流体力学与传热Heterogeneous-Flow-and-Separation.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,3.Heterogeneous Flow and Separation,3.1 Flow Past Immersed Objects,Definition of Drag Coefficient for Flow Past Immersed Objects,1.Introduction and types of drag,The flow of fluids outside immersed bodies appears in many chemical engineering applications and other processing applications.,For example settling,drying and filtration,and so on.,2.Drag coefficient,Correlations of the geometry and flow characteristics for solid objects suspended in fluid are similar in concept and form to the friction factor-Reynolds number correlation given for flow inside conduits.,In flow through pipes,the friction factor was defined as the ratio of the drag force per unit area to the product of fluid density and velocity head.,The Reynolds number for a particle in a fluid is defined as,From dimensional analysis,the drag coefficient of a smooth solid in an incompressible fluid depends upon a Reynolds number and the necessary shape ratios.For a given shape,3.1-2,Drag coefficients of typical shapes,For each particular shape of object and orientation of the object with respect to the direction of flow,a different relation of,C,D,versus Re exists.,Correlations of drag coefficient versus Reynolds number are shown in figure.,Combining Eqs.(3.1-1)and(3.1-3)and solving for,C,D,the drag coefficient predicted by Stokes law is,3.1-4,The variation of,C,D,with Re is quite complicated because of the interaction of the factors that control skin drag and form drag.,For a sphere,as the Reynolds number is increased beyond the Stokes law range,separation occurs and a wake is formed.,In the region of Re about 110,3,to 210,5,the drag coefficient is approximately constant for each shape and,C,D,=0.44 for a sphere.,In the theoretical approach used,the packed column is regarded as a bundle of crooked tubes of varying cross-sectional area.,The theory developed in Chapter 1 for single straight tubes is used to develop the results for the bundle of crooked tubes.,2.Laminar flow in packed beds,Certain geometric relations for particles in packed beds are used in the derivations for flow.The void fraction,in a packed bed is defined as,where,a,is the ratio of total surface area in the bed to total volume of bed(void volume plus particle volume),Since(1-,)is the volume fraction of particles in the bed,The average interstitial velocity in the bed is,u,and is related to the superficial velocity,u,based on the cross section of the empty container by,3.1-9,To determine the equivalent channel diameter,D,e,the surface area for,n,parallel channels of length,L,is set equal to the,surface-volume ratio,times the,particle volume,S,0,L,(1-,),.,3.1-6,where,S,0,is the cross-sectional area of the bed,The void volume in the bed is the same as the total volume of the,n,channels,3.1-7,Combining Eqs.(3.1-6)and (3.1-7)gives an equation for,D,e,3.1-8,For flow at very low Reynolds numbers,the pressure drop should vary with the first power of the velocity and inversely with the square of the channel size,in accordance with the Hagen-Poiseulli equation for laminar flow in straight tubes.,or,3.1-11,The true,L,is larger because of the tortuous path.Experimental data give an empirical constant of 150 for 72,Equation(,3.1-12,)is called the Blake-Kozeny equation for laminar flow,void fractions less than 0.5,effective particle diameter,D,p,and,R,e,p,1000,the drag coefficient is approximately constant at 0.40 to 0.45,and lets,(,3.2-19,),so,the equation is,C,D,=0.44,Equation,(,3.2-19,),is Newton,law and applies only for fairly large particles falling in gases or low-viscosity fluids.,Criterion for settling regime,To identify the range in which the motion of the particle lies,the velocity term is eliminated from Reynolds number by substituting,u,t,from,Eq.(,3.2-16,),to give,for the Stokes law range,（,3.2-21,）,Re=,K,3,/18,.Re1.0,to provide a convenient,criterion,K,let,（,3.2-22,）,Then,from Eq(,3.2-21,),Re=,K,3,/18,.Setting Re=1.0 and solving gives,K,=2.6,.,If the size of the particle is known,K,can be calculated from Eq(,3.2-22,).,If,K,so calculated is less than 2.6,Stokes law applies.,Substitution for,u,t,from,Eq.(,3.2-19,),shows that for the Newtons law range,Re=1.75,K,1.5,.,In the range between Stokes law and Newtons law(,2.6K68.9,),the terminal velocity is calculated from Eq(,3.2-13,)using a value of,C,D,found by trial from Fig,.,Setting this equal to 1000 and solving gives,K,=68.9.,Thus if,K,is greater than 68.9,Newtons law applies.,problem1,Settling of a spherical particle in a air is followed Stokes law,if the temperature changes from 25 to 50,the terminal velocity will();,if settling in liquid,the terminal velocity will(),The terminal velocity is the velocity that the acceleration that a particle moves through the fluid approaches to(),problem2,A single spherical particle settling freely in the fluid and it is laminar flow,when the particle diameter increases,the terminal velocity,u,will,;when the viscosity of fluid increases,u,will,;if the fluid is a gas,what happens to,u,if the temperature increase?,As shown by equations,(,3.2-16,)and,(,3.2-19,),the terminal velocity,u,t,varies with the square of diameter of particle in the()range,whereas in the()range it varies with 0.5 power of the diameter of particle,For a given packed bed,Blake-,Kozeny,equation indicates that the flow is()to the pressure drop and()proportional to the fluid viscosity.,1,、试计算直径为,30m,的球形石英颗粒（其密度为,2650kg/m,3,），在,20,水中和,20,常压空气中的自由沉降速度。,To calculate the terminal velocity of a spherical quartz particle,30 m in diameter and 2650kg/m,3,in density,settling in the water and the air at the temperature of 20,respectively.,Solution:,d,=30m,、,s,=2650kg/m,3,（,1,）,=1.0110,-3,Pas and,=998kg/m,3,for water at t=20,check,It is followed stokes,low,ut,=8.0210-4m/s,（,2,）,=1.8110,-5,Pas,=1.21kg/m,3,for,the air at t=20,Assuming that the type of flowing is followed stokes law,check,u,t,=7.18,10,-2,m/s,。,