Chapter 11 Hydrology

• Hydrologic cycle

• Infiltration

• Rainfall hyetograph, Duration, Intensity, return period

• Effective or Excess rain (ph index, Horton)

• Runoff hydrograph

• Runoff hydrograph for gauged stations (unit hydrograph)

• Synthetic hydrograph for ungauged stations – Rational method

– SCS curve number method

Rainfall Availability and Associated

Growth in Water Resources Engineering

Projects Worldwide

11.1 The Hydrologic Cycle

1 0 0 P r e c i p i t a t i o n o n l a nd

Infiltration

Water table

Groundwater flow

1 Groundwater discharge

38 Surface discharge

61 Evaporation from land

39Moisture over land

385Precipitation

on ocean

424 Evaporationfrom ocean

Surface runoff

Impervious strata

Groundwater Recharge

Precipitation

Snow melt

P

Runoff

Runoff

Evap

ET Evap

Streams

Lake Reservoir

GW

Atmospheric Moisture

Precipitation

Water on Surface Overland Flow

Channel

Flow

The Hydrologic Cycle

Ground Water Ground

Water

Flow

Ocean

Reservoir

Atmosphere

Evaporation

Evapotranspiration

Evaporation

Major Hydrologic Processes

• Precipitation (measured by radar or rain gage)

• Evaporation or ET (loss to atmosphere)

• Infiltration (loss to subsurface soils)

• Overland flow (sheet flow toward nearest stream)

• Streamflow (measured flow at stream gage)

• Ground water flow and well mechanics

• Water quality and contaminant transport (S & GW)

Diagram Showing Two Watersheds

(Catchments)

The Watershed or Basin

• Area of land that drains to a single outlet and

is separated from other watersheds by a

drainage divide.

• Rainfall that falls in a watershed will generate

runoff to that watershed outlet.

• Topographic elevation is used to define a

watershed boundary (land survey or LIDAR)

• Scale is a big issue for analysis

Figure 11.3 Watershed delineation: (a) peak point identification

Figure 11.3 (continued) Watershed delineation: (b) circumscribing boundary. Source: Adapted from

the Natural Resource Conservation Service

(www.nh.nrcs.usda.gov/technical/WS_delineation.html).

Watershed Response

Precipitation over the area

Portion Infiltrates the soil

Portion Evaporates or ET back

Remainder - Overland Flow

Overland flow - Channel flow

Final Hydrograph at Outlet

Reservoir

Tributary

Natural

stream

Urban

Concrete

channel

Q

T

11.2 Precipitation

Watershed area = 240 km2

Arithmetic Mean Method

• The simplest of all is the Arithmetic Mean

Method, which taken an average of all the

rainfall depths

The Theissen polygon method

• This method, first proposed by Thiessen in 1911, considers the representative area for each rain gauge:

• 1. Joining the rain gauge station locations by straight lines to form triangles

• 2. Bisecting the edges of the triangles to form the so-called “Thiessen polygons”

• 3. Calculate the area enclosed around each rain gauge station bounded by the polygon edges (and the catchment boundary, wherever appropriate) to find the area of influence corresponding to the rain gauge. (areas of influence of each rain gauge)

Calculation

The Isohyetal method • This is considered as one of the most accurate methods, but it is dependent

on the skill and experience of the analyst. The method requires the plotting of isohyets as shown in the figure and calculating the areas enclosed either between the isohyets or between an isohyet and the catchment boundary. The areas may be measured with a planimeter if the catchment map is drawn to a scale.

• Area I = 40 km2

• Area II = 80 km2

• Area III = 70 km2

• Area IV = 50 km2

• Total catchment area = 240 km2

• The areas II and III fall between two isohyets each. Hence, these areas may be thought of as corresponding to the following rainfall depths:

• Area II : Corresponds to (10 + 15)/2 = 12.5 mm rainfall depth

• Area III : Corresponds to (5 + 10)/2 = 7.5 mm rainfall depth

• For Area I, we would expect rainfall to be more than 15mm but since there is no record, a rainfall depth of 15mm is accepted. Similarly, for Area IV, a rainfall depth of 5mm has to be taken.

calculation

Area I = 40 km2

Area II = 80 km2

Area III = 70 km2

Area IV = 50 km2

Hydrologic Theory

• One of the principal objectives in

hydrology is to transform rainfall that has

fallen over a watershed area into flows to

be expected in the receiving stream.

• Losses must be considered such as

infiltration or evaporation (long-term)

• Watershed characteristics are important

Watershed Characteristics

Size

Slope

Shape

Soil type

Storage capacity

Reservoir

Divide

Natural

stream

Urban

Concrete

channel

1 mile

11.4 The Watershed Response -

Hydrograph

• As rain falls over a watershed area, a certain portion will infiltrate the soil. Some water will evaporate to atmosphere.

• Rainfall that does not infiltrate or evaporate is available as overland flow and runs off to the nearest stream.

• Smaller tributaries or streams then begin to flow and contribute their load to the main channel at confluences.

• As accumulation continues, the streamflow rises to a maximum (peak flow) and a flood wave moves downstream through the main channel.

• The flow eventually recedes or subsides as all areas drain out.

S e p 8 3

J u n 7 6

A p r 7 9

M a r 9 2

M a r 9 7

2 5 , 0 0 0

3 0 , 0 0 0

5 , 0 0 0

1 0 , 0 0 0

1 5 , 0 0 0

2 0 , 0 0 0

3 6 9 1 2 1 5 1 8 2 1 2 4 T i m e , h r s

Flo

w, cfs

Measured Flow at Main St Gage

29,000 cfs

Time, hrs

Figure 11.11 Schematic of a stream gauging station. Source: USGS, 2008.

HOW IS FLOW MEASURED?

Figure 11.12 Procedure for discharge measurement with current meter. Source: USGS, 2008.

11.5 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

Infiltration and excess rain R

ain

fall

or

infiltra

tion r

ate

(m

m/h

r, in/h

r)

Infiltration capacity rate curve

Actual Infiltration rate

Excess rainfall rate

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

Hydrologic Analysis

RUNOFF

HYDROGRAPHS

A

VR

V

R

dttQR

)(

STORM WATER HYDROGRAPHS

• Graphically represent runoff rates vs. time at watershed outlet

or any selected points of interest in a watershed). and

occurred as a result of rainfall.

• Design Engineers are interested in calculating

– Peak runoff rates (Qp)

– Volume of runoff VR

• Measured hydrographs for the maximum probable storm are

best But not often available

• The volume under the effective rainfall hyetograph is equal to the volume of surface-direct runoff.

• Methods are available to develop a Unit hydrograph or

“synthetic” hydrograph for watersheds that can be used to

determine Qp and VR.

HYDROGRAPH COMPONENTS

• Qp is the maximum flow rate on the hydrograph

• tp (time to peak) is the time from the start of they

hydrograph to qp.

• tb (base time) is the total time duration of the

hydrograph.

HYDROGRAPH COMPONENTS

• Tc (time of concentration) time it takes water to flow from the hydraulically most remote point in a watershed to the watershed outlet

• TL(lag time) is the average of the flow times from all locations in the watershed and can be estimated as the length of time from the center of mass of the first effective rainfall block, to the peak of the runoff hydrograph.

• If each block of effective rainfall has a duration of D

2

DTT Lp

Typical hydrograph (separation of base flow)

• Stream flow hydrographs have two components 1. Direct runoff resulted from a specific rainfall hyetograph 2. base flow from ground water seepage

• Base or groundwater that flow can be separated from direct runoff

Rainfall-Runoff Relationships

• Gauged and ungauged watersheds

• Gauged watersheds

– Watersheds where data on precipitation,

streamflow, and other variables are available

• Ungauged watersheds

– Watersheds with no data on precipitation,

streamflow or other variables.

Unit Hydrograph (HU)

• Unit Hydrograph are developed for gauged

watersheds where data on precipitation,

streamflow, and other variables are

available

• For Ungauged watersheds with no data on

precipitation, streamflow or other variables

we use synthetic hydrograph

03/02/2006

S e p 8 3

J u n 7 6

A p r 7 9

M a r 9 2

M a r 9 7

2 5 , 0 0 0

3 0 , 0 0 0

5 , 0 0 0

1 0 , 0 0 0

1 5 , 0 0 0

2 0 , 0 0 0

3 6 9 1 2 1 5 1 8 2 1 2 4 T i m e , h r s

Flo

w, cfs

Measured Flow at Main St Gage

29,000 cfs

Time, hrs

Unit Hydrograph Theory

• Direct runoff hydrograph resulting from a unit depth of excess rainfall (1.0 in or 1.0 cm) occurring uniformly on a watershed at a constant rate for a specified duration.

• Unit pulse response function of a linear hydrologic system

• Can be used to derive runoff from any excess rainfall R on the watershed of similar duration D or series of excess rain.

Unit hydrograph assumptions

• Assumptions

– Excess rainfall has constant intensity during

duration

– Excess rainfall is uniformly distributed on

watershed

– Base time of runoff is constant

– Ordinates of unit hydrograph are proportional

to total runoff (linearity)

– Unit hydrograph represents all characteristics

of watershed (lumped parameter) and is time

invariant (stationarity)

Typical Unit Hydrograph

Linearity and superposition of the Unit

Hydrograph

Time

(h)

Gross Precipitation

(GRH)

(cm/h)

0 - 1 0.5

1 - 2 2.5

2 - 3 2.5

3 - 4 0.5

Unit Hydrographs - Example

Obtain a Unit Hydrograph for a basin of 315 km2 of area using the gross

rainfall hyetograph and streamflow hydrograph data tabulated below.

Time

(h)

Observed

Hydrograph

(m3/s)

0 100

1 100

2 300

3 700

4 1000

5 800

6 600

7 400

8 300

9 200

10 100

11 100

Cont…

• Separate the baseflow from the observed streamflow hydrograph in order to obtain the Direct Runoff Hydrograph (DRH). For this example, use the horizontal line method to separate the baseflow. From observation of the hydrograph data, the streamflow at the start of the rising limb of the hydrograph is 100 m3/s.

• Compute the volume of Direct Runoff. This volume must be equal to the volume of the Effective Rainfall Hyetograph (ERH).

Cont….

• Volume of Direct Runoff Hydroraph VR = 200+600+900+700+500+300+200+100) m3/s (3600) s = 12'600,000 m3

• Express VR in equivalent units of depth: VDRH in equivalent units of depth = VDRH/Area of the basin = 12'600,000 m3/(315000000 m2) = 0.04 m = 4 cm: R = 4 cm

• Excess rain (4 cm) = Volume of runoff under the hydrograph

• R = CP (C is the runoff constant =4/6=0.67)

• Obtain a Unit Hydrograph by normalizing the DRH. Normalizing implies dividing the ordinates of the DRH by the VR in equivalent units of depth (4 cm).

Table of calculation Time (h) Observed

Hydrograph

(m3/s)

Direct Runoff

Hydrograph

(DRH)

(m3/s)

Unit Hydrograph

(m3/s/cm)

0 100 0 0

1 300 200 50

2 700 600 150

3 1000 900 225

4 800 700 175

5 600 500 125

6 400 300 75

7 300 200 50

8 200 100 25

9 100 0 0

10 100 0 0

Duration, f-index , Excess or Effective

rainfall heytograph (ERH)

• Determine the duration D of the ERH associated with the UH obtained in . In order to do this:

– Determine the volume of losses, VLosses which is equal to the difference between the gross depth of rainfall, VGRH, and the depth of the direct runoff hydrograph, VR .

• VLosses = VGRH - VR = (0.5 + 2.5 + 2.5 +0.5) cm/h 1 h - 4 cm = (6-4)= 2 cm

– Compute the f -index equal to the ratio of the volume of losses to the rainfall duration, tr. Thus,

f-index = VLosses/tr = 2 cm / 4 h = 0.5 cm/h – Determine the ERH by subtracting the infiltration (f -index)

from the GRH:

Duration of the Unit hydrograph

• As observed in the table, the duration of the effective rainfall hyetograph is 2 hours. Thus, D = 2 hours, and the Unit Hydrograph obtained above is a 2-hour Unit Hydrograph.

Time Gross

Precipitatio

n (GRH)

f-index

(cm/h)

Excess rainfall

(cm/h)

(h) (cm/h)

0 - 1 0.5 -0.5 0

1-2 2.5 -0.5 2

2-3 2.5 -0.5 2

3-4 0.5 -0.5 0

Application of UH • Once a UH is derived, it can be used/applied

to find direct runoff and stream flow

hydrograph from other storm events whose

effective rainfall hyetographs can be

represented as a sequence of uniform

intensity (rectangular) pulses each of

duration D.

• using the principles of superposition and

proportionality

Example Application of UH

• Using the UH obtained in A., predict the total streamflow that would be observed as a result of the following ERH

• Determine the volume of each ERH pulse, Pm, expressed in units of equivalent depth:

• Use superposition and proportionality principles:

Time

(h)

Effective

Precipitation

(ERH) (cm/h)

Rainfall

depth

Pm

(cm)

0 - 2 0.5 1.0

2 - 4 1.5 3.0

4 - 6 2.0 4.0

6 - 8 1.0 2.0

1 2 3 4 5 6 7

Time

(h)

UH

(m3/s/cm)

P1*UH

(m3/s)

P2*UH

(m3/s)

P3*UH

(m3/s)

P4*UH

(m3/s)

DRH

(m3/s)

Total

(m3/s)

0 0 0 0 100

1 50 50 50 150

2 150 150 0 150 250

3 225 225 150 375 475

4 175 175 450 0 625 725

5 125 125 675 200 1000 1100

6 75 75 525 600 0 1200 1300

7 50 50 375 900 100 1425 1525

8 25 25 225 700 300 1250 1350

9 0 0 150 500 450 1100 1200

10 75 300 350 725 825

11 0 200 250 450 550

12 100 150 250 350

13 0 100 100 200

14 50 50 150

15 0 0 100

Solution…Cont…

• Columns 2 - 5: Apply the proportionality principle to scale the UH by the actual volume of the corresponding rectangular pulse, Pm. Observe that the resulting hydrographs are lagged so that their origins coincide with the time of occurrence of the corresponding rainfall pulse.

• Column 6: Apply the superposition principle to obtain the DRH by summing up Columns 2 - 5.

• Column 7: Add back the baseflow in order to obtain the Total Streamflow Hydrograph.

Hydrograph

Need for synthetic UH

• UH is applicable only for gauged watershed and

for the point on the stream where data are

measured

• For other locations on the stream in the same

watershed or for nearby (ungauged)

watersheds, synthetic procedures are used.

• Synthetic unit hydrographs provide ordinates of

the unit hydrograph as a function of tp, Qp and a

mathematical or empirical shape description.

Synthetic UH

• Synthetic hydrographs are derived by

– Relating hydrograph characteristics such as

peak flow, base time etc. with watershed

characteristics such as area and time of

concentration.

– Using dimensionless unit hydrograph

– Based on watershed storage

11.6 SCS procedures

Excess Rainfall

• When stream flow hydrograph is not available, the SCS Curve Number Approach is used to determine excess rain that becomes runoff from only rainfall data.

• It also used to construct runoff hydrograph or Unit hydrograph called synthetic hydrograph

• Qp and VR can then be determined for engineering design.

SCS CURVE NUMBER APPROACH

• By far the most popular method.

• Combines initial abstractions and infiltration

losses and estimates rainfall excess as:

SPwhen

SP

SPR 2.0

8.0

2.02

.,,25425400

,,101000

mminSPRforCN

S

inchesinSPRforCN

S

P: Cumulative rainfall (mm, in)

S: soil moisture storage deficit

at time of rainfall (mm, in)

R: runoff (mm, in); begins only

when P > 0.2 S)

CN: Curve number

CN CURVE NUMBER

• A parameter that combines soil type and land use to estimate runoff potential.

• Based on the Hydrologic Soil Group (HSG), land use and condition.

• Range between 0 and 100. The greater the curve number, the greater the potential for RO.

• SCS classified more than 4000 soils into four general HSG (A, B, C, and D)

• Based on soils minimum infiltration rate when the soil is bare and after prolonged wetting.

• In general A have the highest infiltration capacity and lowest runoff potential (sandy soils) and D have lowest infiltration rates and highest runoff potential (clay soils)

• Curve numbers for various land uses ranging from cultivated land to industrial and residential districts.

• Impervious areas and water surfaces are assigned curve numbers of 98-100.

Antecedent Moisture Conditions

Table 11.4 SCS Runoff Curve Numbers for AMC II

CURVE NUMBERS

weighted CN for Mixed Land Uses and Soil

group

• An area weighted CN is used when the

area considered is for mixed land uses and

Hydrological Soil Group.

i

ii

A

CNACN

CREATING AN EFFECTIVE RAINFALL

HYETOGRAPH

• Calculate the accumulated P for each time step from a rainfall hyetograph.

• Calculate the appropriate weighted CN.

• Calculate S using Equation (11.4).

• Find 0.2S.

• For each time step where the accumulated P > 0.2 S calculate the accumulated R using Equation (11.3).

• Find the incremental R at each time step.

• Plot the incremental R vs. time.

EXAMPLE PROBLEM

• Given: – Precipitation (P) = 4.04 in. – A watershed that has:

• 35% cultivated with a D soil group • 30% meadow with a B soil group • 35% thin forest with a C soil group

• Required: – Calculate the surface runoff (excess

rainfall)

35% Cultivated

HSG = D

30% Meadow

HSG = B

35% Thin Forest

HSG = C

Watershed with Land Use % and HSGs

Listed

Cont…EXAMPLE PROBLEM

1. Find the curve numbers Use HSG % CN* Cultivated D 35 91 Meadow B 30 58 Thin Forest C 35 77 *Table 5.1 text (reference is important)

2. Calculate a weighted CN

Weights based on % area CNavg = 0.35(91) + 0.30(58) + 0.35(77) CN avg = 76.2 = 76

Cont…EXAMPLE PROBLEM

3. Calculate the S term S = 1000 / CN – 10 = (1000 / 76) – 10 S = 3.16 in.

4. Check to see if P > 0.2S 0.2S = 0.2(3.16) = 0.63 in. P > 0.2S

5. Calculate surface runoff (R) R = [(P - 0.2S)^2] / (P + 0.8S) R = [(4.04 – 0.2(3.16)]2 / [4.04 + ((0.8)3.16)] R = 1.77 in.

For a rainfall event = 4.04 in. on the given watershed with average soil moisture conditions

SCS Triangle hydrograph

• If R is calculated for a given storm in A

watershed of area A, then a synthetic

triangular hydrograph can be constructed

• Only two parameters are needed

– Peak discharge Qp, and

– Time to peak Tp and Recession time Tr which

usually depend on Tp.

SCS Triangular Hydrograph

• RA = volume under

the curve

Time (hr)

Q (

m3/s

; cfs

)

Qp

Tp Tr = 1.67 Tp

D

TL

Excess rain (R) of duration D

p

p

p

p

p

T

RAQ

T

RAQ

QTpRA

75.0

67.2

2

)67.2(2

1

Tp time to peak

• Now R is find from the SCS procedure

• Tp is usually a function of the longest

travel time in a watershed, called time of

concentration Tc

Tp = 0.67 Tc

Time of Concentration (tc)

• The time it takes flow to move from the most hydraulically remote point in a watershed to the watershed outlet – The distance from the hydraulically most remote point

to the outlet is called the hydraulic length

• tc is the sum of flow times for the various flow segments as the water travels to the watershed outlet – Overland flow + shallow channel flow + open channel

flow

• Travel time for each segment depends on length of travel and flow velocity

Hydraulically most

remote point in the

watershed

TIME of Concentration, Tc

• SCS Equation to calculate time of

Concentration • L = hydraulic length of watershed (feet)

• S = curve number parameter (inches)

• Y = average land slope of the watershed (%)

• Tc = time lag (hours)

5.0

7.08.0

1140

)1(

Y

SLTc

SCS Unit Hydrograph

Time to Peak and Duration

• Duration of rainfall excess should be taken 1/5 to 1/3 Tp.

• Base time, Tb

– Tb=2.67 Tp.

– Some use Tb = 5Tp

– Some use Tb = ∞

pb

cp

cL

c

Lp

TT

TT

TT

TD

DTT

67.2

67.0

6.0

133.0

2

SCS Synthetic Unit hydrograph

• Synthetic UH in which

the discharge is

expressed by the ratio

of Q to Qp and time by

the ratio of T to Tp

• If peak discharge and

lag time are known, UH

can be estimated.

p

pt

AQ

R

75.0

1

SCS unit hydrograph

Example • Construct a 10-min SCS

UH. A = 3.0 km2 and Tc =

1.25 h

smQp

hT

hD

TT

b

Lp

/49.7833.0

)3(08.2

22.2)833.0(67.2

833.02/166.0)25.1(6.02

3

Multiply y-axis of SCS

hydrograph by Qp and x-axis

by Tp to get the required UH,

or construct a triangular UH

q

t

0.833 h

2.22 h

7.49

m3/s.cm

EXAMPLE Solution:

– HSG = D / Commercial T. 5.1 CN = 95

• S = 0.53 in. – Assume AMC = II R = 1.96 in. of runoff

– Find points to develop the unit hydrograph

• tl = 0.75 hr (45 min)

• tp = 1.25 hr (75 min)

• tb = 3.33 hr (200 min)

• qp = 302 cfs / 1 in. of runoff – Plot unit hydrograph

– Check area under the triangle 1 in.

Q (

cfs)

T(min)

600

100

300

150 tb = 200 tp = 75 50

200

400 qp = 302.5 cfs

Volume under triangle = (302.5 cfs x 4,500 sec) / 2 +

[(302.5 x (12,000 – 4,500 sec)] / 2 = 1,812,000 ft3

Surface runoff depth = 1,812,000 ft3 / 21,780,000 ft2 =

0.08 ft = 1.0 in. ok

Check runoff depth?

EXAMPLE

Solution:

qp 2.5” rain = 302.5 cfs x 1.96 in. of SRO

qp 2.5” rain = 592.9 cfs

Plot storm hydrograph

Check area under the triangle 1.96

in.

T(min)

Q (cfs)

Surface runoff depth

= 1.96 in. ok

600

100

300

150 tb = 200 tp = 75 50

200

400

qp = 592.9 cfs

Volume under triangle

= 3,557,400 ft3

11.8 The Rational Method

• Empirical method developed in 1851 to estimate Qp (m3/s, cfs). many limitations but still widely used

• Qp = peak flow (m3/s; cfs) • C = coefficient (dimensionless) • i = rainfall intensity (mm/hr; in/hr) with duration (D) = tc

– tc = time of concentration

• A = drainage area (ha; ac) • 1.008 is conversion factor from ac-in/hr to cfs. • 0.0028 is conversion factor ha-mm/hr to m3/s

CiAQ

CiAQ

p

p

008.1

0028.0

Rational Method

• Rationale of rational method: – If rain falls steadily across the entire watershed

long enough, Peak Q will increase until it equal the

average rate of rain times the basin area (adjust

by a coefficient to account for infiltration)

– the entire watershed contribution to runoff occurs

when rainfall duration (D) = Time of Concentration

tc

– If D ≠ tc , runoff less than when D = tc

– Qp is at a maximum when D = tc

“Rational Method” Limitations

• Reasonable for small watersheds

• Assumes constant and uniform rainfall and constant infiltration

• The runoff coefficient is not constant during a storm

• No ability to predict flow as a function of time (only peak flow)

• Only applicable for storms with duration longer than the time of concentration

• Runoff frequency = rainfall frequency used • Peak flow equation only • Cannot be used to predict time to the peak (tp) and

should not be used to develop hydrographs

pQ CiA=

< 80 ha; 200 acre

Time of Concentration (Tc):

Kirpich

• Tc = time of concentration [min]

• L = “stream” or “flow path” length [ft]

• h = elevation difference between basin

ends [ft]

385.0

36

h

L 10 x 3.35

ct

Watch those units!

Time of Concentration Contd.

• Another name for Tc is gathering time. Tc

can be related to catchment area, slope

etc. using the Kirpich equation:

Tc = 0.02 L 0.77 S – 0.385

• Tc is the time of concentration (min);

• L is the maximum length of flow (m);

• S is the watershed gradient (m/m).

Time of Concentration (Tc):

Hatheway

• Tc = time of concentration [min]

• L = “stream” or “flow path” length [ft]

• S = mean slope of the basin

• N = Manning’s roughness coefficient (0.02

smooth to 0.8 grass overland)

47.0

3

2

S

nLtc

ESTIMATING THE TIME

PARAMETERS

• Time of concentration (tc)

• For some areas, we can sum the time for

various flow segments as the water flows

toward the watershed outlet.

• Segments

– Overland flow

– Shallow channel flow

– Flow in open channels.

n

1ii

ic

v

Lt

Time of Concentration Contd.

• With Tc obtained for the catchment, decide on a

return period.

• For small conservation works, return period is

assumed as 10 years.

• With the Tc and assumed return period, get an

intensity value from the Intensity-Duration curve

derived for the area.

Intensity-Durtaion-Frequency Curve for

Madaba

Rational Method

• Runoff Coefficient (C) – Difficult to accurately determine – Must reflect factors such as: interception, infiltration,

surface detention, antecedent moisture conditions

– Studies have shown that C is not a constant • C increases with wetter conditions

– Table 5.5 contains a range of values for C

– Compute average C for composite areas • Area weighted basis

• Cavg = SCiAi / SAi (same method used with Curve

Numbers)

Runoff Coefficients for the Rational Method

“Rational Formula” - Review

• Estimate tc

• Pick duration of storm = tc

• Estimate point rainfall intensity based on

synthetic storm

• Convert point rainfall intensity to average area

intensity

• Estimate runoff coefficient based on land use

• Calculate Qp

pQ CiA=

Why is this the max flow?

“Rational Formula” - Fall Creek

10 Year Storm

• Area = 126 mi2 = 126 x 640= 80640 acre =

326 km2

• L ­ 15 miles ­ 80,000 ft

• H ­ 800 ft (between Beebe lake and hills)

• tc = 274 min = 4.6 hours

• 6 hr storm = 2.5” or 0.42”/hr

• Area factor = 0.87 therefore i = 0.42 x 0.87 =

0.36 in/hr

tc

3.35 x 106 L3

h

0.385

Area correction

“Rational Formula” - Fall Creek 10 Year

Storm

• C ­ 0.25 (moderately steep, grass-covered

clayey soils, some development)

• Qp = 1.008CiA

• QP = 7300 ft3/s (200 m3/s)

• Empirical 10 year flood is approximately 150

m3/s

2

2 640126

36.025.0008.1

mi

acremi

hr

inQp

Rational Formula Example

• Suppose it rains 0.25” in 30 minutes (i =

0.5 in/hr) on Fall Creek watershed (A =

126 mi2)and runoff coefficient is 0.25.

What is the peak flow?

2

2 640126

5.0

25.025.0008.1

mi

acremi

h

inQp

smcfsQp /1150650,40 3

Peak flow in record was 450 m3/s. What is wrong?

Method not valid for storms with duration less than tc.