Excess Rainfall Reading for today’s material:

Sections 5.3-5.7

Slides prepared by V.M. Merwade

Quote for today (contributed by Tyler Jantzen)

"How many times it thundered before Franklin took the hint! Nature is always hinting at us. It hints over and over again. And suddenly we take the hint.“

Robert Frost

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

-index

-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

-index method

M

mmd tRr

1

• 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 5. ERH = Rm - t

interval timerunoffdriecttongcontributi

rainfallofintervals#indexPhi

rainfall observedrunoffdirect ofdepth

t

M

Rr

m

d

ExampleTime Observed

Rain Flow

in 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

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

M

mmd tRr

in27.0t

in54.0

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

t=0.27

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

PPee is runoff volume, P is is runoff volume, P is precipitation volume, Fprecipitation volume, Faa is continuing is continuing abstraction, and Iabstraction, and Iaa is the is the sum of initial losses sum of initial losses (depression storage, (depression storage, interception, ET)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 Pe

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)

• 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

CNS

a

ea IP

PSF

in5.05.2*2.02.0 SIa

aae FIPP )0.2(

)5.0(5.2)(

)(

PP

SIPIPSF

a

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 Fa

0 0 0 -

1 0.2 0.2 -

2 0.9 0.5 0.34

3 1.27 0.5 0.59

4 2.31 0.5 1.05

5 4.65 0.5 1.56

6 5.29 0.5 1.64

7 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

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