流體力學(xué)答案

1、第一章第一章作業(yè)作業(yè) 1.2 1.6 1.19Newtons Law of Viscosity(牛頓內(nèi)摩擦定律牛頓內(nèi)摩擦定律) ： is dynamic viscosity， N s/m2 (Pa s) Kinematic viscosity (運(yùn)動(dòng)粘度運(yùn)動(dòng)粘度) (m2/s)specific weight (重度重度) , specific gravity (比重比重) S , modulus of elasticity(體積彈性模量體積彈性模量) K)2 . 1 (dyduAF)s(Pa1 . 0P1秒帕/ddpK第二章第二章例例2.1，2.4。作業(yè)作業(yè) 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(無(wú)液氣壓表無(wú)液氣壓表) 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 作業(yè)作業(yè) 3.9, 3.16, 3.19path of a particle(跡線跡線) ，Streamline (流線流線)，stream tube(流管流管) continuity equation (連續(xù)性方程連續(xù)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斷面平均速度斷面平均速度Energy equation for an incompressible fluid(不可壓縮流不可壓縮流體的能量方程，實(shí)際流體總流的伯努利方程體的能量方程，實(shí)際流體總流的伯努利方程)Liner-momentum equation(線動(dòng)量方程線動(dòng)量方程)1 22211221122m N/N22pVpVzzlossesgggg21xxxFQ VV21yyyFQ VV壓力壓力能能位置勢(shì)能位置勢(shì)能動(dòng)能動(dòng)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(開始的開始的) 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間的損失，以管道出口為出口和管道出口之間的損失，以管道出口為0截面，泵出口（表壓計(jì)所在截面）截面，泵出口（表壓計(jì)所在截面）為為g截面，在此兩截面應(yīng)用能量方程。截面，在此兩截面應(yīng)用能量方程。) By application of continuity equation(連續(xù)性方程連續(xù)性方程) 3.16 圖圖3.38中顯示的水泵系統(tǒng)中，當(dāng)泵的入口處開始?xì)庋〞r(shí)，壓強(qiáng)為中顯示的水泵系統(tǒng)中，當(dāng)泵的入口處開始?xì)庋〞r(shí)，壓強(qiáng)為34.47kPa。計(jì)算這種運(yùn)行條件下從水庫(kù)到泵的管道長(zhǎng)度，管道的損失表。計(jì)算這種運(yùn)行條件下從水庫(kù)到泵的

10、管道長(zhǎng)度，管道的損失表示為示為(V12/2g)(0.03L/D)。泵供給流體的馬力（功率）是多少？百分之多少。泵供給流體的馬力（功率）是多少？百分之多少的功率用來(lái)克服流動(dòng)損失？氣壓計(jì)（測(cè)出的是當(dāng)?shù)卮髿鈮海┑淖x數(shù)是的功率用來(lái)克服流動(dòng)損失？氣壓計(jì)（測(cè)出的是當(dāng)?shù)卮髿鈮海┑淖x數(shù)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以水平面為以水平面為2截面，泵入口為截面，泵入口為1截面，在兩截面應(yīng)用能量方程。截面，在兩截面應(yīng)用能量方程。gVDL.Hloss203021lossoopHzgVgpHgp120200lossopHzgVH122pgQHPplossH/HV1V0以水面為為以水面為為2截面，管道出口為截面，管道出口為0截面，在兩截面應(yīng)用能量方程。截面，在兩截面應(yīng)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 圖圖3.40中，當(dāng)系統(tǒng)中流量是中，當(dāng)系統(tǒng)中流量是30 l/s，泵的效率是，泵的效率是80時(shí)，時(shí)，軸的功率是多少？不考

13、慮泵的損失，系統(tǒng)的損失是軸的功率是多少？不考慮泵的損失，系統(tǒng)的損失是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圖圖3.45中，計(jì)算能維持葉片處于靜止?fàn)顟B(tài)的分力中，計(jì)算能維持葉片處于靜止?fàn)顟B(tài)的分力Fx, Fy。Q0=10 l/s; =1000kg/m3; V0=120m/s.第四章第四章例例4.8； 作業(yè)作業(yè) 4.1, 4.2, 4.13，4.20, 4.26Reynolds number(雷諾數(shù)雷諾數(shù)) round tubes (圓管圓管) Laminar flow(層流層流): Re2000ulRe VDVDRe有壓差，相對(duì)運(yùn)動(dòng)的平行平板間縫隙流動(dòng)有壓差，相對(duì)運(yùn)動(dòng)的平行平板間縫隙流動(dòng) 其中其中a為兩平行平板的間距，為兩平行平板

15、的間距，b為板寬，對(duì)環(huán)形間隙為板寬，對(duì)環(huán)形間隙b=d，l 為板長(zhǎng)，為板長(zhǎng)，U為上板運(yùn)動(dòng)速度為上板運(yùn)動(dòng)速度Laminar flow through parallel platesLaminar flow through circular tube(圓管層流圓管層流)21()()4.2.22Uyduph ayyadl3212babaQUpl2horizontal only4.3.10b32pDVl30)(1212ahpdldUaudyQaEnergy equation (能量方程能量方程)沿程損失沿程損失 f is friction factor(沿程阻力系數(shù)沿程阻力系數(shù))。Minor losse

16、s(局部損失局部損失) （出口數(shù)，入口數(shù)；按出口速（出口數(shù)，入口數(shù)；按出口速度系數(shù)計(jì)算）度系數(shù)計(jì)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?yàn)橐驗(yàn)?a d, a l, 故可把二圓柱表面所形成的圓柱環(huán)形間隙故可把二圓柱表面所形成的圓柱環(huán)形間隙看成是無(wú)限寬平行平板間的壓差流動(dòng)。速度可直接用式看成是無(wú)限寬平行平板間的壓差流動(dòng)。速度可直接用式(4.2.2)。 4.1 4.1 圖圖4.234.23a a中油缸內(nèi)油的表壓強(qiáng)為中油缸內(nèi)油的表壓強(qiáng)為0.15MPa0.15MPa，油的粘度，油的粘度0.1Pa0.1Pas s，柱，柱塞的直徑塞的直徑d=50mm=50mm，柱塞

18、與套筒間的徑向間隙，柱塞與套筒間的徑向間隙=0.05mm=0.05mm，套筒的長(zhǎng)，套筒的長(zhǎng)度度l=300mm=300mm。設(shè)以力。設(shè)以力 F推著柱塞使其以速度推著柱塞使其以速度U U = 0.7 m/s = 0.7 m/s向左運(yùn)動(dòng)，求向左運(yùn)動(dòng)，求油的漏損流量、柱塞側(cè)面流體的剪應(yīng)力和力油的漏損流量、柱塞側(cè)面流體的剪應(yīng)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流動(dòng)狀態(tài)為層流，則V,VDReDVflp22 4.9 對(duì)一個(gè)內(nèi)徑為對(duì)一個(gè)內(nèi)徑為3mm的水平管，在的水平管，在Re = 200時(shí)，流體的時(shí)，流體的 = 60 cP, 比重比重S= 0.83, 求流體每米壓降。求流體每米壓降。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 在直徑在直徑300mm的的 流體為水的管道上的實(shí)驗(yàn)顯示，水銀流體為水的管道上的實(shí)驗(yàn)顯示，水銀水差壓計(jì)在相距水差壓計(jì)在相距L=120m的兩端的壓強(qiáng)差為的兩端的壓強(qiáng)差為280 mm Hg，流量是流量是0.23 m3/s，

22、摩擦系數(shù)，摩擦系數(shù)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只與雷諾數(shù)有關(guān)，與相對(duì)粗糙度只與雷諾數(shù)有關(guān)，與相對(duì)粗糙度/D無(wú)關(guān)，