《流体力学英文版复习课件.ppt》由会员分享，可在线阅读，更多相关《流体力学英文版复习课件.ppt（66页珍藏版）》请在三一办公上搜索。
1、Chapter 1.FLUID PROPERTIES 第一章 流体性质,2023/3/12,In physics,a fluid is a substance that continually deforms under an applied shear stress,no matter how small it is.,流体是一种一受到切力作用（不论多么小）就会连续变形的物体。,Definition of a fluid,2023/3/12,(1)Density（密度）,Density is the ratio of the mass of fluid to its volume.,(1.1
2、2),Specific volume(比容):volume occupied by unit mass.,(1.13),The specific volume is the reciprocal(倒数)of density.,(kg/m3),(m3/kg),2023/3/12,(2)Specific Weight（重度）,Its the weight per unit volume,(1.14),in which is the specific weight of fluid,N/m3;G is the weight of fluid,N.,(1.14a),Or the product of
3、density and acceleration of gravity g.,2023/3/12,(3)Relative density and specific gravity,The relative density(相对密度)RD of a fluid is the ratio of its density to the density of a given reference material.,The reference material is water at 4C i.e.,ref=water.=1000kg/m3,dimensionless quantity无量纲,The sp
4、ecific gravity(比重)SG of a fluid is the ratio of its weight to the weight of an equal volume of water at standard conditions(标准状态).,dimensionless quantity无量纲,2023/3/12,(4)Compressibility（压缩性）,The volume of fluid changes under different pressure.As the temperature is constant,the magnitude of compress
5、ibility is expressed by coefficient of volume compressibility(体积压缩系数)p,a relative variation rate（相对变化率）of volume per unit pressure.,(1.15),The bulk modulus of elasticity(体积弹性模量)K is the reciprocal of coefficient of volume compressibility p.,(1.16),（Pa）,（Pa1）,2023/3/12,A mineral oil in cylinder has a
6、 volume of 1000cm3 at 0.1MN/m2 and a volume of 998 cm3 at 3.1MN/m2.What is its bulk modulus of elasticity?,Example 1.1,Solution:,2023/3/12,Predominant Cause:Cohesion is the cause of viscosity of liquid.Transfer of molecular momentum is the cause of viscosity of gas.,(5)Viscosity,Viscosity is an inte
7、rnal property of a fluid that offers resistance to shear deformation.It describes a fluids internal resistance to flow and may be thought as a measure of fluid friction.The resistance of a fluid to shear depends upon its cohesion(内聚力)and its rate of transfer of molecular momentum(分子动量交换).,2023/3/12,
8、Newtons law of viscosity,Figure 1.5 Deformation resulting from application of constant shear force,In Fig.1.5,a substance is filled to the space between two closely spaced parallel plates（平行板）.The lower plate is fixed,the upper plate with area A move with a constant velocity V,a force F is applied t
9、o the upper plate.,2023/3/12,Experiment shows that F is directly proportional to A and to V and is inversely proportional（反比）to thickness h.,(1.18),If the shear stress is=F/A,it can be expressed as,The ratio V/h is the angular velocity of line ab,or it is the rate of angular deformation of the fluid
10、.,2023/3/12,(1.19),The angular velocity may also be written as du/dz,so Newtons law of viscosity is,The proportionality factor(比例因子）is called the viscosity coefficient（黏性系数，黏度）.,Fluids may be classified as Newtonian or non-Newtonian.Newtonian fluid:is constant.(gases and thin liquids稀液)Non-Newtonian
11、 fluid:is not constant.(thick稠的,long-chained hydrocarbons长链碳氢化合物),2023/3/12,Dynamic viscosity and Kinematic viscosity,The dynamic viscosity（动力黏度）is also called absolute viscosity（绝对黏度）.From（1.19）,SI unit：kg/(ms)or Ns/m2 U.S.customary unit：dynes/cm2（达因秒/厘米2）cgs unit：Poise(P,泊).1P=100cP(厘泊)1P=0.1Pas(帕
12、秒),The kinematic viscosity（运动黏度）is the ratio of dynamic viscosity to density.,SI unit：m2/s U.S.customary unit：ft2/s（英尺2/秒）cgs unit：stokes(St,斯).1cm2/s=1St 1mm2/s=1cSt(厘斯),Chapter 2.FLUID STATICS 第二章 流体静力学,2023/3/12,Basic equation of hydrostatics under gravity,Gravity G(G=mg)is the only mass force ac
13、ting on the liquid,fx=0，fy=0，fz=g,Figure 2.4 A vessel containing liquid at rest,rewriting,From(2.5),2023/3/12,For the two points 1 and 2 in the static fluid,For the two points 0 and 1,Figure 2.4 A vessel containing liquid at rest,The pressure at a point in liquid at rest consists of two parts:the su
14、rface pressure,and the pressure caused by the weight of column of liquid.,2023/3/12,Physical meaning,z the position potential energy per unit weight of fluid to the base level；,p/g the pressure potential energy(压强势能)per unit weight of fluid.,Geometrical meaning,z the position height or elevation hea
15、d(位置水头),p/gthe pressure head(压力水头)per unit weight of fluid,Sum of the position head(位置水头)and pressure head is called the hydrostatic head(静水头),also known as the piezometric head(测压管水头).,2023/3/12,local atmospheric pressure(当地大气压)pa,absolute pressure(绝对压强)pabs,gauge pressure(表压,计示压强)=relative pressur
16、e,vacuum pressure(真空压强,真空度)pv,or suction pressure(吸入压强),also called negative pressure(负压强),relative pressure(相对压强)p,It is usually measured in the height of liquid column,such as millimeters of mercury(mmHg,毫米水银柱),denoted by hv.,Related Pressures,2023/3/12,Local atmospheric pressure p=pa,Complete vac
17、uum pabs=0,Absolute pressure,Vacuumpressure,Gaugepressure,Figure 2.6 absolute pressure,gauge pressure and vacuum pressure,p,Absolute pressure,2,ppa,1,ppa,O,Relationship Graph,2023/3/12,To avoid any confusion,the convention is adopted throughout this text that a pressure is in gauge pressure unless s
18、pecifically marked abs,with the exception of a gas,which is absolute pressure unit.,Attention:,2023/3/12,Differential manometermnmit(差压计),used to measure the differences in pressure for two containers or two points in a container.,Structure：,Measurement principle：,Figure 2.10 Differential manometer,
19、2,A,A,pA,1,A,B,h,h1,B,B,pB,h2,A=B=1,For two same air,A=B=0,2023/3/12,EXAMPLE 2.1,A pressure measurement apparatus without leakage and friction between piston and cylinder wall is shown in Fig.2.11.The piston diameter is d=35mm,the relative density of oil is RDoil=0.92,the relative density of mercury
20、 is RDHg=13.6,and the height is h=700 mm.If the piston has a weight of 15N,calculate the value of height difference of liquid h in the differential manometer.,1,1,pa,h,RDoil=0.92,RDHg=13.6,d,h,Figure 2.11 Pressure measurement apparatus,piston,pa,2023/3/12,The pressure on the piston under the weight,
21、From the isobaric surface 1-1 the equilibrium equation is,solving for h,Solution:,Chapter 3.FLUID FLOW CONCEPTS&BASIC EQUATIONS第三章 流体流动概念和基本方程组,2023/3/12,The space pervaded(弥漫,充满)the flowing fluid is called flow field(流场).,velocity u,acceleration a,density,pressure p,temperature T,viscosity force Fv
22、,and so on.,Motion parameters:,2023/3/12,Steady flow and unsteady flow,For steady flow(定常流),motion parameters independent of time.u=u(x,y,z)p=p(x,y,z)Steady flow may be expressed as,The motion parameters are dependent on time,the flow is unsteady flow(非定常流).,u=u(x,y,z,t),p=p(x,y,z,t),2023/3/12,A pat
23、h line(迹线,轨迹线)is the trajectory of an individual fluid particle in flow field during a period of time.Streamline(流线)is a continuous line(many different fluid particles)drawn within fluid flied at a certain instant,the direction of the velocity vector at each point is coincided with（与一致）the direction
24、 of tangent at that point in the line.,Path line and streamline,path line,Streamline,2023/3/12,Cross section,flow rate and average velocity,1.Flow section(通流面)The flow section is a section that every area element in the section is normal to mini-stream tube or streamline.The flow section is a curved
25、 surface（曲面）.If the flow section is a plane area,it is called a cross section(横截面).,2023/3/12,The amount of fluid passing through a cross section in unit interval is called flow rate or discharge.,weight flow rate,volumetric flow rate,mass flow rate,(3.7),For a total flow,2.Flow rate(流量),2023/3/12,3
26、.Average velocity(平均速度),Figure 3.6 Distribution of velocity over cross section,The velocity u takes the maximum umax on the pipe axle and the zero on the boundary as shown in Fig.3.6.,The average velocity V according to the equivalency of flow rate is called the section average velocity(截面平均速度).,Acc
27、ording to the equivalency of flow rate,VA=A udA=Q,therewith,2023/3/12,3.3.2 Control volume,Control volume(cv,控制体)is defined as an invariably hollow volume or frame fixed in space or moving with constant velocity through which the fluid flows.The boundary of control volume is called control surface(c
28、s,控制面).,For a cv:1)its shape,volume and its cs can not change with time.2)it is stationary in the coordinate system.(in this book)3)there may be the exchange of mass and energy on the cs.,2023/3/12,System vs Control Volume,2023/3/12,3.4.1 Steady flow continuity equation of 1D mini stream tube,The ne
29、t mass inflow,dM=(lu1dA12u2dA2)dt,For compressible steady flow dM=0,lu1dA1=2u2dA2,If incompressible,l=2=,u1dA1=u2dA2,The formula is the continuity equation for incompressible fluid,steady flow along with mini-stream tube.,(3.23),2023/3/12,3.4.2 Total flow continuity equation for 1D steady flow,lmV1A
30、1=2mV2A2(3.24),For incompressible fluid flow,is a constant.,Q1=Q2 or V1A1=V2A2,Making integrals at both sides of Eq.(3.23),Integrating it,The total flow continuity equation for the incompressible fluid in steady flow.,2023/3/12,3.5.2 Bernoullis equation,Eq.(3.8)is used for ideal fluid flows along a
31、streamline in steady.Bernoullis equation can be obtained with an integral along a streamline:,(m2/s2)(3.29),This is an energy equation per unit mass.,It has the dimensions(L/T)2 because mN/kg=(mkgm/s2)/kg=m2/s2.,The meanings,u2/2 the kinetic energy per unit mass(mu2/2)/m.,p/the pressure energy per u
32、nit mass.gz the potential energy per unit mass.,Eq.(3.29)shows that the total mechanical energy per unit mass of fluid remains constant at any position along the flow path.,2023/3/12,The Bernoullis equation per unit volume is,(N/m2)(3.30),Because the dimension of u2/2 is the same as that of pressure
33、,it is called dynamic pressure(动压强).,The Bernoullis equation per unit weight is,(mN/N,or,m)(3.31),For arbitrary two points 1 and 2 along a streamline,(3.32),2023/3/12,The mechanical energy per unit weight over the section in gradually varied flow is,Let hw be the energy losses per unit weight of flu
34、id from 1-1 to 2-2,the Bernoullis equation for a total flow is,Let hs be the shaft work per unit weight of fluid,the Bernoullis equation for a real system is,(3.45),(3.44),3.7.2 The Bernoullis equation for the real-fluid total flow,2023/3/12,A centrifugal water pump(离心式水泵)with a suction pipe(吸水管)is
35、shown in Fig.3.20.Pump output is Q=0.03m3/s,the diameter of suction pipe d=150mm,vacuum pressure that the pump can reach is pv/(g)=6.8 mH2O,and all head losses in the suction pipe hw=1mH2O.Determine the utmost elevation(最大提升)he from the pump shaft to the water surface on the pond.,EXAMPLE 3.6,2023/3
36、/12,Solution,1)two cross sections and the datum plane are selected.,The sections should be on the gradually varied flow,the two cross sections here are:1)the water surface 0-0 on the pond,2)the section 1-1 on the inlet of pump.Meanwhile,the section 0-0 is taken as the datum plane,z0=0.,2023/3/12,2)t
37、he parameters in the equation are determined.,The pressure p0 and p1 are expressed in relative pressure(gauge pressure).,，and,So,V0=0.,Let=1,hw0-1=1 mH2O,，and,，and,2023/3/12,3)calculation for the unknown parameter is carried out.,By substituting V0=0，p0=0，z0=0，p1=pv，z1=he，1=1 and V1=1.7 into the Ber
38、noullis equation,i.e.,The values of pv/(g)and V1 are substituted into above formula and it gives,namely,2023/3/12,In the actual flow the velocity over a plane cross section(横截平面)is not uniform,or,If the un-uniform in velocity over the section is taken into account,Eq.(3.48)may be rewritten as,is mom
39、entum correction factor(动量修正系数).=4/3 in laminar flow for a straight round tube.=1.021.05 in turbulent flow and it could be taken as 1.,or,or,(3.49),(3.50),3.8 THE LINEAR-MOMENTUM EQUATION,Chapter 4.FLUID RESISTANCE 4.流体阻力,2023/3/12,hw head losses(水头损失)comprises of friction losses and minor losses.,1
40、.Friction loss(沿程损失,摩擦水头损失)hl In the flow through a straight tube with constant cross section,the energy loss increases linearly in the direction of flow and the loss is called friction loss.,2.Minor loss(次要损失)or Local loss(局部损失)hm When the shape of flow path changes,such as section enlargement and
41、so on,it will give rise to a change in the distribution of velocity for the flow.The change results in energy loss,which is called minor loss or local loss.,4.1 REYNOLDS NUMBER,2023/3/12,Reynolds number Re is used to describe the characteristic of flow.,2023/3/12,The discharge（流量）passing through a f
42、ixed cross section is,4.2 LAMINAR BETWEEN PARALLEL PLATES,2023/3/12,It can be written as,Suppose the diameter of circular tube is d.Substituting known conditions above to the discharge equation,the discharge of circular tube is,Hagen-Poiseuille(哈根-泊肃叶)equation.,So the average velocity of laminar flo
43、w in the circular tube is,4.3 LAMINAR FLOW THROUGH CIRCULAR TUBE,2023/3/12,Fluid flows in a 3-mm-ID horizontal tube.Find the pressure drop per meter.=60 cP,RD=0.83,at Re=200.,Example 4.2,Solution:,2023/3/12,In the Fig.4.7,p is friction loss of laminar flow in the tube between sections 1-1 and 2-2.,L
44、et,the coefficient of friction loss(沿程摩擦系数，沿程损失系数),obtains,Darcy-Weisbch(达西-韦斯巴赫)Equation,4.3.3 Frictional loss for laminar flow in horizontal circular tube,It is used to calculate the frictional pressure drop for laminar flow or turbulent flow in horizontal circular tube.For the convenience of appl
45、ication,it can be written friction loss:,2023/3/12,The losses just occur locally,which called as minor energy losses(局部能量损失)or minor pressure losses,denoted by pm.It usually is expressed as the following equation.,4.6.5 Minor resistance,:the minor loss coefficient or local loss coefficient,V:the vel
46、ocity over the cross section,:the mass density,2023/3/12,It is that the friction loss is the only factor for energy losses in fluid flow.In the simple pipe problem,assuming that the fluid is incompressible,it involves six parameters,Q,L,D,hw,.In general,the length of pipe,L,kinematic viscosity of fl
47、uid,and absolute roughness of pipe,may be determined already.,So,simple pipe problem can be classified into three types:(1)solution for pressure drop hw with Q,L,D;(2)solution for discharge Q with L,D,hw;(3)solution for diameter D with Q,L,hw.,Table 4.5 Simple pipe problem,“Simple pipe problem”,2023
48、/3/12,m is the number of the frictional resistance,n is the number of the minor resistance,the subscripts i and j express the ith and the jth segment of pipeline.,The major work in simple pipe problem is to solve hW.,2023/3/12,EXAMPLE 4.6,A horizontal pipeline is used to supply land with water as sh
49、own in Fig.4.22.The component parts of pipeline are as following:three pipes of length 30m,12m,and 60m;two standard elbows A and B;and a globe valve(fully open)C.If the elevation is H=10 m and the inside diameter of clean cast iron pipe is 150mm,determine:1)the discharge in the pipe;2)the head loss
50、when Q=60 L/s.,Figure 4.22 Horizontal pipeline used to irrigate a cropland,2023/3/12,Solution,The entrance loss coefficient is 0.5,loss coefficient of elbow 0.9,and loss coefficient of the globe valve 10.Bernoullis equation is applied between section 1-1 and section 2-2,including all the losses,Chap
