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QU NTIT TIVEHYDROLOGYChapter 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
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3.1BAS
IN
RECHAR
GE
AND
RU
NOFF
Basin Recharge
Interception together with depression
storage and soil moisture
Direct Runoff
Overland flow and interflow
Influent Streams Intermittent streams (can go dry
because of time elapses between rain)
Effluent Streams
Groundwater
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THREEPATHSTO
A
STREAM
FORWATERNOTWITHHELDASBASIN
RECHARGE
RainOverland
Flow
Nearest
Channel
Rain InterflowNearest
Channel
Rain Percolation Groundwater/SoilMoisture
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HYDROGRAPHANALYSIS
N = Ad0.2
N = Number of days
for recovery after
the peak
Ad= drainage area
in square miles
http://localhost/var/www/apps/conversion/tmp/scratch_4/PPT/Hydrograph_Analysis.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_4/PPT/Hydrograph_Analysis.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_4/PPT/Hydrograph_Analysis.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_4/PPT/Hydrograph_Analysis.ppt8/12/2019 Quantitative hydrology REPORT.pptx
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E
STIMATIN
GVOLUME
OF
RU
NOFF
Theoretical Formula
R = P L G
Where:R -Runoff
P - Precipitation
L - Basin Recharge
G - Groundwater Accretion
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In the design of storm drains and water-control projects, runoff volume iscommonly assumed to be a percentage of rainfall
R = kP
Where:
R Runof
k runoff coefficient
P - Precipitation
Table 3.1 Values of Runoff Coefficients k for
various surfaces
Urban Residential
Simple Houses
Garden ApartmentsCommercial and Industrial
Parks
Asphalt or concretepavement
0.20
0.300.90
0.05-0.30
0.85-1.0
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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.
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EX
A
MP
L
E
Step 1:Assess SiteConditions
In this example we will use a 200 ft2 patio
Step 2:ObtainRunoff
Coefficient
Using the provided table (Table 1), look up the runoffcoefficient that most closely resembles your site. In this caseit is 0.98
Step 3: Dothe Math
Volume Runoff = Surface Area x Runoff Coefficient xRainfall Depth
Volume Runoff = 200ft2 x 0.98 x 0.083ft = 16.3ft3
Note: Make sure that Surface Area and Rainfall Depth are in the sameunits. It doesnt matter what you use, just stay consistent measurements infeet or meters are generally easiest.
Step 4:Convert ifNecessary
Most people have trouble thinking about water volume incubic feet so we will convert to gallons multiplying by7.48gal/ft3. Volume Runoff = 16.3ft3 x 7.48 gal/ft3 =121.gallons
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INFILTR TIONPPT\Infiltration.ppt
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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-1Precipitation index on
previous day
b ranges from 0.85-0.95
When there is no rain for t days,
PaN-1= PaNbt
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RUNOFFFROM
SNOW
3.9 Physics of SnowmeltFactors Affecting Snowmelt
Solar Radiation
Depends on Reflectivity orAlbedo
Heat from warm Air
Turbulence resulting to
speedy wind bringing large
quantities of warm air
Rainfall
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Heat from warm Air
Heat ofFusion of Ice
Heat ofCondensation
of Water
Therefore,
1073/144=7.5 units
Thus, in 1
unit ofmoisture
on snow,
7.5 of
water will
melt
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Rainfall
Where:
Ms amount of melt in inches or millimeter
P - Rainfall or precipitationTw Wet-bulb temperature
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SNOWMELTINBASINSWITHLITTLE
RANGEINELEVATION
Degree-day factors
defined as a departure of 1 degree in mean dailytemperature above 32F.
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 thetotal 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.
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E
xa
m
p
l
e
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 3F 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 44F
Solution: With a temperature of 44F 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) = 39.5F
And the average degree days above 32F 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 sfd
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Did I make myself clear?
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Maraming Salamat!
Kristian Carlo M. BolaBS IN CIVIL ENGINEERING VA
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