Excess Rainfall Reading for today’s material:

Sections 5.3-5.8

Excess rainfall • Rainfall that is neither retained on the land

surface nor infiltrated into the soil• Graph of excess rainfall versus time is

called excess rainfall hyetograph• Direct runoff = observed streamflow -

baseflow• Excess rainfall = observed rainfall -

abstractions• Abstractions/losses – difference between

total rainfall hyetograph and excess rainfall hyetograph

f-index

• f-index: Constant rate of abstraction yielding excess rainfall hyetograph with depth equal to depth of direct runoff

• Used to compute excess rainfall hyetograph when observed rainfall and streamflow data are available

f-index method

M

mmd tRr

1

f

• Goal: pick t, and adjust value of M to satisfy the equation

• Steps1. Estimate baseflow2. DRH = streamflow

hydrograph – baseflow3. Compute rd, rd =

Vd/watershed area4. Adjust M until you get a

satisfactory value of f5. ERH = Rm - ft

interval timerunoffdriecttongcontributi

rainfallofintervals#indexPhi

rainfall observedrunoffdirect ofdepth

t

M

Rr

m

d

f

ExampleTime Observed

Rain Flowin cfs

8:30 203

9:00 0.15 246

9:30 0.26 283

10:00 1.33 828

10:30 2.2 2323

11:00 0.2 5697

11:30 0.09 9531

12:00 11025

12:30 8234

1:00 4321

1:30 2246

2:00 1802

2:30 1230

3:00 713

3:30 394

4:00 354

4:30 303

0

2000

4000

6000

8000

10000

12000

7:30 PM 9:00 PM 10:30 PM 12:00 AM 1:30 AM 3:00 AM 4:30 AM 6:00 AM

Time

Stre

amflo

w (c

fs)

0

0.5

1

1.5

2

2.5

No direct runoff until after 9:30And little precip after 11:00

Have precipitation and streamflow data, need to estimate losses

Basin area A = 7.03 mi2

Example (Cont.)• Estimate baseflow (straight line method)

– Constant = 400 cfs

0

2000

4000

6000

8000

10000

12000

7:30 PM 9:00 PM 10:30 PM 12:00 AM 1:30 AM 3:00 AM 4:30 AM 6:00 AM

Time

Stre

amflo

w (c

fs)

baseflow

Time Observed Direct Runoff

Rain Flow in cfs cfs

8:30 0.15 203 9:00 0.26 246 9:30 1.33 283

10:00 2.2 828 428 10:30 2.08 2323 1923 11:00 0.2 5697 5297 11:30 0.09 9531 9131 12:00 11025 10625 12:30 8234 7834

1:00 4321 3921 1:30 2246 1846 2:00 1802 1402 2:30 1230 830 3:00 713 313 3:30 394 4:00 354 43550 4:30 303

Example (Cont.)• Calculate Direct

Runoff Hydrograph– Subtract 400 cfs

Total = 43,550 cfs

Example (Cont.)• Compute volume of direct runoff

37

3

11

1

11

1

ft10*7.839

/sft 550,43*hr5.0*s/hr3600

n

nn

nd QttQV

• Compute depth of direct runoff

in80.4ft4.0

ft5280*mi03.7ft10*7.839

22

37

AV

r dd

Example (Cont.)• Neglect all precipitation intervals that occur

before the onset of direct runoff (before 9:30)• Select Rm as the precipitation values in the 1.5

hour period from 10:00 – 11:30

)5.0*3*08.220.233.1(80.41

f

f

M

mmd tRr

in27.0tf

in54.0f

in80.4dr

Example (Cont.)

0

2000

4000

6000

8000

10000

12000

7:30 PM 9:00 PM 10:30 PM 12:00 AM 1:30 AM 3:00 AM 4:30 AM 6:00 AM

Time

Stre

amflo

w (c

fs)

0

0.5

1

1.5

2

2.5

ft=0.27

Time Observed Direct Runoff

Excess Rainfall

Rain Flow in cfs cfs in

8:30 0.15 203 9:00 0.26 246 9:30 1.33 283

10:00 2.2 828 428 1.06 10:30 2.08 2323 1923 1.93 11:00 0.2 5697 5297 1.81 11:30 0.09 9531 9131 12:00 11025 10625 12:30 8234 7834

1:00 4321 3921 1:30 2246 1846 2:00 1802 1402 2:30 1230 830 3:00 713 313 3:30 394 4:00 354 43550 4:30 303

SCS method• Soil conservation service (SCS) method is an

experimentally derived method to determine rainfall excess using information about soils, vegetative cover, hydrologic condition and antecedent moisture conditions

• The method is based on the simple relationship that Pe = P - Fa – Ia

Pe is runoff depth, P is precipitation depth, Fa is continuing abstraction, and Ia is the sum of initial losses (depression storage, interception, ET)

Time

Prec

ipit

atio

n

pt

aI aF

eP

aae FIPP

Abstractions – SCS Method• In general

• After runoff begins

• Potential runoff

• SCS Assumption

• Combining SCS assumption with P=Pe+Ia+Fa

Time

Prec

ipit

atio

n

pt

aI aF

eP

aae FIPP

StorageMaximumPotentialSnAbstractioContinuing

nAbstractioInitialExcess Rainfall

Rainfall Total

a

a

e

FIPP

PPe

SFa

aIP

a

eaIPP

SF

SIP

IPP

a

ae

2

SCS Method (Cont.)• Experiments showed

• So

SIa 2.0

SPSPPe 8.0

2.0 2

0

1

2

3

4

5

6

7

8

9

10

11

12

0 1 2 3 4 5 6 7 8 9 10 11 12Cumulative Rainfall, P, in

Cum

ulat

ive

Dir

ect R

unof

f, Pe

, in

10090807060402010

• Surface– Impervious: CN =

100– Natural: CN < 100

100)CN0Units;American(

101000

CN

S

100)CN30Units;SI(

25425400

CNCN

S

SCS Method (Cont.)• S and CN depend on antecedent rainfall

conditions• Normal conditions, AMC(II)• Dry conditions, AMC(I)

• Wet conditions, AMC(III)

)(058.010)(2.4)(IICN

IICNICN

)(13.010)(23)(IICN

IICNIIICN

SCS Method (Cont.)• SCS Curve Numbers depend on soil conditions

Group Minimum Infiltration Rate (in/hr)

Soil type

A 0.3 – 0.45 High infiltration rates. Deep, well drained sands and gravels

B 0.15 – 0.30 Moderate infiltration rates. Moderately deep, moderately well drained soils with moderately coarse textures (silt, silt loam)

C 0.05 – 0.15 Slow infiltration rates. Soils with layers, or soils with moderately fine textures (clay loams)

D 0.00 – 0.05 Very slow infiltration rates. Clayey soils, high water table, or shallow impervious layer

Example - SCS Method - 1• Rainfall: 5 in. • Area: 1000-ac• Soils:

– Class B: 50%– Class C: 50%

• Antecedent moisture: AMC(II)• Land 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

– Parking lots, etc.: 14%

Example (SCS Method – 1, Cont.)

Hydrologic Soil Group

B C

Land use % CN Product % CN Product

Residential (30% imp cover)

20 72 14.40 20 81 16.20

Residential (65% imp cover)

6 85 5.10 6 90 5.40

Roads 9 98 8.82 9 98 8.82

Open land: good cover 4 61 2.44 4 74 2.96

Open land: Fair cover 4 69 2.76 4 79 3.16

Parking lots, etc 7 98 6.86 7 98 6.86

Total 50 40.38 50 43.40

8.8340.4338.40 CNCN values come from Table 5.5.2

Example (SCS Method – 1 Cont.)

• Average AMC

• Wet AMC3.92

8.83*13.0108.83*23

)(13.010)(23)(

IICNIICNIIICN

in25.393.1*8.05

93.1*2.058.0

2.0 22

SPSPPe

in93.1108.83

1000 S

8.83CN

in13.483.0*8.05

83.0*2.058.0

2.0 22

SPSPPe

in83.0103.92

1000S

101000 CN

S

Example (SCS Method – 2)• Given P, CN = 80, AMC(II)• Find: Cumulative abstractions and excess rainfall hyetograph

Time (hr)

Cumulative

Rainfall (in)

Cumulative Abstractions (in)

CumulativeExcess Rainfall

(in)

Excess RainfallHyetograph (in)

P Ia Fa Pe0 01 0.22 0.93 1.274 2.315 4.656 5.297 5.36

Example (SCS Method – 2)• Calculate storage• Calculate initial abstraction• Initial abstraction removes

– 0.2 in. in 1st period (all the precip)– 0.3 in. in the 2nd period (only part

of the precip)• Calculate continuing abstraction

in50.21080

1000101000 CN

S

a

ea IP

PSF

in5.05.2*2.02.0 SIa

aae FIPP )0.2(

)5.0(5.2)(

)(

PP

SIPIPS

Fa

aa

in34.0)0.29.0(

)5.09.0(5.2hr)(2 aF

Time (hr)

CumulativeRainfall (in)

P

0 0

1 0.2

2 0.9

3 1.27

4 2.31

5 4.65

6 5.29

7 5.36

Example (SCS method – 2)• Cumulative abstractions can now be calculated

Time (hr)

Cumulative

Rainfall (in)

Cumulative Abstractions (in)

P Ia Fa0 0 0 -1 0.2 0.2 -2 0.9 0.5 0.343 1.27 0.5 0.594 2.31 0.5 1.055 4.65 0.5 1.566 5.29 0.5 1.647 5.36 0.5 1.65

)0.2()5.0(5.2

PPFa

Example (SCS method – 2)• Cumulative excess rainfall can now be calculated• Excess Rainfall Hyetograph can be calculated

Time (hr)

CumulativeRainfall

(in)

Cumulative Abstractions (in)

CumulativeExcess Rainfall (in)

Excess RainfallHyetograph (in)

P Ia Fa Pe

0 0 0 - 0 0

1 0.2 0.2 - 0 0

2 0.9 0.5 0.34 0.06 0.06

3 1.27 0.5 0.59 0.18 0.12

4 2.31 0.5 1.05 0.76 0.58

5 4.65 0.5 1.56 2.59 1.83

6 5.29 0.5 1.64 3.15 0.56

7 5.36 0.5 1.65 3.21 0.06

aae FIPP

Example (SCS method – 2)• Cumulative excess rainfall can now be calculated• Excess Rainfall Hyetograph can be calculated

Time (hr)

CumulativeRainfall

(in)

Cumulative Abstractions (in)

CumulativeExcess Rainfall (in)

Excess RainfallHyetograph (in)

P Ia Fa Pe

0 0 0 - 0 0

1 0.2 0.2 - 0 0

2 0.9 0.5 0.34 0.06 0.06

3 1.27 0.5 0.59 0.18 0.12

4 2.31 0.5 1.05 0.76 0.58

5 4.65 0.5 1.56 2.59 1.83

6 5.29 0.5 1.64 3.15 0.56

7 5.36 0.5 1.65 3.21 0.06

aae FIPP

0 1 2 3 4 5 6 7

Excess Rainf al lRainf al l

0

0.5

1

1.5

2

2.5

Time (hour)

Rainf al l (in) Rainf al l Hyetogr aphs

Time of Concentration• Different areas of a

watershed contribute to runoff at different times after precipitation begins

• Time of concentration– 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 watershed

Isochrones: boundaries of contributing areas with equal time of flow to the watershed outlet

Stream ordering• Quantitative way of studying

streams. Developed by Horton and then modified by Strahler.

• Each headwater stream is designated as first order stream

• When two first order stream combine, they produce second order stream

• Only when two streams of the same order combine, the stream order increases by one

• When a lower order stream combines with a higher order stream, the higher order is retained in the combined stream