Chapter-3 1 Fluid in Motion

合集下载

流体动力学基本方程

4
τ 21 = c21kl
∂ul ∂u ∂u = c2121 1 = µ 1 ∂xk ∂x2 ∂x2
′ = c21 ′ kl τ 21
′ ∂ul′ ∂u1 ∂u ′ ′ = c2121 =µ 1 ′ ′ ′ ∂xk ∂x2 ∂x2
′ x1 x2
x1
′ x2
cijkl 是四阶张量，考察变换
′ = β im β jnτ mn = β im β jn cmnpq τ ij ∂uq ∂x p = β im β jn cmnpq β kp β lq ∂ul′ ′ ∂xk
——能量方程
二、动能方程
G G G G G dV G G G dV 将动量方程 ρ = ρF + ∇ ⋅ P 两边同时点积 V 得： ρV ⋅ = ρF ⋅ V + V ⋅ (∇ ⋅ P) dt dt G G G G dV 1 d (V ⋅ V ) 1 dV 2 而V ⋅ ，故有动能定理 = = dt 2 dt 2 dt
。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。 §4.本构方程数学预备：二阶张量的坐标变换记 ∇V = E ，将坐标系旋转，从原坐标系 o-xyz 到旋转后的坐标系 o-x′y′z ′ ，二阶张量 E 的张量元满足变换：
Chapter 3
流体动力学基本方程
§1 质量连续性方程（质量守恒方程）一．体系和控制体。体系（物质体）：流体团无论运动到哪，如何变形，总由同一批流体质点组成。控制体：流场中一个确定的子空间，大小、形状、位置都固定。有流体质点不停出入。二．通量的概念和 Reynolds 输运方程质量通量：单位时间内穿过曲面 s 的质量

漆安慎力学

output rces
10-7
Archimedes' Principle
Archimedes' Principle When a body is partially or wholly immersed in a fluid, a buoyant force from the surrounding fluid acts on the body. The force is directed upward and has a magnitude equal to the weight of the fluid that has been displaced by the body. Apparent Weight in a Fluid
10-4
Fluids at Rest
The pressure increases with depth in water. The pressure decreases with altitude in atmosphere. Set up a vertical y axis in a tank of water with its origin at the surface. Consider an imaginary column of water. y1 and y2 are the depths below the surface of the upper and lower column fases, respectively. Three forces act on the column: F1 acts at the top of the column;
m V
( uniform density )
(10-2)

工程流体力学03第三章

For only one inlet and one outlet According to continuity
mout min m
d (mV ) s F m(Vout Vin ) dt
2－out, 1－ in

Fx m(V 2 x V 1x)
y
Fy m(V 2 y V 1 y )
3.2 Basic Physical Laws of Fluid Mechanics
All the laws of mechanics are written for a system, which state what happens when there is an interaction between the system and it’s surroundings. If m is the mass of the system Conservation of mass(质量守恒) Newton’s second law Angular momentum First law of thermodynamic
1-D in & out steady RTT
mV
dmV V dm
(linear momentum)
momentum perunit mass
flux
d (mV ) s ( i AiV iV i )out ( i AiV iV i)in (mi V i )out (mi V i )in i i dt i i
s
t t+dt
t
t+dt
: any property of fluid (m, mV , H , E)
d dm

chapter 3 Basics of fluid flow(revision)

大连理工大学Dalian University of Technology Fluid mechanics Chapter 3 Basics of Fluid Flow2.1 Lagrangian and Eulerian descriptions of fluid motionLagrangian descriptionLagrangian description of fluid flow tracks the position and velocity of individual particles.The path of a fluid particle is given by the vector r (t), and can be expressed in terms Cartesian coordinates as()()()()r t x t i y t j z t k=++Velocity field:(,,,)(,,,)(,,,)x a b c t u t y a b c t v t z a b c t w t ∂⎧=⎪∂⎪∂⎪=⎨∂⎪∂⎪=⎪∂⎩acceleration field :222222(,,,)(,,,)(,,,)x a b c t u t y a b c t v t z a b c t w t ⎧∂=⎪∂⎪∂⎪=⎨∂⎪⎪∂=⎪∂⎩大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow2.1 Lagrangian and Eulerian Descriptions offluid motionEulerian descriptionsA flow domain or control volume is defined by which fluid flows in and out.This method to describe fluid motion is to imagine an array of “windows”in the flow field, and have information for the velocity of fluid particles that pass each window for all timeVelocity field:(,,,)(,,,)(,,,)u u x y z t v v x y z t w w x y z t =⎧⎪=⎨⎪=⎩大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.3 Streamlines, pathlines and streaklinesStreamlinesA Streamline is a curve that is everywhere tangent to the instantaneous local velocity vector.Consider an arc lengthstreamline must be parallel to the local velocity vectorGeometric arguments results in the equation for a streamlinedr dxi dyj dzk =++V ui vj wk=++dr dx dy dz V u v w ===大连理工大学Dalian University of Technologystreamlines大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowPathlinesA Pathline is the actual path traveled by an individual fluid particle over some time period.Same as the fluid particle's material position vectorParticle location at time t:Particle Image Velocimetry (PIV) is a modern experimental technique to measure velocity field over a plane in the flow field.()()()(),,particleparticle particle xt y t z t starttstart t x x Vdt=+∫大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowExample 3.1A velocity field is given by where V 0and l areconstant, at what location in the flow field is the speed equal to V 0? Make a sketch of the velocity field in the first quadrant(x>0, y>0) by drawing arrows representing the fluid velocity at representative locations()0()V V l xi yj =−Solution:The fluid speed V, is()()1/21/2222220VV u v w x y l=++=+The speed is V=V 0at any location on the circle of radius l centered centered at the origin()1/222x y l+=The direction of the fluid velocity00tan V y l v y u V x l xθ−−===理工大学Dalian University of Technology)Stream function大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowExample 3.3Water flowing from the oscillating slit shown in figure produces a velocity field given bywhere u 0, v 0and ωare constants, Thus, the y component ofvelocity remains constant (v =v 0) and the x components of velocity at y=0 coincides with the velocity of the oscillating sprinkler head [u=u 0sin(wt) at y=0].()000sin V u t y v i v jω=−+⎡⎤⎣⎦(a ) determine the streamline that passes the through the origin at t=0; at t=π/2ω.(b) Determine the pathline of the particle that was at the origin at t=0; at t=π/2ω.(c) Discuss the shape of the streakline that pass through the origin大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowExample 3.3Solution: (streamline)The streamline are given by the solution of()000sin v dy vdx u u t y v ω==−⎡⎤⎣⎦In which the variables can be separated and the equation integrated to give()000sin u t y v dy v dxω−=⎡⎤⎣⎦∫∫()0000cos y u v t v x Cv ωω⎡⎤⎛⎞−=+⎢⎥⎜⎟⎝⎠⎣⎦x=0,y=0,t=0----------Æ00C u v ω=00cos 1u y x v ωω⎡⎤⎛⎞=−⎢⎥⎜⎟⎝⎠⎣⎦x=0,y=0,t=π/2ω----------ÆC=00cos 2u y x v πωω⎛⎞=−⎜⎟⎝⎠大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowExample 3.31020sin C x u t C v ω⎡⎤⎛⎞=−+⎢⎥⎜⎟⎝⎠⎣⎦Solution: (pathline)The pathline of a particle can beobtained from the velocity field and the definition of the velocity ()000sin dx dyu t y v and v dtdtω=−=⎡⎤⎣⎦The y equation can be integrated to givethe y coordinate of the pathline as 01y v t C =+0110000sin sin v t C C dxu t u dt v v ωω⎡⎤⎛⎞⎛⎞+=−=−⎢⎥⎜⎟⎜⎟⎝⎠⎝⎠⎣⎦X=y=0,t=0----ÆC 1=C 2=000x and y v t==X=y=0,t=π/2ω----ÆC 1=-πv 0/2ωC 2=-πu 0/2ω0022x u t and y v t ππωω⎛⎞⎛⎞=−=−⎜⎟⎜⎟⎝⎠⎝⎠0v y xu =大连理工大学Dalian University of TechnologySolution: (streakline)大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.4 One-, two, and three-dimensional flowsGenerally, a fluid flow is a rather complex three-dimensional, time-dependent phenomenon.It is possible to make simplify assumptions that allow a much easier understanding of the problem without sacrificing needed accuracy. One of these simplifications involves approximating a real flow as a simpler one-or two-dimensional flow.three-dimensional flow(,,,)V V x y z t ui vj wk==++大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowTwo-dimensional flowIn many situations one of the velocity components may be small (in some sense) relative to the two othercomponents.In situations of this kind it may be reasonable to neglect the smaller component and assume two dimensional flow. (,,)V V x y t ui vj==+Stream filament: the fluid in the stream tube大连理工大学Dalian University of Technologyfinite flow cross大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.5 Some fundamental conception for fluid flow Uniform flow: is one in which the velocity is the same on both magnitude and direction at a given instant at every point in the flow.This strict definition of uniform flow can have little meaning for the flow a real fluid where the velocity varies acrosssection. But when the size and shape of cross section areconstant along the length of channel under consideration. We say the flow is uniform.0V n∂=∂acceleration 大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.5 Some fundamental conception for fluid flow Non-uniform flow:is one in which the velocity is variable at a given instant at every point in the flow.the flow cross section is curved surface.The size and shape of flow cross section varies along the direction of fluid flow The mean velocity is variable along the direction of fluid flow大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.5 Some fundamental conception for fluid flow Gradually varied flow:i s one in which the curvature radius of streamline is large.Since the streamlines is almost parallel, the angle between the streamlines is small. With small remove acceleration and the inertia force can be neglected.大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.5 Some fundamental conception for fluid flow Rapidly varied flowThe angle between the streamlines is large.With larger curvature radius With larger remove acceleration and the inertia force can not be neglected.大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.6 Flow rate and mean velocityFlow rate: the quantity of fluid flowing per unit time across any section.Volume flow rate (discharge) : SI unit, cubic meters per second (m 3/s)Mass flow rate: SI unit, kilogram per second (kg/s) Weight flow rate: SI unit, kilonewtons per second (kN/s)The volume flow rate passing through the element of area dA is Element Mean velocity ()cos (cos )'dQ u dA u dA u dA udA θθ==== i dA’is projection of dA on the plane normalto the direction of u. 大连理工大学Dalian University of Technology Fluid mechanics Chapter 3 Basics of Fluid Flow3.6 Flow rate and mean velocityMass flow rate and the weight flow rateThe flow rate in a real fluidmQ ρ= G Q γ=A A Q dQ udA AV===∫∫A A m dQ udA AV Q ρρρρ====∫∫ AG gmudA AV Q γγγ====∫ u : is the time mean velocity throughan infinitesimal area dA.V: is mean velocity over the entire sectional area A大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.7 Equation of continuityFluid system and control volumeFluid system: a continuous mass of fluid that always contains the same fluid particle.¾The mass of a system is constant¾The shape of the system, and so the boundaries, may change with time.¾The size and shape of system is entirely optional.大连理工大学Dalian University of TechnologyCV character of this term大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowFluid flow from the fire extinguisher tank. Discuss the differences between dB sys /dt and dB cv /dt if B represent mass.Example 3.4Solution:0sys sys sys d dV dB m dt dt dtρ⎛⎞⎜⎟⎜⎟∂⎝⎠===∫0CVCV CV d dV dB m dt dt dtρ⎛⎞⎜⎟∂⎝⎠==<∫大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.7 Equation of continuityConsider a steam tube shown in figureNo fluid can leave or enter the stream tube except at the ends.The fixed volume between the two end sections is a control volume, of volume V.out in CV cv c s v ys B B DB B t t t Dt ∂∂∂=+−∂∂∂CV CV CV dm d V V t dt t ρρ∂==∂∂222222out cvdB A V dt A V dt dt ρρ==111111in cv dB AV dt AV dt dt ρρ==111222CV V AV A V tρρρ∂=−∂111222111222AV A V m AV A V gm G ρργγ===== For steady flowThe mass flow rate in isSo the mass is accumulating in the tank atthe rate of 14.5 kg/s大连理工大学Dalian University of Technology Fluid mechanics Chapter 3 Basics of Fluid FlowA 10-cm jet of water issues from a 1-m diameter tank, as shown. Assume that the velocity in the jet is (2gh)1/2m/s. How long it take for the water surface in the tank to drop from h 0=2m to h f =0.5m?Example 3.6Solution:We will let the control surfacesurround the volume of liquidwithin the tank at all times, so thatthe control volume will decrease insize as time passes.大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.8 Energy equation for steady flow ofincompressible fluids along a streamlineConsider a stream tube, two cross sectional areas d A 1,d A 2at right angles to the streamline, elevation of d A 1,d A 2are Z 1and Z 2, Velocity of the fluid are u 1, u 2in section 1 and section 2,respectively. And the pressure is p 1 and p 2Assumption:The flow is steady;The fluid is isothermal,incompressibleand inviscid.Apply work energy theoremto the fluid system within thestream tube between the twoends.大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.8 Energy equation for steady flow ofincompressible fluids along a streamlineIncrease of mechanical energyThe increase of mechanical energy can be obtained by comparing the spaces that the fluid system occupy before and after dt.Before dt, the space occupied by fluid is 1-1’+1’-2After dt, the space occupied by fluid is 1’-2+2-2’ the volume 1’-2 is the same before and after dt , so thepotential energy and the kinetic energy is invariable.Hence, The increase of the mechanicalenergy of the fluid system is onlyassociated with the increase due tothe new volume occupied 2-2’anddecrease duo to the volume 1-1’left inthe fluid system.大连理工大学Dalian University of Technology Fluid mechanics Chapter 3 Basics of Fluid Flow3.8 Energy equation for steady flow ofincompressible fluids along a streamlineThe volume of 1-1’,2-2’is dQdtMass is Increase of kinetic energy:Increase of potential energy Applying work energy theoremdQdt dQdt gγρ=22222212112222u u u u dQdt mu dQdt g g g γγ⎛⎞⎛⎞−==−⎜⎟⎜⎟⎝⎠⎝⎠()21dQdt Z Z γ−()()2221122122u u p p dQdt dQdt dQdt Z Z g g γγ⎛⎞−=−+−⎜⎟⎝⎠2211221222p u p u Z Z g gγγ++=++Bernoulli equation大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowPitot TubeStatic pressureIn a flowing fluid, we call the fluid pressure p the static pressure because it is the pressure that an instrument would measure if it were static with respect to the fluid, i.e., moving with the fluid. We measure it with piezometer tubes and other devices that attempt to minimize disturbance to the flow. To measure the static pressure in a flow field, we use a static tube. in this device the pressure is transmitted to a gage or manometer through piezometric holes that are evenly space around the circumference of the tube.Holes大连理工大学Dalian University of Technologyobtainable along a given streamline.大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowPitot TubeStatic headTotal head Small holes on bothsides of outer tube Stagnation point Writing Bernoulli equation alongthe streamline ab22a b p p u g γγ=+2a b p p u g γ−=a b v p p h γ−=2vu gh ϕ=Coefficient of instrument: because rarely are the piezometer holes located in precisely the correct position toindicate the true value of p /γ大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.9 Pressure in fluid flowFor the uniform flow, the inertial force disappears, now let us consider how pressure varies over a cross section of flow in a uniform conduit. The figure shows a small prism of the flowing fluid.Forces parallel to the direction of motion, namely the pressure and friction forces and the weight component, must balance out if the flow is steady and parallel.N N The projection of friction forces tothe axis N-N is zero, so the forcesalong N-N, namely pressure andweight component, must balance.12cos 0p A Ay p A γα+−=1212p p Z Z γγ+=+大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.9 Pressure in fluid flowThe pressure variation over a cross section perpendicular to the streamline show that in any plane perpendicular to the direction of parallel and steady flow the pressure varies according to thehydrostatic law. The average pressure is then the pressure at the centroid of such an area.In other words, on a horizontal axis through the conduit andperpendicular to its centerline the pressure is everywhere the same. Since the velocity is higher near the center than near the walls, it follows that the local energy head is also near the center.The height of liquid column are the sameThe streamline is almost parallel forthe gradually varied flow, so it can bedealt with as uniform flow大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowExample 3.7Fluid (water) )flowing in a inclined pipe, the pressure of point A measured by U-tube manometer using mercury is shown in figure. Determine the pressure of point ASolution:Since the point A and point B locate in the same cross section,the hydrostatic law can be satisfied in the cross section.'0.60.3A p γγ+=20.3980713.60.6980734.23/A p kN m =××−×=Note that although point E , D and A locateon the same horizontal plane, the pressureof point E and D are not equal to point A.1Ap ⎜⎝∫A ∫2A大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.10 Energy equation for steady flow of incompressible(b) Kinetic energy2323322222Q A A Au u V dQ dA u dA V dA Q g g g g g γγαγγαγ====∫∫∫∫3333A A A u dA u dA V AV dAα==∫∫∫Kinetic energycorrection factor V: mean velocity in the cross section23311111222A A u dA V V Q g g dA g αγγαγ==∫∫23322222222A A u dA V V Qg g dA g αγγαγ==∫∫大连理工大学Dalian University of Technology Fluid mechanics Chapter 3 Basics of Fluid Flow3.10 Energy equation for steady flow of incompressible(c) Energy loss'1212l l Qh dQ h dQγγ−−=∫Represents the negative work done by viscousfriction per unit time when the fluid across thesection flows from section 1 to section 222111222121222l p V p V Z Z h g g ααγγ−⎛⎞⎛⎞++=+++⎜⎟⎜⎟⎝⎠⎝⎠p 1, Z 1: pressure and elevation for the same point in cross section 1p 2, Z 2: pressure and elevation for the same point in cross section 2Both are absoluteor gage pressure2g大连理工大学Dalian University of TechnologyMass conservation123Q Q Q =+22111222121222l p V p V Z Z h ggααγγ−++=+++22333111131322l p V p V Z Z h ggααγγ−++=+++Generally we can write an energy equation that will fulfill conditions 4and 5. if there are two unknowns in the equation then we must also use the continuity equation.Example 3.8Solution:(1)Choose section 1-1 and 2-2 as flow cross section=2g大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowExample 3.8Z 1=4m, Z 2=0, p 1/γ=0, p 2/γ=021102V gα≈222222V V ggαα=21232l V h g−=2240000322V V g gα++=+++2 4.43/V g m s==()233.140.14.430.0348/4Q VA m s×==×=Similarly: p M =4.904kN/m 2大连理工大学Dalian University of Technology大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowVenturi meter1122(1)Write the Bernoulli equationbetween section (1) and (2), we have2211220022p V p V g gγγ++=++22122122p p V V h g gγγ−=−=∆(2) According to the continuity equation, we have22112V d V d ⎛⎞=⎜⎟⎝⎠422111222d V V h d g g⎛⎞−=∆⎜⎟⎝⎠141221g h V d d ∆=⎛⎞−⎜⎟⎝⎠22111412112441g h Q V d d d d ππ∆==⎛⎞−⎜⎟⎝⎠Which can be measured by experimental method大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.12 hydraulic grade line and energy lineDefinitionHydraulic grade line: if we connected a series of piezometers all along the pipe, the liquid would rise in them to various levels along what is called the hydraulic grade line (HGL) (piezometric line). 22p V H Z gγ=++Elevation headvelocity headPiezometric headTotal engy headStatic pressure headThe z portion of the piezometric head represents the elevation of point A. the vertical distance from point A to the corresponding point on the HGL represent p/γin the flow at point A 大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.12 hydraulic grade line and energy lineEnergy line: is above the HGL by a distance V 2/2g. That is the linedrawn through the pitot-tube liquid surfaces.For the flow of an ideal fluid, the energy line is horizontal. For a real fluid, the energy line must slope downward in the direction of flow because of head loss due to fluid friction大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.12 hydraulic grade line and energy lineHow to draw hydraulic grade lines and energy lines(1) By definition, the EGL is positioned above the HGL by an amount v 2/2g. Thus if the velocity is zero, as in a lake or reservoir , the HGL and EGL will coincide.(2)Head loss for flow in a pipe or channel always means that the EGL will slope downward in the direction of flow, the only exception to this rule occurs when a pump supplies energy to the flow, then an abrupt rise in the EGL (and the HGL) occurs from the upstream side to the downstream side of the pump.Energy addeda大连理工大学Dalian University of Technologygradual expansion such as at the outlet.Energy removed Gradual expansion大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.12 hydraulic grade line and energy lineHow to draw hydraulic grade lines and energy lines(4) if the outlet to a reservoir is an abrupt expansion, all the kinetic energy is lost. Thus the EGL drops an amount v 2/2g at the outlet. (5) If a flow passagechanges diameter,such as in a nozzle or by means of a change in pipe size, the velocity therein will also change, hence, the distance between the EGLand HGL will change, and the slope on the EGL will change.大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.12 hydraulic grade line and energy lineHow to draw hydraulic grade lines and energy lines(6) For steady flow in a pipe that has uniform physicalcharacteristics (diameter, roughness, shape, and so on) along its length, the head loss per unit length will be constant. thus the slope of the EGL and HGL will be constant along the length of pipe.(7)If the HGL fall below the pipe, then p/γis negative, indicating subatmospheric pressure.And so on, energy line:1-a-b-b0-c Hydraulic grade line: 1-a’-b’-bo’-c’大连理工大学Dalian University of Technology11221222l p Z p Z p γγ−++=+++But for the gases flow, we must consider the difference ofatmospheric pressure when the specific weight of gases is not equal to that of atmosphere, specially, in the case which the elevation difference is very large.大连理工大学Dalian University of Technology'11a p p p =+()'2221a a p p p Z Z γ=−+−Pressure difference221211221212()22a a a l V V p p Z p p Z Z Z p ρργγγ−+++=+−−+++γa : the specific weight of atmosphere2212121212()()22a l V V p Z Z p p ρργγ−++−−=++Static pressure Kinetic pressure Elevation pressure21()()s a p p Z Z γγ=+−−22q V p p ρ=+221()()2z a Vp p Z Z ργγ=++−−Potential pressure Stagnation pressure Total pressure大连理工大学Dalian University of Technology(2)if the density of the gas ρ=0.8kg/m . (ρa =1.3kg/m )大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowExample 3.10solution(1): writing energy equationbetween the section A and C, Because the density of gas ρ=ρa =1.2kg/m 3, thus22C A lA C V p p ρ−=+22129.8 1.29 1.222C C V V ×=×+××2117.6/12 4.43/C V m s=×=2314.430.0348/4Q d m sπ=×××=The pressure of point B: Writing energy equation between the section B and C2222B lB CV V p p ρρ−+=+22219 1.221.2 1.2222B V V V p ××××=×++224.5 1.252.92/2B V p N m =××=大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowExample 3.10(2): if the density of the gas ρ=0.8kg/m 3, writing energyequation between the section A and C,2()()2C A a A C lA C V p Z Z p ργγ−+−−=+22129.8(1.20.8)9.8400.890.822C C V V ×+−××=×+××8.28/B V m s=30.065/Q m s=Τhe pressure of point B22190.8(1.20.8)9.82045/22B B V p N m =×××−−××=z p ρ=−Zero pressure lineD Unsteady flow∑The rate of change oraccumulation of momentum within the fixed control volumeThe rates at which momentum leaves the control volumeThe resultant force acting on a fluid mass is equal to the rate of change of momentum of the fluid mass大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.15 The momentum principleIn the case of steady flow, conditions withinthe control volume do not change, soIf we select a control volume so that the control surface is normal to the velocity where it cuts the flow, and lets us specify that the velocity is constant where it cuts across the control surface.()()out in cvcvd d d mV mV F tdt=−∑()11111d d m dt dtmVm V V == ()22222d d m dt dt mVm V V ==Steady flow2211222111F V m V Q V V mQ ρρ=−=−∑大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid Flow3.15 The momentum principleSince the flow is steady, from continuity, we have121122mm m Q Q ρρ==== According to the vector relation21out inV V V V V ∆=−=−()()()21F V V V V mQ Q ρρ=∆=∆=−∑ The direction of must be the same as that of the velocity change Represents the vectorial summation of all forces acting on the fluid mass in the control volume, including:•Gravity, shear forces•Pressure forces including those exerted by fluid surrounding the fluid mass under consideration as well as the pressure forces exerted by the solid boundaries in contact with the fluid massF ∑F ∑ρAV(2(2V Qβyx大连理工大学Dalian University of TechnologyFluid mechanics Chapter 3 Basics of Fluid FlowExample 3.12Energy equation: in free-surface flow such as this where the streamlines are parallel, the water surface is coincident with the hydraulic grade line.221122121222l p V p V Z Z hg g γγ−++=+++22122129.8129.81V V +=+××Continuity equation:122313V V ×=×1 2.56/V m s=2 5.11/V m s=2112215.34/Q AV A V m s===4) Free-body diagram of the fluid within the control volume5) Applying momentum equation()1221x F F F Q V V ρ−−=−4.91x F kN =。

运动会作文精彩时刻英文

运动会作文精彩时刻英文1. The crowd erupted in cheers as the sprinters dashed towards the finish line. The tension in the air was palpable as the athletes pushed their bodies to the limit, their determination shining through their sweat-soaked faces.2. In the long jump event, the athletes soared through the air like graceful birds. The audience held their breath as each competitor leaped, their bodies suspended in mid-air for what seemed like an eternity. The thud of their landing echoed through the stadium, followed by acollective gasp or applause from the spectators.3. The relay race was a display of teamwork and coordination. The baton passed from one runner to another in a seamless motion, each athlete giving their all to maintain the lead. The crowd cheered on their favorite teams, their voices blending together in a symphony of support.4. The high jump event showcased the athletes' agility and precision. With a burst of energy, they soared over the bar, defying gravity for a split second. The bar wobbled precariously with each attempt, adding to the suspense. The crowd erupted in cheers as each successful jump was cleared.5. The shot put event was a display of raw strength and power. The athletes heaved the heavy metal ball with all their might, their muscles bulging with effort. The thud of the shot hitting the ground reverberated through the stadium, earning applause and admiration from the spectators.6. The hurdles race tested the athletes' speed and agility. With each hurdle cleared, they seemed to defy the laws of physics, their bodies soaring over the obstacles effortlessly. The crowd erupted in cheers as each athlete crossed the finish line, their determination shiningthrough their exhausted yet triumphant faces.7. The javelin throw event was a spectacle of precisionand technique. The athletes launched the javelin into the air, their bodies twisting and turning in a graceful motion. The crowd watched in awe as the javelin sailed through the sky, landing with a satisfying thud in the designated area.8. The pole vault event was a display of bravery and skill. The athletes propelled themselves into the air,using a long pole as leverage. The gasps from the audience were audible as they cleared the bar, their bodies arching over it with precision. The crowd erupted in applause, acknowledging their incredible achievement.9. The discus throw event showcased the athletes' strength and finesse. With a powerful spin, they released the discus into the air, their bodies following throughwith a fluid motion. The crowd watched in anticipation as the discus spun through the air, landing with a resounding thud in the designated area.10. The marathon race was a test of endurance and determination. The athletes pushed their bodies to thelimit, their strides steady and determined. The crowd linedthe streets, cheering on the runners as they passed by, their voices providing a much-needed boost of motivation.11. The closing ceremony was a celebration of the athletes' achievements. The stadium was filled with a sense of pride and camaraderie as the medal winners stood on the podium, their national anthems playing in the background. The crowd erupted in applause, their cheers echoing through the night sky, as the curtain fell on a successful and memorable sports event.。

离心泵CFD应用---Fluent计算

图（19）
注：一般的清水离心泵分析可不设置重力
d 11、边界条件设置 UnRegistere 点击 Define → Boundary Conditions，如图（21）
图（20）
图（21）
1）流体区域设置在这个模型里面流体区域有三块，fluid-in(进口)、fluid-run(叶轮)、fluid-out （蜗壳）
is fluid-run,说明这个 wall 是属于叶轮，我们要把它设置为旋转的。
在 wall motion 下选择 moveing wall，如图（35）
g 由于它是相对于叶轮静止的，所以选择 Rotational。此时要查看旋转轴 Z 是否为 1，如
UnRe 果不是的话要设置为 1，其余为 0。如图(36)
图（38） 13、残差设置
点击 Solve→Momitors→Residual,在 Options 在勾选 Plot，点击 OK。如图（39）
图（39）
d 14、初始化设置
点击 Solve→Initialization,在 Compute From 下选择 in，点击 Init→OK。如图（40）
UnRegistere图（40）
15、保存文件点击 Write→Case&Data→保存路径点击 Autosave→在 Autosave Case/Data File Frequency 里面分别填 1，这样的话每计算完一次都会保存数据，在 File Name 下面给文件重新起个名字。对于稳态计算建议保存频率取小一点，可以计算 10 次之后保存一次。而对于非稳态计算可以取大频率。如图（41）
UnR图图（（4e）5）gistered
注：1）创建网格分界面要在 check 网格之前，否则 check 网格时就会出错。 2）在创建网格分界面之后你会发现在 Boundary Conditions 里面多了四个边界（wall-13、wall-25、wall-27、wall-28），如图（6）

Chapter_3

13
Chapter 3
Hydraulic Pumps
3.2 Piston Pumps
3.2.1 Valve shaft radial piston pumps 3.2.2 Swash plate axial piston pumps 3.2.3 Bent axis axial piston pumps
3.1.1 Operating principle 1,2-Non-return valve 3-Spring 4-Cylinder 5-Piston 6-Eccentric wheel
Fig. 3-1 Operating principle displacement pumps
4
Chapter 3
(1) Input power pr (2) Output power P (3) Volumetric efficiency ηv
8
Chapter 3
Hydraulic Pumps
(4) Mechanical efficiency ηm
(5) Overall efficiency η and mechanical efficiency ηm
9
Chapter 3
3.1.3
Hydraulic Pumps
Performance curves
Fig.3-2 shows a performance curve of a hydraulic pump, in which the abscissa represents working pressure p of the hydraulic pump and the ordinate the volumetric efficiencyηv (or the practical flow rate q), overall efficiencyη and input powerＰr.

Chapter 2 Concept and definition

20:34:17
Fundamentals MHM Transfer
1
Fluid
A fluid is defined as a substance which deforms continuously under the action of shear stress.
Liquids and gases
d (V ) 0 dt
Unsteady flow:If the flow at a point varies with
time.
d (V ) 0 dt
Stationary，at rest
Dynamic or motion
20:34:18
Fundamentals MHM Transfer
12
Incompressible flow
The three fundamental laws of physics. Lagrangian representation. Eularian representation. Continuum. Give an example of a compressible fluid. Fluid property and flow property Gradient and dot product
20:34:18
Fundamentals MHM Transfer
11
Steady and unsteady flows
Steady flow: The flow conditions are independent
of time at every point in the flow fields
(2) The macroscopic properties of a continuum are consider to vary smoothly from point to point in the fluid.

流体力学2流体静力学详解

Absolute, Gage, and Vacuum Pressures
Absolute Gage Pressure
Vacuum Esc Pressure
02 - 3
• Actual pressure at a give point is called the absolute pressure. • Most pressure-measuring devices are calibrated to read zero in the
• Normal stress on any plane is pressure (+ for compression)
px DAsinq
W
Dz
q
DAcosq
pz DAcosq
rg(DAcosq Dz)/2
Fx 0 : Fz 0 :
( pnDA)sinq px (DAsinq ) 0
p z DA cosq
• Integrate dp/dz =
z patm
p2
p1
dp
z2
z1
dz
Liquids : or
p2 p1 (z2 z1)
z2
z1
p2
p1
or
p z constant
z ➢ The pressure increases linearly with
the depth of the liquid.
pa = 101,350 Pa
M = 133,100 N/m3
hM
pa M
0.761 m 761 mm
➢ That’s why the pressure is sometimes said to be 761 mmHg or 29.96 inHg.

1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。