CE 374 K – HydrologyCE 374 K Hydrology
Runoff Processes
Daene C. McKinney
WatershedWatershed
• Watershed– Area draining to a stream
– Streamflow generated by water entering surfacewater entering surface channels
– Affected by• Physical vegetative and• Physical, vegetative, and climatic features
• Geologic considerations
• Stream PatternsStream Patterns
– Dry periods• Flow sustained from groundwater (baseflow)groundwater (baseflow)
StreamflowStreamflow
• Atmospheric Atmospheric Moisture
RainSnowpWater– Evapotranspiration
– Precipitation
Interception
RainSnow
Evaporation
Throughfall and Stem Flow
Energy
Precipitation
• Subsurface Water– Infiltration
Snowpack
Surface
Snowmelt
Pervious ImperviousWatershedBoundary
– Groundwater
• Surface Water Soil Moisture
Groundwater
EvapotranspirationInfiltration
Overland Flow
Percolation
Groundwater
Streams and LakesEvaporation
Flow
Groundwater Flow
Channel FlowRunoff
Shoal Creek Flood ‐ 1981Shoal Creek Flood 1981
18000
20000 0
1 (in)
12000
14000
16000
18000(c
fs)
1
2
3
Prec
ip
Precipitation
4000
6000
8000
10000
Runo
ff Streamflow
0
2000
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0
Time (hr)
Streamflow HydrographStreamflow Hydrograph
Peak
Centroid of Precipitation
Basin Lag
Peak
Time of Rise
sch
arge
, QD
i
BaseflowRecession
InflectionPoint
BaseflowRecession
Baseflow
TimeBeginning of Direct Runoff
End of Direct Runoff
Volume of Storm RunoffVolume of Storm Runoff
• Depends on several factors• Depends on several factors– Large watersheds – previous storm events
S ll t h d i d d t f i– Small watersheds – independent of previous storm
R i f ll il bl f ff 3 t• Rainfall available for runoff – 3 parts– Direct runoff
– Initial loss (before direct runoff begins)
– Continuing loss (after direct runoff)
Baseflow SeparationBaseflow Separation
• Depletion of groundwaterDepletion of groundwater during this period
• Continuity equationy q
)()( tQtIdtdS
−=
char
ge, Q ktteQtQ /)(
0 0)( −−=
dteQdS ktt /)(0 0−−−=
)()( tkQtS =D
isc
BaseflowRecession
at time flowat time flow)(
00 tQttQ
==
Time
][constant decay 00Tk
Q=
Baseflow Separation TechniquesBaseflow Separation Techniques
• Straight – line methodStraight line method– Draw a horizontal line segment (A‐B) from beginning of runoff to intersection with recession curve ch
arge
, QD
isc
A B
Direct Runoff
Time
Baseflow
Baseflow Separation TechniquesBaseflow Separation Techniques
• Fixed Base MethodFixed Base Method– Draw line segment (A – C) extending baseflow
20
char
ge, Q
recession to a point directly below the hydrograph peak
– Draw line segment (C‐D)
2.0AN =
Dis
c
A B
D
C
Draw line segment (C D) connecting a point N time periods after the peak
Direct Runoff
Time
CBaseflow2.0AN =
Baseflow Separation TechniquesBaseflow Separation Techniques
• Variable Slope Methodp– Draw line segment (A‐C) extending baseflow recession to a pointrecession to a point directly below the hydrograph peakD li t (B E) ch
arge
, Q
InflectionPoint
– Draw line segment (B‐E) extending baseflow recession backward to a
i t di tl b l thD
isc
A BE
Direct Runoff
point directly below the inflection point
– Draw line segment (C‐E)Time
C
Baseflow
Loss Estimation: Phi – Index MethodLoss Estimation: Phi Index Method
• Effective (excess) rainfall – Rainfall that is not retained or
infiltrated– Becomes direct runoff
Excess rainfall hyetograph
( )∑ Δ−==
M
mmd tRr
1φ
– Excess rainfall hyetograph (excess rainfall vs time)
• Losses (abstraction )– Difference between total and
rainfall observedrunoffdirect ofdepth
==
Rr
m
d
Difference between total and excess rainfall hyetographs
• Phi – Index– Constant rate of loss yielding runoffdirecttongcontributi
rainfallofintervals#indexPhi
==
Mφ
y gexcess rainfall hyetograph with depth equal to depth of direct runoff indexinterval
interval timerunoffdirecttongcontributi
==Δ
mt
Example – Phi Index method Time Observed
Rain Flow
in cfs
8:30 20312000 0
0 5
Total Rainfall HyetographShoal Creek 1981
8:30 203
9:00 0.15 246
9:30 0.26 283
10:00 1.33 8288000
10000
)
0.5
1
1.5
2
Direct Runoff Hydrograph
Find: Constant rate of 10:30 2.2 2323
11:00 0.2 5697
11:30 0.09 9531
12:00 11025
6000St
ream
flow
(cfs 2.5 loss yielding excess
rainfall hyetograph with depth equal to depth of direct runoff
12:30 8234
1:00 4321
1:30 22462000
4000S
2:00 1802
2:30 1230
3:00 713
3:30 394
07:30 PM 9:00 PM 10:30 PM 12:00 AM 1:30 AM 3:00 AM 4:30 AM 6:00 AM
Time
4:00 354
4:30 303
No direct runoff until after 9:30And little precip after 11:00
Basin area A = 7.03 mi2
Example – Phi Index Method (Cont.)Example Phi Index Method (Cont.)• Estimate baseflow (straight line method)
f– Constant = 400 cfs
baseflow
Example – Phi Index Method (Cont.)Example Phi Index Method (Cont.)• Calculate Direct Runoff Hydrograph
– Subtract 400 cfsDate Time
Rainfall Streamflow Time cfs m3in cfs 1/2 hr
24-May 8:30 PM 0.15 2039:00 PM 0 26 246
Observed Direct Runoff
9:00 PM 0.26 2469:30 PM 1.33 283 428 770,400
10:00 PM 2.2 828 1 1,923 3,461,40010:30 PM 2.08 2323 2 5,297 9,534,60011:00 PM 0.2 5697 3 9,131 16,435,80011:30 PM 0.09 9531 4 10,625 19,125,000
25-May 12:00 AM 11025 5 7,834 14,101,20012:30 AM 8234 6 3,921 7,057,8001:00 AM 4321 7 1,846 3,322,8001:30 AM 2246 8 1,402 2,523,6002:00 AM 1802 9 830 1,494,0002:30 AM 1230 10 313 563 400
Volume of direct runoff
2:30 AM 1230 10 313 563,4003:00 AM 713 113:30 AM 3944:00 AM 3544:30 AM 303 Total 7.839E+07
in80.4ft5280*mi03.7
ft10*7.83922
37===
AVr d
d Depth of direct runoff
Example – Phi Index Method (Cont.)Example Phi Index Method (Cont.)
• Neglect all precipitation intervals that occur beforeNeglect all precipitation intervals that occur before the onset of direct runoff (before 9:30)
• Select Rm as the precipitation values in the 1.5 m p phour period from 10:00 – 11:30
( ) ( )50*33
φφ ∑∑ ΔM
RtR( ) ( )
)50*3*082202331(804
5.0*311
φ
φφ ∑ −=∑ Δ−=== m
mm
md RtRr
)5.0*3*08.220.233.1(80.4 φ−++=
in54.0=φ
in27.05.0*54.0 ==Δtφ
Example – Phi Index Method (Cont.)Example Phi Index Method (Cont.)
10000
12000 0
0.5
1
1.5
φΔt=0.27
Effective Runoff Hyetograph
6000
8000
mflo
w (c
fs)
2
2.5
2000
4000Stre
am
07:30 PM 9:00 PM 10:30 PM 12:00 AM 1:30 AM 3:00 AM 4:30 AM 6:00 AM
Time
SCS Curve Number MethodSCS Curve Number Method
• SCS Curve Number (CN) method( )– estimates excess precipitation as a function of
• cumulative precipitation
• soil cover
• land use, and
• antecedent moisture
• Developed for small basins (< 400 sq. mi.)p ( q )
• Classify soils into four types
• Simple to use
• Converts basin storage into something simpler and more manageable (a “curve number” CN)
Losses – SCS MethodLosses SCS Method• Total rainfall separated into
3 parts: n3 parts:– Direct runoff
– Continuing Loss
I iti l L Pre
cip
itat
ion
eP
aae FIPP ++=
– Initial Loss
• SCS Assumption
P
aI aFRunoffActualStorageActual
RainfallTotal=P
TimeptRunoffPotential
RunoffActualStoragePotential
StorageActual=
eea PPIP −− )(
• Solve for Rainfall ExcessLC i i
LossInitial(Runoff) RainfallExcess
Rainfall Total
==
a
e
FIPP
aeeaIPS −
=)(
StorageWatershedMaximumSLossContinuing
==aF( )
SIPIPPa
ae +−
−=
2
SCS Method (Cont.)SCS Method (Cont.)
• Experiments showed SurfaceSurfaceExperiments showed
• So
SIa 2.0=Impervious: CN = 100Impervious: CN = 100Natural: CN < 100Natural: CN < 100
CN, Curve NumberSo( )
SPSPPe 8.0
2.0 2
+−
=
8
9
10
11
12
f, Pe
, in
10090807060
4
5
6
7
8at
ive
Dir
ect R
unof
f 60402010
100)CN0Units;American(
101000
<<
−=CN
S
0
1
2
3
4
Cum
ula100)CN0Units;American( <<
100)C30iS(
25425400−= CN
CNS
00 1 2 3 4 5 6 7 8 9 10 11 12
Cumulative Rainfall, P, in
100)CN30Units;SI( <<
SCS Method (Cont.)SCS Method (Cont.)
• CN depends on previous (antecedent) rainfallCN depends on previous (antecedent) rainfall
• Normal conditions, AMC(II)
• Dry conditions AMC(I) )(2.4)( IICNICN =Dry conditions, AMC(I)
• Wet conditions, AMC(III)
)(058.010)(
IICNICN
−=
)(23)( IICNIIICN =Wet conditions, AMC(III)
55--day antecedent rainfall (in)day antecedent rainfall (in)
)(13.010)(
IICNIIICN
+
55 day a tecede t a a ( )day a tecede t a a ( )
AMC GroupAMC Group Dormant seasonDormant season Growing seasonGrowing season
II < 0.50< 0.50 < 1.4< 1.4
IIII 0.5 0.5 ---- 1.11.1 1.4 1.4 –– 2.12.1
IIIIII > 1.1> 1.1 > 2.1> 2.1
SCS Method (Cont.)SCS Method (Cont.)
• CN depends on soil conditionsCN depends on soil conditions
GroupGroup Minimum Infiltration Minimum Infiltration Rate (in/hr)Rate (in/hr)
Soil typeSoil typeRate (in/hr)Rate (in/hr)
AA 0.3 0.3 –– 0.450.45 High infiltration rates. Deep, well drained High infiltration rates. Deep, well drained sands and gravelssands and gravels
BB 0 150 15 –– 0 300 30 Moderate infiltration rates ModeratelyModerate infiltration rates ModeratelyBB 0.15 0.15 0.300.30 Moderate infiltration rates. Moderately Moderate infiltration rates. Moderately deep, moderately well drained soils with deep, moderately well drained soils with moderately coarse texturesmoderately coarse textures
CC 0.05 0.05 –– 0.150.15 Slow infiltration rates. Soils with layers, Slow infiltration rates. Soils with layers, y ,y ,or soils with moderately fine textures or soils with moderately fine textures
DD 0.00 0.00 –– 0.050.05 Very slow infiltration rates. Clayey soils, Very slow infiltration rates. Clayey soils, high water table, or shallow impervious high water table, or shallow impervious layer layer
Example ‐ SCS Method (1)Example SCS Method (1)
• Rainfall: 5 in. • Area: 1000‐acres• Soils:
– Class B: 50%– Class C: 50%
• Antecedent moisture: AMC(II) ‐ Normal• Land useLand use
– Residential • 40% with 30% impervious cover• 12% with 65% impervious cover
– Paved roads: 18% with curbs and storm sewers– Open land: 16%
• 50% fair grass cover• 50% good grass cover• 50% good grass cover
– Parking lots, etc.: 14%
Example (SCS Method 1, Cont.)Example (SCS Method 1, Cont.)Hydrologic Soil GroupHydrologic Soil Group
BB CC
Land useLand use %% CNCN ProductProduct %% CNCN ProductProduct
R id i l (30% i )R id i l (30% i ) 2020 7272 14 4014 40 2020 8181 16 2016 20Residential (30% imp cover)Residential (30% imp cover) 2020 7272 14.4014.40 2020 8181 16.2016.20
Residential (65% imp cover)Residential (65% imp cover) 66 8585 5.105.10 66 9090 5.405.40
RoadsRoads 99 9898 8.828.82 99 9898 8.828.82
Open land: good coverOpen land: good cover 44 6161 2.442.44 44 7474 2.962.96
Open land: Fair coverOpen land: Fair cover 44 6969 2.762.76 44 7979 3.163.16
Parking lots, etcParking lots, etc 77 9898 6.866.86 77 9898 6.866.86
TotalTotal 5050 40.3840.38 5050 43.4043.40
8.8340.4338.40 =+=CN
Example (SCS Method 1 Cont.)Example (SCS Method 1 Cont.)
• Average AMC 101000−=
CNS8.83=CNAverage AMC CN
22
in93.1108.83
1000=−=S
( ) ( ) in25.393.1*8.05
93.1*2.058.0
2.0 22=
+−
=+−
=SP
SPPe
• Wet AMC ( ) ( ) in13.483.0*8.05
83.0*2.058.0
2.0 22=
+−
=+−
=SP
SPPe
Example (SCS Method 2)Example (SCS Method 2)• Given P, CN = 80, AMC(II)
• Find: Cumulative abstractions and excess rainfall hyetographFind: Cumulative abstractions and excess rainfall hyetograph
Time Time (hr)(hr)
CumulativeCumulativeRainfall (in)Rainfall (in)
Cumulative Cumulative Abstractions (in)Abstractions (in)
CumulativeCumulativeExcess Rainfall (in)Excess Rainfall (in)
Excess RainfallExcess RainfallHyetograph (in)Hyetograph (in)
PP IaIa FaFa PePe00 0011 0 20 211 0.20.222 0.90.933 1.271.2744 2.312.3155 4.654.6566 5.295.2977 5.365.36
Example (SCS Method – 2)Example (SCS Method 2)
• Calculate storage in50.21080
1000101000=−=−=
CNSg
• Calculate initial abstraction
• Initial abstraction removes
80CNin5.05.2*2.02.0 === SIa
– 0.2 in. in 1st period (all the precip)
– 0.3 in. in the 2nd period (only part of the precip)
• Calculate continuing abstraction from SCS
Time Time (hr)(hr)
CumulativeCumulativeRainfall (in)Rainfall (in)
PPCalculate continuing abstraction from SCS method equations
00 00
11 0.20.2
22 0.90.9)5.0(5.2)( −− PIPSF a33 1.271.27
44 2.312.31
55 4.654.65
)0.2()(
)( +=
+−=
PSIPF
a
aa
66 5.295.29
77 5.365.36
Example (SCS method –2, Cont.)Example (SCS method 2, Cont.)
• Cumulative abstractions can now be calculated
Time (hr)
CumulativeRainfall (in)
Cumulative Abstractions (in)
)0.2()5.0(5.2
+−
=P
PFaP I F
0 0 0 -1 0.2 0.2 -2 0 9 0 5 0 34
)0.2( +PP aI aF
2 0.9 0.5 0.343 1.27 0.5 0.594 2.31 0.5 1.055 4 65 0 5 1 56
in34.0)0.29.0(
)5.09.0(5.2hr)(2 =+−
=aF
5 4.65 0.5 1.566 5.29 0.5 1.647 5.36 0.5 1.65
Example (SCS method 2, Cont.)Example (SCS method 2, Cont.)
• Cumulative excess rainfall can now be calculated
• Excess Rainfall Hyetograph can be calculated
Time Cumulative Cumulative Cumulative Excess Rainfall(hr) Rainfall
(in)Abstractions (in) Excess Rainfall (in) Hyetograph (in)
P aI aF eP ePΔ0 0 0 - 0 01 0.2 0.2 - 0 02 0.9 0.5 0.34 0.06 0.063 1.27 0.5 0.59 0.18 0.124 2.31 0.5 1.05 0.76 0.585 4.65 0.5 1.56 2.59 1.836 5.29 0.5 1.64 3.15 0.567 5.36 0.5 1.65 3.21 0.06
Example (SCS method 2, Cont.)Example (SCS method 2, Cont.)
Rainfall (in) Rainfall Hyetographs
2
2.5
1
1.5
0
0.5
Excess RainfallRainfall
0
0 1 2 3 4 5 6 7Time (hour)
Time of ConcentrationTime of Concentration
• Different areas of a watershed contribute toDifferent areas of a watershed contribute to runoff at different times after precipitation beginsbegins
• Time of concentrationTi t hi h ll t f th t h d b i– Time at which all parts of the watershed begin contributing to the runoff from the basin
Time of flow from the farthest point in the– Time of flow from the farthest point in the watershed