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