Quantitative hydrology REPORT.pptx

March 30, 2018 | Author: Hermano Pule | Category: Surface Runoff, Drainage Basin, Rain, Snow, Water And The Environment


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QUANTITATIVE HYDROLOGYChapter 3 3.1 Basin Recharge and Runoff 3.2 Hydrograph Analysis Estimating Volume of Runoff 3.3 Runoff Coefficients 3.4 Infiltration 3.5 Infiltration Indices 3.6 Rainfall-runoff Correlations 3.7 Moisture-accounting Procedures 3.8 Long-period runoff relations Runoff from Snow 3.9 Physics of Snowmelt 3.10 Snowmelt Computation 3.1 BASIN RECHARGE AND RUNOFF  Basin Recharge • Interception together with depression storage and soil moisture  Direct Runoff • Overland flow and interflow  Effluent Streams • Groundwater  Influent Streams • Intermittent streams (can go dry because of time elapses between rain) Rain Overland Flow Nearest Channel THREE PATHS TO A Rain Interflow Nearest Channel STREAM Rain Percolation Groundwater/Soil Moisture FOR WATER NOT WITHHELD AS BASIN RECHARGE N = Ad0.2 N = Number of days for recovery after the peak Ad = drainage area in square miles HYDROGRAPH ANALYSIS Theoretical Formula R=P–L–G Where: R -Runoff P - Precipitation L - Basin Recharge G - Groundwater Accretion ESTIMATING VOLUME OF RUNOFF  In the design of storm drains and water-control projects, runoff volume is commonly assumed to be a percentage of rainfall Table 3.1 Values of Runoff Coefficients k for various surfaces R = kP Where: R – Runof k – runoff coefficient P - Precipitation Urban Residential Simple Houses Garden Apartments Commercial and Industrial Parks Asphalt or concrete pavement 0.20 0.30 0.90 0.05-0.30 0.85-1.0 Table 1: Runoff Coefficients Soil Groups A and B are sandier and Soil Groups C and D are more clayey. These soil classifications would be found in a county soil survey available at any Soil and Water Conservation District office or North Carolina Cooperative Extension center. Step 1: Assess Site Conditions • In this example we will use a 200 ft2 patio  Step 2: Obtain Runoff Coefficient • Using the provided table (Table 1), look up the runoff coefficient that most closely resembles your site. In this case it is 0.98 • Volume Runoff = Surface Area x Runoff Coefficient x Rainfall Depth • Volume Runoff = 200ft2 x 0.98 x 0.083ft = 16.3ft3 • Note: Make sure that “Surface Area” and “Rainfall Depth” are in the same units. It doesn’t matter what you use, just stay consistent – measurements in feet or meters are generally easiest. Step 3: Do the Math • Most people have trouble thinking about water volume in Step 4: cubic feet so we will convert to gallons multiplying by Convert if 7.48gal/ft3. Volume Runoff = 16.3ft3 x 7.48 gal/ft3 = Necessary 121.gallons E X A M P L E INFILTRATION PPT\Infiltration.ppt 3.6 Rainfall-Runoff Correlation Plot of average rainfall versus resulting runoff PaN = bPaN-1 + PN PaN – Atecedent-Precipitation Index at the end of Nth day PaN-1 – Precipitation index on previous day b – ranges from 0.85-0.95 When there is no rain for t days, PaN-1 = PaNbt RUNOFF FROM SNOW 3.9 Physics of Snowmelt  Factors Affecting Snowmelt  Solar Radiation Depends on Reflectivity or Albedo  Heat from warm Air •Turbulence resulting to speedy wind bringing large quantities of warm air Rainfall  Heat from warm Air Heat of Condensation of Water Heat of Fusion of Ice Therefore, 1073/144= 7.5 units Thus, in 1 unit of moisture on snow, 7.5 of water will melt  Rainfall Where: Ms – amount of melt in inches or millimeter P - Rainfall or precipitation Tw – Wet-bulb temperature Degree-day factors  defined as a departure of 1 degree in mean daily temperature above 32°F. Depth of water melted from the snow in inches or millimeter per degree-day may be determined by dividing the volume of stream flow produced by melting snow within a given time period by the total degree days for the period. Usually ranging from 0.05-0.15in/degree-F with an average value of 0.08in/degree-F Ranges from 2-7mm/degree-C day. SNOWMELT IN BASINS WITH LITTLE RANGE IN ELEVATION The area-elevation distribution in a basin on the board. The average snow line is at 5000 ft and the temperature index station is at 6000 ft. Assume a temperature decrease of 3°F per 1000ft increase in elevation and a degree-day factor of 0.10. Compute the snowmelt in second-foot days for a day when the mean daily temperature at the index station is 44°F Solution: With a temperature of 44°F at 6000 ft the freezing level is at 6000 + {[(44-32)/3]*1000} = 10,000 ft The area between the snowline (5000 ft) and the freezing level is 305 sq.mi. from the figure, the average temperature over this area is: 0.5(47+32) = 39.5°F And the average degree days above 32°F is 39.5 – 32 = 7.5 degree days The total melt therefore: 7.5*0.10*305 = 229 sq.mi. inches 26.9*229 = 6150 E x a m p l e sfd Did I make myself clear? Maraming Salamat! Kristian Carlo M. Bola BS IN CIVIL ENGINEERING VA
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