估計(jì)水塔的水流量

1、估計(jì)水塔的水流量1、問題提出：某地區(qū)用水管理機(jī)構(gòu)需要對居民的用水速度（單位時(shí)間的用水量） 和日總用水量進(jìn)行估計(jì)?，F(xiàn)有一居民區(qū)，其自來水是由一個(gè)圓柱形水塔提供，水塔高12.2m，塔的直徑為17.4m。水塔是由水泵根據(jù)水塔中的水位自動加水，一般水泵每天工作兩次，按照設(shè)計(jì)，當(dāng)水塔中的水位降至最低水位，約8.2m時(shí)，水泵自動啟動加水；當(dāng)水位升高到最高水位，約10.8m時(shí)，水泵停止工作。 表1給出的是某一天的測量數(shù)據(jù)，測量了28個(gè)時(shí)刻的數(shù)據(jù)，但由于水泵正向水塔供水，有三個(gè)時(shí)刻無法測到水位（表中用表示），試建立數(shù)學(xué)模型，來估計(jì)居民的用水速度和日用水量。 表1 水塔中水位原始數(shù)據(jù) 2、問題分析：日用水量用水

2、速度每個(gè)時(shí)刻水塔中水的體積3、模型假設(shè)：影響水從水塔中流出的流量的唯一因素是公眾對水的傳統(tǒng)要求；水塔中的水位、氣候條件、溫度變化等不影響水流量的大??；水泵充水速度水塔的水流量與水泵狀態(tài)獨(dú)立；恒定，且遠(yuǎn)大于水塔的水流速度；水流量曲線是一條連續(xù)光滑的曲線；表1數(shù)據(jù)是準(zhǔn)確的；4、模型的建立與求解：（1）、 水塔中水的體積其中， ，（r為底面半徑,d為水面高度）（2）在Matlab命令窗口直接運(yùn)行(不包括未知三點(diǎn))>>t=0,0.921,1.843,2.949,3.871,4.978,5.900,7.006,7.928,8.967,10.954,12.032,12.954,13.875,1

3、4.982,15.903,16.826,17.931,19.037,19.959,20.839,22.958,23.880,24.986,25.908;>>v=2301.1,2254,2213.3,2169.8,2135.8,2095.9,2065.4,2027.1,1994.6,1954.6,2572.9,2496.8,2427.8,2362.7,2295.4,2237.3,2182.9,2121.3,2059.7,2005.3,1954.6,2572.9,2518.4,2462.0,2420.7;>> scatter(t,v)得到水塔中水體積的散點(diǎn)圖（3）在Matl

4、ab中編寫腳本文件（不包括未知三點(diǎn)）采用數(shù)值微分的一階微商的兩點(diǎn)公式（末位處近似為sd(n)=sd(n-1)t=0,0.921,1.843,2.949,3.871,4.978,5.900,7.006,7.928,8.967,10.954,12.032,12.954,13.875,14.982,15.903,16.826,17.931,19.037,19.959,20.839,22.958,23.880,24.986,25.908;v=2301.1,2254,2213.3,2169.8,2135.8,2095.9,2065.4,2027.1,1994.6,1954.6,2572.9,2496.8

5、,2427.8,2362.7,2295.4,2237.3,2182.9,2121.3,2059.7,2005.3,1954.6,2572.9,2518.4,2462.0,2420.7;for i=1:9 sd(i)=abs(v(i+1)-v(i)/(t(i+1)-t(i);endsd(10)=sd(9);for i=11:20 sd(i)=abs(v(i+1)-v(i)/(t(i+1)-t(i);endsd(21)=sd(20);for i=22:24 sd(i)=abs(v(i+1)-v(i)/(t(i+1)-t(i);endsd(25)=sd(24);scatter(t,sd)hold o

6、nplot(t,sd)得到水塔中水流速度的散點(diǎn)圖及光滑圖（4）預(yù)測水塔中的未知流速1在Matlab中運(yùn)行腳本文件（不包括未知三點(diǎn)）：采用數(shù)值微分的一階微商的兩點(diǎn)公式（末位處近似為sd(n)=sd(n-1)t=0,0.921,1.843,2.949,3.871,4.978,5.900,7.006,7.928,8.967,10.954,12.032,12.954,13.875,14.982,15.903,16.826,17.931,19.037,19.959,20.839,22.958,23.880,24.986,25.908;v=2301.1,2254,2213.3,2169.8,2135.8,

7、2095.9,2065.4,2027.1,1994.6,1954.6,2572.9,2496.8,2427.8,2362.7,2295.4,2237.3,2182.9,2121.3,2059.7,2005.3,1954.6,2572.9,2518.4,2462.0,2420.7;for i=1:9 sd(i)=abs(v(i+1)-v(i)/(t(i+1)-t(i);endsd(10)=sd(9);for i=11:20 sd(i)=abs(v(i+1)-v(i)/(t(i+1)-t(i);endsd(21)=sd(20);for i=22:24 sd(i)=abs(v(i+1)-v(i)/(

8、t(i+1)-t(i);endsd(25)=sd(24);sd得到速度（不包括未知三點(diǎn)）sd = Columns 1 through 9 51.1401 44.1432 39.3309 36.8764 36.0434 33.0803 34.6293 35.2495 38.4986 Columns 10 through 18 38.4986 70.5937 74.8373 70.6840 60.7949 63.0836 58.9382 55.7466 55.6962 Columns 19 through 25 59.0022 57.6136 57.6136 59.1106 50.9946 44.

9、7939 44.79392采用拉格朗日插值法估計(jì)未知三點(diǎn)的速度:在Matlab命令窗口直接運(yùn)行>> x0=7.928,8.967;>> y0=38.4968,38.4968;>> lglr3(x0,y0,9.981)ans = 38.4968>> x0=8.967,9.981;>> y0=38.4968,38.4968;>> lglr3(x0,y0,10.925)ans = 38.4968>> x0=19.959,20.839;>> y0=57.6136,57.6136;>> lglr3

10、(x0,y0,22.015)ans = 57.61363在Matalb中運(yùn)行腳本文件t=0,0.921,1.843,2.949,3.871,4.978,5.900,7.006,7.928,8.967,9.981,10.925,10.954,12.032,12.954,13.875,14.982,15.903,16.826,17.931,19.037,19.959,20.839,22.015,22.958,23.880,24.986,25.908;sd=51.1401,44.1432,39.3309,36.8764,36.0434,33.0803,34.6293,35.2495,38.4986,38.4968,38.4968,38.4986,70.5937,74.8373,70.6840,60.7949,63.0836,58.9382,55.7466,55.6962,59.0022,57.6136,57.6136,57.6136,59.1106,50.9946,44.7939,44.7939;scatter(t,sd)hold onplot(t,sd)得到水塔中水流速度的散點(diǎn)圖及光滑圖（new）(4)a、通過曲線擬合，擬合出上述函數(shù)（f1） b、通過數(shù)值積分（梯形，辛普森）求出用水量(f2)5、模型檢驗(yàn)：應(yīng)該另外測試一批數(shù)據(jù)檢驗(yàn)?zāi)Ｐ?f1,f2)6、模型分析