黏性流体的一维定常流动-习题答案.docx

1、【例题解析】第六章 黏性流体的一维定常流动V例 6-1管道直径 d = 100mm ， 输送水的流量 q = 0.01 m3/s ，水的运动黏度 n = 110 -6 m2/s， 求水在管中的流动状态？若输送n = 1.14 10 -4 m2/s的石油， 保持前一种情况下的流速不变，流动又是什么状态？解：（1）雷诺数Re = VdnV = 4qVpd 2= 4 0.013.14 0.12= 1.27 （ m/s）故水在管道中是紊流状态。Re = 1.27 0.1 = 1.27 105 2000110 -6（2）Re = Vd =n1.27 0.11.14 10 -4= 1114 2000故油在

2、管中是层流状态。例 6-2有一长方形风道长 l = 40m， 截面积 A = 0.50.8m2 ，管壁绝对粗糙度 = 0.19mm，输送t = 20的空气，流量 qV解：平均流速= 21600m3/h，试求在此段风道中的沿程损失。当量直径V = qVA=21600= 15 (m/s)3600 0.5 0.8de =2hb h + b= 2 0.5 0.8 = 0.615 (m)0.5 + 0.820空气的运动黏度n = 1.6310-5m2/s，密度= 1.2kg/m3。雷诺数Re = Vden= 15 0.615 = 565950 1.6310 -5相对粗糙度查莫迪曲线图得l = 0.016

3、5 = 0.19 = 0.00031d e615沿程损失Vl2hf = l= 0.016540 152= 12.3 (m 空气柱)de 2g0.6152 9.806沿程压强损失pf = hf rg = 12.3 9.806 1.2 = 144.8 (Pa)例 6-3 如图 6-1 所示，水平短管从水深 H = 16m 的水箱中排水至大气中，管路直径d1 = 50mm， d 2 = 70mm，阀门阻力系数z 门 = 4.0，只计局部损失，不计沿程损失，并认为水箱容积足够大，试求通过此水平短管的流量。解： 列截面 00 和 11 的伯努利方程V 2H + 0 + 0 = 0 + 0 + 1 +（z

4、+ z+ zV 2+ z ） 12g入由表查得z 入=0.5， z 扩1 =0.24， z 缩2 =0.30，故扩1缩2门 2g12gH1+ z 入 + z 扩1 + z 缩2 + z 门V1 =11+ 0.5 + 0.24 + 0.30 + 4.02 9.806 16= 7.2 （m/s）通过水平短管的流量q = V p d 2 = 7.2 p 0.052 = 0.01413 （m3/s）V1 4 14图 6-1 例 6-3 示意图图 6-2 例 6-4 示意图311例 6-4有一并联管道如图 6-2 所示。若已知 qV = 300m /h， d = 100mm， l = 40m，d 2 =

5、 50mm，l2 = 30m，d 3 = 150mm，l3 = 50m，l1 = l2 = l3 = 0.03。试求各支管中的流量 qV 1 、qV 2 、 qV 3 及并联管道中的水头损失。解：根据并联管道的流动规律因此有qV = qV 1 + qV 2 + qV 3hf1 = hf2 = hf3l V 2l V 2l V 2（a）111l 11 = l 22 = l 33 d1 2gd 2 2gd 3 2g401 4q20.030.1 V 1 2 9.806 p 0.12 3014q 2= 0.030.05 V 22 9.806 p 0.052 5014q 2= 0.030.15 V 32

6、 9.806 p 0.152 9898.98qV 1 = 238873.79qV 2 = 1634.8qV 3所以qV 2 = 0.2qV1 （b）将式（b）代入式（a）中qV 3= 2.46qV 1 qV = qV 1 + 0.2qV 1 + 2.46qV 1qV = 3.66qV 1所以qV 1= qV3.66= 300 = 81.97 （m3/h）3.66V 2V 1q= 0.2q= 0.2 81.97 = 16.36 （m3/h）V 3V 1q= 2.46q= 2.46 81.97 = 201.65 （m3/h）并联管道的水头损失l V 24014 81.97 2hf(a-b) = h

7、f1 = l1 11 = 0.03 = 5.15 （mH2O）d1 2g0.12 9.806 3600p 0.12 SVa例 6-5如图 6-3 所示，虹吸管总长l =21m，坝顶中心前管长l =8m。管内径 d =0.25m， 坝顶中心与上游水面的高度差 h1 =3.5m，二水位落差 h2 =4m。设沿程阻力系数l =0.03，虹吸管进口局部阻力系数V 1 =0.8，出口局部阻力系数V 2 =1，三个 45折管的局部阻力系数均为0.3，试求虹吸管的吸水流量 q 。若当地的大气压强 p =105Pa，水温t =20，所对应的水的密度=998kg/m3，水的饱和压强 p =2.42103Pa，试

8、求最大吸水高度。图 6-3 例 6-5 示意图解：对上下游液面列伯努利方程，得pa = -pa l V 2所以流速V =rgh2 + rg + l d +2gh2l l + Vd2 9.8 40.03 21 + (0.8 +1+ 0.3 3)0.25=V 2g= 3.875 m/sp2qV = 0.25 4 3.875 = 0.19 m3/s对上游液面和坝顶中心列伯努利方程，得pa =p2V 2 l V 2rgh1 + rg + 2g + l d+ V 1 2g+ V 1 2g = 1 +1+ l l + Vpa - p2rg= h + 1+ l l1 d V 2hd1 h2l l +Vd当

9、p2 = p S ，可得最大吸水高度h1 pa - pSrg1+ l l + V1- dlh2 =(100 - 2.42) 103 -998 9.81+ 0.032180.25+ (0.8 + 0.3) 4 = 7.63 ml d + V0.03+ (0.8 +1+ 0.3 3)0.25可见最大吸水高度为 7.63m，现吸水高度为 3.5m，远小于最大吸水高度，所以不会出现水的汽化，故虹吸作用不会被破坏。一般而言，虹吸管的吸水高度不得超过 7m。例 6-6 某厂有一条铸铁输水管线，如图 6-4 所示，长 500m，直径 200mm，流量为 100L/s，水源为 5m深井；进口有一滤网（ =5.

10、2），管路中有 10 个 90弯头（ =0.48），两个闸门（ =0.08），运动黏度系数n =1.3mm2/s，绝对粗糙度 =1.3mm，2 截面与水泵轴线在同一水平面上。求水泵所需压头。解：在 1 和 2 截面间建立能量方程V 2H = 5 + 2g + hWV = 4qVpd 2= 4 100 10-33.14 0.22= 3.18 m/s由Re = Vd = 4.9 105 ， = 0.0065 ，可由莫迪图查得沿程nd损失系数l = 0.033 ，则图 6-4 例 6-6 示意图l V 2500 3.182hW = V + l d = 5.2 + 10 0.48 + 2 0.08 +

11、 0.033= 47.8 m2g0.2 2 9.8 i代入原式可得所需泵的压头为V 23.182H = 5 + 2g + hW= 5 + 47.8 = 53.3 m2 9.8例 6-7 输水钢管直径 d =1000m，壁厚d =20mm，长 l =800m，钢的弹性模量 E =2 1010Pa，管内水的流速v =1.2m/s试求突然关闭阀门所引起的水击压强和水击周期。0解：水的弹性模量 E = 2 109 Pa，则水击波的传播速度为E0 r1 + E0 dE dc = 1154.7 m/s水击压强p = rcv = 1.3856 106 Pah水击周期T = 4l / c = 2.77 sEx

12、ample E6-1 Oil, withr = 900 kg/m3 andn = 0.00001 m2/s, flows at 0.2m3/s through 500m of 200 mm-diameter cast iron pipe( e = 0.26 mm). Determine ( a ) the head loss and ( b ) the pressure drop if the pipe slopes down at 10in the flow direction.Solution:First compute the velocity from the known flow r

13、ate:V = QpR 2=0.2p (0.1) 2= 6.4 m/sThen the Reynolds number isRe = Vd = 6.4 0.2 = 128000 Thendn0.00001e = 0.26 = 0.0013d200Enter theMoody chart on the right ate / d = 0.0013(you will have to interpolate), andmove to the left to intersect with Re=128000. Read f 0.0225 . Then the head loss is=L V 2 =

14、500 6.42=mAns.( a )h fThereforefd 2g0.02250.22 9.81117h = p + z - z = p + L sin10ofrg12rgforp = rg (h - 500 sin 10o )= rg(117 - 87)= 900 9.81 30 = 265000 PaAns.( b )Example E6-2Water,r = 1.94slugs/ft 3 andn = 0.000011 ft3/s, is pumped between tworeservoirs at 0.2 ft 3 /s through 400 ft of 2-in-diame

15、ter pipe and several minor losses, as shown inFig.E6-1. The roughness ratio ise / d = 0.001.Compute the pump horsepower required.Fig.E6.1 schematic view of example E6-2Solution: Write the steady flow energy equation between sections 1 and 2, the two reservoir surfaces:pV 2pV 2 1 + 1 + z1 = ( 2 + 2 +

16、 z 2 ) + h f + h j - hprg2grg2gWhere hpis the head increase across the pump. But sincep1 = p2andV1 = V2 0 , solvefor the pump head:hp = z 2- z1+ h f+ h j= 120 - 20 V+22g( lL +d K )(1)Now with the flow rate known, calculateV = qVA=0.21 ( 2 ) 2= 9.17 ft/s412Now list and the minor loss coefficients:Los

17、sKSharp entrance Open globe valve 12-in bend Regular 90elbowHalf-closed gate valve Sharp exit0.56.90.250.952.71.0 K = 12.3Calculate the Reynolds number and pipe friction factor:9.17( 2 )Re d= Vdv= 12 = 1390000.000011Fore / d = 0.001,from the Moody chart readl = 0.0216.Substitute into Eq.(1):=9.17 2

18、0.0246(400)h100 +p2 32.22+12.3 12=100 + 84 = 184 ftpump head The pump must provide a power to the water ofV PP = rgq h = 1.94(32.3)1bf/ft 3 (0.2ft 3 / s)(184ft) = 2300ft 1bf/sThe conversion factor is 1 hp=550ft1bf/s. ThereforeP = 2300 = 4.2550hpAns.Allowing for an efficiency of 70 to 80 percent, a p

19、ump is needed with an input of about 6 hp.【习题答案】6-1 V1 = 1.526 m/s， p1 = 28139.3 Pa； p2 = 1119.9 kPa6-2 (1)Re = 281332紊流；（2） Re = 1581层流6-36-4m = 0.023Re1 = 2PasRe 236-5 h f = 8.8 m 油柱6-6 h f = 5.5 mH2O6-7 qv max = 8.25 L/s6-8 qv = 4.25 m3/s6-9 qv = 40.1 L/s6-10p f= 6.49 Pa6-11 h = 0.23 m6-12 （1） p

20、A= pa+ 2 rg(h - 1) ； h = 1mh + l1 + l3（2） qv= 0.00556； h = 1m6-13 a = 1.0586-14 （1）Vmax = 0.36 m/s；(2)h f = 0.423 m 原油柱；(3) t = 5.18 10-4 rp6-15 （1） h f 1 = p ；（2） h f 1 =h f 24h f 226-16qv = 0.105 m3/s6-17H = 7.37 m6-18pv = 24152.97 Pa6-19 H = 9.32 m6-20 hw = 10.79 mH2O6-21 p2 = 100567.923 Pa6-22 H

21、 = 25.7 m6-23 qv = 12.17 m3/s6-24z 2 = 8.653 ，理论值z 2 = 9 （或z 1 = 0.5408 ，理论值z 1 = 0.5625 ）6-25 D1 = 210 mm6-26 qv = 0.068 m3/s6-27 p1 = 55035.82 Pa6-28d = 0.6105 m6-29qv1 =2qv286-30 向上流动： pe = -39075 Pa；向下流动： pe = -26475 Pa6-31 H = 2.367 m6-32qv1 = 0.0283 m3/s； qv2 = 0.0967 m3/s； h f= 2.72 mH2O6-33p f= 231.67 Pa6-34 qV = 5.89 L/s6-35 ph = 1514.7 kPa6-36 p1 = 2517.44 kPa6-37hw1-2 = 0.051 mH2O； hw1-3 = 7.95 mH2O