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新编工程流体力学课后答案++第三章+流体动力学基础.doc

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新编工程流体力学课后答案++第三章+流体动力学基础.doc

上传人:iris028 2019/11/21 文件大小:70 KB

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新编工程流体力学课后答案++第三章+流体动力学基础.doc

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文档介绍::,u,u,u,uxxxxa,,u,u,uxxyz,t,x,y,z,2,2u,2u,2,22t,2x,2y,t,y,z,,,,xy,2,6t,4x,2y,2z,34,u,u,u,uyyyya,,u,u,uyxyz,t,x,y,z,1,u,u,1,t,y,z,t,x,z,,,,yx,1,x,y,2z,3,u,u,u,uzzzza,,u,u,uzxyz,t,x,y,z,1,u,u,1,2t,2x,2y,t,y,z,,,,xz,1,t,x,2y,z,112222a,a,a,a,:3265a,uy,3xyu,xy,xy,32(1)xxy13225a,,yu,y,yy33222a,a,a,(2)二元流动(3)恒定流(4):187h77u,,yhmax7,,,,QudAubdyyubh,,max0max,,1A088h,,7hQ7v,,umaxA8v122,,,,2dv2,,v,v,2,,:,,12,,d1,,,,:Hvv1v23dd3d21qq21,23(1)Q,dv,,Q,q,,q,Q,(2)v,,,d14Q2v,,,:渠中:Q,vbh,3m/s,2m,1m,6ms11,32管中:Q,Q,Q,,v,,d21244Q2d,,,:,2Bv,,,,,6msA2,,则22pv306AAH,,,,,,,,h,,,,,,,,:,水将从B点流向A点。H,HBA22pvpvAABB或:z,,,z,,,hABw,,g2gg2g解得水头损失为:,水将从B点流向A点。h,,:1,1、2,2、3,3为等压面。11由左边:p,p,,gh11左ph22331h由右边:p,p,,gh22右***h2水且:p,p,,ghA21pp2ppu右左max因为:zz,,,,,gg,g222则:,,u,p,p右左max,2,p,p,gh,h,,,,,,,2121,2,,gh,,ghppp,,,2,,,,,,ghpp,u,:v,,,vA,,,,p或直接由公式:求。,,u,2g,1hmaxp,,,,,:12Q12mmv2,,,h11ssA23,,,,,2以过2—2断面渠底处的水平面为基准面0—0,对1—1和2—2过水断面列能量方程:22vv,,1122Lsin,h,,0,h,,h,w122g2g122hLsinhhvv,,,,,,,,,,,,w1211222g122,2000,,2,,2,,,,,2,,:1(1)阀门关闭时,对1,1和2,2列能量方程:,0,0,0,,10,012121得:H,5m12v(题意理解一:h,2是整个管路的水头损失)w2g阀门开启时,对1,1和3,3列能量方程:22vvH02,,,12g2g5v,,2g,,,223(2)Q,dv,,,,,(题意理解二:是至压力表断面的管路水头损失)h,2w2g阀门开启时,对1,1和2,2列能量方程:22vvH22,,,12g2g2v5,2,32gv,2g,,,,223Q,dv,,,,,(2):1dd2212v,()v,4v2110d1dQ0221ppp,p1212(z,),(z,),12,,,ggghp,,,13600,850Hgoil,h,hpp850,oil,15h,,1和2,2断面列能量方程:22pvpv1122,,,,g2g,g2g22v,v15221,v,,15,,,,2Q,vd,,11452g2,,,222K,d,,,,(m/s)或:(),1(),,,K,h,,,,,:p1,,