中国石油大学(华东)工程流体力学答案5

2014/6/315-5H11110002500.025LmDmml===2225002000.026LmDmml===25/qLs=25/QLs=132222312214425100.796/3.140.24()450101.019/3.140.25QQqQvmsDQqvmsDpp--=+´´===´+´´===´解：求图示串联管路按长管计算所需的水头H=?5-5H11110002500.025LmDmml===2225002000.026LmDmml===25/qLs=25/QLs=12227.4(0.03)2ffHhhmvmg=+==232111112222222101.0190.02520.2519.65.305000.7960.02620.219.62.10ffLvhDgmLvhDgmll==´´===´´=UnRegistered2014/6/325-6.Q=100L/s,确定Q1,Q2及AB间的水头损失()()()2111121212221232221333232211123122332111621621622.561.660.8051.8050.1/0.055/,0.045/4.29fffffffLQhDgQQLhhhhDgQQLhDgQQQQQQQQQmsQmsQmshmlplplpü=ïïï-ï==+ýïï-ï=ïþ=-Þ==+=====解：11110002500.025LmDmml===1Q2229003000.024LmDmml===3333002500.025LmDmml===Q2Q5-7n图中一根管长L1=30m，直径d1=50mm，另一根管长L2=50m，直径d2=100mm，阀门的局部阻力系数ζ=3，假设沿程阻力系数λ1=0.04和λ2=0.03。试求流量在两并联管路中是如何分配的？并联管路中水头损失又为多少？12QQQ=+332510m/sQ-=´解：并联管2需考虑局部水头损失：L1,d1L2,d2Q22122ffvhhgz=[phone]()22LvLvdgdgllz=+UnRegistered2014/6/335-7332510m/sQ-=´L1,d1L2,d2Q2221544222142130500.04(0.033)0.050.10.1240.1180.054.61QQQQQQ=+×=×=2211221222112244()()()LQLQddddllzpp=+312510/5.61Q-=´3314.4610m/sQ-=´33220.5410m/sQ-=´23211152521163080.04(4.4610)6.3220.059.83.14fLhQmdglp-==´=´5-8.A20mB19mCD5m20005202522111521313201162500160.10.02510.60.2519.63.14161510.620.034/0.066/fCDfCDADfACfCDfAChLQhdgmzzhhLQhdgQmsQQQmslplp=´==´=++=-==-=解：分支管路,串联，先求＝CD管由A、B联合供水，Q0=100L/s,求Q1，Q2，d2。UnRegistered2014/6/345-8.A20mB19mCD5m22225222163.420.242BDfBCfCDfBCzzhhLQhmdgdmlp=++===5-9121211257,0.86,630,50,2,1)??2)10305,?DmmscStzzmpatpatQikmDmmQQn===-======-=原油，求变径12124.7531.750.25524(502)10306070.860.093/0.024641.8/,Re7693910,6070.012510ffffppzzhhmhDQmsLQvDvmsDhiLggnpn+=++-´=+==========´解：1)设变径前后均为水力光滑区流态符合UnRegistered2014/6/355-9440.251.7514.754.751311410102)0.0246()6070.0996/7.1%fhQDDQmsQQQn´=+==-=5-13保证自流式虹吸管中液体流量，只计沿程水头损失，已知：保证层流，d应为多少？l/2处A断面上的极限真空为hV=5.4m水柱，输油管在贮油池油面以上的最大允许高度hmax=？33423110/,2,44,10/,900/Qmshmlmmskgmnr--=====h1h1/2l,dA1/2l,dUnRegistered2014/6/365-13解：1)按层流计算，对两池液面1-1,2-2列能量方程33423110/,2,44,10/,900/Qmshmlmmskgmnr--=====212flvhhdgl==h1h1/2l,dA1/2l,d144.152Qlhdn==34414.1510104420.055dm--=´´´´=112222124644Re2lQhddgp=5-132)hV=5.4m水柱,对1-1,3-3列能量方程33423110/,2,44,10/,900/Qmshmlmmskgmnr--=====2max02Afpvhhgg=+++2max5.40.4214.990.919.6hm=--=h1h1/2l,dA1/2l,d11223321max22Aphvhgg=---32244100.42/3.140.055Qvmsdp-´===´UnRegistered2014/6/375-16hvH231.50.75220.00363/4vhmHHmdQgHmspm=====解：d=0.03m,圆柱形外管嘴。管嘴内的真空度1.5m，求Q。5-17n已知x=4.8m,y=2m,H=3mn求：孔口的流速系数221gtytvxc==9.84.87.513/222cgvxmsyÞ==´=´gvHc2)1(2孔x+=7.5130.98229.83cvgHf===´´流速系数Hp0yx解：UnRegistered2014/6/38d1=0.008m,d2=0.01m,d3=0.006m,H=1.2m，h=0.025m,求Q、h1、h2。3211232111122222233343321120.98,0.82,0.8224242()41.13510/0.16,0.270.11QQQdQghdQghdQgHhQmshmhmhhmmmmpmpmpm-========+=´==-=解：ABCd2d1d3h1h2Hh5-185-19n水从封闭水箱上部直径d1=30mm的孔口流至下部，然后经d2=20mm的圆柱行管嘴排向大气中，流动恒定后，水深h1=2m，h2=3m，水箱上的压力计读数为4.9MPa，求流量Q和下水箱水面上的压强p2，设为定常流。已知：6.01=m20.82m=ccUnRegistered2014/6/395-19解：经过孔口的流量=经过管嘴的流量÷÷øöççèæ-++gpm21112111224ppdhgdQ＝÷÷øöççèæ+gpm22222224phgdQ＝Pap421034.4´=smphgdQQ/1011.32433222222-´=÷÷øöççèæ+=gpm＝cc5-20n圆台形容器如图所示，液体经过容器底部直径d的小孔口流出，求容器排空及排出液体深度的1/2时所需的时间。12020cm,10cm,10cm,0.5cm,0.62DDhdm=====02004ThDQdtdhp=-òò解：流出的体积=容器减少的体积。设dt时间液面下降dh，液面直径为D，容器排空时间为T。224dQghpm=0220012ThDTdtdhdghm==òòD2D1h0hdhdUnRegistered2014/6/3105-2012020cm,10cm,10cm,0.5cm,0.62DDhdm=====D2D1h0hdhd022012hDTdhdghm=òθD12220()DDDDhDhh-=+=+02220()12hDhTdhdghm+=ò021/23/25/22202142[2]352hTDhDhhdgm=++5/25/25/22142[20.10.10.1]172.04s350.620.00519.6T=´+´+´=´´5-2012020cm,10cm,10cm,0.5cm,0.62DDhdm=====D2D1h0hdhdθD00021/23/25/2/222/22142[2]352hhhTDhDhhdgm=++02/220023355200001[2(/2)42(/8)(/32)]35hTDhhdDhhhhm=-+-+-172.04sT=04/22100.005681.610.620.519.6hTs==´UnRegistered