尾矿库排水系统调洪演算

尾矿库排水系统调洪演算调洪演算的目的是依据既定的排水系统确定所需的调洪库容及泄洪流量。对确定的来水过程线，排水构筑物愈小，所需调洪库容就愈小。一、数解法〔一〕对于洪水过程线可概化为三角形，且排水过程线可近似为直线的简洁状况，其调洪库容和泄洪流量之间的关系可按公式〔1〕确定。tVtq=Qp(1-Wp

)〔1〕式中 所需排水构筑物的泄流量，米3/秒；Qp——设计频率P3/秒；V——3；tW——频率为P3。p〔二泄水量与尾矿库线，从而定出泄流量和调洪库容。尾矿库内任一时段△t的水量平衡方程式如公式〔2〕如下。s z

s

)△t=V-Vz z

(2)2 △t- 2式中Q、Q——时段始、终尾矿库的来洪流量，米3/秒；s zq、q——时段始、终尾矿库的泄洪流量，米3/秒；s zV、V——时段始，终尾矿库的蓄洪量，米3。z s令Q=1/2(Q+Q)，将其代入公式〔3〕，整理后得：s z1 1V+ - q△t) (3)zz s sz2 2—调洪库〔Vt〕之间的关系绘出q-V+〔1/2〕q△t和q-V-〔1/2〕q△t输助曲线备查。1：某尾矿库初期坝装满时，水面面积F＝2.52，陆面面积F＝s 11.52，L＝0.81公里，E＝385公里/公里，J＝0.2，N＝0.2，0 0 0N=0.08，mp＝2.0，μ＝1毫米/秒，S＝137.5毫米/小时，n＝0.55，s p 1n=0.75，试求p=2%的设计洪水过程线。2取△t取△t＝1小时，设计雨量H24p1。分段类别主雨峰段段号1t〔小时〕H24p的时程安排计算t1－nHi〔毫米〕Hti－Hti1HR〔毫米〕－雨峰段雨峰段11.01.00137.5137.5137.5231.31618143.521.75351.4962062512.5471.627224189.0591.732238.514.57.256111.822250.5126.07121.8622565.55.58131.90261.55.55.59141.9352664.54.510151.9727155.011162.0027544.012172.03279.54.54.513182.06279.544.014192.09283.544.015202.115287.53.53.516212.14229144.017222.16629533.018232.191301.53.53.519242.2153053.53.52中公式，无视地下汇流，则：取等流时面积块数n＝1iv＝mJ1/3Q1/4＝3.73×0.21/3×Q1/4＝2.18Q1/4计算与绘制坡面汇流及河槽汇流关心曲线如图1所示。坡面汇流、河槽汇流及水面降水3所示。依据表中末栏的计算数据绘制的洪水过程如图2所示。表2 汇流关心曲线计算公式图1 坡面汇流与河槽汇流关心曲线Qg=＋Q〔米00000000Qg=＋Q〔米00000000000000秒〕013.512.51.251.251.250.50.210.110.110.110.092.432.5223.512.52.52.03.251.50.630.420.330.440.362.432.7933.012.02.250.754.02.00.830.730.370.810.662.082.7444.013.02.50.54.52.30.960.900.241.050.852.783.6353.512.52.750.454.952.71.131.050.201.251.072.433.5064.013.02.750.055.02.71.131.130.061.311.122.783.9074.013.03.00.35.32.91.211.170.051.361.162.783.9484.513.53.250.355.653.11.291.250.091.451.243.134.3794.013.03.250.155.83.31.371.330.091.541.312.784.08105.014.03.50.206.03.41.421.400.091.631.393.484.87114.513.53.750.356.353.71.541.480.091.721.473.134.60125.514.54.00.306.654.01.671.610.141.861.593.825.41135.514.54.50.507.154.31.791.730.142.001.713.825.53146.015.04.750.457.64.71.961.880.172.171.854.176.02157.2516.255.630.938.545.52.292.130.282.452.095.047.13169.018.07.131.6310.166.92.882.590.502.952.526.258.771712.5111.59.752.8513.019.84.092.49－0.032.922.498.6811.11821.75120.7516.136.3319.3418.07.505.803.316.235.1015.1020.219137.51136.578.630.6379.9494.039.1523.3318.2324.4624.5095.50120.02021.75120.7578.63－15.3764.6073.030.4034.7810.2834.7436.8015.1051.92112.5111.516.13－56.877.734.82.0016.20－20.6014.1413.008.6821.6229.018.09.754.9512.689.53.962.98－10.024.123.516.259.76坡面汇流〔毫米〕河槽汇流〔米2/秒〕历时时段雨水面降水〔小时〕HR〔毫米〕μ Qs=KsHRI －Qi1 Mi Qi Bμ I=kLQi －Qi1 Mi Qi 3— － 〔米/秒〕237.2516.257.13－2.3710.31237.2516.257.13－2.3710.317.052.943.45－0.064.063.465.048.50246.015.05.63－1.428.895.82.422.68－0.783.282.804.176.9725－6.802.090.810.331.38－1.421.861.5901.5926－1.80.290.1110.050.19－1.400.460.3900.3927－0.29000.1800.03－0.300.100.0900.0928－0.090.010.0100.0129－0.01000030数 ；〔2〕水面降雨的流量过程线是台阶形的，而陆面出流过程线是折线形的〔图2〕。两者同时程直接相加的结果是洪峰前的流量少算了一些的平均流量 与水面降水流量Q同时程相加，即用〔 ＋Q〕△t进展调洪计算。S S2：

图2 P＝2％设计洪水过程线1－河槽汇流；2－水面降水；3－陆面与水面集合出流过程线某尾矿库按澄清距离的要求确定最低水位为400.00如图3所示。试按〔例1〕的来水过程线和泄流量与库水位关系曲线，用数解法进展调洪演算。图3 尾矿库调洪库容曲线解：将求得的不同库水位时的泄流量及相应的调洪库容于表41、2、3栏，取△t＝360秒计算V＋ q△t和V－ q△t列于5、6栏。依据计算结果绘出关心曲线如图4所示。HqVq△tV＋q△tV－q△tHqVq△tV＋q△tV－q△t4000000040113.64380024500683001930040221.995300394001347005590040326.81546004830020290010630040431.42217005650027820016520040534.72965006250035900023400040638.23800006880044880031120040741.35000007440057440042560040844.366500079800744800585200其次按公式〔3〕列表进展调洪演算将例1求得的洪水过程线Q－t数据列于表5的1、2栏，取△t＝3600秒计算 △t列于4栏。其余各栏计算如下：表5 尾矿库的调洪演算tQ〔3/秒〕〔3/秒〕Q△t〔3〕V＋q△tq〔3/秒〕V－q△t001.265000012.532.63456045600.91128022.732.7059480107602.15301032.683.189750127602.55356043.683.5811400149602.99418053.483.6812900170803.42476063.883.90513240180003.60503073.934.1314050190803.82532084.334.2314850202304.03565094.134.50515200208504.175850104.884.7316220220704.416220114.585.02517000232204.646520125.475.4718100246204.926920135.475.7719700266205.327470146.076.5820790282605.657910157.097.74523620315306.318830168.409.7727800366307.33102301711.1315.8235100453309.07126801820.5060.25570006968013.502108019120.087.0521700023808029.51319802054.136.8931400044598038.33080802119.6814.5213290044098038.1303980229.358.925230035628035.0230280238.497.683210026238030.5152580246.874.212760018018026.088480251.550.971510010158018.036780260.40.23490402708.0511270270720119902.4334028033400.67930299300.19250302500.057031700.013432340.01033000～1时段始点〔时序t＝0〕△t＝0，∵ △t＝0，qo＝O，Vo＝0,Vo± q0△t＝00～1时段终点〔时序t＝1〕，按公式〔3〕：1 1 0 0 V＋ q△t＝ △t＋〔V－ q△t〕＝ △t＝45603查关心曲线得q＝0.911 1 0 0 1 米3/秒及V－ q△t＝1280米3/秒，将这些数据填入表中时序1的5、6、7栏，这些数据也即1～2时段始点的数据：1 1～2时段终点〔时序t=2〕：2 2 1 V＋ q△t＝ △t＋〔V－ q△t〕＝9480＋1280＝107603，查关心曲线得2 2 1 2 2 q＝2.15米3/秒及V－ q△t＝3010米3，将这些数据填入表中时序2的、5、6、7栏，2 2 x这些数据也即时段2～3始点的数据……，以不同些计算至Vx＋ q△t＝0为止。x5计算可得最大泄流量消灭于时序20，qm＝38.33/V＋ q△t确定，即Vt＝〔V＋ q△t〕－ q△t＝445980－ 38.3×3600＝3770303。二、图解法〔一〕对于来水过程线可概化为三角形，而排水过程线不能近似为直线的状况，图解方法如下：1、将三解形洪水过程线绘于第一象限，尾矿库的调洪库容与泄流量关系曲线绘于其次5所示。图5 调洪演算图解图2、从原点O向左取oa=洪水总量W=1/2Qpt(t为洪水过程总历时)；3、自三解形顶点b向左作水平线，与y轴相交于c，连接acVt-q曲线相交于d；4d点向右作水平线与退水过程线相交于e，则e的纵坐标即为所求和qm；5d点向下作垂线，与x轴相交于h，则oh即为所需的调洪库容。〔二〕q与调洪库容Vt之间的关系求出各个然后按下述进展图解。图6 调洪演算图解图1、自o点向左取ob=0a(a为时段中点)，再从b点向上作垂线与q-ΦC，自C向右作水平线与过第一时段总点t1的垂线相交于dd的纵坐标即为所求的第一时段终点的泄流量q1；2Ccf=def向上作垂线与q-Φ线相交于g