Hydrophobicity and Unit Hydrographs in Watershed Management
Learn about the impact of dry soil on infiltration during floods and explore the concept of synthetic unit hydrographs for watershed analysis. Discover how hydrophobicity affects soil properties and the computation of standard unit hydrographs using Snyder's method.
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Presentation Transcript
Heard on the news about California Floods Infiltration was less because the soil was dry. Why? Green-Ampt method says that drier soil has higher But Hydrophobicity (especially after fires) http://geography.swansea.ac.uk/hydrophobicity/ soil_hydrophobicity.htm K( ) 0 for dry soil due to residual water being disconnected ? = ? 1 +? ? ?
Synthetic Unit Hydrographs A unit hydrograph is intended to quantify the unchanging characteristics of the watershed The synthetic unit hydrograph approach quantifies the unit hydrograph from watershed attributes Table 8.4.1 gives the steps to compute Snyder s Synthetic Unit Hydrograph 1/3 2/3
Example 8.4.1 A watershed has a drainage area of 5.42 mi2; the length of the main stream is 4.45 mi, and the main channel length from the watershed outlet to the point opposite the center of gravity of the watershed is 2.0 mi. Using Ct = 2.0 and Cp = 0.625, determine the standard synthetic unit hydrograph for this basin. What is the standard duration? Use Snyder s method to determine the 30- min unit hydrograph parameter. 1/3 2/3
Result (4.05,570) (5.97,427.5) (7.47,285) (3.09,427.5) (2.37,285) (14.1,0) (0,0)
Example 8.4.1 Snyder's Synthetic Unit Hydrograph - Result A watershed has a drainage area of 5.42 mi2; the length of the main stream is 4.45 mi, and the main channel length from the watershed outlet to the point opposite the center of gravity of the watershed is 2.0 mi. Using Ct = 2.0 and Cp = 0.625, determine the standard synthetic unit hydrograph for this basin. What is the standard duration? Use Snyder s method to determine the 30- min unit hydrograph parameter. Follow the procedure of table 8.4.1 L = main channel length = 4.45 mi Lc = length to point opposite centroid = 2.0 mi A = watershed area = 5.42 mi2 ??= ?1??? ?? ??= ??/5.5 = 0.7 ? ???= ??+ 0.25 ?? ?? = 3.85 + 0.25 0.5 0.7 = ?.? ?? 0.3 ? = 1 2 4.45 20.3= 3.85 ? ?2??? ??? ???= = 640 0.625 5.42/3.8 = ??? ??? Widths ?75= ?75 440 570/5.421.08= 2.88 ? 770 570/5.421.08= 5.04 ? ??? 1.5 ?50 ?75= 25815.42 2.88 = 14.1 ? 1.08= (4.05,570) ???/? ?50 ?50= 1.08= ???/? (3.09,427.5) (5.97,427.5) ? ??= 2581 570 1.5 5.04 W75 (2.37,285) (7.41,285) W50 1/3 2/3 (14.1,0)
S Curves From Mays, 2011, Ground and Surface Water Hydrology
