107第十章明渠流和闸孔出流及堰流10—1某梯形断面粉质粘土渠道中的均匀流动,如图所示。已知渠底宽度b=2.0m,水深h0=1.2m,边坡系数m=1.0,渠道底坡度i=0.0008,粗糙系数n=0.025,试求渠中流量Q和断面平均流速v,并校核渠道是否会被冲刷或淤积.解:A=222200(21.211.2m3.84m)bhmh2221(221.211m5.39m)obhmQ=AC122353iRiAn152233330.00083.845.39m/s3.46m/s0.025v=3.46m/s0.90m/s3.84QA校核:由表10-3查得粉质粘土最大允许流速为1m/s,因h0=1.2m>1.0m,需乘以系数k=1.25,所以vmax=11.25m/s=1.25m/s,最小允许流速vmin=0.4m/s,满足vmax>v>vmin条件。10—2设有半正方形和半圆形两种过流断面形状的渠道,具有相同的n=0.02,A=1.0m2,i=0.001,试比较它们在均匀流时的流量Q的大小。解:设正方形渠道流量为Q1,半圆形渠道流量为Q2,Q1=1522133111111iACRiAn15222332222222iQACRiAn其中AAAnnii212121,,,212321()QQ由于212Ah,12Ah,114482AhA又22222π22,,ππ2π2rAAArrA210.894,2232193.0)89.0(QQQ1152522233133331110.0011(81)m/s0.8m/s0.02iQAn3320.8m/s0.86m/s0.93Q10—3某梯形断面渠道中的均匀流动,流量Q=20m3/s,渠道底宽b=5.0m,水深h0=2.5m,边坡系数m=1.0,粗糙系数n=0.025,试求渠道底坡i。解:15222335233()iQnQAinA由知108222200(52.512.5)m18.75mAbhmh22021(522.511)m12.07mbhm25233200.025()0.000418.7512.07i0.00030.01i可保证正常的排水条件,且不必人工加固。10—4为测定某梯形断面渠道的粗糙系数n值,选取L=1500m长的均匀流段进行测量。已知渠底宽度b=10m,边坡系数m=1.5,水深h0=3.0m。两断面的水面高差z=0.3m,流量Q=50m3/s,试计算n值。解:12fzzzh,0.30.00021500ffhiJJL由均匀流基本方程可得152233inAQc-=,式中222(103.01.53.0)m43.5mA=??2(1023.011.5)m20.82m1522330.000243.520.820.0250n10—5某梯形断面渠道底坡i=0.004,底宽b=5.0m,边坡系数m=1.0,粗糙系数n=0.02,求流量Q=10m3/s时的正常水深h0及断面平均流速0v。解:1155222333223()1(21)hbmhiiQAnnbhm将已知值代入,由试算法可得00.76mh=222200(50.7610.76)m4.38mAbhmh010m/s2.28m/s4.38QvA10—6某梯形断面渠道中的均匀流,流量Q=3.46m3/s,渠底坡度i=0.0009,边坡系数m=1.5,粗糙系数n=0.02,正常水深h0=1.25m,试设计渠底宽度b。解:1523002230()(21)hbmhiQACRinbhm将已知值代入,由试算法可得0.80mb=。10—7设有一块石砌体矩形陡槽,陡槽中为均匀流。已知流量32.0m/sQ=,底坡09.0i,粗糙系数020.0n,断面宽深比0.20hbh,试求陡槽的断面尺寸0h及b。解:20,0002.0,22,4,hbbhAhhh代入均匀流基本方程得522330012(2)(4)Qnhhi将已知值代入,解得:00.43m20.43m0.86mhb==?,10—8试拟定某梯形断面均匀流渠道的水力最优断面尺寸。已知边坡系数5.1m,粗109糙系数025.0n,底坡002.0i,流量33.0m/sQ=。解:由水力最佳条件知2202(1)2(11.51.5)0.61hbmmh2200000.612.11bhAbhmhh,200214.22bhmh将已知值代入均匀流基本方程得320352021)22.4()11.2(002.0025.00.3hh解得:01.09m0.611.09m0.66mhb,10—9需在粉质粘土地段上设计一条梯形断面渠道。已知均匀流流量Q=3.5m3/s,渠底坡度i=0.005,边坡系数m=1.5,粗糙系数n=0.025,试分别按(1)不冲允许流速vmax及(2)水力最优条件设计渠道断面尺寸,并确定采用哪种方案设计的断面尺寸和分析是否需要加固。解:(1)由表10-3查得粉质粘土max1.0m/sv=2220000max3.53.51.51.0QAmbhmhbhhv====+=+(1)由221332max1vcRiiAnc-==,1222330.0051.03.50.025c-=?,16.66mc=200213.6116.66bhmbhc=++=+=(2)联解(1)、(2)式得:010.22m15.87mhbì=??í?=?07.68m11.06mhbì=??í?=-?(舍去)(2)()()22021211.51.50.61hbmmhb==+-=+-=,00.61bh=2222000000.611.52.11Abhmhhhh=+=+=,20000210.613.614.22bhmhhhc=++=+=5213321QiAnc-=,52123320013.50.005(2.11)(4.22)0.025hh-=?1023300.931467h-=,8300.93h=,00.97mh=,0.610.97m0.59mb=?2202.111.99mAh==max3.5m/s1.76m/s1.0m/s1.99QvvA===>=。按不冲允许流速计算结果水深太浅,故按水力最优条件设计断面尺寸,需加固,可采用干...