流体力学英文版1

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流体力学(Fluid

流体力学(Fluid
上 篇
流体力学(Fluid Mechanics) 流体力学(Fluid Mechanics)

第一章 绪 论(Introduction)
流体的定义、 第一节 流体的定义、特征及连续介质假设
(Definition, Characteristics and Continuum Hypothesis of a Fluid) 1、定义:能够流动的物质。 、定义:能够流动的物质。 可以定义为: 根据其力学特征(mechanical characteristics)可以定义为: 受到任何微小剪切力(shear force)的作用都能够导致其连续 变形的物质。 变形的物质。 的总称。 流体是液体(liquid)和气体(gas)的总称。 和气体 的总称
p =
lim
A→ 0
P A
τ =
lim
A→ 0
T A
第四节 流体力学模型
理想流体模型: 理想流体模型:不考虑流体的黏性 黏性流体模型: 黏性流体模型:考虑流体的黏性 不可压缩流体模型: 不可压缩流体模型:不考虑流体的压缩性 可压缩流体模型: 可压缩流体模型:考虑流体的压缩性
dF X dF Y dF Z X = lim , Y = lim , Z = lim dv → m dv → m dv → m m m m
G X X = = 0 m G Y Y = = 0 m G Z Z = = g m
在重力场中: 在重力场中:
二、表面力(surface force)
定义:指作用在流体中所取某部分流体体积表面上的力。 定义:指作用在流体中所取某部分流体体积表面上的力。 表面力可分解成两个分力: 表面力可分解成两个分力: 法向力P -----与流体表面垂直 与流体表面垂直 切向力τ---与流体表面相切。 切向力τ---与流体表面相切。 与流体表面相切

流体力学英文版第一章绪论—Introuduction

流体力学英文版第一章绪论—Introuduction
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Expression of fields
Noห้องสมุดไป่ตู้es
Vector fields: Velocity field, V = (V(r, t)) = (u(r, t), v(r, t)). For a streamline, it satisfies dr × V = 0.
The mechanics is built on the base of the corresponding engineering problem, physical models, and mathematic models or algorithms.
Deformation is also important in mechan-
Content
1. Curse overview 2. Field 3. Indicial notation 4. Fluid properties
Notes
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Curse overview
Notes
Fundamentals of Fluid mechanics
Chapter I Physical properties of fluid mechanics Field, Indicial notation, fluid properties. Chapter II Fluid statics Fluid is either at rest or moving without shearing stresses. Chapter III Fluid kinematics The description and visualization of its motion. Chapter IV Fluid dynamics The analysis of the specific forces necessary to produce the motion: Idea flow and viscous flow. Chapter V Dimensional analysis (Option) Buckingham Π theorem to conduct dimensional analysis.

工程流体力学英文原版

工程流体力学英文原版

工程流体力学英文原版Engineering Fluid Mechanics: An Introduction.Engineering fluid mechanics is a crucial discipline within the field of engineering that deals with the study of fluids and their interactions with solid boundaries. It is a fundamental branch of physics and engineering that finds applications in various fields such as civil, mechanical, aerospace, and chemical engineering. The study of fluid mechanics involves the understanding of fluid properties, fluid statics, fluid dynamics, and fluid control.1. Fluid Properties.Fluids are substances that continuously deform under the application of shear stress. They lack a fixed shape and take the shape of the container in which they are contained. Fluids can be classified as liquids or gases, depending on their state and properties. Liquids have adefinite volume but no fixed shape, while gases expand tofill the available space.Some important fluid properties include density, viscosity, compressibility, and surface tension. Density is the mass per unit volume of a fluid. Viscosity representsthe internal friction of a fluid and affects its flow behavior. Compressibility describes how a fluid responds to changes in pressure, while surface tension arises from the intermolecular forces at the fluid's surface.2. Fluid Statics.Fluid statics deals with the behavior of fluids at rest, or in equilibrium. It involves the study of pressure distribution in fluids, buoyancy, and hydrostatics.Pressure is a force per unit area acting perpendicular tothe surface, and it is a fundamental quantity in fluid mechanics. Buoyancy is the upward force exerted by a fluid on an immersed object, and it is responsible for thefloating of objects on water. Hydrostatics deals with the equilibrium of fluids under the influence of gravity andother external forces.3. Fluid Dynamics.Fluid dynamics is concerned with the motion of fluids and the forces acting on them. It involves the study of fluid flow, fluid mechanics equations, and fluid control. Fluid flow can be laminar or turbulent, depending on the velocity and other fluid properties. Laminar flow is smooth and orderly, while turbulent flow is chaotic and irregular.The fundamental equations of fluid dynamics include the conservation of mass, momentum, and energy. The conservation of mass states that the rate of change of mass within a control volume is equal to the net mass flow rate into the volume. The conservation of momentum relates the forces acting on a fluid element to its acceleration, while the conservation of energy accounts for the conversion of energy forms within a fluid system.4. Fluid Control.Fluid control involves the manipulation and manipulation of fluid flow using pumps, valves, and other devices. Pumps are used to increase the pressure or flow rate of a fluid, while valves are used to control the direction or amount of fluid flow. Other devices such as nozzles, diffusers, and turbines are also employed to modify fluid flow characteristics.In conclusion, engineering fluid mechanics is a crucial discipline that deals with the study of fluids and their interactions with solid boundaries. It involves the understanding of fluid properties, fluid statics, fluid dynamics, and fluid control. This knowledge is essentialfor engineers to design, analyze, and optimize fluid systems in various engineering applications.。

流体力学(英文)

流体力学(英文)

2021/8/11
5
Experimental Projects (1)
In-class (Required) (Total hours: 6)
E1 Hydrostatics (1 hour) E2 Velocity measurement with Pitot tube (1 hour) E3 Flow measurement with Venturi tube (1 hour) E4 Verification of Bernoulli equation (1 hour) E5 Reynolds experiment (1 hour) E6 Measurement of friction loss coefficient for circular pipe
Water and air are researched mostly in fluid mechanics. Besides these, fluid also includes vapor as work medium in turbine, lube oil, water which mixing with mud and sand, blood, melting metal and the gas with complicated components coming from burning, plasmas under high temperature condition, etc.
B2 Boundary layer theory and flow past body (4 hours)
Part C (Total hours: 4) Extended contents
C1 Resaerch progress in Fluid Mechancis (1 hour) C2 Modern flow measurement technologies (1 hour) C3 Introduction to numerical simulation of turbulence flow

流体力学英文版课后习题答案

流体力学英文版课后习题答案

1.1 What will be the (a) the gauge pressure and (b) the absolute pressure of water at depth 12m below the surface? ρwater = 1000 kg/m 3, and P atmosphere = 101kN/m 2. Solution:Rearranging the equation 1.1-4gh p p a b ρ+=Set the pressure of atmosphere to be zero, then the gauge pressure at depth 12m below the surface iskPa gh p p a b 72.1171281.910000=⨯⨯+=+=ρAbsolute pressure of water at depth 12mkPa Pa gh p p a b 72.2182187201281.91000101000==⨯⨯+=+=ρ1.3 A differential manometer as shown in Fig. is sometimes used to measure small pressure difference. When the reading is zero, the levels in two reservoirs are equal. Assume that fluid B is methane (甲烷), that liquid C in the reservoirs is kerosene (specific gravity = 0.815), and that liquid A in the U tube is water. The inside diameters of the reservoirs and U tube are 51mm and 6.5mm , respectively. If the reading of the manometer is145mm., what is the pressure difference over the instrument In meters of water, (a) when the change in the level in the reservoirs is neglected, (b) when the change in the levels in the reservoirs is taken into account? What is the percent error in the answer to the part (a)? Solution :p a =1000kg/m 3 p c =815kg/m 3 p b =0.77kg/m 3 D/d=8 R=0.145mWhen the pressure difference between two reservoirs is increased, the volumetric changes in the reservoirs and U tubesR d x D 2244ππ=(1)soR D d x 2⎪⎭⎫⎝⎛= (2)and hydrostatic equilibrium gives following relationshipg R g x p g R p A c c ρρρ++=+21 (3)sog R g x p p c A c )(21ρρρ-+=- (4)substituting the equation (2) for x into equation (4) givesg R g R D d p p c A c )(221ρρρ-+⎪⎭⎫⎝⎛=- (5)(a )when the change in the level in the reservoirs is neglected,()Pag R g R g R D d p p c A c A c 26381.98151000145.0)()(221=⨯-=-≈-+⎪⎭⎫⎝⎛=-ρρρρρ(b )when the change in the levels in the reservoirs is taken into account()Pa g R g R D d gR g R D d p p c A c c A c 8.28181.98151000145.081.9815145.0515.6)()(22221=⨯-+⨯⨯⨯⎪⎭⎫ ⎝⎛=-+⎪⎭⎫⎝⎛=-+⎪⎭⎫⎝⎛=-ρρρρρρerror=%=7.68.2812638.281-1.4 There are two U-tube manometers fixed on the fluid bed reactor, as shown in the figure. The readings of two U-tube manometers are R 1=400mm ,R 2=50mm, respectively. The indicating liquid is mercury. The top of the manometer is filled with the water to prevent from the mercury vapor diffusing into the air, and the height R 3=50mm. Try to calculate the pressure at point A and B .Solution: There is a gaseous mixture in the U-tube manometer meter. The densities of fluids are denoted by Hg O H g ρρρ,,2, respectively. The pressure at point A is given by hydrostaticequilibriumFigure for problem 1.4g R R g R g R p g H g O H A )(32232+-+=ρρρg ρis small and negligible in comparison with H g ρand ρH2O , equation above can besimplifiedc A p p ≈=232gR gR H g O H ρρ+=1000×9.81×0.05+13600×9.81×0.05=7161N/m²1gR p p p H g A D B ρ+=≈=7161+13600×9.81×0.4=60527N/m1.5 Water discharges from the reservoir through the drainpipe, which the throat diameter is d. The ratio of D to d equals 1.25. The vertical distance h between the tank A and axis of the drainpipe is 2m. What height H from the centerline of the drainpipe to the water level in reservoir is required for drawing the water from the tank A to the throat of the pipe? Assume that fluid flow is a potential flow. The reservoir, tank A and the exit of drainpipe are all open to air.Solution:Bernoulli equation is written between stations 1-1 and 2-2, with station 2-2 being reference plane:2222222111u gz p u gz p ++=++ρρ Where p 1=0, p 2=0, and u 1=0, simplification of the equation22u Hg =1The relationship between the velocity at outlet and velocity u o at throat can be derived by the continuity equation:22⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛D d u u o 22⎪⎭⎫⎝⎛=d D u u o 2Bernoulli equation is written between the throat and the station 2-2 3Combining equation 1,2,and 3 givesSolving for H H=1.39m1.6 A liquid with a constant density ρ kg/m 3 is flowing at an unknown velocity V 1 m/s through a horizontal pipe of cross-sectional area A 1 m 2 at a pressure p 1 N/m 2, and then it passes to a sectionof the pipe in which the area is reduced gradually to A 2 m 2 and the pressure is p 2. Assuming no friction losses, calculate the velocities V 1 and V 2 if the pressure difference (p 1 - p 2) is measured.Solution :In Fig1.6, the flow diagram is shown with pressure taps to measure p 1 and p 2. From the mass-balance continuity equation , for constant ρ where ρ1 = ρ2 = ρ,222200uu p =+ρ()===144.281.92100081.910002125.11112442-⨯⨯⨯--⎪⎭⎫ ⎝⎛==ρρgh d D u Hg2112A A V V =For the items in the Bernoulli equation , for a horizontal pipe, z 1=z 2=0Then Bernoulli equation becomes, after substituting 2112A A V V =for V 2, ρρ22121211212020p A A V p V ++=++ Rearranging,2)1(21212121-=-A A V p p ρ ⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫⎝⎛-12221211A Ap p V ρ=Performing the same derivation but in terms of V 2,⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫⎝⎛--21221212A A p p V ρ=1.7 A liquid whose coefficient of viscosity is µ flows below the critical veloc ity for laminar flowin a circular pipe of diameter d and with mean velocity V . Show that the pressure loss in a length of pipeL p ∆ is 232d V μ. Oil of viscosity 0.05 Pas flows through a pipe of diameter 0.1m with a average velocity of 0.6m/s.Calculate the loss of pressure in a length of 120m.Solution :The average velocity V for a cross section is found by summing up all the velocities over the cross section and dividing by the cross-sectional area1From velocity profile equation for laminar flow2 substituting equation 2 for u into equation 1 and integrating3rearranging equation 3 gives1.8. In a vertical pipe carrying water, pressure gauges areinserted at points A and B where the pipe diameters are 0.15m and 0.075m respectively. The point B is 2.5m belowA and when the flow rate down the pipe is 0.02 m 3/s, the pressure at B is 14715 N/m 2 greater than that at A. Assuming the losses in the pipe between A and B can beexpressed as gV k 22where V is the velocity at A, find thevalue of k .If the gauges at A and B are replaced by tubes filled with water and connected to a U-tube containing mercury of relative density 13.6, give a sketch showing how the levels in the two limbs of the U-tube differ and calculate the value of this difference in metres. Solution:d A =0.15m; d B =0.075m z A -z B =l =2.5m Q =0.02 m 3/s,p B -p A =14715 N/m 2Figure for problem 1.8⎰⎰==RR rdr uR udA A V 020211ππ⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛--=22014R r R L p p u L μ2032D L pp V L μ-=232dV L p μ=∆Pa d VL p 115201.01206.005.0323222=⨯⨯⨯==∆μs m d QV V d Q AA AA /132.115.0785.002.044222=⨯===ππs m d QV V d Q BB BB /529.4075.0785.002.044222=⨯===ππWhen the fluid flows down, writing mechanical balance equation222222AB B B A A AV kV g z p V g z p +++=++ρρ 213.1253.4100014715213.181.95.2222k++=+⨯ k 638.0260.10715.14638.0525.24++=+=k 0.295making the static equilibriumgR g x g l p g R g x p H g A B ρρρρρ+∆++=+∆+()()mm ggl p p R gH A B 7981.91260081.910005.214715-=⨯⨯⨯-=---=ρρρ1.9.The liquid vertically flows down through the tube from the station a to the station b , then horizontally through the tube from the station c to the station d , as shown in figure. Two segments of the tube, both ab and cd ,have the same length, the diameter and roughness. Find:(1)the expressions ofgp ab ρ∆, h fab , g p cdρ∆ and h fcd , respectively.(2)the relationship between readings R 1and R 2 in the U tube.Solution:(1) From Fanning equationFigure for problem 1.92Vl h fab λ=andsoFluid flows from station a to station b , mechanical energy conservation giveshence2from station c to station d hence3From static equationp a -p b =R 1(ρˊ-ρ)g -l ρg 4 p c -p d =R 2(ρˊ-ρ)g 5 Substituting equation 4 in equation 2 ,thentherefore6Substituting equation 5 in equation 3 ,then7Thus R 1=R 222V d l h fcd λ=fcdfab h h =fab bah p p +=+ρρlg fab b a h p p =+-lg ρfcd d c h pp +=ρρfcd dc h p p =-ρfab h g l g R =+--'lg 1ρρρρ)(gR h fab ρρρ-'=1g R h fcd ρρρ-'=21.10 Water passes through a pipe of diameter d i=0.004 m with the average velocity 0.4 m/s, as shown in Figure.1) What is the pressure drop –∆P when water flows through the pipe length L =2 m, in m H 2O column?2) Find the maximum velocity and point r at whichit occurs.3) Find the point r at which the average velocity equals the local velocity.4)if kerosene flows through this pipe ,how do the variables above change ?(the viscosity and density of Water are 0.001 Pas and 1000 kg/m 3,respectively ;and the viscosity and density of kerosene are 0.003 Pas and 800 kg/m 3,respectively )solution: 1)1600001.01000004.04.0Re =⨯⨯==μρudfrom Hagen-Poiseuille equation1600004.0001.024.0323222=⨯⨯⨯==∆d uL P μ m g p h 163.081.910001600=⨯=∆=ρ 2)maximum veloc ity occurs at the center of pipe, from equation 1.4-19 max0.5Vu =so u max =0.4×2=0.8m3)when u=V=0.4m/s Eq. 1.4-172max 1⎪⎪⎭⎫ ⎝⎛-=w r r u u5.0004.01max2=⎪⎭⎫⎝⎛-u V r = m r 00284.071.0004.05.0004.0=⨯==4) kerosene:427003.0800004.04.0Re =⨯⨯==μρudPa pp 4800001.0003.01600=='∆='∆μμFigure for problem 1.10m g p h 611.081.98004800=⨯=''∆='ρ1.12 As shown in the figure, the water level in the reservoir keeps constant. A steel drainpipe (with the inside diameter of 100mm) is connected to the bottom of the reservoir. One arm of the U-tube manometer is connected to the drainpipe at the position 15m away from the bottom of the reservoir, and the other is opened to the air, the U tube is filled with mercury and the left-side arm of the U tube above the mercury is filled with water. The distance between the upstream tap and the outlet of the pipeline is 20m.a) When the gate valve is closed, R=600mm, h=1500mm; when the gate valve is opened partly, R=400mm, h=1400mm. The friction coefficient λ is 0.025, and the loss coefficient of the entrance is 0.5. Calculate the flow rate of water when the gate valve is opened partly. (in m³/h)b) When the gate valve is widely open, calculate the static pressure at the tap (in gauge pressure, N/m²). l e /d ≈15 when the gate valve is widely open, and the friction coefficient λ is still 0.025.Solution :(1) When the gate valve is opened partially, the water discharge isSet up Bernoulli equation between the surface of reservoir 1—1’and the section of pressure point 2—2’,and take the center of section 2—2’ as the referring plane, then∑+++=++21,2222121122—f h p u gZ p u gZ ρρ (a )In the equation 01=p (the gauge pressure)222/396304.181.910004.081.913600m N gh gR p O H H g =⨯⨯-⨯⨯=-=ρρFigure for problem 1.12021=≈Z uWhen the gate valve is fully closed, the height of water level in the reservoir can be related to h (the distance between the center of pipe and the meniscus of left arm of U tube).gR h Z g H g O H ρρ=+)(12(b )where h=1.5mR=0.6mSubstitute the known variables into equation b2222_1,113.22)5.01.015025.0(2)(66.65.110006.013600V VV K d l h mZ c f =+⨯=+==-⨯=∑λ Substitute the known variables equation a9.81×6.66=2213.21000396302V V ++the velocity is V =3.13m/s the flow rate of water ish m V d V h /5.8813.312.0436004360032=⨯⨯⨯=⨯=ππ2) the pressure of the point where pressure is measured when the gate valve is wide-open.Write mechanical energy balance equation between the stations 1—1’ and 3-3´,then∑+++=++31,3233121122—f h p V gZ p V gZ ρρ (c )since m Z 66.61=311300p p u Z =≈= 2223_1,81.4 2]5.0)151.035(025.0[ 2)(V V V K d l l h c e f =++=++=∑λinput the above data into equation c ,9.8122V 81.4266.6+=⨯Vthe velocity is: V =3.51 m/sWrite mechanical energy balance equation between thestations 1—1’and 2——2’, for the same situation of water level∑+++=++21,2222121122—f h p V gZ p V gZ ρρ(d )since m Z 66.61=2121003.51/0(page pressure Z u u m sp =≈≈=)kg J V K d l hc f /2.26251.3)5.01.015025.0(2)(222_1,=+⨯=+=∑λinput the above data into equation d ,9.81×6.66=2.261000251.322++p the pressure is: 329702=p1.14 Water at 20℃ passes through a steel pipe with an inside diameter of 300mm and 2m long. There is a attached-pipe (Φ60⨯3.5mm) which is parallel with the main pipe. The total length including the equivalent length of all form losses of the attached-pipe is 10m. A rotameter is installed in the branch pipe. When the reading of the rotameter is2.72m 3/h, try to calculate the flow rate in the main pipe and the total flow rate, respectively. The frictional coefficient of the main pipe and the attached-pipe is 0.018 and 0.03, respectively.Solution : The variables of main pipe are denoted by a subscript1, and branch pipe by subscript 2.The friction loss for parallel pipelines is 2121S S s f f V V V h h+==∑∑The energy loss in the branch pipe is22222222u d l l he f ∑∑+=λ In the equation03.02=λsm u d ml l e /343.0053.04360072.2053.01022222=⨯⨯===+∑πinput the data into equation c kg J hf /333.02343.0053.01003.022=⨯⨯=∑The energy loss in the main pipe is333.022111121===∑∑u d l h hf f λSo s m u /36.22018.023.0333.01=⨯⨯⨯=The water discharge of main pipe ish m V h /60136.23.043600321=⨯⨯⨯=πTotal water discharge ish m V h /7.60372.26013≈+=1.16 A Venturimeter is used for measuring flow of water along a pipe. The diameter of the Venturi throat is two fifths the diameter of the pipe. The inlet and throat are connected by water filled tubes to a mercury U-tube manometer. The velocity of flow along the pipe is found to beR 5.2 m/s, where R is the manometer reading in metres of mercury. Determine the loss of headbetween inlet and throat of the Venturi when R is 0.49m. (Relative density of mercury is 13.6). Solution:Writing mechanical energy balance equation between the inlet 1 and throat o for V enturi meterf o o hg z V p g z V p +++=++22121122ρρ 1 rearranging the equation above, and set (z 2-z 1)=xf o oh xg V V p p ++-=-22121ρ2from continuity equationFigure for problem 1.1611221125.625V V dd V V o o =⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛= 3 substituting equation 3 for V o into equation 2 gives()fff f oh xg R h xg Rh V h xg V V p p ++=++=+=++-=-94.1185.203.1903.19206.3922121211ρ4from the hydrostatic equilibrium for manometerg x g R p p H g o ρρρ+-=-)(1 5substituting equation 5 for pressure difference into equation 4 obtainsf Hgh xg R gx g R ++=+-94.118)(ρρρρ 6rearranging equation 6kg J R R R R gR h Hg f /288.267.494.11861.12394.118)(==-=--=ρρρ1.17.Sulphuric acid of specific gravity 1.3 is flowing through a pipe of 50 mm internal diameter. A thin-lipped orifice, 10mm, is fitted in the pipe and the differential pressure shown by a mercury manometer is 10cm. Assuming that the leads to the manometer are filled with the acid,calculate (a)the weight of acid flowing per second, and (b) the approximate friction loss in pressure caused by the orifice.The coefficient of the orifice may be taken as 0.61, the specific gravity of mercury as 13.6, and the density of water as 1000 kg/m 3Solution: a)2.0501010==D D =⨯-=-=-81.9)130013600(1.0)(21g R p p H g ρρ()sm p p D D C V o /63.231.461.056.1861.0130081.9)130013600(1.022.0161.0214214102=⨯=≈⨯-⨯-=-⎪⎪⎭⎫ ⎝⎛-=ρs kg V D m /268.0130063.201.0442220=⨯⨯⨯==πρπb) approximate pressure drop=⨯-=-=-81.9)130013600(1.0)(21g R p p H g ρρ12066.3Papressure difference due to increase of velocity in passing through the orificePa D D V V V V p p o8.44882)2.01(63.213002242412222212221=-=⎪⎪⎭⎫⎝⎛-=-=-ρρpressure drop caused by friction lossPa p f 5.75778.44883.12066=-=∆2.1 Water is used to test for the performances of pump. The gauge pressure at the discharge connection is 152 kPa and the reading of vacuum gauge at the suction connection of the pump is24.7 kPa as the flow rate is 26m 3/h. The shaft power is 2.45kw while the centrifugal pump operates at the speed of 2900r/min. If the vertical distance between the suction connection and discharge connection is 0.4m, the diameters of both the suction and discharge line are the same. Calculate the mechanical efficiency of pump and list the performance of the pump under this operating condition. Solution:Write the mechanical energy balance equation between the suction connection and discharge connection2_1,2222121122f H gp g u Z H g p g u Z +++=+++ρρwherem Z Z 4.012=-(Pa 1052.1(Pa 1047.22_1,215241≈=⨯=⨯-=f H u u pressure gauge p pressure gauge p ))total heads of pump is m H 41.1881.9100010247.01052.14.055=⨯⨯+⨯+= efficiency of pump isN N e /=ηsince kW g QH N e 3.1360081.9100041.18263600=⨯⨯⨯==ρN=2.45kWThen mechanical efficiency%1.53%10045.23.1=⨯=η The performance of pump isFlow rate ,m³/h 26 Total heads ,m18.41 Shaft power ,kW 2.45 Efficiency ,%53.12.2 Water is transported by a pump from reactor, which has 200 mm Hg vacuum, to the tank, in which the gauge pressure is 0.5 kgf/cm 2, as shown in Fig. The total equivalent length of pipe is 200 m including all local frictional loss. The pipeline is φ57×3.5 mm , the orifice coefficient of C o and orificediameter d o are 0.62 and 25 mm, respectively. Frictionalcoefficient λ is 0.025. Calculate: Developed head H of pump, in m (the reading R of U pressure gauge in orifice meter is 168 mm Hg)Solution: Equation(1.6-9)Mass flow rates kg S V m o o /02.21000025.0414.312.42=⨯⨯⨯==ρ 2) Fluid flow through the pipe from the reactor to tank, the Bernoulli equation is as followsfor V 1=V 2f H z gp p H +∆+-=ρ12 ∆z=10msm Rg D d C V f /12.444.69375.062.01000)100013600(81.9168.025025162.02144000=⨯=-⨯⨯⎪⎭⎫ ⎝⎛-=-⎪⎭⎫⎝⎛-=ρρρ)(Pa p 7570710013.17602001081.95.054=⨯⨯+⨯⨯=∆ ∆p/ρg=7.7mThe relation between the hole velocity and veloc ity of pipeFriction losssoH=7.7+10+5.1=22.8m2.3 . A centrifugal pump is to be used to extract water from a condenser in which the vacuum is 640 mm of mercury, as shown in figure. At the rated discharge, the net positive suction head must be at least 3m above the cavitation vapor pressure of 710mm mercury vacuum. If losses in the suction pipe accounted for a head of 1.5m. What must be the least height of the liquid level in the condenser above the pump inlet? Solution :From an energy balance,WhereP o =760-640=120mmHgP v =760-710=50mmHgUse of the equation will give the minimum height H g as2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half, what will the new flowrate be ?• Density of acid 1840kg/m 3• V iscosity of acid 25×10-3PasSolution:Velocity of acid in the pipe:sm D d V V /12112.42200=⎪⎭⎫ ⎝⎛⨯=⎪⎭⎫ ⎝⎛=mgu d l f H f 1.581.92105.0200025.02422=⨯⨯==NPSH H gp p H f vog ---=ρmNPSH H gp p H f vo g 55.335.181.9100081.913600)05.012.0-=--⨯⨯⨯-=---=(ρs m d m d mpipe of area tional cross flowrate volumetric u /32.3025.01840785.03785.04sec 222=⨯⨯===-=ρπρReynolds number:6109102532.31840025.0Re 3=⨯⨯⨯==-μρud from Fig.1.22 for a smooth pipe when Re=6109, f=0.0085 pressure drop is calculated from equation 1.4-9kg J u d l f ph f /450232.3025.0600085.042422=⨯==∆=ρkPa p 5.8271840450=⨯=∆or friction factor is calculated from equation1.4-25kgJ u d l u d l f ph f /426232.3025.0606109046.042Re 046.042422.022.02=⨯⨯⨯==∆=--=ρkPa p 84.7831840426=⨯=∆if the pressure drop falls to 783.84/2=391.92kPa8.18.12.12.038.12.12.022.0012.089.1079`2025.060102518401840046.042046.042Re 046.043919202u u u dl u d l p p =⎪⎭⎫⎝⎛⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛⨯⨯⨯==∆='∆----ρμρρ=sos m u /27.236.489..1079012.03919208.18.1==⨯=new mass flowrate=0.785d 2u ρ=0.785×0.0252×2.27×1840=2.05kg/s2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half on assumption that the change of friction factor is negligible, what will the new flowrate be ?Density of acid 1840kg/m 3V iscosity of acid 25×10-3Pa Friction factor 32.0Re 500.00056.0+=f for hydraulically smooth pipeSolution:Write energy balance equation:f h gu z g p H g u z g p +++=+++2222222111ρρ gu d l g p h H f 22λρ=∆== 342=ρπu ds m d u /32.31840025.014.3124322=⨯⨯=⨯=ρπ 611510251840025.032.3Re 3=⨯⨯⨯=-0087.061155.00056.0Re 500.00056.032.032.0=+=+=f92.4681.9232.3025.0600087.04222=⨯⨯==∆==g u d l g p h H f λρΔp=46.92×1840×9.81=847.0kpa2.6 The fluid is pumped through the horizontal pipe from section A to B with the φ38⨯2.5mm diameter and length of 30 meters, shown as figure. The orifice meter of 16.4mm diameter is used to measure the flow rate. Orifice coefficient C o =0.63. the permanent loss in pressure is 3.5×104N/m 2, the friction coefficient λ=0.024. find:(1) What is the pressure drop along the pipe AB?(2)What is the ratio of power obliterated in pipe AB to total power supplied to the fluid when the shaft work is 500W , 60%efficiency? (The density of fluid is 870kg/m 3)solution :∑+++=+++f AA A A AA h u p g z w u p g z 2222ρρρλρ22p u d l h p p f BA ∆+==-∑ 247.0334.162=⎪⎭⎫⎝⎛=A A o()()s m gR C u /5.8870870136006.081.9297.063.02247.01200=-⨯⨯=''--=ρρρ∴u = (16.4/33)2×8.5=2.1m/s∴242/76855105.321.2033.030870024.0m N h p p f B A =⨯+⨯==-∑ρ(2)W u d p Wm 1381.2033.0785.0768554Ne 22=⨯⨯⨯=∆==ρπρ sothe ratio of power obliterated in friction losses in AB to total power supplied to the fluid%%=461006.0500138⨯⨯AcknowledgementsMy deepest gratitude goes first and foremost to Professor aaa , my supervisor, for her constant encouragement and guidance. She has walked me through all the stages of the writing of this thesis. Without her consistent and illuminating instruction, this thesis could not havereached its present form.Second, I would like to express my heartfelt gratitude to Professor aaa, who led me into the world of translation. I am also greatly indebted to the professors and teachers at the Department of English: Professor dddd, Professor ssss, who have instructed and helped me a lot in the past two years.Last my thanks would go to my beloved family for their loving considerations and great confidence in me all through these years. I also owe my sincere gratitude to my friends and my fellow classmates who gave me their help and time in listening to me and helping me work out my problems during the difficult course of the thesis.My deepest gratitude goes first and foremost to Professor aaa , my supervisor, for her constant encouragement and guidance. She has walked me through all the stages of the writing of this thesis. Without her consistent and illuminating instruction, this thesis could not havereached its present form.Second, I would like to express my heartfelt gratitude to Professor aaa, who led me into the world of translation. I am also greatly indebted to the professors and teachers at the Department of English: Professordddd, Professor ssss, who have instructed and helped me a lot in the past two years.Last my thanks would go to my beloved family for their loving considerations and great confidence in me all through these years. I also owe my sincere gratitude to my friends and my fellow classmates who gave me their help and time in listening to me and helping me work out my problems during the difficult course of the thesis.。

流体力学英语词汇翻译(1)

流体力学英语词汇翻译(1)

流体力学英语词汇翻译(1)流体动力学 fluid dynamics连续介质力学 mechanics of continuous media介质 medium流体质点 fluid particle无粘性流体 nonviscous fluid, inviscid fluid连续介质假设 continuous medium hypothesis流体运动学 fluid kinematics水静力学 hydrostatics液体静力学 hydrostatics支配方程 governing equation伯努利方程 Bernoulli equation伯努利定理 Bernonlli theorem毕奥-萨伐尔定律 Biot-Savart law欧拉方程 Euler equation亥姆霍兹定理 Helmholtz theorem开尔文定理 Kelvin theorem涡片 vortex sheet库塔-茹可夫斯基条件 Kutta-Zhoukowski condition 布拉休斯解 Blasius solution达朗贝尔佯廖 d'Alembert paradox雷诺数 Reynolds number施特鲁哈尔数 Strouhal number随体导数 material derivative不可压缩流体 incompressible fluid质量守恒 conservation of mass动量守恒 conservation of momentum能量守恒 conservation of energy动量方程 momentum equation能量方程 energy equation控制体积 control volume液体静压 hydrostatic pressure涡量拟能 enstrophy压差 differential pressure流[动] flow热电偶高温计 thermocouple pyrometer 流线 stream line流面 stream surface流管 stream tube迹线 path, path line流场 flow field流态 flow regime流动参量 flow parameter流量 flow rate, flow discharge涡旋 vortex涡量 vorticity涡丝 vortex filament涡线 vortex line涡面 vortex surface涡层 vortex layer涡环 vortex ring涡对 vortex pair涡管 vortex tube涡街 vortex street卡门涡街 Karman vortex street马蹄涡 horseshoe vortex对流涡胞 convective cell卷筒涡胞 roll cell涡 eddy涡粘性 eddy viscosity环流 circulation环量 circulation速度环量 velocity circulation偶极子 doublet, dipole驻点 stagnation point总压[力] total pressure总压头 total head静压头 static head总焓 total enthalpy能量输运 energy transport速度剖面 velocity profile库埃特流 Couette flow单相流 single phase flow单组份流 single-component flow均匀流 uniform flow非均匀流 nonuniform flow二维流 two-dimensional flow三维流 three-dimensional flow准定常流 quasi-steady flow非定常流 unsteady flow, non-steady flow 暂态流 transient flow周期流 periodic flow振荡流 oscillatory flow分层流 stratified flow无旋流 irrotational flow有旋流 rotational flow轴对称流 axisymmetric flow不可压缩性 incompressibility不可压缩流[动] incompressible flow浮体 floating body定倾中心 metacenter阻力 drag, resistance减阻 drag reduction表面力 surface force表面张力 surface tension毛细[管]作用 capillarity来流 incoming flow自由流 free stream自由流线 free stream line外流 external flow进口 entrance, inlet出口 exit, outlet扰动 disturbance, perturbation分布 distribution传播 propagation色散 dispersion弥散 dispersion附加质量 added mass ,associated mass 收缩 contraction镜象法 image method无量纲参数 dimensionless parameter 几何相似 geometric similarity运动相似 kinematic similarity动力相似[性] dynamic similarity平面流 plane flow势 potential势流 potential flow速度势 velocity potential复势 complex potential复速度 complex velocity流函数 stream function源 source汇 sink速度[水]头 velocity head拐角流 corner flow空泡流 cavity flow超空泡 supercavity超空泡流 supercavity flow空气动力学 aerodynamics低速空气动力学 low-speed aerodynamics 高速空气动力学 high-speed aerodynamics 气动热力学 aerothermodynamics亚声速流[动] subsonic flow跨声速流[动] transonic flow超声速流[动] supersonic flow锥形流 conical flow楔流 wedge flow叶栅流 cascade flow非平衡流[动] non-equilibrium flow细长体 slender body细长度 slenderness钝头体 bluff body钝体 blunt body翼型 airfoil翼弦 chord薄翼理论 thin-airfoil theory构型 configuration后缘 trailing edge迎角 angle of attack失速 stall脱体激波 detached shock wave波阻 wave drag诱导阻力 induced drag诱导速度 induced velocity临界雷诺数 critical Reynolds number前缘涡 leading edge vortex附着涡 bound vortex约束涡 confined vortex气动中心 aerodynamic center气动力 aerodynamic force气动噪声 aerodynamic noise气动加热 aerodynamic heating离解 dissociation地面效应 ground effect气体动力学 gas dynamics稀疏波 rarefaction wave热状态方程 thermal equation of state喷管 Nozzle普朗特-迈耶流 Prandtl-Meyer flow瑞利流 Rayleigh flow可压缩流[动] compressible flow可压缩流体 compressible fluid绝热流 adiabatic flow非绝热流 diabatic flow未扰动流 undisturbed flow等熵流 isentropic flow匀熵流 homoentropic flow兰金-于戈尼奥条件 Rankine-Hugoniot condition 状态方程 equation of state量热状态方程 caloric equation of state 完全气体 perfect gas拉瓦尔喷管 Laval nozzle马赫角 Mach angle马赫锥 Mach cone马赫线 Mach line马赫数 Mach number马赫波 Mach wave当地马赫数 local Mach number冲击波 shock wave激波 shock wave正激波 normal shock wave斜激波 oblique shock wave头波 bow wave附体激波 attached shock wave激波阵面 shock front激波层 shock layer压缩波 compression wave反射 reflection折射 refraction散射 scattering衍射 diffraction绕射 diffraction出口压力 exit pressure超压[强] over pressure反压 back pressure爆炸 explosion爆轰 detonation缓燃 deflagration水动力学 hydrodynamics液体动力学 hydrodynamics泰勒不稳定性 Taylor instability盖斯特纳波 Gerstner wave斯托克斯波 Stokes wave瑞利数 Rayleigh number自由面 free surface波速 wave speed, wave velocity 波高 wave height波列 wave train波群 wave group波能 wave energy表面波 surface wave表面张力波 capillary wave规则波 regular wave不规则波 irregular wave浅水波 shallow water wave深水波 deep water wave重力波 gravity wave椭圆余弦波 cnoidal wave潮波 tidal wave涌波 surge wave破碎波 breaking wave船波 ship wave非线性波 nonlinear wave孤立子 soliton水动[力]噪声 hydrodynamic noise 水击 water hammer空化 cavitation空化数 cavitation number空蚀 cavitation damage超空化流 supercavitating flow水翼 hydrofoil水力学 hydraulics洪水波 flood wave涟漪 ripple消能 energy dissipation海洋水动力学 marine hydrodynamics 谢齐公式 Chezy formula欧拉数 Euler number弗劳德数 Froude number水力半径 hydraulic radius水力坡度 hvdraulic slope高度水头 elevating head水头损失 head loss水位 water level水跃 hydraulic jump含水层 aquifer排水 drainage排放量 discharge壅水曲线 back water curve压[强水]头 pressure head过水断面 flow cross-section明槽流 open channel flow孔流 orifice flow无压流 free surface flow有压流 pressure flow缓流 subcritical flow急流 supercritical flow渐变流 gradually varied flow急变流 rapidly varied flow临界流 critical flow异重流 density current, gravity flow堰流 weir flow掺气流 aerated flow含沙流 sediment-laden stream降水曲线 dropdown curve沉积物 sediment, deposit沉[降堆]积 sedimentation, deposition沉降速度 settling velocity流动稳定性 flow stability不稳定性 instability奥尔-索末菲方程 Orr-Sommerfeld equation 涡量方程 vorticity equation泊肃叶流 Poiseuille flow奥辛流 Oseen flow剪切流 shear flow粘性流[动] viscous flow层流 laminar flow分离流 separated flow二次流 secondary flow近场流 near field flow远场流 far field flow滞止流 stagnation flow尾流 wake [flow]回流 back flow反流 reverse flow射流 jet自由射流 free jet管流 pipe flow, tube flow内流 internal flow热电偶高温计 thermocouple pyrometer 拟序结构 coherent structure猝发过程 bursting process表观粘度 apparent viscosity运动粘性 kinematic viscosity动力粘性 dynamic viscosity泊 poise厘泊 centipoise厘沱 centistoke剪切层 shear layer次层 sublayer流动分离 flow separation层流分离 laminar separation湍流分离 turbulent separation分离点 separation point附着点 attachment point再附 reattachment再层流化 relaminarization起动涡 starting vortex驻涡 standing vortex涡旋破碎 vortex breakdown涡旋脱落 vortex shedding压[力]降 pressure drop压差阻力 pressure drag压力能 pressure energy型阻 profile drag滑移速度 slip velocity无滑移条件 non-slip condition壁剪应力 skin friction, frictional drag 壁剪切速度 friction velocity磨擦损失 friction loss磨擦因子 friction factor耗散 dissipation滞后 lag相似性解 similar solution局域相似 local similarity气体润滑 gas lubrication液体动力润滑 hydrodynamic lubrication 浆体 slurry泰勒数 Taylor number纳维-斯托克斯方程 Navier-Stokes equation 牛顿流体 Newtonian fluid边界层理论 boundary later theory边界层方程 boundary layer equation边界层 boundary layer附面层 boundary layer层流边界层 laminar boundary layer湍流边界层 turbulent boundary layer温度边界层 thermal boundary layer边界层转捩 boundary layer transition边界层分离 boundary layer separation边界层厚度 boundary layer thickness位移厚度 displacement thickness。

流体力学英文版1

流体力学英文版1
(2) Recovery – For a solid, if the elastic limit is not exceeded, the deformation disappears when the force is removed; – A fluid continues to flow as long as the force is applied and will not recover its original form when the force is removed.
若连续介质不适用,应如何处理呢? 13
1.3 Dimension and Units 单位与量纲
Dimensions are properties that can be measured, e.g. length, velocity, area, volume, acceleration etc. 可以测量的性质叫量纲。
3
Figure 1.1 Behavior of (a) solid and (b) fluid, under the action of a constant shear
solid
fluid
与时间无关
与时间有关
4
Main differences between the behavior of solids and
It breaks down whenever the mean free path of the molecules (平 均分子自由行程,10−7 )approximately becomes the same magnitude order as the smallest significant characteristic dimension (特征长度) of the problem.

流体力学_英文课件第1章

流体力学_英文课件第1章

∂T q = −k ∂n
q: heat flux in n direction per unit area
k: coefficient of thermal conductivity T: temperature n: direction of heat transfer
1.3 The Fluid as a Continuum (连续介质 连续介质) 连续介质
Turbulence Famous experiment on transition Reynolds Number
20th century
Ludwig Prandtl (1875-1953)
Boundary layer theory(1904)
To be the single most important tool in modern flow analysis.
Shear stress
y
U U
du τ∝ dy
Velocity gradient
u(y)
x
du τ =µ dy
The velocity gradient is comparable to deformation.
This kind of linear fluid is called Newtonian fluid. (牛顿流体 牛顿流体) 牛顿流体
陆利蓬 李秋实 王洪伟 景晓东 邹正平 李志平 (4-5 班) (1-3班)
Grades:
30% homework + 70% final exam
Contents
Chapter 1………Introduction (6 hours) Chapter 2………Fluid Statics (4 hours) Chapter 3………Integral Relations (12 hours) Chapter 4………Differential Relations (8 hours) Chapter 5………Boundary Layer (6 hours) Chapter 6………Flow Compressibility (2 hours) Chapter 7………Vorticity (4 hours) Summary (2 hours)

流体力学基础知识(英文)

流体力学基础知识(英文)

1
Fundamentals of Ship Science MSc Course SESS6001 University of Southampton Inviscid Fluid An inviscid fluid or ideal fluid is a simplified conceptual idealisation of a real fluid. An inviscid fluid cannot support shear. Hence in the vicinity of solid boundaries the ‗no-slip‘ boundary condition of a real fluid is not realisable. This implies that continuity of fluid velocity across such boundaries is limited to the normal velocity component as ‗slippage‘ is possible in the tangential directuations of fluid motion (Lagrangian and Eulerian) are only dependent upon pressure gradient and body or gravitational influences and not upon the real fluid shear influences due to fluid viscosity. These alternative forms of motion equations are derived from first principles in a later section. Incompressible Fluid An incompressible fluid is one that neither gains nor losses mass in a selected volume V bounded by the surface S . Schematically we can think of this situation as corresponding to number of particles flowing into V corresponds to number of particles flowing out of V subject to no mass change within V . n ρv . n dS = normal mass flux from the elemental surface dS.

最新流体力学英文版课后习题答案

最新流体力学英文版课后习题答案

1.1 What will be the (a) the gauge pressure and (b) the absolute pressure of water at depth 12mbelow the surface? ρwater = 1000 kg/m 3, and P atmosphere = 101kN/m 2.Solution:Rearranging the equation 1.1-4gh p p a b ρ+=Set the pressure of atmosphere to be zero, then the gauge pressure at depth 12m below the surfaceiskPa gh p p a b 72.1171281.910000=⨯⨯+=+=ρAbsolute pressure of water at depth 12mkPa Pa gh p p a b 72.2182187201281.91000101000==⨯⨯+=+=ρ1.3 A differential manometer as shown in Fig. is sometimes used to measure small pressuredifference. When the reading is zero, the levels in two reservoirs are equal. Assume that fluid B ismethane (甲烷), that liquid C in the reservoirs is kerosene (specific gravity = 0.815), and thatliquid A in the U tube is water. The inside diameters of the reservoirs and U tube are 51mm and6.5mm , respectively. If the reading of the manometer is145mm., what is the pressure differenceover the instrument In meters of water, (a) when the change in the level in the reservoirs isneglected, (b) when the change in the levels in the reservoirs is taken into account? What is thepercent error in the answer to the part (a)?Solution :p a =1000kg/m 3 p c =815kg/m 3 p b =0.77kg/m 3 D/d=8 R=0.145mWhen the pressure difference between two reservoirs is increased, the volumetric changes in the reservoirs and U tubesR d x D 2244ππ= (1) so R D d x 2⎪⎭⎫ ⎝⎛= (2) and hydrostatic equilibrium gives following relationshipg R g x p g R p A c c ρρρ++=+21 (3)sog R g x p p c A c )(21ρρρ-+=- (4)substituting the equation (2) for x into equation (4) givesg R g R D d p p c A c )(221ρρρ-+⎪⎭⎫ ⎝⎛=- (5) (a )when the change in the level in the reservoirs is neglected,()Pa g R g R g R D d p p c A c A c 26381.98151000145.0)()(221=⨯-=-≈-+⎪⎭⎫ ⎝⎛=-ρρρρρ(b )when the change in the levels in the reservoirs is taken into account()Pa g R g R D d g R g R D d p p c A c c A c 8.28181.98151000145.081.9815145.0515.6)()(22221=⨯-+⨯⨯⨯⎪⎭⎫ ⎝⎛=-+⎪⎭⎫ ⎝⎛=-+⎪⎭⎫ ⎝⎛=-ρρρρρρ error=%=7.68.2812638.281- 1.4 There are two U-tube manometers fixed on the fluid bed reactor, as shown in the figure. The readings of two U-tube manometers are R 1=400mm ,R 2=50mm, respectively. The indicating liquid is mercury. The top of the manometer is filled with the water to prevent from the mercury vapor diffusing into the air, and the height R 3=50mm. Try to calculate the pressure at point A and B .Solution: There is a gaseous mixture in the U-tube manometer meter. The densities of fluids are denoted by Hg O H g ρρρ,,2, respectively. The pressure at point A is given by hydrostatic equilibriumFigure for problem 1.4g R R g R g R p g Hg O H A )(32232+-+=ρρρg ρis small and negligible in comparison with Hg ρand ρH2O , equation above can be simplifiedc A p p ≈=232gR gR Hg O H ρρ+=1000×9.81×0.05+13600×9.81×0.05=7161N/m²1gR p p p Hg A D B ρ+=≈=7161+13600×9.81×0.4=60527N/m1.5 Water discharges from the reservoir through the drainpipe, which the throat diameter is d. The ratio of D to d equals 1.25. The vertical distance h between the tank A and axis of the drainpipe is 2m. What height H from the centerline of the drainpipe to the water level in reservoir is required for drawing the water from the tank A to the throat of the pipe? Assume that fluid flow is a potential flow. The reservoir, tank A and the exit ofdrainpipe are all open to air.Solution: Bernoulli equation is written between stations 1-1 and 2-2, with station 2-2 being reference plane: 2222222111u gz p u gz p ++=++ρρ Where p 1=0, p 2=0, and u 1=0, simplification of the equation22u Hg =1The relationship between the velocity at outlet and velocity u o at throat can be derived by the continuity equation:22⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛D d u u o 22⎪⎭⎫ ⎝⎛=d D u u o 2 Bernoulli equation is written between the throat and the station 2-23 Combining equation 1,2,and 3 givesSolving for HH=1.39m1.6 A liquid with a constant density ρ kg/m 3 is flowing at an unknown velocity V 1 m/s through a horizontal pipe of cross-sectional area A 1 m 2 at a pressure p 1 N/m 2, and then it passes to a section of the pipe in which the area is reduced gradually to A 2 m 2 and the pressure is p2. Assuming no friction losses, calculate the velocities V 1 and V 2 if the pressure difference (p 1 - p 2) is measured. Solution :In Fig1.6, the flow diagram is shown with pressure taps to measure p 1 and p 2. From the mass-balance continuity equation , for constant ρ where ρ1 = ρ2 = ρ,222200u u p =+ρ()===144.281.92100081.910002125.11112442-⨯⨯⨯--⎪⎭⎫ ⎝⎛==ρρg h d D u Hg2112A A V V = For the items in the Bernoulli equation , for a horizontal pipe,z 1=z 2=0Then Bernoulli equation becomes, after substituting 2112A A V V = for V 2, ρρ22121211212020p A A V p V ++=++ Rearranging,2)1(21212121-=-A A V p p ρ ⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫ ⎝⎛-12221211A A p p V ρ=Performing the same derivation but in terms of V 2,⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--21221212A A p p V ρ=1.7 A liquid whose coefficient of viscosity is µ flows below the critical velocity for laminar flow in a circular pipe of diameter d and with mean velocity V . Show that the pressure loss in a length of pipe L p ∆ is 232d V μ. Oil of viscosity 0.05 Pas flows through a pipe of diameter 0.1m with a average velocity of 0.6m/s.Calculate the loss of pressure in a length of 120m.Solution :The average velocity V for a cross section is found by summing up all the velocities over the cross section and dividing by the cross-sectional area1From velocity profile equation for laminar flow2 substituting equation 2 for u into equation 1 and integrating3rearranging equation 3 gives1.8. In a vertical pipe carrying water, pressure gauges areinserted at points A and B where the pipe diameters are0.15m and 0.075m respectively. The point B is 2.5m belowA and when the flow rate down the pipe is 0.02 m 3/s, thepressure at B is 14715 N/m 2 greater than that at A.Assuming the losses in the pipe between A and B can be expressed as g V k 22where V is the velocity at A, find the value of k .If the gauges at A and B are replaced by tubes filled with water and connected to a U-tube containing mercury of relative density 13.6, give a sketch showing how the levels in the two limbs of the U-tube differ and calculate the value of this difference in metres.Solution:d A =0.15m; d B =0.075mz A -z B =l =2.5mQ =0.02 m 3/s,p B -p A =14715 N/m 2 Figure for problem 1.8 ⎰⎰==R R rdr u R udA A V 020211ππ⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛--=22014R r R L p p u L μ2032D L p p V L μ-=232d V L p μ=∆Pa d VL p 115201.01206.005.0323222=⨯⨯⨯==∆μs m d QV V d Q AA AA /132.115.0785.002.044222=⨯===ππsm d Q V V d Q BB BB /529.4075.0785.002.044222=⨯===ππWhen the fluid flows down, writing mechanical balance equation222222AB B B A A A V k V g z p Vg z p +++=++ρρ213.1253.4100014715213.181.95.2222k ++=+⨯k 638.0260.10715.14638.0525.24++=+=k 0.295making the static equilibriumgR g x g l p g R g x p Hg A B ρρρρρ+∆++=+∆+()()mm g g l p p R g H A B 7981.91260081.910005.214715-=⨯⨯⨯-=---=ρρρ1.9.The liquid vertically flows down through the tube from thestation a to the station b , then horizontally through the tube fromthe station c to the station d , as shown in figure. Two segments ofthe tube, both ab and cd ,have the same length, the diameter androughness.Find:(1)the expressions of g p ab ρ∆, h fab , g pcdρ∆and h fcd , respectively.(2)the relationship between readings R 1and R 2 in the U tube.Solution:(1) From Fanning equationFigure for problem 1.92V l h fab λ=andsoFluid flows from station a to station b , mechanical energy conservation giveshence2from station c to station dhence3From static equationp a -p b =R 1(ρˊ-ρ)g -l ρg 4p c -p d =R 2(ρˊ-ρ)g5 Substituting equation 4 in equation 2 ,thentherefore6Substituting equation 5 in equation 3 ,then7ThusR 1=R 222V d l h fcd λ=fcdfab h h =fab b a h p p+=+ρρlg fab b a h p p =+-lg ρfcddc h pp +=ρρfcd d c h p p =-ρfabh g l g R =+--'lg 1ρρρρ)(g R h fab ρρρ-'=1g R h fcd ρρρ-'=21.10 Water passes through a pipe of diameter d i=0.004 m with the average velocity 0.4 m/s, as shown in Figure.1) What is the pressure drop –∆P when water flows through the pipe length L =2 m, in m H 2O column?2) Find the maximum velocity and point r at which it occurs.3) Find the point r at which the average velocityequals the local velocity. 4)if kerosene flows through this pipe ,how do thevariables above change ?(the viscosity and density of Water are 0.001 Pasand 1000 kg/m 3,respectively ;and the viscosityand density of kerosene are 0.003 Pas and 800kg/m 3,respectively )solution:1)1600001.01000004.04.0Re =⨯⨯==μρud from Hagen-Poiseuille equation1600004.0001.024.0323222=⨯⨯⨯==∆d uL P μ m g p h 163.081.910001600=⨯=∆=ρ 2)maximum velocity occurs at the center of pipe, from equation 1.4-19max 0.5V u = so u max =0.4×2=0.8m3)when u=V=0.4m/s Eq. 1.4-172max 1⎪⎪⎭⎫ ⎝⎛-=wr r u u 5.0004.01max2=⎪⎭⎫ ⎝⎛-u V r = m r 00284.071.0004.05.0004.0=⨯== 4) kerosene:427003.0800004.04.0Re =⨯⨯==μρud Pa pp 4800001.0003.01600=='∆='∆μμFigure for problem 1.10m g p h 611.081.98004800=⨯=''∆='ρ1.12 As shown in the figure, the water level in the reservoir keeps constant. A steel drainpipe (with the inside diameter of 100mm) is connected to the bottom of the reservoir. One arm of the U-tube manometer is connected to the drainpipe at the position 15m away from the bottom of the reservoir, and the other is opened to the air, the U tube is filled with mercury and the left-side arm of the U tube above the mercury is filled with water. The distance between the upstream tap and the outlet of the pipeline is 20m.a) When the gate valve is closed, R=600mm, h=1500mm; when the gate valve is opened partly, R=400mm, h=1400mm. The friction coefficient λ is 0.025, and the lo ss coefficient of the entrance is 0.5. Calculate the flow rate of water when the gate valve is opened partly. (in m³/h)b) When the gate valve is widely open, calculate the static pressure at the tap (in gauge pressure, N/m²). l e /d ≈15 when the gate valve is widely open, and the friction coefficient λ is still 0.025.Solution :(1) When the gate valve is opened partially, the water discharge isSet up Bernoulli equation between the surface of reservoir 1—1’ and the section of pressure point 2—2’,and take the center of section 2—2’ as the referring plane, then∑+++=++21,2222121122—f h p u gZ p u gZ ρρ (a ) In the equation 01=p (the gauge pressure)222/396304.181.910004.081.913600m N gh gR p O H Hg =⨯⨯-⨯⨯=-=ρρ Figure for problem 1.120021=≈Z uWhen the gate valve is fully closed, the height of water level in the reservoir can be related to h (the distance between the center of pipe and the meniscus of left arm of U tube).gR h Z g Hg O H ρρ=+)(12 (b )where h=1.5mR=0.6mSubstitute the known variables into equation b 2222_1,113.22)5.01.015025.0(2)(66.65.110006.013600V V V K d l h m Z c f =+⨯=+==-⨯=∑λ Substitute the known variables equation a9.81×6.66=2213.21000396302V V ++ the velocity is V =3.13m/sthe flow rate of water is h m V d V h /5.8813.312.0436004360032=⨯⨯⨯=⨯=ππ2) the pressure of the point where pressure is measured when the gate valve is wide-open. Write mechanical energy balance equation between the stations 1—1’ and 3-3´,then∑+++=++31,3233121122—f h p V gZ p V gZ ρρ (c ) since m Z 66.61=311300p p u Z =≈=2223_1,81.4 2]5.0)151.035(025.0[ 2)(V V V K d l l h c e f =++=++=∑λ input the above data into equation c ,9.8122V 81.4266.6+=⨯Vthe velocity is: V =3.51 m/sWrite mechanical energy balance equation between thestations 1—1’ and 2——2’, for the same situation of water level ∑+++=++21,2222121122—f h p V gZ p V gZ ρρ (d )since m Z 66.61=212103.51/0(page pressure Z u u m s p =≈≈=)kg J V K d l hc f /2.26251.3)5.01.015025.0(2)(222_1,=+⨯=+=∑λ input the above data into equationd , 9.81×6.66=2.261000251.322++p the pressure is: 329702=p1.14 Water at 20℃ passes through a steel pipe with an inside diameter of 300mm and 2m long. There is a attached-pipe (Φ60⨯3.5mm) which is parallel with the main pipe. The total length including the equivalent length of all form losses of the attached-pipe is 10m. A rotameter is installed in the branch pipe. When the reading of the rotameter is2.72m 3/h, try to calculate the flow rate in the main pipe and the total flow rate, respectively. The frictional coefficient of the main pipe and the attached-pipe is 0.018 and 0.03, respectively.Solution : The variables of main pipe are denoted by a subscript1, and branch pipe by subscript 2.The friction loss for parallel pipelines is2121S S s f f V V V h h +==∑∑The energy loss in the branch pipe is 22222222u d l l h e f ∑∑+=λ In the equation 03.02=λs m u d ml l e /343.0053.04360072.2053.01022222=⨯⨯===+∑πinput the data into equation ckg J h f /333.02343.0053.01003.022=⨯⨯=∑The energy loss in the main pipe is 333.022111121===∑∑u d l h h f f λ So s m u /36.22018.023.0333.01=⨯⨯⨯= The water discharge of main pipe ish m V h /60136.23.043600321=⨯⨯⨯=π Total water discharge ish m V h /7.60372.26013≈+=1.16 A Venturimeter is used for measuring flow of water along a pipe. The diameter of the Venturi throat is two fifths the diameter of the pipe. The inlet and throat are connected by water filled tubes to a mercury U-tube manometer. The velocity of flow along the pipe is found to beR 5.2 m/s, where R is the manometer reading in metres of mercury. Determine the loss of head between inlet and throat of the Venturi when R is 0.49m. (Relative density of mercury is 13.6). Solution:Writing mechanical energy balance equation between the inlet1 and throat o for Venturi meterf o o hg z V p g z V p +++=++22121122ρρ 1 rearranging the equation above, and set (z 2-z 1)=xf o oh xg V V p p ++-=-22121ρ 2 from continuity equationFigure for problem 1.1611221125.625V V d d V V o o =⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛= 3 substituting equation 3 for V o into equation 2 gives()f f f f oh xg R h xg R h V h xg V V p p ++=++=+=++-=-94.1185.203.1903.19206.3922121211ρ 4from the hydrostatic equilibrium for manometerg x g R p p Hg o ρρρ+-=-)(1 5 substituting equation 5 for pressure difference into equation 4 obtainsf Hgh xg R gx g R ++=+-94.118)(ρρρρ 6 rearranging equation 6 kg J R R R R g R h Hg f /288.267.494.11861.12394.118)(==-=--=ρρρ1.17.Sulphuric acid of specific gravity 1.3 is flowing through a pipe of 50 mm internal diameter. A thin-lipped orifice, 10mm, is fitted in the pipe and the differential pressure shown by a mercury manometer is 10cm. Assuming that the leads to the manometer are filled with the acid,calculate (a)the weight of acid flowing per second, and (b) the approximate friction loss in pressure caused by the orifice.The coefficient of the orifice may be taken as 0.61, the specific gravity of mercury as 13.6, and the density of water as 1000 kg/m 3Solution: a)2.0501010==D D =⨯-=-=-81.9)130013600(1.0)(21g R p p Hg ρρ()s m p p D D C V o /63.231.461.056.1861.0130081.9)130013600(1.022.0161.021*******=⨯=≈⨯-⨯-=-⎪⎪⎭⎫ ⎝⎛-=ρs kg V D m /268.0130063.201.0442220=⨯⨯⨯==πρπb) approximate pressure drop=⨯-=-=-81.9)130013600(1.0)(21g R p p Hg ρρ12066.3Pa pressure difference due to increase of velocity in passing through the orificePa D D V V V V p p o 8.44882)2.01(63.213002242412222212221=-=⎪⎪⎭⎫ ⎝⎛-=-=-ρρ pressure drop caused by friction lossPa p f 5.75778.44883.12066=-=∆2.1 Water is used to test for the performances of pump. The gauge pressure at the discharge connection is 152 kPa and the reading of vacuum gauge at the suction connection of the pump is 24.7 kPa as the flow rate is 26m 3/h. The shaft power is 2.45kw while the centrifugal pump operates at the speed of 2900r/min. If the vertical distance between the suction connection and discharge connection is 0.4m, the diameters of both the suction and discharge line are the same. Calculate the mechanical efficiency of pump and list the performance of the pump under this operating condition.Solution:Write the mechanical energy balance equation between the suction connection and discharge connection 2_1,2222121122f H gp g u Z H g p g u Z +++=+++ρρ wherem Z Z 4.012=-(Pa 1052.1(Pa 1047.22_1,215241≈=⨯=⨯-=f H u u pressure gauge p pressure gauge p ))total heads of pump is m H 41.1881.9100010247.01052.14.055=⨯⨯+⨯+= efficiency of pump is N N e /=ηsince kW g QH N e 3.1360081.9100041.18263600=⨯⨯⨯==ρN=2.45kWThen mechanical efficiency %1.53%10045.23.1=⨯=η The performance of pump is Flow rate ,m³/h26 Total heads ,m18.41 Shaft power ,kW2.45 Efficiency ,%53.12.2 Water is transported by apump from reactor, which has200 mm Hg vacuum, to thetank, in which the gaugepressure is 0.5 kgf/cm 2, asshown in Fig. The totalequivalent length of pipe is200 m including all localfrictional loss. The pipeline isφ57×3.5 mm , the orificecoefficient of C o and orificediameter d o are 0.62 and 25mm, respectively. Frictional coefficient λ is 0.025. Calculate: Developed head H of pump, in m (the reading R of U pressure gauge in orifice meter is 168 mm Hg)Solution:Equation(1.6-9)Mass flow rates kg S V m o o /02.21000025.0414.312.42=⨯⨯⨯==ρ 2) Fluid flow through the pipe from the reactor to tank, the Bernoulli equation is as follows for V 1=V 2f H z gp p H +∆+-=ρ12 ∆z=10ms m Rg D d C V f /12.444.69375.062.01000)100013600(81.9168.025025162.02144000=⨯=-⨯⨯⎪⎭⎫ ⎝⎛-=-⎪⎭⎫ ⎝⎛-=ρρρ)(Pa p 7570710013.17602001081.95.054=⨯⨯+⨯⨯=∆ ∆p/ρg=7.7mThe relation between the hole velocity and velocity of pipeFriction losssoH=7.7+10+5.1=22.8m2.3 . A centrifugal pump is to be used to extract water from a condenser in which the vacuum is 640 mm of mercury, as shown in figure. At the rated discharge, the netpositive suction head must be at least 3m above the cavitation vaporpressure of 710mm mercury vacuum. If losses in the suction pipeaccounted for a head of 1.5m. What must be the least height of the liquid level in the condenser above the pump inlet?Solution :From an energy balance,WhereP o =760-640=120mmHgP v =760-710=50mmHgUse of the equation will give the minimum height H g as2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half, what will the new flowrate be ?• Density of acid 1840kg/m 3• Viscosity of acid 25×10-3 PasSolution:Velocity of acid in the pipe:s m D d V V /12112.42200=⎪⎭⎫ ⎝⎛⨯=⎪⎭⎫ ⎝⎛=m g u d l f H f 1.581.92105.0200025.02422=⨯⨯==NPSH H g p p H f v og ---=ρm NPSH H g p p H f v o g 55.335.181.9100081.913600)05.012.0-=--⨯⨯⨯-=---=(ρs m d m d mpipe of area tional cross flowrate volumetric u /32.3025.01840785.03785.04sec 222=⨯⨯===-=ρπρReynolds number:6109102532.31840025.0Re 3=⨯⨯⨯==-μρud from Fig.1.22 for a smooth pipe when Re=6109, f=0.0085 pressure drop is calculated from equation 1.4-9kg J u d l f ph f /450232.3025.0600085.042422=⨯==∆=ρ kPa p 5.8271840450=⨯=∆ or friction factor is calculated from equation1.4-25kg J u d l u d l f ph f /426232.3025.0606109046.042Re 046.042422.022.02=⨯⨯⨯==∆=--=ρkPa p 84.7831840426=⨯=∆ if the pressure drop falls to 783.84/2=391.92kPa8.18.12.12.038.12.12.022.0012.089.1079`2025.060102518401840046.042046.042Re 046.043919202u u u d l u d l p p =⎪⎭⎫ ⎝⎛⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛⨯⨯⨯==∆='∆----ρμρρ= so s m u /27.236.489..1079012.03919208.18.1==⨯= new mass flowrate=0.785d 2u ρ=0.785×0.0252×2.27×1840=2.05kg/s2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half on assumption that the change of friction factor is negligible, what will the new flowrate be ?Density of acid 1840kg/m 3Viscosity of acid 25×10-3 Pa Friction factor 32.0Re 500.00056.0+=f for hydraulically smooth pipe Solution:Write energy balance equation:f h gu z g p H g u z g p +++=+++2222222111ρρ gu d l g p h H f 22λρ=∆== 342=ρπu ds m d u /32.31840025.014.3124322=⨯⨯=⨯=ρπ 611510251840025.032.3Re 3=⨯⨯⨯=- 0087.061155.00056.0Re 500.00056.032.032.0=+=+=f 92.4681.9232.3025.0600087.04222=⨯⨯==∆==g u d l g p h H f λρ Δp=46.92×1840×9.81=847.0kpa2.6 The fluid is pumped through the horizontal pipe from section A to B with the φ38⨯2.5mm diameter and length of 30 meters, shown as figure. The orifice meter of 16.4mm diameter is used to measure the flow rate. Orifice coefficient C o =0.63. the permanent loss in pressure is3.5×104N/m 2, the friction coefficient λ=0.024. find:(1) What is the pressure drop along the pipe AB?(2)What is the ratio of power obliterated in pipe AB to total power supplied to the fluid when the shaft work is 500W, 60%efficiency? (The density of fluid is 870kg/m 3 )solution :∑+++=+++f A A A A AA h u p g z w u p g z 2222ρρ ρλρ022p u d l h p p f BA ∆+==-∑247.0334.162=⎪⎭⎫ ⎝⎛=A A o()()s m gR C u /5.8870870136006.081.9297.063.02247.01200=-⨯⨯=''--=ρρρ ∴u = (16.4/33)2×8.5=2.1m/s∴242/76855105.321.2033.030870024.0m N h p p f B A =⨯+⨯==-∑ρ(2)W u d p Wm 1381.2033.0785.0768554Ne 22=⨯⨯⨯=∆==ρπρsothe ratio of power obliterated in friction losses in AB to total power supplied to the fluid%%=461006.0500138⨯⨯。

流体力学(英文版)

流体力学(英文版)

Fluid MechanicsCourse Code: 83165000Course Name: Fluid MechanicsCourse Credit: 3Course Duration: The 3rd SemesterTeaching Object: Undergraduate Students in Space SciencePre-course:Advanced Mathematics, General PhysicsCourse Director: Shi Quanqi Lecturer Philosophical DoctorCourse Introduction:Fluid Mechanics is a foundation course which is used extensively in the engineering .At the same time, it is also a basic course in learning magnetohydrodynamics (MHD)。

The courseincludes the concept of the continuum, kinematics, the hydrodynamics Equations, the similarity theory and dimensional analysis, and the dynamics of gas flow. Application examples in fields of power source, dynamics, Aviation, machinery, space science and daily life will be introduced. When the students finish this course, they will be able to understand the basic concept, the general theory and the analytical method in Fluid Mechanics, and will know its application in the Space Sciences, engineering, and our daily life.Course Examination:Students’ Final Scores = Scores of Ordinary Tests * 30% + Scores of the Final Exam * 70%;Scores of ordinary tests vary according to students’ performance in class and homework.;The final exam will be open-book.Appointed Teaching Materials:[1] Ding Zurong,Fluid Mechanics,Beijing:China Higher Education Press,2005.Bibliography:[1]. Wu Wangyi,Fluid Mechanics,Beijing:Peking University Press,2004.。

fluids mechanics 流体力学

fluids mechanics 流体力学
2 v v T I( v) with = 0 3
(19-9c)


Rectangular coordinates (x, y, z) Normal stresses
2 v xx 2 x v x 3 v y 2 yy 2 v 3 y 2 v zz 2 z v z 3
r-component
2 v v v r v 2 v v v v r vr r r r r r sin r t r p 1 2 1 sin r 1 r rr 2 g r r r r r sin r sin r
(19-11d) (19-11e) (19-11f)
Spherical coordinates (r, , ) Normal stresses v 2 rr 2 r v r 3 1 v v r 2 2 v r 3 r
(19-12d) (19-12e) (19-12f)

Dv p + 2 v + g Dt
Rectangular coordinates (x, y, z )
x-component
2 v v v v v 2 v x 2 v x p x v x x v y x v z x 2x g x x y z x y2 z 2 t x
2 v zz 2 z v z 3 1 1 v v z rv r v r r r z
(19-11a) (19-11b) (19-11c) (19-5b)

流体力学英文版复习

流体力学英文版复习

2013-8-23
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(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.
AV F h
If the shear stress is =F/A, it can be expressed as
(1.18)
V h
The ratio V/h is the angular velocity of line ab, or it is the rate of angular deformation of the fluid.
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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. The energy per unit weight of fluid can be also
2013-8-23
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(1) Density (密度)
Density is the ratio of the mass of fluid to its volume.
m (kg/m3 ) ρ V
(1.12)
Specific volume (比容) : volume occupied by unit mass.

流体力学英文版复习

流体力学英文版复习
Solution:
p K V / V 6 6 3.110 0.110 (998 1000) /1000 1.5 109 (Pa)
7
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(5) Viscosity
Viscosity is an internal property of a fluid that offers resistance to shear deformation. It describes a fluid's 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).
dp
The bulk modulus of elasticity ( 体积弹性模量 ) K is the reciprocal of coefficient of volume compressibility p.
dp K p dV / V
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(Pa)
(1.16)
6
Example 1.1 A mineral oil in cylinder has a volume of 1000cm3 at 0.1MN/m2 and a volume of 998 cm3 at 3.1MN/m2. What is its bulk modulus of elasticity?
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 to the upper plate.

流体力学-1

流体力学-1

M孔的孔面平行与流线。两处的压强差可从 U形管中液面的高度差测得,即
P A PM 1 2
v
2
gh
v
2 gh

式中h是U形管中液面的高度差, 是U形管中液体的密度 。
例题 水在截面不同的水平管中作稳定流动,出口处 的截面积为管的最细处的3倍。若出口处的流速为 2m/s,问最细处的压强为多少?若在此最细处开一 小孔,水会不会流出来? 解:由连续性方程S1 v1 = S2 v2 ,得S2 = 6(m/s)
Q SB
0(m/s) . 12 12 2 10
20 ( m/s) 4 60 10 0 . 12
vB
又根据柏努利方程可得 PB = PA+ρ vA2-ρ vB2-ρ ghB = 9.8× 2 2×105+ ×1000×122 – 2
1 1 ×1000× 202 –1000× 2
= 5.24×104 (N/m2)
三、柏努利方程的应用
1、汾丘里流量计(Venturimeter)
液体 中间逐渐缩小 稳定流动。 S1、P1、v1 和S2、P2、v2
P1 1 2
v1 P2
2
1 2
v2
2
S1 v1 = S2 v2 P1-P2=ρ gh
解得:
v1 S
2 gh
2
S1
第五章
液体的流动
• 流体(fluid)
气体和液体
•流动性 流体各部分之间极易发生相对移动, 因而没有固定的形状。 •流体力学(fluid mechanics)
流体静力学 (hydrostatics) 研究静止流体规律的
学科,如阿基米德原理、帕斯卡原理等。

蔚山大学讲义:流体力学(英语)

蔚山大学讲义:流体力学(英语)

L ECTURE 1F UNDAMENTAL C ONCEPTS,D EFINITIONS IN F LUID M ECHANICS1. Solid and FluidA solid can resist a shear stress by a static deformation; a fluid cannot.Any shearing stress applied to a fluid, no matter how small, will result in motion of that fluid. Shearing stress is created whenever a tangential force acts on a surface. The fluid moves and deforms continuously as long as the shear stress is applied. As a corollary, we can say that a fluid at rest must be in a state of zero shear stress, a state often called the hydrostatic stress condition in structural analysis.2. ContinuumAs far as we know, fluids are aggregations of molecules, widely spaced for a gas, closely spaced for a liquid. The distance between molecules is very large compared with the molecular diameter. The molecules are not fixed in a lattice but move about freely relative to each other. Thus fluid density, or mass per unit volume, has no precise meaning because the number of molecules occupying a given volume continually changes. This effect becomes unimportant if the unit volume is large compared with, say, the cube of the molecular spacing, when the number of molecules within the volume will remain nearly constant in spite of the enormous interchange of particles across the boundaries. If, however, the chosen unit volume is too large, there could be a noticeable variation in the bulk aggregation of the particles.The density of a fluid is best defined asThe limiting volume is about 10-9mm3for all liquids and for gases at atmospheric pressure. Most engineering problems are concerned with physical dimensions much larger than this limiting volume, so that density is essentially a point function and fluid properties can be thought of as varying continually in space. Such a fluid is called a continuum, which simply means that its variation in properties is so smooth that the differential calculus can be used to analyze the substance.3. Stress TensorIn continuum mechanics, stress is a measure of the average force per unit area of a surface within a deformable body, and therefore has units of N/m2(Pa), dynes/cm2, psi, etc. Stress depends on two quantities that are both vectors, namely force and the orientation of the surface. This means that stress is actually a second order tensor, which is the generalization of a vector. The stress tensor is denoted by σ, and its components are denoted by .• i: refers to the plane on which the stress acts, or more specifically, to the orientation of the outward normal to that plane.• j: refers to the direction in which the force acts.Thus, refers to a stress created by a force acting in the y-direction on a surface with a normal in the x-direction. The stress tensor has 9 components:[][][]• i≠j: shear stress.• i = j: normal stress. Considered positive when the force and the normal vector to the surface point in the same direction. That means that tensile stresses arepositive and compressive stresses are negative•the stress tensor is symmetric, so that4. Lagrangian and Eulerian DescriptionsLagrangian approach : described as a material-following approach. For example, if a fluid particle was at position x0at time t=0, then its velocity would be expressed as v(t; x0), where the x0in brackets serves to identify the specific fluid particle that we are referring to. The Lagrangian approach is conceptually simple, but turns out to be very difficult in practice because all fluid particles are moving so that it is hard to keep track of which particles are being referred to.Eulerian approach: described the fluid velocity as a function of position, x. This is analogous to looking into a flow and measuring the velocity at a given point x. Even if the point is fixed, the velocity that we measure represents different fluid particles at each instant in time, since the fluid flowing past x is always changing.5. Velocity FieldVelocity is a vector function of position and time and thus has three components u, v, and w, each a scalar field in itself:̂ ̂ ̂6. PressurePressure is the (compression) stress at a point in a static fluid. Next to velocity, the pressure p is the most dynamic variable in fluid mechanics. Differences or gradients in pressure often drive a fluid flow, especially in ducts. In low-speed flows, the actual magnitude of the pressure is often not important, unless it drops so low as to cause vapor bubbles to form in a liquid.7. ViscosityWhen a fluid is sheared, it begins to move at a strain rate inversely proportional to a property called its coefficient of viscosity . Consider a fluid element sheared in one plane by a single shear stress . The shear strain angle will continuously grow with time as long as the stress is maintained.In the limit of infinitesimal changes, this becomesγNewtonian fluid: a fluid whose relationship between shear stress and shear rate of strain is linear and passes through the originNon-Newtonian fluid: a fluid whose flow properties are not described by a single constant value of viscosityShear rateγ̇Apparent or effective viscosity can be represented by Ostwald-de Waele power law:()If n < 1 : Psudoplastic (shear thinning)n > 1 : Dilatant (shear thickening)n = 1 : Newtonian8. Steady & UnsteadyIf no flow variables are changing as a function of time in the Eulerian approach, then the flow is said to be steady. In other words, ifthen, it is steady flow.In this case,However, the fluid particle has acceleration and this tells that9. Streamlines, Streaklines and PathlinesStreamline: a line everywhere tangent to the velocity vector at a given instant.Streakline: the locus of particles which have earlier passed through a prescribed point.Pathline: the actual path traversed by a given fluid particle. Can be found by a time exposure ofa single marked particle moving through the flowTimeline: a set of fluid particles that form a line at a given instant.The streamline is convenient to calculate mathematically, while the other three are easier to generate experimentally. Note that a streamline and a timeline are instantaneous lines, while the pathline and the streakline are generated by the passage of time.In steady flow, streamlines, pathlines, and streaklines are identical.10. Diffusion and ConvectionDiffusion: a passive process that relies on random collisions between molecules. For example, if you very gently put a drop of cream into a still cup of coffee you will see it slowly spread out, even though you are not stirring the coffee. This is due to water molecules in the coffee crashing into the cream and causing the molecules of cream to spread within the coffee. One of the characteristic features of diffusion is that the time t it takes for a molecule to be transported a distance L is given by~L2ℌwhere ℌ is known as the diffusion coefficient. The diffusion coefficient depends onboth the species being transported and the material that it is being transported in, and isa measure of how quickly the substance can. You will notice from the above expressionthat the diffusion time gets very large when the distance L gets large. The rule of thumb is that diffusion is very fast over short distances, but very inefficient over longer distances, and that diffusion of a small substance is faster than diffusion of a large substance. Diffusion is typically responsible for transport processes where the length scale is tens of microns or less.Convection: a transport process due to the motion of a fluid. The transported species gets carried along with the moving fluid. Convection is at work if you stir your coffee up to mix in the cream. It is much more efficient than diffusion over large distances, and is the mechanism of transport in the cardiovascular and respiratory systems. The time scale of the convection process can be estimated by~Luwhere u is the convection velocity。

流体力学(英文版)

流体力学(英文版)

流体力学(英文版)Fluid MechanicsCourse Code: 83165000Course Name: Fluid MechanicsCourse Credit: 3Course Duration: The 3rd SemesterTeaching Object: Undergraduate Students in Space Science Pre-course:Advanced Mathematics, General PhysicsCourse Director: Shi Quanqi Lecturer Philosophical Doctor Course Introduction:Fluid Mechanics is a foundation course which is used extensively in the engineering .At the same time, it is also a basic course in learning magnetohydrodynamics (MHD)。

The course includes the concept of the continuum, kinematics, the hydrodynamics Equations, the similarity theory and dimensional analysis, and the dynamics of gas flow. Application examples in fields of power source, dynamics, Aviation, machinery, space science and daily life will be introduced. When the students finish this course, they will be able to understand the basic concept, the general theory and the analytical method in Fluid Mechanics, and will know its application in the Space Sciences, engineering, and our daily life.Course Examination:Students’ Final Scores = Scores of Ordinary Tests * 30% + Scores of the Final Exam * 70%;Scores of ordinary tests vary according to students’ performance in class and homework.;The final exam will be open-book.Appointed Teaching Materials:[1] Ding Zurong,Fluid Mechanics,Beijing:China HigherEducation Press,2005.Bibliography:[1]. Wu Wangyi,Fluid Mechanics,Beijing:Peking University Press,2004.。

流体力学英语词汇翻译(1)

流体力学英语词汇翻译(1)

流体力学英语词汇翻译(1)流体动力学 fluid dynamics连续介质力学 mechanics of continuous media介质 medium流体质点 fluid particle无粘性流体 nonviscous fluid, inviscid fluid连续介质假设 continuous medium hypothesis流体运动学 fluid kinematics水静力学 hydrostatics液体静力学 hydrostatics支配方程 governing equation伯努利方程 bernoulli equation伯努利定理 bernonlli theorem毕奥-萨伐尔定律 biot-savart law欧拉方程 euler equation亥姆霍兹定理 helmholtz theorem开尔文定理 kelvin theorem涡片 vortex sheet库塔-茹可夫斯基条件 kutta-zhoukowski condition 布拉休斯解 blasius solution达朗贝尔佯廖 d'alembert paradox雷诺数 reynolds number施特鲁哈尔数 strouhal number随体导数 material derivative不可压缩流体 incompressible fluid质量守恒 conservation of mass动量守恒 conservation of momentum能量守恒 conservation of energy动量方程 momentum equation能量方程 energy equation控制体积 control volume液体静压 hydrostatic pressure 涡量拟能 enstrophy压差 differential pressure流[动] flow流线 stream line流面 stream surface流管 stream tube迹线 path, path line流场 flow field流态 flow regime流动参量 flow parameter流量 flow rate, flow discharge 涡旋 vortex涡量 vorticity涡丝 vortex filament涡线 vortex line涡面 vortex surface涡层 vortex layer涡环 vortex ring涡对 vortex pair涡管 vortex tube涡街 vortex street卡门涡街 karman vortex street 马蹄涡 horseshoe vortex对流涡胞 convective cell卷筒涡胞 roll cell涡 eddy涡粘性 eddy viscosity环流 circulation环量 circulation速度环量 velocity circulation偶极子 doublet, dipole驻点 stagnation point总压[力] total pressure总压头 total head静压头 static head总焓 total enthalpy能量输运 energy transport速度剖面 velocity profile库埃特流 couette flow单相流 single phase flow单组份流 single-component flow均匀流 uniform flow非均匀流 nonuniform flow二维流 two-dimensional flow三维流 three-dimensional flow准定常流 quasi-steady flow非定常流 unsteady flow, non-steady flow 暂态流 transient flow周期流 periodic flow振荡流 oscillatory flow分层流 stratified flow无旋流 irrotational flow有旋流 rotational flow轴对称流 axisymmetric flow不可压缩性 incompressibility不可压缩流[动] incompressible flow浮体 floating body定倾中心 metacenter阻力 drag, resistance减阻 drag reduction表面力 surface force表面张力 surface tension毛细[管]作用 capillarity来流 incoming flow自由流 free stream自由流线 free stream line外流 external flow进口 entrance, inlet出口 exit, outlet扰动 disturbance, perturbation分布 distribution传播 propagation色散 dispersion弥散 dispersion附加质量 added mass ,associated mass 收缩 contraction镜象法 image method无量纲参数 dimensionless parameter 几何相似 geometric similarity运动相似 kinematic similarity动力相似[性] dynamic similarity平面流 plane flow势 potential势流 potential flow速度势 velocity potential复势 complex potential复速度 complex velocity流函数 stream function源 source汇 sink速度[水]头 velocity head拐角流 corner flow空泡流 cavity flow超空泡 supercavity超空泡流 supercavity flow空气动力学 aerodynamics低速空气动力学 low-speed aerodynamics 高速空气动力学 high-speed aerodynamics 气动热力学 aerothermodynamics亚声速流[动] subsonic flow跨声速流[动] transonic flow超声速流[动] supersonic flow锥形流 conical flow楔流 wedge flow叶栅流 cascade flow非平衡流[动] non-equilibrium flow细长体 slender body细长度 slenderness钝头体 bluff body钝体 blunt body翼型 airfoil翼弦 chord薄翼理论 thin-airfoil theory构型 configuration后缘 trailing edge迎角 angle of attack失速 stall脱体激波 detached shock wave波阻 wave drag诱导阻力 induced drag诱导速度 induced velocity临界雷诺数 critical reynolds number前缘涡 leading edge vortex附着涡 bound vortex约束涡 confined vortex气动中心 aerodynamic center气动力 aerodynamic force气动噪声 aerodynamic noise气动加热 aerodynamic heating离解 dissociation地面效应 ground effect气体动力学 gas dynamics稀疏波 rarefaction wave热状态方程 thermal equation of state喷管 nozzle普朗特-迈耶流 prandtl-meyer flow瑞利流 rayleigh flow可压缩流[动] compressible flow可压缩流体 compressible fluid绝热流 adiabatic flow非绝热流 diabatic flow未扰动流 undisturbed flow等熵流 isentropic flow匀熵流 homoentropic flow兰金-于戈尼奥条件 rankine-hugoniot condition 状态方程 equation of state量热状态方程 caloric equation of state完全气体 perfect gas拉瓦尔喷管 laval nozzle马赫角 mach angle马赫锥 mach cone马赫线 mach line马赫数 mach number马赫波 mach wave当地马赫数 local mach number 冲击波 shock wave激波 shock wave正激波 normal shock wave斜激波 oblique shock wave头波 bow wave附体激波 attached shock wave 激波阵面 shock front激波层 shock layer压缩波 compression wave反射 reflection折射 refraction散射 scattering衍射 diffraction绕射 diffraction出口压力 exit pressure超压[强] over pressure反压 back pressure爆炸 explosion爆轰 detonation缓燃 deflagration水动力学 hydrodynamics液体动力学 hydrodynamics泰勒不稳定性 taylor instability盖斯特纳波 gerstner wave斯托克斯波 stokes wave瑞利数 rayleigh number自由面 free surface波速 wave speed, wave velocity 波高 wave height波列 wave train波群 wave group波能 wave energy表面波 surface wave表面张力波 capillary wave规则波 regular wave不规则波 irregular wave浅水波 shallow water wave深水波 deep water wave重力波 gravity wave椭圆余弦波 cnoidal wave潮波 tidal wave涌波 surge wave破碎波 breaking wave船波 ship wave非线性波 nonlinear wave孤立子 soliton水动[力]噪声 hydrodynamic noise 水击 water hammer空化 cavitation空化数 cavitation number空蚀 cavitation damage超空化流 supercavitating flow水翼 hydrofoil水力学 hydraulics洪水波 flood wave涟漪 ripple消能 energy dissipation海洋水动力学 marine hydrodynamics 谢齐公式 chezy formula欧拉数 euler number弗劳德数 froude number水力半径 hydraulic radius水力坡度 hvdraulic slope高度水头 elevating head水头损失 head loss水位 water level水跃 hydraulic jump含水层 aquifer排水 drainage排放量 discharge壅水曲线 back water curve压[强水]头 pressure head过水断面 flow cross-section明槽流 open channel flow孔流 orifice flow无压流 free surface flow有压流 pressure flow缓流 subcritical flow急流 supercritical flow渐变流 gradually varied flow急变流 rapidly varied flow临界流 critical flow异重流 density current, gravity flow堰流 weir flow掺气流 aerated flow含沙流 sediment-laden stream降水曲线 dropdown curve沉积物 sediment, deposit沉[降堆]积 sedimentation, deposition沉降速度 settling velocity流动稳定性 flow stability不稳定性 instability奥尔-索末菲方程 orr-sommerfeld equation 涡量方程 vorticity equation泊肃叶流 poiseuille flow奥辛流 oseen flow剪切流 shear flow粘性流[动] viscous flow层流 laminar flow分离流 separated flow二次流 secondary flow近场流 near field flow远场流 far field flow滞止流 stagnation flow尾流 wake [flow]回流 back flow反流 reverse flow射流 jet自由射流 free jet管流 pipe flow, tube flow内流 internal flow拟序结构 coherent structure猝发过程 bursting process表观粘度 apparent viscosity运动粘性 kinematic viscosity动力粘性 dynamic viscosity泊 poise厘泊 centipoise厘沱 centistoke剪切层 shear layer次层 sublayer流动分离 flow separation层流分离 laminar separation湍流分离 turbulent separation分离点 separation point附着点 attachment point再附 reattachment再层流化 relaminarization起动涡 starting vortex驻涡 standing vortex涡旋破碎 vortex breakdown涡旋脱落 vortex shedding压[力]降 pressure drop压差阻力 pressure drag压力能 pressure energy型阻 profile drag滑移速度 slip velocity无滑移条件 non-slip condition壁剪应力 skin friction, frictional drag 壁剪切速度 friction velocity磨擦损失 friction loss磨擦因子 friction factor耗散 dissipation滞后 lag相似性解 similar solution局域相似 local similarity气体润滑 gas lubrication液体动力润滑 hydrodynamic lubrication 浆体 slurry泰勒数 taylor number纳维-斯托克斯方程 navier-stokes equation 牛顿流体 newtonian fluid边界层理论 boundary later theory边界层方程 boundary layer equation边界层 boundary layer附面层 boundary layer层流边界层 laminar boundary layer湍流边界层 turbulent boundary layer温度边界层 thermal boundary layer边界层转捩 boundary layer transition边界层分离 boundary layer separation边界层厚度 boundary layer thickness位移厚度 displacement thickness。

流体力学中科大 英文版

流体力学中科大 英文版

Distinction between a gas and a liquid (2)
● VAPOR: – Gas whose T and P very near the liquid phase
– Steam is a vapor, state near that of water
● GAS: – Super-heated vapor, far away from liquid phase – Volume of gas or vapor greatly affected by ΔT and ΔP ● Thermodynamics
– Irrigation -- sedimentation
– Meteorology -- oceanography
第12页,共57页。
Scope of Fluid Mechanics (2)
Used in the design of:
– Water supply system – Dam spillways – Shock absorbers, brakes – ships, submarines – Aircrafts, rockets
-- breakwaters, marinas -- computer disk drives
-- pumps, HVAC systems -- artificial organs
Yatchs
hang gliders
第13页,共57页。
As fish habitat...
第14页,共57页。
For reptile habitat...
第37页,共57页。
Historical development (2) ● Newton (1642-1727): laws of motion, law of
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11
Significance of continuum concept
As a consequence of the continuum assumption, each fluid property is assumed to have a definite value at every point in space. Therefore fluid properties such as density, temperature, velocity, and so on, are considered to be continuous functions of position and time, and this lead to a field* description of fluid flow. In particular the velocity field is given by
Instead, molecular theory may be used for above cases
8
How to represent flow property at a point in a continuum
For example, the density at point C
C
When δ V is shrank (收缩) to a very small size, it then will
δ V*~10−9 mm3
10
In a continuum, volume of a fluid particle or element satisfy (连续介质中的流体质点、流体微团满足下 列条件):
Large enough in microscope(微观足够大), Small enough in macroscope(宏观足够小), i.e. 10−9 mm3 of air at standard conditions contains approximately 3× 107 molecules.
ML−1T −2
16
z The Principle of Dimensional Homogeneity 量纲一致性原则
In engineering and science, all equations must be dimensionally homogeneous, that is, each additive term in an equation must have the same dimensions.
如果对空间区域的每个点, 都对应某个物理量的一个确 定的值,则称在上确定了该量的一个场。物理量是数量
r r 还是向量决定其为数量或者向量场。 V=V(x,y,z,t)
12
2
NOTE: 连续介质概念对下列场合适用吗?
1. 稀薄气体 2. 微小空间(尺度)的流动
“Continuum concept” breaks down whenever the mean free path of the molecules (分子自由行程) approximately becomes the same magnitude order (相 同数量级) as the smallest significant characteristic dimension (特征长度) of the problem.
5
Problem: Fluids under a shear stress must be in a kind of motion
-------- deformation. How to keep the fluid at rest ? (FW p4)
ANS: Fluid takes shape of container!
6
1
1.2 Continuum concept
连续介质概念
All fluids are composed of molecules in constant motion. However in most engineering applications we are interested in the average or macroscopic effects that we ordinarily perceive and measure. We thus treat a fluid as an infinitely divisible substance, a continuum, thus from macro point of view, we do not need to concern with the behavior of individual molecules. This is the so-called continuum concept in classical fluid mechanics.
若连续介质不适用,应如何处理呢? 13
1.3 Dimension and Units 单位与量纲
Dimensions are properties that can be measured, e.g. length, velocity, area, volume, acceleration etc. 可以测量的性质叫量纲。
It breaks down whenever the mean free path of the molecules (平 均分子自由行程,10−7 )approximately becomes the same magnitude order as the smallest significant characteristic dimension (特征长度) of the problem.
represent a fluid particle/element, and this volume is termed
elemental volume.
9
ρ
=
δm
lim
δV →δV
*
δ
V
For all liquids and for all gases at atmospheric pressure, the value of lower limiting volume is about
or secondary units
导出单位
They are derived units from combination of the above primary ones (units, dimensions or quantities)
For instance, a force has a dimension and SI unit
3
Figure 1.1 Behavior of (a) solid and (b) fluid, under the action of a constant shear
solid
fluid
与时间无关
与时间有关
4
Main differences between the behavior of solids and
– Basic flow analysis Techniques, control volume and system, Eulerian and Lagrangian Descriptions
– Thermodynamic properties of a Fluid
– Viscosity and other secondary properties (Newtonian/nonNewtonian; no-slip condition, turbulence; surface tension )
在经典流体力学中,只考虑分子平均或宏观 作用,不考虑单独分子的性能 。
7
Validity of continuum concept
The concept of a continuum is the basis of classical fluid mechanics. The continuum assumption is valid in treating the behavior of fluids under normal conditions.
Chapter 1 Introduction
概论
Main Subjects
In this chapter, we focus on the following concepts and
definitions:
– Definition of a fluid;
– Continuum concept;
– Units and Dime源自sions;fluids under an applied force
(1) Time-dependency – For a solid, the strain(应变)/deformation is a function of the applied stress and independent of the time over which the force is applied, providing that the elastic limit is not exceeded; – For a fluid, the strain is dependent on time.
(2) Recovery – For a solid, if the elastic limit is not exceeded, the deformation disappears when the force is removed; – A fluid continues to flow as long as the force is applied and will not recover its original form when the force is removed.
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