流体力学答案

1、第一章第一章作业作业 1.2 1.6 1.19Newtons Law of Viscosity(牛顿内摩擦定律牛顿内摩擦定律) ： is dynamic viscosity， N s/m2 (Pa s) Kinematic viscosity (运动粘度运动粘度) (m2/s)specific weight (重度重度) , specific gravity (比重比重) S , modulus of elasticity(体积弹性模量体积弹性模量) K)2 . 1 (dyduAF)s(Pa1 . 0P1秒帕/ddpK第二章第二章例例2.1，2.4。作业作业 2.3, 2.6, 2.9, 2.

2、17331000kg/m ,9806N/mwwwphg-h0hppaFree surface: y=0, p=pagywaterairh is measured vertically from free-liquid 2.6 Bourdon gage A inside a pressure tank reads 12 psi. Another bourdon gage B outside the pressure tank and connected with it reads 20 psi, and an aneroid barometer(无液气压表无液气压表) reads 30 in H

3、g. What is the absolute pressure measured by A in inches of mercury? Solution: wHg312 psi20 psi+30 in Hg12201220in Hg30 in Hg30 in Hg=95.12 in Hg62.413.612AabsAgagebarpppS21 psi1 lb/in3362.4lb/ft62.4lb/inwater2.9 In Fig.2.6c S1 = 0.86, S2 =1.0, h1= l50 mm, h2= 90mm. Find pA in millimeters of mercury

4、 gage. If the barometer reading is 720 mm, what is pA in meters of water absolute?Solution:mmHg34. 56 .13/6 .72O)(mmH6 .7286. 090115021221ShShhAO)(mH8646. 91000) 1/6 .13(34.725(mmHg)34.72534. 57207202AAhp2.17 Neglecting the weight of the container find (a) the force tending to lift the circular top

5、CD, and (b) the compressive load on the pipe wall at A-A. Solution: (a)221144wFpAdhd Sgh第三章第三章例例3.14, 3.16, 3.19 作业作业 3.9, 3.16, 3.19path of a particle(迹线迹线) ，Streamline (流线流线)，stream tube(流管流管) continuity equation (连续性方程连续性方程), (for incompressible, steady flow)Bernoulli equation（伯努利方程）（伯努利方程）1 122Q

6、AVAVg2 g2 z22222111vpzvp断面平均速度断面平均速度Energy equation for an incompressible fluid(不可压缩流不可压缩流体的能量方程，实际流体总流的伯努利方程体的能量方程，实际流体总流的伯努利方程)Liner-momentum equation(线动量方程线动量方程)1 22211221122m N/N22pVpVzzlossesgggg21xxxFQ VV21yyyFQ VV压力压力能能位置势能位置势能动能动能3.9 Neglecting losses, determine the discharge in Fig.3.35. So

7、lution:2222112122zp gVzp gV2221120000)(Vghgh2221121)(VghShS221806. 93048. 0)41375. 0(V22112211222VVgz pgz p1223.16 The pumping system shown in Fig.3.38 must have pressure of 34.47kPa in the discharge line when cavitation is incipient(开始的开始的) at the pump inlet. Calculate the length of pipe from the r

8、eservoir to the pump for this operating condition if the loss in this pipe can be expressed as (V12/2g)(0.03L/D). What horsepower is being supplied to the fluid by the pump? What percent of this power is being used to overcome losses? Barometer reads 762 mm Hg. Solution:By neglecting the loss betwee

9、n the gage and the outlet, (忽略泵忽略泵出口和管道出口之间的损失，以管道出口为出口和管道出口之间的损失，以管道出口为0截面，泵出口（表压计所在截面）截面，泵出口（表压计所在截面）为为g截面，在此两截面应用能量方程。截面，在此两截面应用能量方程。) By application of continuity equation(连续性方程连续性方程) 3.16 图图3.38中显示的水泵系统中，当泵的入口处开始气穴时，压强为中显示的水泵系统中，当泵的入口处开始气穴时，压强为34.47kPa。计算这种运行条件下从水库到泵的管道长度，管道的损失表。计算这种运行条件下从水库到泵的

10、管道长度，管道的损失表示为示为(V12/2g)(0.03L/D)。泵供给流体的马力（功率）是多少？百分之多少。泵供给流体的马力（功率）是多少？百分之多少的功率用来克服流动损失？气压计（测出的是当地大气压）的读数是的功率用来克服流动损失？气压计（测出的是当地大气压）的读数是762 mm Hg。 V1V0ggooVAVAgggogoVVVAAV4)50/100()/(222022ggoogpVpVzzgggg0gzzpressureabsolutein2030200211210gVDL.zgVgpgpvpage12104523Pa.pv5010013251760763.p12030210V/Hgp

11、pg.DLvV1V0303216216()(34.47 100)8.575(m/s)151015ogVpp(m/s)953. 0575. 8)150/50()/(211ooVAAV以水平面为以水平面为2截面，泵入口为截面，泵入口为1截面，在两截面应用能量方程。截面，在两截面应用能量方程。gVDL.Hloss203021lossoopHzgVgpHgp120200lossopHzgVH122pgQHPplossH/HV1V0以水面为为以水面为为2截面，管道出口为截面，管道出口为0截面，在两截面应用能量方程。截面，在两截面应用能量方程。3.19 Determine the shaft horsep

12、ower for an 80 percent efficient pump to discharge 30 l/s through the system of Fig.3.40. The system losses, exclusive of pump losses, are 12 V2/2g, and H=16m.Solution:2321501030)/.(AQV).(VgHgHpower27112168069100021222800.QgHPpowerpower 3.19 图图3.40中，当系统中流量是中，当系统中流量是30 l/s，泵的效率是，泵的效率是80时，时，轴的功率是多少？不考

13、虑泵的损失，系统的损失是轴的功率是多少？不考虑泵的损失，系统的损失是12 V2/2g。并且并且 H=16m。2211221222212000002powerlosspowerpVpVzHzHggVHHg 3.31 Calculate the force components Fx, Fy needed to hold the stationary vane of Fig.3.45. Q0=10 l/s; =1000kg/m3; V0=120m/s. Solution: Linear momentum equation:0202140 Q.QQQQAv dvdvtFcsxcvxxVQcosVQc

14、osVQFx00201xF00201sinVQsinVQFyQ2 3.31图图3.45中，计算能维持叶片处于静止状态的分力中，计算能维持叶片处于静止状态的分力Fx, Fy。Q0=10 l/s; =1000kg/m3; V0=120m/s.第四章第四章例例4.8； 作业作业 4.1, 4.2, 4.13，4.20, 4.26Reynolds number(雷诺数雷诺数) round tubes (圆管圆管) Laminar flow(层流层流): Re2000ulRe VDVDRe有压差，相对运动的平行平板间缝隙流动有压差，相对运动的平行平板间缝隙流动 其中其中a为两平行平板的间距，为两平行平板

15、的间距，b为板宽，对环形间隙为板宽，对环形间隙b=d，l 为板长，为板长，U为上板运动速度为上板运动速度Laminar flow through parallel platesLaminar flow through circular tube(圆管层流圆管层流)21()()4.2.22Uyduph ayyadl3212babaQUpl2horizontal only4.3.10b32pDVl30)(1212ahpdldUaudyQaEnergy equation (能量方程能量方程)沿程损失沿程损失 f is friction factor(沿程阻力系数沿程阻力系数)。Minor losse

16、s(局部损失局部损失) （出口数，入口数；按出口速（出口数，入口数；按出口速度系数计算）度系数计算）Laminar flow, Re200022fL VhfDg22Vhkgflosseshh64Ref 1 2221122112222pVpVzzlossesgggglaminar flow: =2；turbulent flow: =1 4.1 In Fig 4.23a U = 0.7 m/s. Find the rate at which oil is carried into the pressure chamber by the piston and the shear force and

17、total force F acting on the piston. Solution 1: U=0.7 m/s， a=0.0510 3 m， =1 P =0.1 Pa s 因为因为 a d, a l, 故可把二圆柱表面所形成的圆柱环形间隙故可把二圆柱表面所形成的圆柱环形间隙看成是无限宽平行平板间的压差流动。速度可直接用式看成是无限宽平行平板间的压差流动。速度可直接用式(4.2.2)。 4.1 4.1 图图4.234.23a a中油缸内油的表压强为中油缸内油的表压强为0.15MPa0.15MPa，油的粘度，油的粘度0.1Pa0.1Pas s，柱，柱塞的直径塞的直径d=50mm=50mm，柱塞

18、与套筒间的径向间隙，柱塞与套筒间的径向间隙=0.05mm=0.05mm，套筒的长，套筒的长度度l=300mm=300mm。设以力。设以力 F推着柱塞使其以速度推着柱塞使其以速度U U = 0.7 m/s = 0.7 m/s向左运动，求向左运动，求油的漏损流量、柱塞侧面流体的剪应力和力油的漏损流量、柱塞侧面流体的剪应力和力F F的大小。的大小。 262623120 711 10137505 100 05 102 0 1Uydpuayyadl. yayyyy. 660 15 1001 100 15dpp.dll. 0.000050.00005620060.000050.000052363000.0

19、5( 137505 10)15 100.05(13750)2.73 10 (m /s)23Qdudyyydyyy Solution 2: U=0.7 m/s， a=0.0510 3 m， =1 P =0.1 Pa s3632.73 10 m /s212babaQUpl 1214250 1/s2duUdpaydyadl60.000050.0000513750 10 1014250 1/syyduydy 0.1 142501425 Padudy 262()0.15 100.0514250.05 0.1544294.52433.576328.1 NFpddl 4.9 Determine the pr

20、essure drop per meter of 3-mm-ID horizontal tubing for flow of liquid, = 60 cP, sp gr = 0.83, at Re = 200.Solution:20064Re642000200fReRec流动状态为层流，则V,VDReDVflp22 4.9 对一个内径为对一个内径为3mm的水平管，在的水平管，在Re = 200时，流体的时，流体的 = 60 cP, 比重比重S= 0.83, 求流体每米压降。求流体每米压降。DVfLpgVDLfhpf2222 4.13 A test on 300-mm-diameter pip

21、e with water showed a gage difference of 280 mm on a mercury-water manometer connected to two piezometer rings l20 m apart, The flow was 0.23 m3/s. What is the friction factor? Solution: 4.13 在直径在直径300mm的的 流体为水的管道上的实验显示，水银流体为水的管道上的实验显示，水银水差压计在相距水差压计在相距L=120m的两端的压强差为的两端的压强差为280 mm Hg，流量是流量是0.23 m3/s，

22、摩擦系数，摩擦系数f 是多少？是多少？ (Pa)37330760280101325p42/DQAQV22/VpLDf872222pDVfLpgVDLfhpf 4.20 A 60-mm-diameter smooth pipe 150 m long conveys 10 l/s water at 25 from a water main p=1.6 MN/m2, to the top of a building 25 m above the main, What pressure can be maintained at the top of the building? Solution: p=1.6 106Pa, Q=10 10 3m3/s, l=150m, d=60 10 3m, HT=25m, At 25 C, =997kg/m3, =0.893 10 6m2/s ( =1000kg/m3(at 4 C), =1 10 10 6m2/s (at 20 C).) smooth pipe(光滑管光滑管)：f只与雷诺数有关，与相对粗糙度只与雷诺数有关，与相对粗糙度/D无关，