB)(0hHgpAghvB2ghppBA)(2BABPPvnExplaining (1) Total pressure at point A (Stagnation Pressure -PA )221vppBA PressureDynamicPressureStaticPressureStagnationn(2) Static PressurePBn1) The pressure PB can be measured with a manometer moving at the same velocity as the flow

46、s.n2) If the change of static pressure on the section of pipe is neglected, the height of liquid in the little hole perpendicular to the pipe can represents the static pressure as shown. n(3) Pitot-static pressure tubenThe velocity of flow is usually measured by Pitot-static pressure tube in enginee

47、ring. )(2strealppvpsptnExample 8 Analyze the principle of Venturi tube as shown in the figure. 2222121122pvgzpvgz2121vAAv )(1 2)(2212212AAzgppv)(1 2)(2212212AAzgppAqv流速：流速：体积流量：体积流量：nin which, is caused by viscosity and turbulence of fluid. Usually, =0.980.99)(1 )(22122AAghAqvnExample 9 Determine hm

48、ax of the siphon pipe for saturation pressure pVnAnalysisgphzgvgpzgvStreamlinegphzgvgpzgvStreamlineVseasspeaksurfaceaseassoutletsurface)(22:)(22:max221220svgphgvgpghvVeaemax21221max)(1hppghVaand 思考题4.7 Equation for Steady Motion of a Real Fluid Along a Streamline)d(d)d21(cosd)d21()d)(d(vvvAvsLLwAppA

49、ApppA1212)(ttvvmFsgALvgvzgpdddd)2()2(22222111gvzgpgALSgvzgpgALShf)2()2(22222111gvzgphgvzgpf4.8 Energy Equation for Pipe Flow n1) For one-dimensional, incompressible and steady flow, 2222221211Qgzvhgzvh4.8.1 Comparison between the energy Equation and Bernoullis Equation In which, -Quantity of heat pe

50、r fluid mass from the outside. Qn2) And for ideal fluid, according to the Bernoullis equation,) (221222221121QIIpgzvpgzv2222112122pgzvpgzvnTherefore, for viscous, one-dimensional, incompressible and steady flow,in which, Losses1-2 are caused by viscous friction.212222112122Lossespgzvpgzv4.8.2 Modifi

51、ed Coefficient of Kinetic Energy and Modified Coefficient of Momentum (1) Modified coefficient of kinetic energynConsidering the influences of velocity distribution on the section, In which, va -average velocity 2222a2a2vqvAvvdAvmaAAvdAAv1a (2) Modified coefficient of momentumSimilarly, (3) Coeffici

52、ents of pipe flow 2222amAvqvdAvamaAAvqAvdAvvdAv22Laminar flowTurbulent flow21.051.124/31.011.05CoefficientsDatumtypesFlow4.8.3 Bernoullis Equation for Total Flow nRegarding a segment of pipe as a control volume, for viscous, incompressible and steady flow, nin which， hwhead loss. Specially, for a vi

53、scous fluid in pipe, w22222s1121122hgpzgvhgpzgvaajfwhhhnhsgotten energy per unit weight quantity of flow, gvdlh2:LossFrictional2fmgqPhssflowariedvslowlyinfrictionviscoustheof_becausegvh2:LossLocal2jflowariedvrapidlyinenergyofwastetheof_becausen Pseffective shaft power, caused by pump, blower and tur

54、bomachine, etc.4.9 Momentum Equation nAccording to the law of momentum conservation ( Newtons second law)TimeUnitperMomentuminIncreaseTimeUnitperMomentumEffluentTimUnitperMomentumInfluentForceSurfaceForceMasseCVCSCSCSsCVdVvtvdAvnvdAvndAfdVf2112)()(CSCSCVCSsCVvdAvnvdAvndVvtdAfdVf1212CSnCSnCVCSsCVdAvv

55、dAvvdVvtdAfdVf1212CSnCSnCSsCVdAvvdAvvdAfdVf For a steady flow 1) Specially, for a steady flow with inlets and outlets zmzmszfzymymsyfyxmxmsxfxvqvqFFvqvqFFvqvqFF112211221122Avqqaveragen,averageaveragem in which, 2) For one dimensional steady flow )()()(121212zzmszfzyymsyfyxxmsxfxvvqFFvvqFFvvqFFExplai

56、ningThe equations above can be applied to only the inertial system or the coordinate system moving at uniform linear velocity because of being derived from Newtons second law.nExample 10 In the following Fig.，the inflow and outflow radii are 25 and 15cm, respectively; the inflow and outflow angles w

57、ith respect to the horizontal line (1and2) are 450 and 300, respectively; q is 50 L/s; the area average inlet and outlet pressure are 8.5 and 5.83kPa; and the total fluid weight in the pipe is 2.0N. Find the horizontal and vertical force required to hold the pipe in place.nSolutionThe momentum excha

58、nge and force vectors are drawn in the following figure.The magnitudes of Fp1 and Fp2 are calculated asThe magnitudes of v and qm are calculated as follows111ApFp222ApFp111Aqvvn222Aqvvnqqqqmmm21)sinsin(sinsin)coscos(coscos1122221111222211vvqFWFFvvqFFFmpypmpxpnAccording to the momentum equation, the

59、components of F are calculated asn.nSubstitution of parameters and solving for Fx and Fy givesnThe negative signs indicate that the assumed directions for Fx and Fy are incorrect.)N(2 .963)N(4 .801yxFF4.2 Moment-of-Momentum Equation 4.2.1 Moment-of-Momentum EquationnAccording to the law of moment of momentum,TimeUnitperMomentinIncreaseTimeUnitperMomentEffluentTimeUnitperMomentInfluentForceSurfaceofMomentForceMassofMomentCVCSCSsCSCVdVvrtdAvnvrdAvnvrdAfrdVfr)()(21dAvnvrdVvrtdAfrdVfrCSCVsCSCV)(dAvvrdAvvrdVvrtdAfrdVfrnCSnCSCVsCSCV1212nFor a steady flownThe moment-of-momentum equation