Estimating peak flow for lake Wheeler farm test watershed Watershed parameters: AREA - main watershed upstream of the pipe: 24 acre - with exp. forested area and small SW strip added: 24+8+9=41 acres (all of this contributes to the ponding area) - smaller western watershed that drains into upper pipe: 15 acre (may contribute to ponding, see photo) - total that may go into the ponding area : 56 acre LENGTH and ELEVATION max length of flow in feet L = 2000ft elevation difference H = 430ft - 365ft = 65ft L/H = 2000/65 = 30.7 mean slope 3.4 deg TIME TO CONCENTRATION tc = 0.0078 * 2000^0.77 * 30.7^0.385 = 2.71 * 3.73 = 10.13min RAINFALL and PEAK FLOW 10 year storm: 6in/24hours, if uniformly distributed in time: 0.25in/hr BUT rainfall rarely follows this pattern, rather it would be high intensity during a short period of time and then lower intensity for hours. Given that our tc is only 10 minutes we need to know what would be the max intensity during the 24hr period, for a duration equal to the "time of concentration" of the watershed 2 year 15 minute rainfall is 1.0inch http://www.intelisolve.com/hydro35_twoyr_15minute_map.htm based on data - see noaa, the rainfall intensity for 15min, 10 year rainfall is 4.92in! (am I reading it correctly?) http://hdsc.nws.noaa.gov/hdsc/pfds/orb/nc_pfds.html -- Rich what did you use for K and b and what did you get for I ? To see how much we actually got during some recent rainfalls, I have requested hourly data for Ernesto and Alberto when we had high daily value (Alberto was 10in) to see what the hourly intensities were. But so far let us use the old data 31. march 2002 with 0.9 inch precipitation between 6-7pm, and lets use C=0.7: qp = C*i*A = 0.7 * 0.9 in/hr1 * 24 acre = 15.12 cfs for the 41 acre area qp=25 cfs for comparison, for the NOAA data (that is if we get 10 year 15min rainfall) qp = 0.7 * 4.92 * 24 = 82 cfs the actual data for Alberto: hourly 1.3in/hr June 14, 10-11am leading to qp = 0.7 * 1.3 * 24 acre = 21.8 cfs per minute 0.33in/10min (rate 1.98 in/hr) June 14, 10.29-10.39 leading to qp = 0.7 * 1.98 * 24 = 33.26 parameters based on lidar data =============================== area = 32.8 acre length L=560m=1837ft elevation difference H 431-369 = 62 tc= 0.0078 * 326 * 3.7 = 9.4min peak flow using rational equation from Haan for 0.9in/hr q = 0.7 * 0.9 * 32.8 = 20.7 cft SEDSPEC NC design manual - rational method for 10year storm qp = 20cfs culvert should be 18inch (45cm) SCS Curve number (for pasture - need to try grass cover) qp=10cfs ? culvert 12inch SIMWE (older simulation done for the previous project) rainfall 0.036m/hr = 1.4in/hr, Mannings n=0.05, infiltration rate = 0 peak flow in the lower parts of concentrated flow (it changes from cell to cell so I am listing it as an interval) qp=0.68 - 1.1 m3/s = 24cfs - 38cfs water depth in concentrated flow area h=30cm=1ft water depth in ponding area h=60cm=2ft I get steady state after 8 minutes I will run it for 4in/hr to see what I get. Now I am really curious what it will be in the real world.