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