华中科技大学流体力学课件Fm8解析
流体力学第八章(Fluid Mechanics)
2nd – Year Fluid Mechanics, Faculty of Engineering and Computing, Curtin University FLUID MECHANICS 230For Second-Year Chemical, Civil and Mechanical EngineeringFLUID MECHANICS LECTURE NOTESCHAPTER 8 FLOW OVER IMMERSED BODY8.1 IntroductionWe have discussed pipe flow in Chapter 6, which is internal flow because the fluid is confined within well-defined boundaries. This chapter deals with external flows, i.e. flows over bodies immersed in the fluid. Typical examples of external flows include the flow of water over a submarine (Figure 8-1a) or a fish (Figure 8-1b), the flow of air over an aircraft (Figure 8-1c).(a)(b) (c)Figure 8-1 Examples; a) submarine; b) fish; c) aircraft;The fluid forces (drag and lift) on the immersed bodies are of important considerations in practice. In this chapter, we will learn the fundamentals of drag and lift, as well as the methods for determining and optimising these forces in engineering applications.8.2 Drag and liftAs shown in Figure 8-2, the forces exerted on the surface of an aerofoil by the fluid include the pressure force (Figure 8-2a) and the viscous force (Figure 8-2b). The results of these forces are the net drag force F D and lift force F L (Figure 8-2c). Taking a small element dA on the aerofoil surface in Figure 8-2c, we can decompose the pressure force (PdA ) and viscous force (dA w τ) into x-direction force dF x and y-direction force dF y , as shown in Figure 8-2d,θτθsin )(cos )(dA PdA dF w x += θτθcos )(sin )(dA PdA dF w y +−=.Integrating dF x and dF y along the surface, we can calculate the drag and lift forces as, dA dA P dF F w x D ∫∫∫+==θτθsin cos (E8-1) ∫∫∫+−==dA dA P dF F w y L θτθcos sin . (E8-2)Equations E8-1 and E8-2 reveals that 1) both shear stress and pressure contribute to the drag and lift. In the case of drag, the former is called friction drag ; the later is called pressure drag ;2) to calculate drag and lift, sufficient knowledge are required on three aspects: the body shape as it determines the distribution of θ; the distribution of pressure P and the distribution of shear stress w τ along the surface.Practically, it is very difficult to obtain the distribution of pressure and shear stress. Therefore, E8-1 and E8-2 are generally not very useful. To make it easy for engineers, dimensionless coefficients are often used instead. These dimensionless numbers are called drag coefficient C D and lift coefficient C L , defined as A V F C D D 22ρ= (E8-3) A V F C L L 22ρ= (E8-4) where A is projected (or frontal) area, i.e. the project area of the body in the flow direction.Figure 8-2 Forces on an aerofoil [1]: a) pressure force; b) viscous force; c). resultant drag andlift; d) pressure and viscous forces on a small surface area dA in (c), plotted in an x-y coordination system.8.3 The boundary layer8.3.1 Hydraulician and hydrodynamicistFor a fundamental understanding on the cause of drag and lift, we must have a good understanding on the boundary layer concept. Let’s start with some historical background on two groups of fluid mechanicians who developed two different approaches in dealing with problems of fluid mechanics by late 19th century.One is the group of hydraulicians, focused on experiments and attempted to generalise useful design equations from experimental data. This group developed the filed of experimental hydraulics, delivering empirical solutions with little theoretical content. The other is the group of hydrodynamicists, who focused on differential equations describing flows and tried to apply them to practical problems. In order to solve these differential equations, the fluid was assumed to have zero viscosity and constant density. This group developed the filed of theoretical hydrodynamics, seeking pure theoretical solutions based on ideal-fluid flows.The ideal-fluid solutions of hydrodynamicists agreed well with the observations of flows did not involve solid surface, e.g. tides, however did not agree with observed behaviours in the problems that concerned the hydraulicians, e.g, flow over immersed bodies. The following are two typical examples.(1) Flow over a thin plateFigure 8-3 illustrates the differences between the ideal solutions by the hydrodynamicists and the experimental observations by the hydraulicians for flow over a thin plate. For an ideal-fluid flow (Figure 8-3a), the fluid is inviscid, there is no friction between the fluid and the surface of the thin plate. Therefore, the fluid will maintain its free stream velocity when it flows over the thin plate. However, for a real-fluid flow, the interaction between the viscous fluid with the surface leads to a velocity gradient near the surface of the plate although the fluid far away from the plate still maintains its free stream velocity.(2) Flow over a circular cylinderFigure 8-5 shows the patterns of ideal and real flow over a circular cylinder. In absence of viscous effects, the wall shear stress is zero therefore the streamlines are symmetrical (see Figure 8-6a). The fluid coasts from the front (point A), pasts the top (point C) and then reaches to the rear (point F) of the cylinder.However, the experimental observations by the hydraulicians are very different from the above analysis by the hydrodynamicists, as shown in Figure 8-4b. A real-fluid flow cannot coast along the cylinder surface down to point F. Instead, flow separation occurs from the surface at a point between point C and F, leading to the formation of turbulent wake in the downstream (see Figure 8-4b). The friction between the viscous fluid and the cylinder surface leads to inevitable energy loss so that the flow does not have sufficient kinetic energy to travel along the surface down to the rear of the cylinder.From the above examples, it seems that the hydrodynamicists calculated what can NOT be observed while the hydraulicians observed what can NOT be calculated [2]. By early 20th century, the hydraulicians and hydrodynamicists had completely gone to live in their own worlds. The hydraulicians continued to solve their problems by trial and error based on experiments while the hydrodynamicists kept publishing academic papers based on mathematics with little bearing on engineering problems.(a) Ideal flow over a thin plate(b) Real flow over a thin plateFigure 8-3 Ideal (a) and real (b) flow over a thin plate(a) Ideal flow a cylinder (b) Real turbulent flow over a cylinderFigure 8-4 Ideal (a) and real (b) flow over a cylinder8.3.2 The boundary layer conceptThe revolutionary thinking was finally due in 1904. The two fields of theoretical hydrodynamics and experimental hydraulics were united by a German professor, Ludwig Prandtl (1857-1953). Prandtl proposed the concept of the boundary layer , which has brought the hydrodynamics and hydraulics together and laid the foundation of modern fluid mechanics.F A CTurbulentwakeThe key concept of the boundary layer is to conceptually divide the flow into two regions: •The boundary layer, the region close to solid surfaces, the effects of viscosity are too large to be ignored. Within the boundary layer, ideal-fluid flow is unsatisfactory and a set of boundary layer equations should be used.•Free stream region, the region outside the boundary layer, the effects of viscosity is small and can be neglected. In this region, Ideal-fluid flow is satisfactory.At the edge of boundary layer, the pressures and velocities of the two regions should be matched. Prandtl arbitrarily suggested the boundary layer be considered that region in which the x component of the velocity, v, is less than 0.99 times of the free-stream velocity, V, as shown in Figure 8-5.Figure 8-5 Thickness of the boundary layerThe introduction of the boundary layer concept is revolutionary in seamlessly uniting the fields of theoretical hydrodynamics and experimental hydraulics. The boundary layer concept is commonly accepted as the foundation of modern fluid mechanics, because it has •clarified numerous unexplained phenomena;•provided a much better intellectual basis for discussing complicated flows;•become a standard idea in minds of fluid mechanicians;•brought analogous ideas in heat and mass transfer, generally with useful results. However, it should be noted that•the division of the flow filed by the boundary layer concept does NOT correspond any physically obvious boundary;•the edge of boundary layer does NOT correspond to any sudden change in the flow but rather corresponds to an arbitrary definition;•even with this simplification the calculation is still difficult, and in general only approximate mathematical solutions are possible.Let’s use the boundary layer concept to explain the patterns of flow over a thin plate and flow over a circular cylinder at various Reynolds’ numbers.(1). Flow over a thin plate (Figure 8-6) [1]At a very low Reynolds number, e.g. Re = 0.1 (Figure 8-6a), the viscous force is more important than the inertial force. The viscous effects are therefore strong and the plate affects the flow considerably in a wide range (see the grey area) in all directions. Consequently, there is an extensively wide range around the plate in which the streamlines deflected considerably. The boundary layer is very thick. Outside the boundary layer is the free stream region where viscous effects are no long important so that the ideal-flow solutions is applicable.At a moderate Reynolds number, e.g. Re = 10 (Figure 8-6b), the region in which the viscous effects are important becomes much smaller. The boundary layer is much thinner. The streamlines over the plate only deflected somewhat.At a large Reynolds number, e.g. Re = 107 (Figure 8-6c), the flow is dominated by inertial force. The viscous effects are negligible anywhere except in a thin boundary layer close to the surface of the plate and the wake region. As the boundary layer is very thin, the flow streamlines are largely unaffected except slightly deflected near and within the boundary layer.Figure 8-6 Patterns of flow over a thin plate [1] at (a) a low Reynolds number; (b) a moderate Reynolds number and (c) a large Reynolds numberFigure 8-7 Patterns of flows over a circular cylinder [1] at (a) a low Reynolds number; (b) a moderate Reynolds number and (c) a large Reynolds number2). Flow over a circular cylinder (Figure 8-7) [1]At a low Reynolds number, e.g. Re = 0.1 (Figure 8-7a), flow past a cylinder is dominated by the viscous force. The viscous effects influence a large portion of the flow field, stretching to several diameters in any direction of the cylinder. However, the flow can still coast along the surface of the cylinder slowly. Such flow is sometime called creeping flow, which has streamlines essentially symmetric about the centre of the cylinder.At a moderate Reynolds number, e.g. Re = 50 (Figure 8-7b), the inertial force becomes more important. The region ahead of the cylinder in which the viscous effects are important becomes much smaller. The viscous effects are convected downstream and the flow loses its symmetry. The flow inertia dominates so that it does not coast along the surface down to the rear of the body, resulting in the formation of flow separation bubbles behind the cylinder.At a large Reynolds number, e.g. Re = 105 (Figure 8-6c), the flow is dominated by the inertialforce. The region affected by the viscous forces is forced further downstream, leading to theformation of a very thin boundary layer on the front portion of the cylinder. The boundary lay can be laminar or turbulent, depending on Reynolds number. Due to the strong inertia force, the boundary layer separation occurs from the cylinder, leading to the formation of a turbulent wake region extending far downstream. In the free stream region outside the boundary layer and the wake region, the velocity gradient is zero and the fluid flows as if it were inviscid.8.4 Drag force and streamlined bodies8.4.1 Drag forceIn 1710, Isaac Newton (cited in [2]) dropped hollow spheres from the inside of the dome of St Paul’s Cathedral in London and measured their rate of fall. He calculated that the drag force F D on a sphere and concluded that the following equation holds 242222V D V A F D ρπρ== where A is called projected (or frontal) area – the projected area seen by a person lookingtoward the object from a direction parallel to the upstream (see Figure 8-8).Figure 8-8 Drag force on a free-fall sphereIn other words, Isaac Newton thought that 122=V A F D ρ. However, subsequent experimental investigations by many other researchers found that the above formula must be modified by a coefficient C D in the right-hand side, i.e. D D C V A F =22ρ This is essentially same as the definition of drag coefficient C D in E8-3. Normally, drag coefficient C D is not 1. We can calculate the drag force of the flow on an immersed body as (E8-5)Therefore, when calculate drag force using E8-5, we have accumulated the effects of all rest factors into a single coefficient, C D , i.e. drag coefficient. One can see that the drag coefficient C D in E8-5 plays a similar role of the friction factor f in E6-22. The key difference between E8-5 and E6-22 is that in the case of pipe flow, the geometry of pipes (with different length and diameter) varies while all spheres have the same shape.8.4.2 Drag coefficient chartAccording to E8-1 and the discussion in Section 8-2, the drag coefficient of an object will be a function of Reynolds number, the shape and the surface properties of the object. We haverties)face_prope ,shape,sur Φ(C D Re = (E8-6)Practically, finding the exact function of E8-6 is extremely difficult, if not impossible. Therefore, for engineering applications, what we need is a drag coefficient chart, similar to Moody chart used for determining friction factor in pipe flows. Figure 8-9 shows the dragcoefficient as a function of Reynolds number for a smooth sphere and a smooth cylinder.Figure 8-9 Drag coefficient of a smooth sphere and a smooth circular cylinder [1].EX8-1: An exampleWe caught two breams in two consecutive casts. If we pulled the strings at the same speed, explain why we felt it was much harder to pull the big bream?Solutions:Figure 8-10The projected area of the big bream, A bf , is a much bigger than that of the small bream A sf . As both fishes are bream, we can reasonable assume that the two fishes have similar shape and surface characteristics. As both fishes were caught in consecutive casts in the same water area and we pulled the string at the same speed V , the water flows over the two fishes have same Reynolds numbers. Therefore, the drag coefficients can be taken as same. According to E8-5, the water would have induced much higher drag force on the big bream when we pulled the string.8.4.3 Pressure dragEquation E8-1 in Section 8.2 clearly shows that the drag consists of two components, friction drag and pressure drag. For flow over an immersed circular cylinder, as shown in Figure 8-7c, the fluid friction within the boundary layer certainly leads to the friction drag. In this section, let’s have a detailed analysis on the cause of pressure drag.Let’s start with inviscid flows, as shown in Figure 8-11. In absence of viscosity, the flow will coast along the cylinder surface and the streamlines are symmetrical. Based on theoretical hydrodynamics (further details can be found in Chapter 6 of Reference [1]), the distribution of pressure and velocity along the surface of the cylinder are )sin 41(21220θρ−+=V P P (E8-7) θsin 2V V fs = (E8-8) which are plotted in Figures 8-11b and 8-11c.Figure 8-11c indicates that for an ideal fluid (0=μ and ρ is constant), the fluid velocity along the surface varies from 0=fs V at the very front and rear (stagnation point A and F) of the cylinder to the maximum of V V fs 2= at the bop (point C) and bottom of the cylinder.Figure 8-11 Ideal flow over a circular cylinder [1] (a) streamlines for the ideal flow; (b) pressure distribution of the ideal flow; (c) fluid velocity distribution on the cylinder surfaceFigure 8-12 Real flow over a circular cylinder [1] (a) boundary layer separation; (b)distribution of pressure coefficient.Similarly, as shown in Figure 8-11b, the pressure distribution is also symmetrical about thevertical midplane of the cylinder, varying from a minimum of 2023V P ρ−at the top orbottom of the cylinder to a maximum of 2021V P ρ+. The decrease in pressure in thedirection of flow along the front half of the cylinder is termed as favourable pressure gradientwhile the increase in pressure in the direction of flow along the rear half of the cylinder istermed adverse pressure gradient . In absence of viscous effects, the fluid travelling from thefront to the back of the cylinder coasts down the “pressure hill” (see Figure 8-5b) from°=0θat point A to °=90θat point C and then back up to the hill to °=180θ(from point C topoint F ). Therefore, there is only energy exchange between kinetic energy and pressureenergy without any energy loss.However, the experimental observations are very different from the above theoreticalanalysis, as shown in Figure 8-12. At a large Reynolds number, the flow forms the boundarylayer on the surface of the cylinder and cannot coast along the cylinder surface down to therear stagnation point F. The boundary layer separation occurs at point D on the rear surface, (a)(b)leading to the formation of turbulent wakes in the downstream (see Figure 8-12a). The separation of the boundary layer can be explained by the pressure distribution in Figure 12b. Due to the friction between the viscous fluid and the cylinder surface, energy loss is inevitable so that after pass through point C, the fluid does not have enough kinetic energy to climb the pressure hill up to point F which sits on the top of pressure hill therefore the boundary layer separates at point D in Figure 12a.Figure 8-12b also indicates that the location of separation therefore the width of the turbulent wake and the pressure distribution on the surface depend on the nature of the boundary layer. Compared with a laminar boundary layer, a turbulent boundary layer has more kinetic energy and momentum so that it can flow further around the cylinder, resulting in a narrower wake, less drag, corresponding to a decrease in drag coefficient from point D to E in Figure 8-10. Because of the boundary layer separation, the average pressure on the front half of the cylinder is significantly greater than that on the rear half. This leads to the development of a large pressure drag. Under turbulent conditions, the friction drag is insignificant compared with the pressure drag, as discussed blow.Figure 8-13 Two objects of significant different size that have the same drag force [1]: (a) a circular cylinder - a blunt body, C D = 1.2; (b) a streamlined strut C D = 0.12.Table 8-1 Drag coefficients of various bodies8.4.4 Streamlined bodiesIt is clear now that the drag force developed on an object immersed in turbulent fluid isdominantly pressure drag as a result of boundary layer separation. Therefore, we can optimisethe body shape of the object to minimise the boundary layer separation hence reduce pressuredrag. This requires us to design streamlined bodies .Figure 8-13 shows the significance of body streamlining. The streamlined strut has a sizemuch bigger than the circular cylinder. However, the two bodies have the same drag forces.The boundary layer separation on the streamlined strut has been postponed to the tail of thebody so that pressure drag is minimised compared with that on a circular cylinder, which is ablunt body. It is interesting to revisit the shapes of submarine, fish and aerofoil in Figures 8-1and 8-2, we would appreciate that these bodies are actually all streamlined.8.4.5 Drag coefficient for various objectsThe drag coefficient information for a wide range of objects is available in the literature [1,2].Some of this information is listed in Table 8-1.8.4.6 Drag coefficient at low Reynolds number (Re < 1)At Re < 1, the flow over an immersed body is called creeping flow , dominated by viscouseffects. The drag coefficient gives the 1/Re dependence. Table 8-2 shows the drag coefficientsfor various bodies at Re < 1.Table 8-2 Drag coefficient of various bodies at Re < 1, from Reference [1]8.6 Terminal velocityIf an object in a body of fluid is free to move subject only to gravity, or perhaps centrifugalforces in certain circumstances, then it accelerates to a particular velocity at which the tractiondeveloped and the other forces on the body balance. This velocity is called terminal velocity .Generally, the acceleration is very rapid so that the whole process can be treated as the objecttravelling at the terminal velocity.Figure 8-14a illustrates the concept of terminal velocity of a sphere settling in a fluid. Figure8-14b shows the force analysis of the sphere. When the sphere travels at the terminal velocityV t , the force is balanced. V V V V Re4.20Re6.13Re0.24Re2.22Therefore, as shown in Figure 8-14b, the gravity force F W , drag force F D and buoyancy forceF B are in balance,B D W F F F += (E8-9)We know thatobject object W gV F ρ= (E8-10)object fluid B gV F ρ= (E8-11) projected t fluid D D A V C F 22ρ= (E8-12) where6)2(3433D D V object ππ== (E8-13)42D A projected π=. (E8-14)Assuming Re < 1, for a sphere (see Table 8-2), the drag coefficient is DV C t fluid fluid D ρμ24Re 24==. (E8-15)Substituting E8-15 into E8-12, we have t fluid t fluid t fluid fluid projected t fluid D D DV D V D V A V C F πμπρρμρ342242222===. (E8-16)This is the famous Stock’s law, which is only valid when Re < 1.(a) Concept of terminal velocity (b) Force analysisFigure 8-14 Terminal velocity of a sphereSubstituting E8-11, E8-13 and E8-16 into E8-9, we have fluidfluid object t gD V μρρ18)(2−= (E8-17)Equation E8-17 is not generally given for calculations so that if not available we are requiredto go through the force analysis for its derivation.Once we have calculated the terminal velocity, one final critical step is to double check Re toensure that 0.1Re <=fluid t fluid D V μρ (E8-18)so that our assumption leading to E8-16 is indeed valid.Solutions:We can use a free-body diagram of a mineral particle, as shown in Figure 8-14b. The mineralparticle moves downward with a constant velocity V t (relative to the moving air flow)that isgoverned by a balance between weight of particle, F W , the buoyancy force of the surroundingair, F B , and the drag of air on the particle, F D . Please Note: it is acceptable that thebuoyancy force of the surrounding air, F B , is neglected.(a) We haveB D W F F F += (E8-19)63p p p p W D g gV F πρρ== (E8-20) 63p air p air B D g gV F πρρ== (E8-21)Assuming Re <1, we have the drag coefficient of an mineral particle pt air air D D V C ρμ24Re 24== (E8-22) Therefore the drag force p t air pt air air p t air D p t air D D V D V D V C D V F πμρμπρπρ3244214212222=== (E8-23) Substitute E8-20, E8-21 and E8-23 into E8-19, we have p air p air p p UD D g D g πμπρπρ36633+= Therefore, the terminal velocity of a mineral particle relative to the flowing air is air p air p t gD V μρρ18)(2−= or airp p t gD V μρ182= if buoyancy is neglected,Therefore, we haves m m Ns m s m m kg m kg gD V air p air p t /029.0)/1079.1(18)1020(/8.9)/29.1/2400(18)(25262332=×××××−=−=−−μρρCheck Reynolds number 0.10418.0/1079.11020/029.0/29.1Re 2563<<=××××==−−m Ns m s m m kg D V air p t air μρ Therefore, the assumption to use E8-22 is valid.(b) We can calculate the air travelling velocity in the vertical tube s m m s L m L D Q A Q V tube air /017.0)05.0(14.3)60min 110001min 0.2(44232=××××===π Therefore, the residence time of a mineral particle in the vertical tube is 3.16/)017.0029.0(75.0=+=+=sm m V V L t air t sReferences[1]. Munson BR, Young DF and Okiishi TH, Fundamentals of Fluid Mechanics, 4th Edition,John Wiley & Sons, Brisbane, 2002.[2]. Noel de Nevers, Fluid Mechanics for Chemical Engineers, 3rd Edition, McGraw-Hill’sChemical Engineering Series, Sydney 2005.。
流体力学第八章-39页PPT精品文档
§ 8.3 理想流体的运动微分方程
二、兰姆运动微分方程式(续)
兰姆运动微 分方程式
vtx
fz
1
p z
v2 ()
z 2
x
(
pF
v2 2
)
2(vz y
vy z
)
y
(
pF
v2 2
)
2(vx z
fy
p
f
z
(x,
fx y,
z)
y
o
x
z
§8.3 理想流体的运动微分方程
一、欧拉运动微分方程式(续)
x轴方向的受力
左面中心受力： (pp dx)dydz
x 2
右面中心受力： (pp dx)dydz
x 2
质量力：
fx
p p dx x 2
fy p
fz fx
p p dx x 2
不可压缩流体的定 常或非定常流动：
vvxvyvz 0 x y z
§8.1 微分形式的连续方程
二、其它形式的连续方程（续）
二维可压缩流体 的定常流动：
x(vx)y(vy)0
二维不可压缩流 体的定常或非定 常流动：
vx vy 0 x y
§8.2 流体微团运动的分解 有旋流动和无旋流动
2 z x
v M y v y v y yy 1 2 ( v y z v z y )z 1 2 ( v x y v y x )x 1 2 ( v y z v z y )z 1 2 ( v x y v y x )x
华中科技大学流体力学课后习题答案完整版
解： v |(1,2) =
v
2 x
+
v
2 y
|(1,2) = 30.41m / s ；
a=
a
2 x
+
a
2 y
|(1,2) =
(∂vx / ∂x ⋅ vx )2 + (∂vy / ∂x ⋅ vx + ∂vy / ∂y ⋅ vy )2 = 167.71m / s2 。
2.4 （1） ax = 35, a y = 15 ；（2）260。
直立部分： P2
=
ρg⎜⎛ h ⎝
+
h ⎟⎞ ⋅ hB 2⎠
=
3 2
ρgh 2 B
方向向左；作用点距离水平面为
yD
=
3 2
h+
Bh3 12 3h 2 ⋅ Bh
=
14 h 9
⇒ L2 = 2h −14h 9 = 4h 9 M 2 = P2 ⋅ L2 = 2ρgh3 B 3
于是关闭闸门所需的力 P 由力矩平衡方程
H2
− h2
设此合力的作用点距底部 x 处，则
( ) R ⋅ x = P1 ⋅ H 3 − P2 ⋅ h 3 = ρgB H 3 − h3 6
将 H = 7.5m
⇒
x
=
H
2 + Hh + h2
3(H + h)
h = 3m B = 5m 代入得 R = 1160KN
x = 2.79m
1.29 解：闸门自动开启，此时压力中心 D 应与 O 点重合；水位超过 H，则压力中心 D 高
解：（1） ax |(2,1) = (∂vx / ∂x ⋅ vx + ∂vx / ∂y ⋅ v y ) |(2,1) = 35 ，
华中科技大学 流体力学第八章_6
8-18，8-20
8.9 自由淹没射流
射流 ━ 由孔口、喷嘴喷射到另一流体区域中的运动流体； 自由淹没射流 ━ 射流喷入同一种无界的环境流体。 一般不考虑重力影响，自由射流不受固体壁面的作 用力，在其内部和边界上压强相等，等于环境流体压强。 射流一般都是湍流。
射流存在于许多工程技术领域中，如：火箭、 喷气式飞机、蒸汽泵、汽轮机、锅炉、燃烧室、化 工混合设备、自动控制射流元件、射流切割设备、 水力采掘设备、消防设备、通风设备、空调、排热 和排气设备，等等。
2 um
0
2 y2 2 exp 2 dy u0 b0理后成为
2 2 um b 2u0 b0
um b0 1.596 b u0
u0 和 b0 都是常数，该式给出射流主体段中任意 截面的半宽 b 与轴线速度 um 之间的关系。
平面射流
1 um x
轴对称射流
1 um x
如果矩形喷口的宽度与圆形喷口的直径相同，在 喷射初速度相同的情况下，矩形喷口喷射得较远。
例 圆截面送风口半径 R0 0.2 m ，送风量 Q0 2 m3/s ，
试求 （1）距送风口 x x0 = 5 m处喷射气流半径R； （2）此处最大风速um。 解 出口速度
第8章完 ！
Q0 2 u0 m/s 15.9 m/s 2 2 R0 0.2
（1）距送风口 x x0 = 5 m 处射流半径
R Cx R0 0.114 x x0 0.2 0.114 5 m 0.77 m
r
R 0.114 x
A0 2R0
O
R(x)
xC x0 3.634R0
流体力学课件PPT课件
注意：恒定流中流线与迹线重合
第27页/共90页
四、流管、流束、元流、总流、过流断面
1.流管
在流场中通过任意不与流线重合的封闭曲线上各 点作流线而构成的管状面。
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2.流束
流管内所有流线的总和。流束可大可小，视流管 封闭曲线而定。
•元流：流管封闭曲线无限小，故元流又称微元流束。 •总流：流管封闭曲线取在流场边界上，总流即为许
x
y方向:
my
(uy ) dxdydz
y
z方向:
mz
(uz ) dxdydz
z
据质量守恒定律：
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单位时间内流进、流出控制体的流体质量差之总和
等于控制体内流体因密度发生变化所引起的质量增
量 即
mx
my
mz
t
dxdydz
将 mx、my、mz 代入上式，化简得：
(ux ) (u y ) (uz ) 0
第54页/共90页
1.伯努利方程的物理意义
• z mgz : 单位重量流体所具有的位能。 mg
•
p
mg
p
/
mg
:
单位重量流体所具有的压能。
•z p :
单位重量流体所具有的势能。
•
u2 2g
1 2
mu
2
/
mg
:
单位重量流体所具有的动能。
第55页/共90页
• z p u2 : 单位重量流体所具有的机械能。
第8页/共90页
§3-1 描述流体运动的方法
一、拉格朗日方法
1.方法概要
着眼于流体各质点的运动情况，研究各质点 的运动历程，并通过综合所有被研究流体质点的 运动情况来获得整个流体运动的规律。
流体力学基础讲解PPT课件
05
流体流动的湍流与噪声
湍流的定义与特性
湍流定义
湍流是一种高度复杂的三维非稳态、带旋转的不规则流动。 在湍流中，流体的各种物理参数，如速度、压力、温度等都 随时间与空间发生随机的变化。
湍流特性
湍流具有随机性、不规则性、非线性和非稳定性等特性。在 湍流中，流体的速度、方向和压力等都随时间和空间发生变 化，形成复杂的涡旋结构。
环境流体流动与环境保护
要点一
环境流体流动
环境中的流体流动对环境保护具有重要影响。例如，大气 中的气流会影响污染物的扩散和迁移，水流会影响水体中 的污染物迁移和沉积等。
要点二
环境保护
通过对环境中的流体流动进行研究和模拟，可以更好地了 解污染物扩散和迁移规律，为环境保护提供科学依据。同 时，通过合理规划和设计流体流动系统，可以有效降低污 染物对环境的影响，保护生态环境。
04
流体流动的能量转换
能量的定义与分类
总结词
能量是物体做功的能力，可以分为机械能、热能、电能等。在流体力学中，主要关注的是机械能中的 动能和势能。
详细描述
能量是物体做功的能力，它有多种表现形式，如机械能、热能、电能等。在流体力学中，我们主要关 注的是机械能，它包括动能和势能两种形式。动能是流体运动所具有的能量，与流体的速度和质量有 关；势能则是由于流体所处位置而具有的能量。
流体流动噪声
流体流动过程中产生的噪声主要包括 机械噪声和流体动力噪声。机械噪声 主要由机械振动和摩擦引起，而流体 动力噪声主要由湍流和流体动力振动 引起。
噪声控制
为了减小流体流动产生的噪声，研究 者们提出了各种噪声控制方法，如改 变管道结构、添加消音器和改变流体 动力特性等。这些方法可以有效降低 流体流动产生的噪声。
华科研究生之有限元课件：FEM-第8讲
微元体温度升 高所需的热量
三个方向传入微 元体的净热量
微元体内热源 产生的热量
ρ ——物体密度 c ——比热，单位质量物体温度升高
一度所需的热量 kx ,ky , kz —— 热传导系数
有限元分析及应用
胡于进
8-1 温度场问题的边界条件
•
整理得：
cρ
∂T ∂t
−
∂ ∂x
(kx
∂T ∂x
)−
∂ ∂y
∂J
e n
∂Ti
⎪⎪ ⎨ ⎪
∂J
e n
∂Tj
⎪ ⎪
∂J
e n
⎫ ⎪ ⎪ ⎪⎪ ⎬ ⎪ ⎪ ⎪
=
k 4A
⎡⎢bi2 ⎢ ⎣⎢
+
ci2
⎪⎩∂Tm ⎪⎭
bibj + cic j b2j + c2j
bibm b j bm
+ +
cicm c jcm
⎤ ⎥ ⎥
⎨⎧⎪TTij
⎫ ⎪ ⎬
=
[H
]e
{T }e
=
0
bm2 + cm2 ⎦⎥ ⎪⎩Tm ⎭⎪
h22
" "
h2n "
hnn
⎥ ⎥ ⎥ ⎥ ⎦
⎪⎪⎨⎪"T2 ⎪⎪⎬⎪ ⎩⎪Tn ⎭⎪
=
⎪⎪⎨⎪"p2 ⎩⎪ pn
⎪⎪ ⎬ ⎪ ⎭⎪
• 对于其他带热源的稳态温度场或三维温度场计算其方 法相似。
有限元分析及应用
[H ]{T} = { p}
胡于进
8-3 热变形的计算
• 当弹性体的温度改变时，体内各部分将随温度变化而产生变 形，这种变形常称为热变形。考虑到弹性体实际工作中都受
华科物院 第7章 流体力学 第1讲 PPT课件
证明第二个特征 隔离法
px y
在流体内部取一直角三
棱柱，边长为 x, y, n
厚度 为z，重力沿y 方向。
p
n
x
g
py
根据牛顿第二定律
pxyz pnz sin max pyxz pnz cos mg may
当流体静止时， ax ay 0
t x0 , y0 ,z0
v g t x0 , y0 ,z0
w h
ay
2 f t 2
2g t 2
x0 , y0 ,z0 x0 , y0 ,z0
az
2h t 2
x0 , y0 ,z0
特点： 牛顿运动定律适用 考察某一确定的流体质元
Tn 可分解为 Tnn和 Ttn
平行于法向 平行于平面
法向应力或 正应力
切应力
Tn Tn
z f
M
nˆ s
nx , ny , nz
y
x
Tn不仅与M点的位置有关，而且与截面的方向 nˆ有关
张量
Tnn
lim
s0
f n sn
Ttn
lim
s0
ft sn
应力与 s 的取向有关
或者：
1s10v1 2s20v2
m t
s2
v2
2
如果所取截面 s1 ， s2 不与流速垂直，其法线与速
度分别成 1， 2 角，则计算投影
1s1
v1
2
s2
v2
m t
流体力学第八章量纲分析和相似原理
直接实验难于进行；对于那些尚未建造的设备，如要设计一座新的水坝、
建造一艘新型舰船，则根本谈不上用实验的方法探索其规律性；直接实
验的方法不适用于大型设备的破坏性试验，如水坝、大型容器等的爆破
试验；此外，直接实验方法常常只能得出个别量之间的规律性关系，难
于抓住现象的全部本质。
第1页
退出 返回
精品ppt课件
位系统，例如在国际单位制中选用的kg，m，s，K为一种基本单位系统。
其余物理量的单位均是导出单位。
（二）量纲
用基本单位系统来表示物理量单位的式子称为该物理量的量纲，用[ ]或
可用该物理l]
或L，质量的量纲为[m]或M，温度的量纲为[T]或，速度的量纲为[l][t]-1
表8.1给出了流体力学一般问题中所涉及的各种物理量的量纲。
（三）有量纲量和无量纲量
具有单位的物理量称为有量纲量，其大小与选择的单位系统有关；没 有单位的物理量称为无量纲量，其大小与选择的单位系统无关。角度在物 理学中是以弧度表示的，平面角定义为对应的弧长除以曲率半径，立体角 定义为对应的曲面面积除以曲率半径的平方，都是没有单位的，所以角度 是无量纲量。
第2页
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精品ppt课件
4
第八章 量纲分析和相似原理
第一节 量纲分析和定理
一、物理量的单位和量纲
（一）单位及单位系统
度量物理量要有单位，如时间t的单位有s，min，hr…等，长度l的单位
有mm，cm，m…等。单位分为基本单位和导出单位。在一般流体力学问
题中时间、长度、质量和温度的单位为基本单位，它们构成一个基本单
（8.2）
式中，为无量纲量，a为量纲不独立的物理量，a1，a2，，ak为量纲
流体力学课件 ppt
流体阻力计算
利用流体动力学方程，可以计算 流体在管道中流动时的阻力，为 管道设计提供依据。
管道优化设计
通过分析流体动力学方程，可以 对管道设计进行优化，提高流体 输送效率，减少能量损失。
流体动力学方程在流体机械中的应用
泵和压缩机性能分析
流体动力学方程用于分析泵和压缩机的性能 ，预测其流量、扬程、功率等参数，为机械 设计和优化提供依据。
适用于不可压缩的流体。
方程意义
描述了流体压强与密度、重力加速度和深度之间的 关系。
Part
03
流体动力学基础
流体运动的基本概念
01
02
03
流体
流体是气体和液体的总称 ，具有流动性和不可压缩 性。
流场
流场是指流体在其中运动 的区域，可以用空间坐标 和时间描述。
流线
流线是表示流体运动方向 的曲线，在同一时间内， 流线上各点的速度矢量相 等。
能量损失的形式
流体流动的能量损失可以分为沿程损失和局部损失两种形式。沿程损失是指流体在流动过程中克服摩擦阻力而损 失的能量，局部损失是指流体在通过管道或槽道的局部障碍物时损失的能量。
Part
05
流体动力学方程的应用
流体动力学方程在管道流动中的应用
稳态流动和非稳态
流动
流体动力学方程在管道流动中可 用于描述稳态流动和非稳态流动 ，包括流速、压力、密度等参数 的变化规律。
变化的流动。
流体动力学基本方程
1 2
质量守恒方程
表示流体质量随时间变化的规律，即质量守恒原 理。
动量守恒方程
表示流体动量随时间变化的规律，即牛顿第二定 律。
3
能量守恒方程
表示流体能量随时间变化的规律，即热力学第一 定律。
华中科技大学流体力学电子档第1章 (打印A4)
工程流体力学讲稿华中科技大学土木工程与力学学院力学系陈应华E-mail第一章绪论§流体与流体力学1.流体的定义：定义：凡不能象固体一样保持其一定形状，并容易流动的物质称为流体。
流体包括液体和气体。
液体的特点：①.液体有一定的容积。
②.在容器中的液体可形成一定的自由表面。
③.液体不容易压缩。
④.没有一定的形状，容易流动。
气体的特点：①.气体没有一定的容积。
②.在容器中的气体不存在自由表面。
③.气体极易压缩。
④.没有一定的形状，容易流动。
液体与气体的共同特点：没有一定的形状，容易流动。
容易流动：流体在任何微小的剪力或拉力的作用下，它们都会发生连续变形（即流动）。
2.流体力学的发展简史：古典流体力学+ 实验水力学→（现代）流体力学（现代）流体力学：理论流体力学工程流体力学（水力学）空气动力学计算流体力学环境流体力学多相流流体力学等等3.流体力学的研究方法：流体力学是研究流体平衡和机械运动的力学规律及其工程应用的一门力学学科。
流体力学的研究方法主要有：理论分析、实验研究和数值计算等。
§连续介质模型流体质点：微观上充分大，宏观上充分小的流体分子团。
比如1cm3的标态水（1atm，20˚C水温）中约含有×1022个水分子。
10-12cm3的标态水中约含有×1010个水分子。
连续介质模型：认为流体是由无任何空隙的流体质点所组成的连续体。
流体的密度、温度等物理量连续分布。
连续介质模型是欧拉在1753年提出的假说。
有了这个模型，我们就可以采用连续函数这一强有力的数学工具来分析流体的流动规律。
连续介质模型的适用范围：常温常压下的气体和液体。
§ 流体的密度及粘性一.流体的密度：1.密度的定义：流体具有维持它原有运动状态的特性，这种特性称为惯性。
表征惯性的物理量是质量。
质量愈大，则惯性愈大。
流体的密度(ρ)： VM ρV V ∆∆=∆→∆'lim ΔV ′可理解为：微观上足够大，宏观上足够小的流体体积。
《流体力学》PPT课件
h
3
流体力学的基础理论由三部分组成: 一是流体处于平衡状态时，各种作用在流体上的力之间关系
的理论，称为流体静力学； 二是流体处于流动状态时，作用在流体上的力和流动之间关
系的理论，称为流体动力学； 三是气体处于高速流动状态时，气体的运动规律的理论，称
为气体动力学。 工程流体力学的研究范畴是将流体流动作为宏观机械运动进
温度 t （℃）
20 20 20 20 20 20 20 20 20 20 20 20 -257 -195 20
密度
（ kg/m3） 998
1026 1149
789 895 1588 1335 1258 678 808 850-958 918
72 1206 13555
相对密度 d
1.00 1.03 1.15 0.79 0.90 1.59 1.34 1.26 0.68 0.81 0.85-0.93 0.92 0.072 1.21 13.58
动 力 黏 度 104
（ P a·s） 10.1 10.6 — 11.6 6.5 9.7 —
14900 2.9
19.2 72 —
0.21 2.8
15.6
2021/1/10
h
14
表1-2
在标准大气压和20℃常用气体性质
气体
空
气
二氧化碳
一氧化碳
氦
氢
密度
（ kg/m3） 1.205 1.84 1.16
h
1
第一节 流体力学的研究对象和任务
目
第二节 流体的主要物理性质
录
第三节 流体的静压强及其分布规律
第四节 流体运动的基本知识
第五节 流动阻力和水头损失
返回
流体力学基础讲解132页PPT
流体力学基础讲解
26、机遇对于有准备的头脑有特别的 亲和力 。 27、自信是人格的核心。
28、目标的坚定是性格中最必要的力 量泉源 之一， 也是成 功的利 器之一 。没有 它，天 才也会 在矛盾 无定的 迷径中 ，徒劳 无功。- -查士 德斐尔 爵士。 29、困难就是机遇。--温斯顿．丘吉 尔。 30、我奋斗，所以我快乐。--格林斯 潘。
71、既然我已经踏上这条道路，那么，任何东西都不应妨碍我沿着这条路走下去。——康德 72、家庭成为快乐的种子在外也不致成为障碍物但在旅行之际却是夜间的伴侣。——西塞罗 73、坚持意志伟大的事业需要始终不渝的精神。——伏尔泰 74、路漫漫其修道远，吾将上下而求索。——屈原 75、内外相应，言行相称。——韩非
谢谢你的阅读
❖ 知识就是财富 ❖ 丰富你的人生
《流体力学》课件
流体力学的应用领域
总结词
流体力学的应用领域与实例
详细描述
流体力学在日常生活、工程技术和科学研究中有广学、石油和天然气工业中的流体输送等。
流体力学的发展历程
总结词
流体力学的发展历程与重要事件
详细描述
流体力学的发展经历了多个阶段，从 早期的水力学研究到近代的流体动力 学和计算流体力学的兴起。历史上， 牛顿、伯努利等科学家对流体力学的 发展做出了重要贡献。
损失计算
根据流体流动的阻力和能量损失，计算流体流动的总损失。
流体流动阻力和能量损失的减小措施
优化管道设计
采用流线型设计，减少流体与 管壁的摩擦。
合理配置局部障碍物
减少不必要的弯头、阀门等， 或优化其设计以减小局部阻力 。
选择合适的管材
选用内壁光滑、摩擦系数小的 管材。
提高流体流速
适当提高流体的流速，可以减 小沿程损失和局部损失。
流体动力学基本方程
连续性方程
表示质量守恒的方程，即单位时间内流出的质量等于单位 时间内流入的质量。
01
动量方程
表示动量守恒的方程，即单位时间内流 出的动量等于单位时间内流入的动量。
02
03
能量方程
表示能量守恒的方程，即单位时间内 流出的能量等于单位时间内流入的能 量。
流体动力学应用实例
航空航天
飞机、火箭、卫星等的设计与制造需要应用 流体动力学知识。
流动方程
描述非牛顿流体的流动规律，包括连续性方程 、动量方程等。
热力学方程
描述非牛顿流体在流动过程中的热力学状态变化。
非牛顿流体的应用实例
食品工业
01
非牛顿流体在食品工业中广泛应用于番茄酱、巧克力、奶昔等
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则正的切应力逆坐标轴方向；
8.1 黏性流体中的应力
流体内一点的应力有九个分量
pxx pxy pxz
P
p yx pzx
p yy pzy
p yz pzz
其中
应力张量
pij p ji 切应力互等定律
由微元体的力矩平衡可证切应力的对称性
8.1 黏性流体中的应力
二、广义牛顿内摩擦定律
定解条件 1、非定常流动的初始场 2、边界条件
在特殊条件下可得到N-S方程的解析解 例如:两平行平板间的定常层流流动
第8章 黏性不可压缩流体的流动
§8.3 N-S方程的解析解 一、斜平面上液膜的定常流动
§4-9 缝隙流动 §4-10 边界层流动、 边界层分离及物体阻力
边界条件
U
u 0, y 0
常微分方程
f 2 f 0
边界条件
f (0) 1 f () 0
用相似性变量
y 2 t
u f
U
相似性解
u 1 2 e2 d 1 erf
U
0
8.3 N-S方程的解析解 例：两平行平板间的定常层流流动
y
U
h o
u（y）
x
h
由于板 压的 强运梯动度产和生压的强流梯动度产生的流动 均质不可压缩,不计质量力
vr=vz =0, v= u(r)
fr = f= fz=0
/= 0
r、方向运动方程
边界条件
u=0，r=R2 外壁面无滑移
u=R1，r=R1
内壁面同速
u2 r
1
dp dr
d2u dr2
1 r
du dr
u r2
0
u
R12
R22 R12
R22
r2 r
8.3 N-S方程的解析解 三、无限大平板在自身平面内启动所带动的流体运动
vu仅uy 是wyuz的函fx数
1
p x
(
2u x2
2u y 2
2u z 2
)
v t
u
v x
v
v y
w v z
fy
1
p y
(
2v x2
2v y 2
2v z 2
)
w u w v w w w t x y z
fz
1
p z
(
2w x2
2w y 2
w z2 )
8.3 N-S方程的解析解 需要求解的方程组成为
意义：1. 建立应力与变形速度的关系
2. 测量速度比测量应力方便
例如: xoy平面内的切应力与角变形速度关系
pxy 2 xy
即有
xy
1 2
( v x
u ) y
角变形速度
pxy
pyx
( v
x
u ) y
8.1 黏性流体中的应力
二、广义牛顿内摩擦定律（流体的本构方程）
牛顿流体正应力
pxx
p
2 3
( u
§8.2 不可压缩黏性流体运动的基本方程
一、Navier-Stokes (N-S)方程
1. 黏性流体微团受力分析
2. 应力形式的运动方程
3. N-S方程
pyy
p y y y
dy 2
8.2 不可压缩黏性流体运动的基本方程
pxy
pyx
( v x
u ) y
2. 应力形式的运动方程
pxx
p
2 3
( u x
第八章 黏性不可压缩流体的流动
8.1 黏性流体中的应力 8.2 不可压缩黏性流体运动的基本方程 8.3 N-S方程的解析解 8.4 边界层的基本概念及基本方程 8.5 平板层流边界层的相似性解 8.6 边界层动量积分方程 8.7 湍流边界层与混合边界层 8.8 边界层分离及物体阻力 8.9 自由淹没射流
v y
w) z
2
u x
du dt
f x
pxx x
pyx y
pzx z
dv dt
f y
pxy x
pyy y
pzy z
dw dt
f z
pxz x
pyz y
pzz z
代入本构关系式后得到运动方程
8.2 不可压缩黏性流体运动的基本方程 3. N-S方程
不可压缩粘性流体的运动方程
du
dt
fx
1
p x
(2u
壁面流体无滑移
二阶偏微分方程 需要两个边界条件
uuU0,, yyhh
y
忽液略体液上面表摩面擦与力板同速
8.3 N-S方程的解析解
习题8-5：解N-S方程求平板间的速度分布
由流动的特性 u 0, w 0, 0, f g
z
t
充分发展流动 二u元、0定，常由连续性方程 v 0
x
u t
u
u x
8.3 N-S方程的解析解
解N-S方程求平板间的速度分布
由流动的特性 u 0, w 0, 0, f 0
z
t
充分发展流动 u二，维0由、连定续常性、方不程计质量力v 0
x
u t
u
u x
vu仅uy 是wyuz的函fx数
1
p x
(
2u x2
2u y 2
2u z 2
)
v t
u
v x
v
v y
p= C v= w= 0, u=u(y, t) /x= /z= 0
x方向运动方程
定解条件
u t
2u y2
t 0, y 0, u 0 t 0, y 0, u U y , u 0
量纲分析 u/U= f (, y, t)
8.3 N-S方程的解析解 三、无限大平板在自身平面内启动所带动的流体运动
第8章 黏性不可压缩流体的流动
§ 8.1 黏性流体中的应力
一、黏性流体中的应力
二、广义牛顿内摩擦定律 应力正方向的规定:
pzz
正的正应应力力沿的作符用号面p外ij （法或向；ij ）
ppxzzx pxx
pzy pyz
pxy pyx
若作用面外i 法应向力沿作坐用面标方轴向方向,
p则yy 正的切j 应应力力沿方坐向标轴方向；
2u
y
2
g
sin
1
p x
p
y
g
cos
p g
cos
y
f
(x)
p
z
0
边界条件
u 0, y 0; u U, y h
已知u 仅是 y 的函数，而 p/ x 仅是x 的函数
2u y2
1
( p x
g sin )
C
u
C 2
y2
C1 y
C2
8.3 N-S方程的解析解 二、无限长同心圆柱面之间的定常流动。
x 2
2u y 2
2u ) z 2
dv
dt
fy
1
p y
(
2v x 2
2v y 2
2v z2 )
dw
dt
fz
1
p z
(
2w x 2
2w y 2
2w z2 )
连续性方程
u v w 0 x y z
8.2 不可压缩黏性流体运动的基本方程 二、求解N-S方程的定解条件
求解N-S方程的途径 1、精确解 2、近似解 3、数值解
x
v y
w) z
2
u x
p yy
p
2 3
( u
x
v y
w) z
2
v y
pzz
p
2 3
( u
x
v y
w) z
2
w z
其中
p
1 3(pxxFra bibliotekp yy
pzz )
不可压缩流体的正应力?
牛顿流体切应力
pxy
pyx
( v
x
u ) y
pzx
pxy
( u
z
w) x
pyz
pzy
( w
y
v ) z
第8章 黏性不可压缩流体的流动