Hydrograph_analysis_2 hydro.pdf

7/23/2019 Hydrograph_analysis_2 hydro.pdf

1/68

Advance Hydrology

Hydrograph analysis


2/68

hydrograph

hydrograph

A hydrograph is a continuous

plot of instantaneous discharge

v/s time.

A hydrograph may be used to

show how the water flow in a

drainage basin (particularly

river runoff) responds to a period

of rain.

hydrograph


3/68

hydrograph

hydrograph

Hydrograph results from

combination of physiographic

and meteorological conditions in

a watershed.

.

Watershed


4/68

hydrograph

hydrograph

Hydrograph represents the

integrated effects of

climate, hydrologic losses,

surface runoff, interflow,

and ground water flow

Hydrologic cycle


5/68

hydrograph

hydrograph

Detailed analysis of

hydrographs is usually

important in

1. Flood damage mitigation,

2. Flood forecasting

3. Design flows for

structures that convey

floodwaters.

Flood forecasting, mitigation,Design of structures


6/68

Hydrograph analysis

Hydrograph analysis

During the rainfall, hydrologic losses

such as infiltration, depression storage

and detention storage must be satisfied

prior to the onset of surface runoff

As the depth of surface detention

increases, overland flow may occur in

portion if a basin Water eventually

moves into small rivulets, small

channels and finally the main stream of

a watershed

Some of the water that infiltrates thesoil may move laterally through upper

soil zones (subsurface storm flow) until

it enters a stream channel

Hydrograph analysis


7/68

Hydrograph analysis

Hydrograph analysis

If the rainfall continues at a

constant intensity for a very long

period, storage is filled at some

point and then an equilibriumdischarge can be reached

In equilibrium discharge the

inflow and outflow are equal

The point P indicates the time at

which the entire discharge area

contributes to the flow

The condition of equilibrium

discharge is seldom observed in

nature, except for very small basins,

because of natural variations in

rainfall intensity and duration

Hydrograph analysis


8/68

Hydrograph analysis

Hydrograph analysis

The typical hydrograph is

characterized by a1. Rising limb

2. Crest

3. Recession curve

The inflation point on thefalling limb is oftenassumed to be the pointwhere direct runoff ends

Hydrograph analysis


9/68

Hydrograph analysis

A flood hydrograph is a graph of two axis,'discharge' and 'time' Plotted on the graphis the amount of discharge over a period oftime.

There are many different factors that canaffect the appearance and shape of ahydrograph. Certain conditions can causethe line on the hydrograph to be tall and

thin and other conditions can cause it to beshort and wide.


10/68

Factors that influence the hydrograph shape and volume

Meteorological factors

Rainfall intensity and

pattern

Areal distribution or rainfallover the basin and

Size and duration of the

storm event

Rainfall intensity and pattern


11/68

Factors that influence the hydrograph shapeand volume

Physiographic or watershedfactors

1.Size and shape of the drainage area

2.Slope of the land surface and main

channel

3.Channel morphology and drainagetype

4.Soil types and distribution

5.Storage detention in the watershed

The size, shape and relief of

the basin are important controls. Water

takes longer to reach the trunk streamin a large, round basin than in does in a

small, narrow one.

Where gradients are steep,

water runs off faster, reaches the river

more quickly and causes a steep rising

limb. Prolonged heavy rain causes more

overland flow than light drizzly rain.

Shape and slope


12/68

Factors that influence the hydrographshape and volume

Human factors

1. vegetation interceptsprecipitation and allowsevaporation to take place

directly into the atmosphereso reducing the amount ofwater available for overlandflow

2. large number ofimpermeable surfaces in

urban areas encourages runoff into gutters and drainscarrying water quickly tothe nearest river.

vegetation


13/68

Hydrograph analysis


14/68

Hydrograph analysis

1. The ascending limb is the first part of

the line on a hydrograph that rises to the peak

discharge.

1. If the gradient is steep, then this can indicatethat the amount of rainfall becoming overland

flow is very high, the result of this is that all the

water reaches the river very quickly and all in a

short period of time, this gives the immediate

steep ascending limb on the hydrograph.


15/68

Hydrograph analysis

2. Peak discharge is the term used to

describe the maximum amount of discharge

from the river over the period of time recorded;

this peak discharge can be high or lowdepending on Climatic factors.


16/68

Hydrograph analysis

3. The descending limb is the last part of the line

on a hydrograph, showing the discharge dropping with

time after the peak discharge. This line can also be

either steep or gentle.

1. If it is steep then it indicates that the river is very

efficient because it takes the excess water from the flood

away quickly and brings the river back to its base flow.

2. If the descending limb is gentle then it could mean the

river is less efficient, it could also mean that the storm

endured over a long period of time and much water is

still being contributed to the river by groundwater flow.


17/68

Base-flow Separation

Discharge which is not associated

with the storm (i.e. from groundwater) is

termed base-flow. Hydrograph or base-

flow separation is performed to determinethe portion of the hydrograph attributable

to base-flow.


18/68



Discharge which is notassociated with thestorm (i.e. fromgroundwater) is termedbase-flow.

Hydrograph or base-flowseparation is performed

to determine the portionof the hydrographattributable to base-flow.


19/68



1.Constant-discharge

method

2.Constant-slope 3.Concave(mostrealistic)

4.Master depletion curve

method (use when the

most accurate model ofhydrograph recessions is

needed).


= 0.83.


20/68


Constant-discharge method:

Assume basef low constant

regardless of stream height

(discharge). project from minimum

value immediately prior to

beginning of storm

hydrograph.

Constant-discharge method:


21/68


Constant-slope

Connect inflection point onreceeding limb of storm

hydrograph to beginning ofstorm hydrograph

Assumes flow from aquifersbegan prior to start of currentstorm, arbitrarily sets it toinflection point

For large watersheds setinflection point at = . ,where N is number of daysafter hydrograph peak, A isDischarge area in km2

Constant-slope


22/68


Concave (most realistic):

Assume baseflow decreaseswhile streamflow increases(i.e. to peak of storm

hydrograph) project hydrograph trend

from minimum dischargevalue immediately prior tobeginning of stormhydrograph to directly

beneath hydrograph peak connect that point to

inflection point on reseedinglimb of storm hydrograph

Concave (most realistic):


23/68


Master depletion curve

Master depletion curvemethod use when the mostaccurate model of hydrograph

recessions is needed. combine data from several

recessions to make generalrecession model.

from this an equation of the

form = can bederived, which gives dischargeat any time t after dischargeis measured.

Master depletion curve


24/68

Effective rainfall

Effective rainfall Effective rainfall

Effective rainfall is

equal to the difference

between total rainfall andactual losses. These losses

consist in interception,

storage within depressions,

infiltration and

evapotranspiration.


25/68

Effective rainfall

The simpler method to determine

rainfall excess include

1. index method

2. Horton infiltration method


26/68

Effective rainfall

The index method

The index is defined as the averageinfiltration capacity that, for a certainrainfall, is considered as a constantvalue in time. Above this infiltrationcapacity the surplus of theprecipitation is considered as effectiverainfall. this index is obtained bydrawing a parallel line to the timeabscissa so that the area of thehyetograph above this line wouldrepresent the effective rainfall. Thisindex integrates all the losses that

occur in the process of runoffformation: interception, retention,evaporation and infiltration. For therainfall values having significantdepths, interception and retention indepressions might be neglected.

The index method


27/68

Effective rainfall

Horton method estimates infiltration with anexponential-type equation that slowly declines intime as rainfall continues and is given by

f= fc + (fofc) e-kt ( when rainfall intensity i>f)where

f = infiltration capacity (in./hr)

fo = initial infiltration capacity (in./hr)

fc = final infiltration capacity (in./hr)

k = empirical constant ()


28/68

Example 1

Rainfall of magnitude 3.8 cm and 2.8 cm

occurring on two consecutive 4-h durations ona catchment area 27 = produced the

following hydrograph of flow at the outlet ofthe catchment. Estimate the rainfall excess

and -index.

Time from startof rain

(h)

-6 0 6 12 18 24 30 36 42 48 54 60 66

flow

(m3/s)

6 5 13 26 21 16 12 9 7 5 5 4.5 4.5


29/68

Example 1

Solution

Base flow separation:

Using Simple straight

line method, N = 0.83 .= 0.83

(27).= 1.6 days = 38.5

h

So the base flow startsat 0thh and ends at the

point (12+38.5)

Hydrograph


30/68

Example 1 (Solution)

DRH ordinates Base flow seperation


31/68


Area of DRH = (6*60*60)[1/2 (8)+1/2 (8+21)+ 1/2 (21+16)+ 1/2 (16+11)+ 1/2

(11+7)+ 1/2 (7+4)+ 1/2 (4+2)+1/2 (2)] = 1.4904 * 106

Run-off depth(P) = Runoff volume/catchmentarea =

1.4904 * 106/27* 106= 0.0552m = 5.52cm =(Rainfall excess)

Total rainfall (P) = 3.8 +2.8 = 6.6cm Duration of rainfall excess = 8h

-index = (P-R)/t = (6.6-5.52)/8 = 0.135cm/h


32/68

Example 2

A storm over a catchment of area 5.0 km2 had a duration of 14hours. The

mass curve of rainfall of the storm is as follows:

If the -index of the catchment is 0.4cm/h, determine the effectiverainfall hyetograph and the volume of direct runoff from the

catchment due to the storm.

Time from start of

storm (h)

0 2 4 6 8 10 12 14

Accumulat ed

rainfall (cm)

0 0.6 2.8 5.2 6.6 7.5 9.2 9.6


33/68


Total effective rainfall = Direct runoff due to storm = area of ER hyetograph ==(0.7+0.8+0.35+0.45)*2 = 4.6 cm

Volume of direct runoff = (4.6/100) * 5.0*(1000)2 = 230000m3


34/68

Unit hydrograph

The unit hydrograph is the direct runoff

hydrograph produced by a storm of given

duration such that the total volume of excess

rainfall is 1 mm. The total volume of directrunoff is also 1 mm.


35/68

35

Unit Hydrograph Theory


36/68

36

Unit Hydrograph Lingo

Duration

Lag Time

Time of Concentration

Rising Limb Recession Limb (falling limb)

Peak Flow

Time to Peak (rise time)

Recession Curve

Separation

Base flow


37/68

37

Graphical Representation

Lagtime

Timeofconcentration

Duration of

excess

precipitation.

Baseflow


38/68

38

Methods of Developing UHGs

From Streamflow Data

Synthetically

Snyder SCS

Time-Area (Clark, 1945)

Fitted Distributions

Geomorphologic


39/68

Unit hydrograph

1.The unit hydrograph is the direct runoff hydrographproduced by a storm of given duration such that the total volume ofexcess rainfall is 1 mm. The total volume of direct runoff is also 1mm.

2. The ordinates of UH indicate the direct runoff flowproduced by the watershed for every millimeter of excess rainfall;

therefore, the units are

/

3. A volume of 1 mm is the amount of water in a 1-mm layeruniformly distributed over the entire watershed area. This volumeis equal to the area under the UH.

4. Storms of different durations produce different UHs evenif the excess rainfall volume is always 1 mm.

5. Longer storms will likely produce smaller peaks andlonger duration in the UH.

6. The duration associated with the UH that of originatingstorm and not the base duration of the UH


40/68

Unit hydrograph

Assumptions

Unit hydrograph theory assumes that watersheds behave as linearsystems. The following are the fundamental assumptions of UH theory.

1. The duration of direct runoff is always the same for uniform-intensity storms of the same duration, regardless of the intensity. This meansthat the time base of the hydrograph does not change and that the intensityonly affects the discharge.

2. The direct runoff volumes produced by two different excessrainfall distributions are in the same proportion as the excess rainfall volume.This means that the ordinates of the UH are directly proportional to the stormintensity. If storm A produces a given hydrograph and Storm B is equal to

storm A multiplied by a factor, then the hydrograph produced by storm B willbe equal to the hydrograph produced by storm A multiplied by the samefactor.


41/68

Unit hydrograph

3. The time distribution of the direct runoff is independent of

concurrent runoff from antecedent storm events. This implies that direct

runoff responses can be superposed. If storm C is the result of adding storms

A and B, the hydrograph produced by storm c will be equal to the sum of the

hydrographs produced by storm A and B.

4. Hydrologic systems are usually nonlinear due to factor such

as storm origin and patterns and stream channel hydraulic properties.

In other words, if the peak flow produced by a storm of a certain

intensity is known, the peak corresponding to another storm (of thesame duration) with twice the intensity is not necessarily equal to

twice the original peak.


42/68

Unit hydrograph

5. Despite this nonlinear behavior, the unit hydrograph concept is

commonly used because, although it assumes linearity, it is a convenient tool

to calculate hydrographs and it gives results within acceptable levels of

accuracy.

6. The alternative to UH theory is kinematic wave theory anddistributed hydrologic method.


43/68

43

Derived Unit Hydrograph

0.0000

100.0000

200.0000

300.0000

400.0000

500.0000

600.0000

700.0000

0.00

00

0.1600

0.32

00

0.4800

0.64

00

0.80

00

0.96

00

1.1200

1.28

00

1.4400

1.60

00

1.76

00

1.92

00

2.08

00

2.24

00

2.4000

2.56

00

2.72

00

2.88

00

3.04

00

3.20

00

3.36

00

3.52

00

3.68

00

Baseflow

Surface

Response


44/68

Estimating Excess Precip.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Time (hrs.)

Precipitation(inches)

Uniform loss rate of

0.2 inches per hour.


45/68

Excess Precipitation

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Excess

Prec.

(inches)

Time (hrs.)

Small amounts ofexcess precipitation at

beginning and end maybe omitted.

Derived unit hydrograph is theresult of approximately 6

hours of excess precipitation.


46/68

Derived Unit Hydrograph

0.0000

100.0000

200.0000

300.0000

400.0000

500.0000

600.0000

700.0000

0.0000 0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000

Total

Hydrograph

Surface

Response

Baseflow


47/68

Example 3

Obtain a Unit Hydrograph for a basin of 282.6 km2of area using the rainfall

and streamflow data tabulated below.

Time

(h)

Observed

Hydrograph (m3/s)

0 160

1 150

2 350

3 800

4 1200

5 900

6 750

7 550

8 350

9 225

10 150

11 140

Time

(h)

Gross Precipitation

(GRH)

(cm/h)

0 - 1 0.25

1 - 2 2.75

2 - 3 2.75

3 - 4 0.25


48/68

Example 3(Solution)

Unit Hydrograph Derivation 1. Separate the base flow 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

base flow. From observation of the hydrograph data, thestreamflow at the start of the rising limb of the hydrograph is 150m3/s. Thus, use 150 m3/s as the base flow.

2. Compute the volume of Direct Runoff. This volume must be equal

to the volume of the Effective Rainfall Hyetograph (ERH).


49/68

Example 3(Solution)

Unit Hydrograph

Time (h) Observed

Hydrograph(m3/s)

Direct Runoff

Hydrograph(DRH) (m3/s)

Unit

Hydrograph(m3/s/cm)

0 160 10 --

1 150 0 0

2 350 200 40

3 800 650 130

4 1200 1050 210

5 900 750 150

6 750 600 120

7 550 400 80

8 350 200 40

9 225 75 1510 150 0 0

11 140 0 0

Unit Hydrograph


50/68

Example 3(Solution)

Thus, for this example:

VDRH= (200+650+1050+750+600+400+200+75) m3/s

(3600)s = 14'130,000 m3

3. Express VDRHin equivalent units of depth:

VDRHin equivalent units of depth = VDRH/Abasin=14'130,000 m3/(282600000 m2) = 0.05 m = 5 cm.

4.Obtain a Unit Hydrograph by normalizing the DRH.Normalizing implies dividing the ordinates of the DRH by theVDRHin equivalent units of depth.


51/68

Example 3(Solution)

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

a) Determine the volume of losses, VLosseswhich is equal to the differencebetween the volume of gross rainfall, VGRH, and the volume of the directrunoff hydrograph, VDRH.

VLosses

= VGRH

- VDRH

= (0.25 + 2.75 + 2.75 +0.25) cm/h *1 h - 5 cm = 1 cm

b) Compute the -index equal to the ratio of the volume of losses to therainfall duration, tr. Thus,

-index = VLosses/tr= 1 cm / 4 h = 0.25 cm/h

c) Determine the ERH by subtracting the infiltration (e.g., -index) from

the GRH:


52/68

Estimating ER

0

0.1

0.2

0.3

0.4

2.75

3.00

4.007

1 2 3 4 5 6 7

Time (hrs.)

Precipitation(cm)

Uniform loss rate of

0.25 Cm per hour.

7 0

Time (hrs.)


53/68

Example 3(Solution)

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

(h)

Gross Precipitation

(GRH)

(cm/h

0-1 0.00

1-2 2.75

2-3 2.75

3-4 0.00


54/68

54

Changing the Duration of the unit hydrograph

Very often, it will be necessary to change the duration ofthe unit hydrograph.

If unit hydrographs are to be averaged, then they must beof the same duration.

Also, convolution of the unit hydrograph with aprecipitation event requires that the duration of the unithydrograph be equal to the time step of the incrementalprecipitation.

The most common methods of altering the duration of aunit hydrograph are:

1. Short duration to long duration

2. S-curve method.

Short duration to long duration


55/68

A UH has a particular duration, D. Unit hydrographs are linear, they can be added, for

example, to generate a hydrograph for a storm 2x, 3x, 4x, etc. longer

1hr UH

2hr UH

The linear property of a UH can be used to generate a UH of a larger duration.

Here two copies of a 1-hour UH are lagged by 1-hr and added, then the ordinates

are divided by 2, to make a 2-hour UH (1 cm over two hours).

However this lagging method is restricted to integer multiples of the original

duration.

Short duration to long duration


56/68

Example 4

6 hr UH from 3 hr UH 6 hr UH from 3 hr UH


57/68

S-Curve

S-curve method

The S-curve method

involves continually lagging

a unit hydrograph by its

duration and adding the

ordinates.

For the present example,

the 6-hour unit hydrograph

is continually lagged by 6

hours and the ordinates are

added.

S-curve

0.00

10000.00

20000.00

30000.00

40000.00

50000.00

60000.00

0 612

18

24

30

36

42

48

54

60

66

72

78

84

90

96

102

108

114

120

Flow(

cfs)

Time (hrs.)


58/68

Scurve

Scurve

S curve method works for

any duration. The first step

is to add a series of UHsof

duration D, each lagged by

time period D, This

corresponds to the runoff

hydrograph from a

continuous rainfall excessintensity of 1/D cm/hour.

Scurve


59/68

Scurve method works for any duration, D

By shifting a copy of the S-curve by Dhours, and subtracting the ordinates, theresulting hydrograph (dashed line - - - - - - ) must be due to rainfall of intensity 1/D cm/hour

that lasts for a duration of Dhours.

To convert the hydrograph (dashed line - - - - -) to a UH, multiply ordinates by

D/D,resulting in a UH of duration D. Dneed NOT be an integral multiple of D.


60/68

D hr UH from D hr UH


61/68

Example 5

The ordinates of a 2-hr hydrograph for a particular basin are given below.

Determine the ordinates of the S-curve hydrograph and therefrom theordinates of the 3-hr unit hydrograph.

Time

(hr)

Discharge

(cumecs)

01

2

3

4

5

67

8

9

10

075

250

300

275

200

10075

50

25

0

Example 5 (solution)


62/68


Scurve

Make a spreadsheet with

the 2-hr UH ordinates, then

copy them in the next

column lagged by D=2

hours.

Keep adding columns until

the row sums are fairly

constant. The sums are the

ordinates of S-curve.

Scurve



63/68


Scurve lagged by 3 hrs

Here are the two copies of

S-curve based on a sum of

2-hr UHs.

The second copy is lagged

by D=3 hours.

Scurve lagged by 3 hrs



64/68


3-hrs unit hydrograph

Here the original UH was for a2-hour storm, and we need aUH for a 3-hour storm.

A 2-hour UH is for a stormwith 1/2 cm/hr for 2 hours

A 3-hour UH is for a stormwith 1/3 cm/hr for 3 hours

1/2 x 2/3 = 1/3

Calculate the difference, thenmultiply by D/D ,here = 2/3,to get your 3 hour UH.

3-hrs unit hydrograph


65/68

Example 6

2hr UH from 4 hr UH2hr UH from 4 hr UH

Example 7


66/68

Example 76 hr UH from 4 hr UH

(Short cut)


67/68

Prediction

This method provides a hydrograph that predicts thebehavior of a flood from a storm of any duration

Thank


68/68

Thank

You

Documents

Hydrograph_analysis_2 hydro.pdf