工程流体力学(孔珑版)第六章-题解

1589327660852第1页共4页第六章管内流动和水力计算液体出流【6-11】加热炉消耗qm=300kg/h的重油，重油的密度ρ=880kg/m3，运动黏度ν=0.000025m2/s。如图6-54所示，压力油箱位于喷油器轴线以上h=8m处，而输油管的直径d=25mm，长度l=30m。求在喷油器前重油的计示压强？[62504Pa]。图6-54习题6-11示意图【解】流速sm0.192915025.08803600/3004441222dqdqAqvmmV2300192.915000025.0025.00.192915vdRe输油管内流动是层流沿程损失m3.91557807.9025.0915.1920.19291530323226422222fgdRelvgvdlRegvdlh以油箱液面为1-1，喷油器前为2-2断面，列写伯努利方程：w222221121122hgpzgvgpzgvaa由于是层流，221；fwhh；01av；vva1；hzz21；01p。w2222hgpgvhPa62489.50.1929153.915578807.988022w2vhhgp【6-16】用新铸铁管输送25℃的水，流量qv=300L/s，在l=1000m长的管道上沿程损失为hf=2m（水柱），试求必须的管道直径。【解】vdqV24124dqvV522222f8242gdlqgdqdlgvdlhVVf2258ghlqdVdqddqvdReVV1442新铸铁管ε=0.25～0.42，25℃水的运动黏度ν=0.897×10-6m2/s。取ε=0.3，λ=0.0175得d=0.5790m1589327660852第2页共4页【6-20】图6-58所示为一突然扩大的管道，其管径由d1=50mm突然扩大到d2=100mm，管中通过流量qV=16m3/h的水。在截面改变处插入一差压计，其中充以四氯化碳（ρ=1600kg/m3），读得的液面高度差h=173mm。试求管径突然扩大处的损失系数，并把求得的结果与按理论计算的结果相比较。图6-58习题6-20示意图【解】(1)按小截面流速计算的损失系数实际值vdqV2412114dqvV，2224dqvVgvgpzgvgpzgv2222212222211121222211222121vpvpv242424221122221221dqpdqpdqVVV41212142412142111421dqppddqVV212412421412214142411811421111ppqddddqppdddVVhlgpghglp21ghpp2154013.0173.0807.910001600100016805.010050181242424124211ghqdddV(2)按小截面流速计算的损失系数理论值5625.010050114411222221222212211ddddAA(3)按大截面流速计算的损失系数实际值vdqV2412114dqvV，2224dqvV1589327660852第3页共4页gvgpzgvgpzgv2222222222211121222222222121vpvpv242424222222221221dqpdqpdqVVV42222142412142111421dqppddqVV212422412422214242412811421111ppqddddqppdddVVhlgpghglp21ghpp218.642173.0807.91000160010001681.015010081242424224122ghqdddV(4)按大截面流速计算的损失系数理论值915010011441222212221222122ddddAA【6-21】自鼓风机站供给高炉车间的空气量qv=120000m3/h，空气温度t=20℃，运动黏度ν=0.0000157m2/s。输气管总长l=120m，其上有五个弯曲半径R1=2.6m的弯曲处和四个弯曲半径R2=1.3m的弯曲处，还有两个闸阀，其局部损失系数ζ=2.5。管壁的绝对粗糙度ε=0.5mm。设输气管中空气的流速v=25m/s，而热风炉进口处的计示压强pie=156896Pa。试求输气管所需的管径d和鼓风机出口处的计示压强poe。【解】vdqV241m1.303253600/12000044vqdV20℃时的空气密度ρ=1.205kg/m3，声速c=343.1m/s。马赫数0.300.0728651.34325cvMa，可以不考虑气体压缩性的影响。管壁的相对粗糙度4-3103.8375303.1105.0d雷诺数6102.074740000157.0303.125vdReRedReb685.0385.0101.8484105.0303.123082308流动处于紊流粗糙管平方阻力区，沿程损失系数1589327660852第4页共4页74.12lg21d1.574374.1105.02303.1lg274.12lg2232d弯曲半径R1=2.6m的弯曲处的局部损失系数0.14556.2303.1163.0131.0163.0131.05.35.311Rd弯曲半径R1=2.6m的弯曲处的局部损失系数0.29533.1303.1163.0131.0163.0131.05.35.322Rd列写能量方程gvgvgvgvdlgpzgvgpzgv2224252222222222122222221112121zz，21vv，21，oe1pp，ie2ppgvdlgpgp2245221ieoePa106004.1225205.15.222953.041455.05303.11209699.1156896224552221ieoevdlpp