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当前位置:首页 > 商业/管理/HR > 人事档案/员工关系 > 【重庆大学出版社·肖明葵】版【水力学习题评讲】第五章
2110112vHHHg2300.120.6229.830.0373/4QAgHms题5.1图21dHHBAC112,5.l水从A箱通过直径为10cm的薄壁孔口流入B水箱,流量系数为0.62,设上游水箱中水面高程H1=3m保持不变。试分别求:(1)A水箱敞开,B水箱中无水时;(2)A水箱敞开,B水箱中的水深H2=2m时;(3)A水箱水面压强为2kpa时,通过孔口的流量。解:(1)此情况为薄壁孔口的自由出流,若v1为孔口前的渐变流断面平均流速,取v1=0,则120vv22112201212vvHHHmg3020.025/QgHms,则:作用水头,题5.1图21dHHBAC11212,vv(2)此种属于薄壁孔口的恒定淹没出流,下游的渐变流过水断面1-1、2-2的断面平均流速,依题意取,分别为上游,(3)以孔口型心所在水平面为基准面,选择渐变流过水断面1-1、2-2,写1-2的伯诺里方程,222112212120()222cAvpvvHHgggg120vv2211220121.2042AvvpHHHmgg3020.0237/QgHms200012210.555.012829.8vHHgm2300.220.6229.85.01280.193/4QgHms题5.3图Hap0.2md5.0mH00.5m/sv5.3有一直径的圆形锐缘薄壁孔口,其中心在上游水面下的深度,孔口前的来流流速,孔口出流为全部完善收缩的自由出流,求孔口出流量Q。0.62,作用水头解:对薄壁小孔口的全部完善收缩的自由出流有:流量系数0.620.82n2111010120.6224dQAgHgH2222020220.8224ndQAgHgH0112HHH022HHl5.6两敞口水箱用一直径为d1=40mm的薄壁孔口连通,如图所示。右侧水箱的底部接一直径为d2=30mm的圆柱形管嘴,长l=0.1m,孔口的上游水深H1=3m,水流保持恒定,求管嘴流量Q2和下游水深H2。,题5.6图d2H1Hl1Q21Qd2解:孔口出流和管嘴出流的流量系数分别为:孔口出流和管嘴出流的流量公式:其中12QQ22121220.622()0.822()44ddgHHgHl21.896Hm22022202320.82240.0363nQAgHdgHm/s题5.6图d2H1Hl1Q21Qd2连续性方程则有代入数据解得:00000wHh2222220.50.650.65122222(0.0350/0.51.31)19.6wfjlvvvvvhhhdgggggvd24,QQAvvd531.162.08dd1.015dm50ml0.0310.50.65w2133m/sQ5.13圆形有压涵管如图所示,管长。上、下游水位差H=3m,各项阻力系数:沿程,进口,弯头,出口。试求当涵管通过流量为时,有压涵管的管径。解:以2-2断面为基准面,写出1-1,2-2之间的液体的伯诺里方程,题5.13图Hl=50m11220.70.025.12虹吸管将河道中的水引入水池,如图所示。钢管总长为30m,直径d=400mm设每一弯头的局部阻力系数为,又.求管中流量和最大真空值。解:本题属淹没出流。以下游液面2-2为基准,作用水头流速系数可直接代公式022HHmcc121210.48300.020.520.710.4wld020.4829.822.985m/svgH222.9850.452219.6vmg232.9850.40.375/4QvAms故流量cc0.48233300032wpvhg2313063(2)22430.452(10.020.520.7)4.8640.4vwpvhgdm由式看出3-3断面为真空。其真空度为从而可得真空值:Pv=P3=47.67kPa.又任取断面3-3,写1→3能量方程(以1-1断面为基准面)0.0125n2211lll123Hz题5.15图集水井钻井5.15用虹吸管自钻井输水至集水井如图所示。虹吸管长12360mllll200mmd,直径,钻井与集水井间的1.5mH。试求虹吸管的流量。恒定水位高差已知选用钢管管道进口、弯头及出口的局部10.5230.541.0,,阻力系数分别为1/61/6110.2()0.0125448.557CRn22889.80.033348.557gC解:00000wHh22222220.50.50.5122222(0.033360/0.20.531)0.6419.6wfjlvvvvvhhhdgggggvv21.50.64,1.53/vvms233.14150.21.5340.0481/QAvms以2-2断面为基准面,写出1-1、2-2间液体的伯努力方程:2211lll123Hz题5.15图集水井钻井30.0628m/sQ12323m17m15m12mhhhl,,,1320.21340.0731,mHm(eeNNQH),各处局部阻力系数沿程阻力系数λ=0.023。求:水泵的扬程及有效功率5.18水泵抽水系统如图所示,流量,管径均为d=200mm,解:以水池水面为基准面0-0,列断面0-0至水箱断面2-2列能量方程,22121200()022mwvvHhhhgg题5.18图31403022l3h1hh2wfjhhh1232(/sin30)20.02(31230)0.221.056fhhlhdgm2212342()(30.210.0731)219.644.2830.874119.6jvhgm120vv22440.06282/0.2Qvmsd因为管中流速题5.18图31403022l3h1hh21.0560.87411.93wfjhhhm3171.9321.93mHmm9.80.062821.9313.5kWeNQH扬程有效功率1.056fhm0.8741jhm2002aawppvHhggg2()2aswppvHhgg解:以1-1断面为基准面写出1-1与2-2之间液体的伯诺里方程:离心水泵压水管吸水管122Hs1题5.19图325m/hQ123.5m1.5mll,320ml18mzsh。吸水管长。压水管长。水泵提水高度6m。试确定水泵的允许安装高度5.19从水池取水,离心泵管路系统布置如题5.19图。水泵流量,水泵最大真空度不超过并计算水泵的扬程H。,QAv22227.00.252221.03(0.0457.5/0.17.00.25)0.57519.6wfjlvvvhhhdgggm21.035.70.57519.65.07sHm2440.00811.03/3.14150.12Qvmsd离心水泵压水管吸水管122Hs1题5.19图5.20用离心泵将湖水抽到水池,流量Q=0.2m3/s,湖面标高▽1=85.0m,水池水面标高▽3=105.0m,吸水管长l1=10m,水泵的允许真空值为4.5m,吸水管底阀局部水头损失系数ζe=2.5,90°弯头局部阻力系数ζw=0.3,水泵入口前的渐变收缩段局部阻水系数ζ=0.1,吸水管沿程阻力系数λ=0.022,压力管道采用铸铁管,其直径d2=400mm,长度l2=1000m,n=0.013,试确定:(1)吸水管的直径d1;(2)水泵的安装高程▽2;(3)带动水泵的动力机械功率。1.0m/sv1440.21.00.2540.5mQdv1d解:(1)吸水管的直径采用经济流速10.5md选标准管径(2)水泵的安装高程安装高程是以水泵的允许真空值来控制的,令水泵轴中心线距湖面高差为hs,则▽2=▽1+hs,其中hs可按下式计算:2112()21014.5(10.0222.50.31)4.280.519.6svlvhhdgm311058520zm21101()(0.0222.50.31)0.1720.519.6walvhmdg(3)带动水泵的动力机械功率:式中:z为提水高度wh为吸水管及压水管道中水头损失之和,吸水管按短管计算,压水管按长管计算。()wpQzhNwah(1)求s21834.2889.28hm水泵轴中心高程0.3489.1759.523wwawphhhm所以总的水头损失为30.7,9.8kN/m()9.80.2(209.523)83kW0.7wQzhN取则动力机械功率为222220.210009.1752.09wpQhlmKwph20.5320.410.4()2.09/40.0134KACRms(2)求压水管的流量模数:222440.21.59m/s0.4pQvd11/6611()(0.1)40.013dCn0.552.41/Cms22889.80.028552.4gC22()210001.59(0.02850)9.2080.419.6pwppplvhdgmm200.1779.229.38wawpHzhhm19.80.229.3882.264kW0.7emNQHN若将压水管按短管计算wph则为(>9.175m)5.21水池A和B的水位保持不变,用一直径变化的管道系统相连接,图所示。管道直径d1=150mm,d2=225mm,管长l1=6m,l2=15m,两水池水面高差为6m,两管道沿程阻力系数λ=0.04,试求通过管道的流量?并绘制总水头线及测压管水头线。解:求Q,设管径为d1的管道断面平均流速为v1,管径为d2的管道断面平均流速为v2题5.21图ld11ld22HBA1122221211244ddQvv21221()2.25vdvd222221112122112000002222vlvvlvHgdggdg进出(-1)222226156(0.52.250.042.251.250.041.0)0.150.2252vg()22.723/vms122.256.127/vvms23220.108/4dQvms122.25vv0.5,1.0进出选渐变流过水断面1-1、2-2,并取2-2断面为基准面0-0,写1-2的伯诺里方程:题5.21图ld11ld22HBA1122222126.1271.915,0.37822*9.82vvmgg222212122.7270.51.9150.958,1.250.5912229.8jjvvhmhmgg2进2231.00.3780.3782jvhmg出21111260.041.9153.0640.15flvhdgm22222150.040.3783.06420.225flvhmdg(二)绘制总水头线及测压管水头线:流速水头,局部水头损失,沿程水头损失,题5.21图ld11ld22BAHhj1hf1hj2hf2hj3222011102220333fHhSlQSlQSlQ26101200,9.029/dmmSsm
本文标题:【重庆大学出版社·肖明葵】版【水力学习题评讲】第五章
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