《水环境模拟》考核题2012答案.doc

2012年春水环境模拟考试题一、计算题1 在一条宽阔略有弯曲的河流中心设有一工业排污口，污水流量为0.15m3/s，污水中含有有害物质的浓度为110ppm，河流水深为3.5米，流速为0.8m/s，摩阻流速u*=0.060m/s，假定污水排入河流后在垂向可立即均匀混合，已知横向扩散系数，试计算排污口下游400m处污染带宽度及断面上最大浓度。设排污口下游400m处断面上允许最大浓度为6ppm，问排污口的排污流量可以增加多少倍（假定排放浓度维持不变）？2 一个均匀河段稳定排放BOD，河水平均流速u=45km/d，水温T=20，始端河水BOD和DO浓度分别为L0=18mg/L，C0=7mg/L，K1=0.500d-1，K2=1.00d-1，计算临界溶解氧和下游处55km的BOD浓度。3 某河道分成四个河段，试进行BOD和DO的一维水质模拟。上游来水流量为Q10=9.5m3/s，L10=2.0mg/L，O10=8mg/L，各河段耗氧系数K1分别为0.2,0.2,0.25,0.3d-1，复氧系数K2均为0.55d-1，水流通过各河段的历时分别为0.6,0.9,1.0,1.0d；各排放口污水流量Qi分别为0.5,0.3,0.4,0.5m3/s；BOD5浓度Li分别为300，200，400，100mg/L；DO浓度Oi均为0.9mg/L；各取水口取水流量Q3i为别为0.2,1.0,0.0,1.0m3/s，饱和溶解氧浓度为10.0mg/L。4 某河段长X=50km，河段u=0.5m/s，已知该河段K1=0.3d-1，K2=0.8d-1，K3=0.18d-1，R=1mg/(L.d)，P=0.5mg/(L.d)，L0=30mg/L，Os=10mg/L，O0=4mg/L。（1）计算该河段BOD和DO的沿程变化；（2）当要求河段内DO不应小于3mg/L时，推求起始断面BOD的去除率。5 （选做题）一个拟建工程向大湖中排放废水q=5000m3/d，BOD5=20mg/L,废水在湖中的扩散角为90o，离排水口1000m处的湖水所受到的影响基本可忽略，湖水BOD基线浓度为2mg/L，溶解氧9mg/L，饱和溶解氧10mg/L，BOD降解的耗氧系数K1=0.1d-1，湖水复氧速率K2=0.15d-1。离排放口500m处为一景点，求该景点附近水域的BOD值。2、 电算题1 设在时间t=0到t=1小时内，因偶然事故均匀排放BOD5，仅河水起始断面L0为10mg/L，河段平均流速为4.5km/h，纵向离散系数E=1.8km2/h，K1=0.014h-1。试用数值法求解BOD5的浓度L(x,t)。要求附计算框图与程序。给出计算结果并给出不同时刻的BOD5沿程分布图。2对一维污染物浓度扩散方程，以时间前差，空间前差建立差分格式，分析其稳定性，导出稳定性条件。三、论述题1 QUAL2E模型考虑了哪些水质参数，简述其原理。2简述湖泊水库富营养化模型的最新进展。四、论文或PPT选做一个1写一篇与水环境模拟、预测相关的论文。2 做一个水环境模拟与预测的PPT。（备注：若发现在作业有雷同现象，以零分计；第一大题第5小题为选做题，做了的对期末考核成绩有提高，请同学们尽量都做。）一、计算题1.Solving：(1) The biggest concentration of section on 400m in the downstream:Cmax=Muh4Exxu=0.151100.83.540.43.50.0604000.8ppm=0.257ppm(2) Because of the linear relation between Cmax and M in the condition of invariability on other parameters, the relational expression can be summarized as M,M=Cmax,Cmax=60.257=23.35 In the factor of the unchanged emitting concentration , the blowdown flow can enlarge about 22 times.2 Solving: (1) Calculating the critical dissolved oxygen concentration: The concentration of oxygen saturation mg/L According to the formula: the critical dissolved oxygen concentration is:Cc=9.0698-181.000.5001.000.5001-1.000.500-19.0698-71811-1.000.500=3.9851mg/L(2) Calculating the BOD concentration in x=50km: According to the formula:the BOD concentration in x=50km is:L55km=L0e-K1xu=18e-0.5005545=9.769mg/L3 Solving:The imformation is summarized in the fig below: (1) Calculating the elements of matrix A B C and D. 0 = exp(-k11t1) = exp(-0.20.6) = 0.8869 1 = exp(-0.20.9) = 0.8353 2 = exp(-0.251.0) = 0.7788 3 = exp(-0.31.0) = 0.7408 0 = exp(-k21t1) = exp(-0.550.6) = 0.7189 1 = exp(-0.550.9) = 0.6096 2 = 3 = exp(-0.551.0) = 0.5769 0=k110-0k21-k11=0.20.8869-0.71890.55-0.2=0.096 1=0.20.8353-0.60960.55-0.2=0.1292=0.250.7788-0.57690.55-0.25=0.1683=0.30.7408-0.57690.55-0.3=0.197 0 = Os(1-0) = 10(1-0.7189) = 2.811 1 = 10(1-0.6096) = 3.904 2 = 3 =10(1-0.5769) = 4.231 a0=Q11-Q31Q210=9.5-0.29.5+0.5-0.20.8869=0.8417a1=9.8-19.8+0.3-10.8353=0.8078a2=9.1-09.1+0.4-00.7788=0.7460a3=9.5-19.5+0.5-10.7408=0.6996b0=Q1Q21=0.59.8=0.05102b1=0.39.1=0.03297b2=0.49.5=0.04211b3=0.59.0=0.05556c0=Q11-Q31Q210=9.5-0.29.5+0.5-0.20.7189=0.6822c1=9.8-19.8+0.3-10.6096=0.5895c2=9.1-09.1+0.4-00.5769=0.5526c3=9.5-19.5+0.5-10.5769=0.5449d0=Q11-Q31Q210=9.5-0.29.5+0.5-0.20.096=0.0911d1=9.8-19.8+0.3-10.129=0.125d2=9.1-09.1+0.4-00.168=0.161d3=9.5-19.5+0.5-10.197=0.186f0=Q11-Q31Q210=9.5-0.29.5+0.5-0.22.811=2.668f1= 9.8-19.8+0.3-13.904=3.775f2=9.1-09.1+0.4-04.231=4.053f3=9.5-19.5+0.5-14.231=3.996g1=a0L10=0.88692.0=1.7738h1=c0O10-d0L10=0.68228-0.09112.0=5.2754 (2) Calculating matrix U V m and n.A=1000-a11000-a21000-a31=1000-0.80781000-0.74601000-0.69961B=b00000b10000b20000b3=0.0510200000.0329700000.0421100000.05556C=1000-c11000-c21000-c31=1000-0.58951000-0.55261000-0.54491D=0000d10000d20000d30=00000.12500000.16100000.1860A-1=1000-0.80781000-0.74601000-0.69961-1=10000.80781000.60260.7460100.42160.52190.69961C-1=1000-0.58951000-0.55261000-0.54491-1=10000.58951000.32580.5526100.17750.30110.54491U=A-1B=10000.80781000.60260.7460100.42160.52190.699610.0510200000.0329700000.0421100000.05556=0.051020000.041210.03297000.030740.024600.0421100.021510.017210.029460.05556V=-C-1DU=-10000.58951000.32580.5526100.17750.30110.5449100000.12500000.16100000.18600.051020000.041210.03297000.030740.024600.0421100.021510.017210.029460.05556=0000-0.00640000-0.01071-0.0053100-0.00954-0.00747-0.007830m=A-1g=10000.80781000.60260.7460100.42160.52190.699611.7738000=1.77381.43291.06890.7478n=C-1BO+C-1f+h+C-1DA-1g=10000.58951000.32580.5526100.17750.30110.544910.0510200000.0329700000.0421100000.055560.90.90.90.9+10000.58951000.32580.5526100.17750.30110.544912.6683.7754.0533.996+5.2754000-10000.58951000.32580.5526100.17750.30110.5449100000.12500000.16100000.186010000.80781000.60260.7460100.42160.52190.699611.7738000=10000.58951000.32580.5526100.17750.30110.544910.045920.029670.037900.05000+7.9433.7754.0533.996-00.2210.2310.199=7.9898.2948.4438.448(3) Calculating the BOD5 and DO in the downsides of each sectionsL2=UL+m=0.051020000.041210.03297000.030740.024600.0421100.021510.017210.029460.05556300200400100+1.77381.43291.06890.7478=17.0820.3932.0527.98mg/LO2=VL+n=0000-0.00640000-0.01071-0.0053100-0.00954-0.00747-0.007830300200400100+7.9898.2948.4438.448=7.996.374.170.96mg/L4. Solving: (1)The analytic solution of D-C BOD-DO model: Where, 0.5m/s=43.2km/d So, (2) If O=-36.29exp-154x+26.18exp-190x=0 ,x=44.456 km . In the other word, the minimum of the DO appears on 44.456km. Ox=44.456 kmL0=6.575-0.1605(L0-2.0833) When Ox=44.456 kmL03 mg/L, L024mg/L. So, the removal rate is at least (30-24)/3