complete ch#6 hydro graph

8/3/2019 Complete Ch#6 Hydro Graph

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HYDROLOGY - BASIC

CONCEPTS



Hydrograph

Introduction

When a storm occurs it contributes to the stream flow.Various instruments are used to document stormcharacteristics,

for example, rain gauges are used to record period anddepth of rainfall, current meter is used to find dischargein streams, etc.

The data thus collected is analyzed for its beneficialuse. The stream flow characteristics are analyzed bywhat is called hydrograph analysis.

Continued..



Hydrograph

The graphical relation between anyhydrological quantity (stage, velocity,discharge, etc.) & the time is known as a

hydrograph. Hydrographs are of three types:

Discharge Hydrograph

Stage Hydrograph Velocity Hydrograph

Continued..



Hydrograph

Discharge Hydrograph It is the graphical representation of discharge against time.

Generally a hydrograph means discharge hydrograph.

Stage Hydrograph It is the graphical representation of stage against time.

Stage hydrograph is useful only for the design of flood-protection works like embankment.

Velocity Hydrograph

It is the graphical representation of velocity against time.



Effective Rainfall

All of the rain usually does not go into thestream but a certain part of it reaches thestream & causes rise in the stream flow while

the remaining part of rain is accounted for invarious forms of precipitation losses.

The portion of rainfall which contributes to

stream flow is called Effective Rainfall.



Basin Lag or Time Lag

A useful index to define the location of thepeak of the hydrograph is the basin lag,which is the time difference between the

center of mass of the rainfall and the centerof the mass of the hydrograph.

Rise of the stream flow depends upon theeffective rainfall and type of the soil, and time

lag depends upon type of area & also uponcharacteristics of rainfall (Figure 1).

Continued..



Basin Lag or Time Lag

Hydrograph

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

0 2 4 6 8 10

Time, t

D i s c h a r g e , Q

DRO Hydrograph

Lag

Figure 1: Time Lag in Surface Runoff



Parts of a Hydrograph

A hydrograph consists of three Parts (Fig 2):

The rising Limb ‘BC’

The crest or peak ‘C’.

The falling Limb or Recession curve ‘CDE’

The shape of the rising limb depends upon thestorm characteristics which are the duration of

rainfall, intensity of rainfall, areal distribution ofthe rainfall, etc.

Continued..




The crest segment is controlled by the stormcharacteristics and the distribution of streamsin the area, which further depends upon

geological structure of the area. The falling limb depends upon the control on

storage release, which itself depends upon

the geological structure.

Continued..




Hydrograph

0.0

1.0

2.0

3.0

4.0

5.0

6.0

0 2 4 6 8 10 12

Time, t


DRO Hydrograph

A

B

C

D

E

Rising Limb

Crest

Falling Limb or

Recession Curve

Fig 2: Parts of Hydrograph



Components of Hydrograph

A hydrograph normally consists of the fourcomponents:

Channel Precipitation

Direct Runoff (DRO)

Inter Flow or Sub-surface Flow

Ground or Base Flow



Hydrograph Analysis

Separation of Hydrograph Components Hydrograph Analysis means dividing total runoff

indicated by the hydrograph into its (above-mentioned) components, of which direct runoff(DRO) and Base Flow are more important.

The separation of hydrograph components isdone by one of the following four methods. Straight Line Method

Fixed Base Length Method

Variable Slope Method

Recession Curve Equation

Continued..



Hydrograph Analysis

Straight Line Method A hydrograph as shown in Figure 3. From

point ‘A’ draw a straight horizontal line whichmeets the falling limb of hydrograph at point‘B’.

Above line ‘AB’ is the direct runoff and belowthis line is the base flow.

The volumes of direct runoff and base flow(i.e. area under the curve) can be calculated byany suitable method.

Continued..



Hydrograph Analysis

The Straight Line Method for separation ofhydrograph has the advantage of producing anextremely long time base for the direct runoffhydrograph.

The time base varies from storm to stormdepending on the flow at the point of rise.

It is a computationally easy method.

Continued..



Hydrograph Analysis

Hydrograph

0.0

1.0

2.0

3.0

4.0

5.0

6.0

0 2 4 6 8 10 12

Time, t


DRO Hydrograph

A B

Horozontal Line from the

point of rise in rising limbof hydrograph

Fig. 3: Separation of Hydrograph Components Continued..



Hydrograph Analysis

Fixed Base Length Method

Let us have the hydrograph shown in Figure 4. From point ‘A’(existing prior to the storm) extend the curve to point ‘B’ which isvertically below the peak.

Now draw a line from point ‘B’ to point ‘C’ on the hydrograph

equal to a distance on X-axis given by equation below:N = Ad

0.2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1

Where,

N = Length of time in days, and

Ad = Drainage area in square miles

This line will meet the falling limb at, say point ‘C’. Join the point‘B’ to ‘C’ by a line ‘BC’.

Area below ‘ABC’ is the base flow and that above it is the directrunoff.

Continued..



Hydrograph Analysis

Hydrograph

0

100

200

300

400

500

600

0 2 4 6 8 10 12

Time, t


DRO Hydrograph

A C

Extended

RecessionCurve

A0.2

B

Figure 4: Separation of Base Flow by Fixed Base Length Method Continued..



Hydrograph Analysis

Variable Slope Method

This method is based on the fact that if water rises in thestream the water level in the stream becomes greater thanthe adjoining ground water table level.

As a result the flow occurs from the stream to thegroundwater and is stored there as bank storage.

Refer to Figure 5, below. Take a point ‘A’ on the previous

recession curve, and the line ‘AE’ by extending the

previous recession curve. Point ‘E’ is under the peak.

Continued..



Hydrograph Analysis

Now take a point ‘B’ on the recession curve andextend it backward. Take a point ‘C’ on recessioncurve where the recession curve is changingslope.

This point is called the point of inflexion. Draw avertical line from the point ‘C’.

It cuts the line drawn backward from point ‘B’, atpoint ‘D’. Join point ‘E’ and point ‘D’.

The area under the line ‘AEDB’ shows the baseflow component

Continued..



Hydrograph Analysis

Hydrograph

0

100

200

300

400

500

600

0 2 4 6 8 10 12

Time, t

D i s c h a r g e ,

Q

DRO Hydrograph

AB

Extended

Recession

Curve Point of Inflexion

E

C

D

Figure 5: Separation of Base Flow by Variable Slope Method Continued..



Hydrograph Analysis

This method is not a well defined one. It depends uponexperience, so it may vary from one expert to another.

In general, Straight Line Method is the simplest and easiestbut also the least accurate.

Fixed Base Method is widely used and gives practicallygood results.

Whereas Variable Slope Method is not commonly used andis difficult but can give better results depending upon the

expertise of the person using it.

Continued..



Hydrograph Analysis

Recession Curve Equation The recession side of the hydrograph is formed of

the surface recession, interflow recession andbase flow recession.

Although the interflow, because of similarcharacteristics, is taken as part of the surfacerunoff in the hydrograph analysis.

There are means to separate the threecomponents on the basis of the exponentialrecession equation of the form.

Continued..



Hydrograph Analysis

The equation is:Q = Q0 e –δt - - - - - - - - - - - - - - - - - - - - - - - - - - 2

Where,Q0 = initial dischargeQ = discharge a time interval afterwards

δ = recession constant The equation can be written in a general form.

Q t+1 = Q t e -δt

Taking log on both sideslog ( Qt+1) = log (Q t e -δt)

log Qt+1 = log Qt - δt log e - - - - - - - - - - - - - - - - - - 3 Graph of this equation for different recession constants ‘δ’ will

give the components of hydrograph.



Factors Affecting Hydrograph Shape

Hydrograph shape depends on climatic andcatchment characteristics.

Among the climatic characteristics the mostimportant are the rainfall intensity, duration and

weather. Among the catchment characteristics the

topography and geological conditions are moreimportant.

These are discussed below: Storm Characteristics Catchment Characteristics



Storm Characteristics

Intensity of Rainfall

The intensity of rainfall has a predominant effecton the shape of the hydrograph.

If other conditions remain the same, a higherintensity storm will produce a rapid rise in thehydrograph and a higher peak than that in case ofa low intensity rainfall.

Such floods give little warning time and so aredangerous.

Continued..




Duration of Rainfall

The duration of rainfall is important if it is greaterthan the time of concentration (defined as the time

of travel from the farthest point in the catchmentarea to the gauging station).

In such a situation, if the rainfall of certain intensityoccurs uniformly, the whole of the catchment area

is contributing runoff at the gauging station and thehydrograph attains its high peak.

Continued..




The runoff will remain constant at the peak rate ifthe rainfall occurs for duration longer than the timeof concentration.

If rainfall of the same intensity occurs for durationless than the time of concentration, thehydrograph will rise to a smaller peak.

Continued..




Weather

The shape of the hydrograph greatly depends on theseasonal distribution of rainfall. During summer losses dueto evaporation may produce a small peak hydrograph.

On the other hand, in winter, losses are small with theresult that even a small intensity storm may produce arelatively rapid rise and high peak of the hydrograph.

Antecedent soil moisture conditions (generally being higher

in winter) are important as for high soil moisture; the stormwill produce more runoff.



Catchment Characteristics

Size of Catchment

The catchment area affects the stream flow in a variety of ways.Even if the intensity and depth of precipitation is assumed to beconstant,

The hydrograph of a smaller catchment rises to its peak and then

recedes more rapidly than that for a larger catchment, becausefor the latter, it takes longer for the runoff to reach the gaugingstation.

The hydrograph of a larger catchment area, therefore, hasbroader base than that of a smaller one. Moreover, the larger thearea, the greater will be the heterogeneity in soil and vegetationcover, with a consequent effect on the stream flow of each part ofthe catchment. The vegetation cover increases the time ofconcentration.

Continued..




The hydrograph of a larger catchment area,therefore, has broader base than that of a smallerone.

Moreover, the larger the area, the greater will bethe heterogeneity in soil and vegetation cover,with a consequent effect on the stream flow ofeach part of the catchment.

The vegetation cover increases the time ofconcentration.

Continued..




Shape of the Catchment Area The shape of the catchment area affects stream flow by

altering the time of concentration and the pattern ofdrainage tributaries.

If the length of the catchment along the main stream is lessthan the width across the same stream, separate runoffpeaks generated by a heavy rainfall are likely to reach thegauging station at the same time from the tributaries, withthe result that the flood peak in the main stream increases.

Thus a hydrograph with a high peak and narrow base isobtained.

Continued..




However, if the length of the catchment along the mainstream is larger than width, the tributaries will be of shorterlength and contribution from each tributary area is likely toreach the gauging stations at intervals.

After an intense storm over the whole of the catchmentarea there will be some lag between the times at which thepeak from each tributary reaches the gauging station.

The hydrograph will therefore, have a low peak andbroader base, because the remote tributaries will continuecontributing for some time even after the rain fall has

stopped.

Continued..




Elevation of the Catchment

The elevation of the catchment also plays animportant role in some hydrologic characteristics

of hydrograph, particularly due to variation intemperature and precipitation.

The temperature reduces with the increase inelevation with the result that above a certain

elevation, the temperature becomes so low that allprecipitation falls as snow.

Continued..




In the Indo-Pakistan Sub-continent, contributionfrom areas, at altitude 15,000 ft. above the meansea level, is mainly due to snowmelt.

The floods from snowmelt are of usually low peakand broader base.

Continued..




Catchment Slope

The slope of the catchment affects stream flow byreducing the rate of infiltration due to increased

speed of water-movement towards drainagechannels.

The steeper the slope the faster will be the flowand rise in the hydrograph. The hydrograph in

such a situation will have higher peak and smallerbase.



Unit Hydrograph

While designing hydraulic structures such as,we are mostly interested in estimatingmaximum probable rate of runoff.

Various methods of estimating maximumprobable rate of runoff are as under:

Frequency Analysis

It may be used provided adequate stream flowrecords are available.

Continued..



Unit Hydrograph

Empirical Formulae

Involving various physical characteristics of the basin i. e.

Q = C I A (Rational Formula) - - - - - - - - - - - - 4

Where,

I = Intensity of rainfall,

C = Constant, and

A = Area of cross-section.

Very detailed rainfall-runoff models have been developed

which estimate losses separately and hence calculate thedirect runoff

Continued..



Unit Hydrograph

Flood Routing

Flood Routing will be discussed in detail later.

Unit Hydrograph Method

Of the four methods of estimating maximumprobable rate of runoff, the unit hydrographmethod is being discussed in detail in thisdiscussion.

Continued..



Unit Hydrograph

Physical characteristics of the basin remainrelatively constant. The variablecharacteristics of the storm cause variation in

the shape of resulting hydrograph. The storm characteristics as discussed

earlier are the rainfall duration, time-intensity

pattern, areal distribution of rainfall and theamount of rainfall.

Continued..



Unit Hydrograph

Rainfall Duration

The volume under unit hydrograph is always the same for agiven catchment and is equal to one unit say one inch ofrunoff (note that volume of one inch of runoff means that

one inch of rainfall excess over the whole catchment. The actual volume will be the catchment area multiplied by

one). Peak value for a lesser duration unit hydrograph ishigher and is sharp, but the peak value for a higherduration unit hydrograph is lower with larger base length.

Continued..



Unit Hydrograph

Areal Distribution of Rainfall

The unit hydrograph is based upon uniformlydistributed rainfall over whole of the catchment.

So the actual flood hydrograph has to beestimated accordingly from the unit hydrographtaking into account the effect of areal distributionof rainfall.



Derivation of Unit Hydrograph

A unit hydrograph derived from a single storm may have someerror, and it is desirable to average the unit hydrograph fromseveral storms of the same duration.

This should not be an arithmetic average of concurrentcoordinates; since, if peaks do not occur at the same time, the

average peak will be lower than many of the individual peaks. The proper procedure is to compute the average of the peak

flows and times to peak. The average unit hydrograph is thensketched to conform to those of the other graphs, passingthrough the computed average peak, and having a volume of

one-inch.

Continued..




Example 1

From the data given in Table 1 below, derive a 6-hour unit hydrograph if this data is obtained from a

six-hours duration storm and the catchment areais 3,200 square kilometers.

Continued..




Date Time Discharge Base Flow

(day) (m³/s) (m³/s)(1) (2) (3) (4)

11-Jan 1st 21 21

12-Jan 2nd 21 21

13-Jan 3rd 21 21

14-Jan 4th 20 20

15-Jan 5th 20 20

16-Jan 6th 20 20

17-Jan 7th 20 20

18-Jan 8th 20 20

19-Jan 9th 521 11

20-Jan 10th 160 21

21-Jan 11th 82 29

22-Jan 12th 63 35

23-Jan 13th 59 47

24-Jan 14th 53 53

25-Jan 15th 47 47

26-Jan 16th 45 45

27-Jan 17th 38 38

T a b l e 1 :

D a t a f o r U n i t h

y d r o g r a p h




Solution

The calculations are shown in Table 2.

The Direct Runoff (DRO) is obtained by subtracting baseflow from discharge values.

The DRO is then multiplied by time interval to get totalvolume. This total volume is divided by catchment area toget depth of effective rainfall in centimeters spread uniformlyover whole catchment.

By definition of unit hydrograph, its ordinates are obtained bydividing ordinates of DRO by effective rain.

Continued..




y g p

Date Time Discharge Base Flow DRO Volume Ordinates of Unit Hydrograph

(day) (m³/s) (m³/s) (m³/s) (x106 m³) (m³/s)

(1) (2) (3) (4) (5)=(3)-(4) (6) (7)=(5)/Eff. Rain

11-Jan 1st 21 21 0 0.00 0.00

12-Jan 2nd 21 21 0 0.00 0.0013-Jan 3rd 21 21 0 0.00 0.00

14-Jan 4th 20 20 0 0.00 0.00

15-Jan 5th 20 20 0 0.00 0.00

16-Jan 6th 20 20 0 0.00 0.00

17-Jan 7th 20 20 0 0.00 0.00

18-Jan 8th 20 20 0 0.00 0.00

19-Jan 9th 521 11 510 44.06 255.00

20-Jan 10th 160 21 139 12.01 69.50

21-Jan 11th 82 29 53 4.58 26.50

22-Jan 12th 63 35 28 2.42 14.00

23-Jan 13th 59 47 12 1.04 6.00

24-Jan 14th 53 53 0 0.00 0.00

25-Jan 15th 47 47 0 0.00 0.00

26-Jan 16th 45 45 0 0.00 0.00

27-Jan 17th 38 38 0 0.00 0.00

Total Volume 64.11 x106 m³

Area 3200 x106 m² Volume 32.05x106 m³

Effective Rain 2.00cm Effective Rain 1.00 cm

T a b l e 2 : D a t a f o r U n i t h y d r o g r a p h




The resulting unit hydrograph and DRO hydrograph are shown in Fig. 6

One-Day Unit Hydrograph

0

100

200

300

400

500

600

0 5 10 15 20

Time (Day)

D i s c h a r g e ( m ³

/ s )

Total Hydrograph DRO Hydrograph One Day Unit Hydrograph

Fig 6 Developing a 6-hr unit hydrograph



Derivation of Hydrograph from a Unit

Hydrograph

Once the unit hydrograph of certain durationis determined, it can be used to compute ahydrograph of the same duration.

In order to derive the hydrograph of sameduration as that of the unit hydrograph, wehave to proceed in reverse to the process

used to derive unit hydrograph

Continued..




Hydrograph

The procedure will be as follows: The unit hydrograph of a certain duration is given.

Derive the ordinates of direct runoff hydrograph bymultiplying the unit hydrograph ordinates byeffective rainfall.

Add base flow to the ordinates of direct runoffhydrograph to get total runoff hydrograph. Thepeak of this hydrograph gives the maximum flow

due to the storm.

Continued..




Hydrograph

Example 2

Given the unit hydrograph of 4-hour duration(Table 3), calculate the ordinates of total runoff

hydrograph of same duration from an excessprecipitation of 10 cm.

Assume constant Base Flow of 14 m3 /sec.

Continued..



f d h f




Hydrograph

Solution

The excess rainfall has been reported as 10cm.The ordinates of unit hydrograph are simply

multiplied by excess rainfall to get ordinates ofDirect Runoff (DRO).

Once the ordinates of DRO are obtained, baseflow is added to it to get Total flow ordinates.

These computations are given in Table 4.

Continued..

D i i f d h f i




Hydrograph

Date Time Time Ordinates ofUnit

Hydrograph

DRO BaseFlow

Total Flow

(Hours) (Hours) (m³/s) (m³/s) (m³/s) (m³/s)

(1) (2) (3) (4) (5)=(4) x ExcessRainfall Depth

(6) (7)=(5)+(6)

1-Jan 12 Noon 0 0.00 0 14 14

24 Midnight 12 3.00 30 14 44

2-Jan 12 Noon 24 17.00 170 14 184

24 Midnight 48 34.00 340 14 354

3-Jan 12 Noon 60 51.00 510 14 524

24 Midnight 72 17.00 170 14 184

4-Jan 12 Noon 84 6.00 60 14 74

24 Midnight 96 0.00 0 14 14Table 4: Derivation of total runoff hydrograph.

D i i f H d h f U i




Hydrograph

Derivation of Total Runoff Hydrograph

0

50

100

150

200

250

300

350

400

450

500

550

600

0 20 40 60 80 100 120

Time (Hours)

F l o w

( m ³ / s )

12 Hour Unit Hydrograph DRO Hydrograph Total Runoff Hydrograph

•The unit hydrograph, DRO hydrograph and total hydrograph is shown in Fig. 7

F i g . 7 : T o t a l R u n o f f H y d r o

g r a p h

S C



S Curve

The S curve is hydrograph obtained by summation of infinitenumber of unit hydrographs of given duration each lagged bytime interval equal to the duration of the given hydrograph.

The S curve can be used to derive unit hydrograph of anydesired duration.

The Derivation of the S curve is very simple. We keep on adding given unit hydrograph ordinates each lagged

by time equal to the duration of given hydrograph from theprevious one till we obtain a constant value after addition or itstarts fluctuating between some values.

S C



S Curve

Example 3 Given is a unit hydrograph of 2 hour duration (column 2 of

Table 5). Derive the 2 hours duration ‘S’ curve.

Time (Hours) Flow m³/s

0 0

2 95

4 164

6 121

8 91

10 54

12 18

14 0

Table 5: Ordinates of 2 -Hour Unit Hydrograph Continued..

S C



S Curve

SOLUTION

First, we derive ‘S’ curve: The calculations are

shown in Table 6. In this table ordinates of unit

hydrograph of 2-hour duration are given in column2. Column 1 is time in hours. Other columns haveordinates in m³/s. Columns 3 to 14 show laggedunit hydrographs. Column 15 is the addition which

is the ‘S’ curve. The ‘S’ curve is shown in Figure 8

Continued..

S C



S Curve(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15)

0 0 0

2 95 0 95

4 164 95 0 259

6 121 164 95 0 380

8 91 121 164 95 0 471

10 54 91 121 164 95 0 525

12 18 54 91 121 164 95 0 543

14 0 18 54 91 121 164 95 0 543

16 0 18 54 91 121 164 95 0 543

18 0 18 54 91 121 164 95 0 543

20 0 18 54 91 121 164 95 0 543

22 0 18 54 91 121 164 95 0 543

24 0 18 54 91 121 164 95 0 543

26 0 18 54 91 121 164 95 543

28 0 18 54 91 121 164

30 0 18 54 91 121

32 0 18 54 91

34 0 18 54

36 0 18

38 0

T a

b l e 6 : U n i t h y d r o g r a p h d a t a a n d

d e r i v a t i o n o f S

c u r v e .



D i i f A D i U i



Derivation of Any Duration Unit

Hydrograph

The procedure for derivation of any requiredduration unit hydrograph from a given-duration unithydrograph is described below:

Derive the S-Curve (as per procedure explained above).

Offset the position of the S-curve for a period equal to thedesired duration of hydrograph to be derived. This is calledthe lagged S-curve.

Subtract the ordinates of lagged S-curve from the original

S-curve at the respective time intervals. The ordinates of unit hydrograph of required duration are

obtained by multiplying the difference obtained in Step 3with the ratio given duration over required duration (i. e.multiply by the ratio = given duration/desired duration)

Continued..

D i i f A D i U i




Hydrograph

Example 4

Given is unit hydrograph of 2-hours duration(Column (2), Table 5 of Example 3). Find 6-hours

unit hydrograph for the same catchment. Solution

The derivation of S curve is given in Example 3.Using that S curve the derivation of 6 hours unit

hydrograph is given below in Table 7.

Continued..

D i ti f A D ti U it




Hydrograph

Time 2-hour UnitHydrographOrdinates

S-Curve Lagged S-Curve

Difference 6 Hour UnitHydrograph Ordinates

(1) (2) (3) (4) (5)=(3)-(4) = (5) x 2 / 6

0 0 0 0 0

2 95 95 95 32

4 164 259 259 86

6 121 380 0 380 127

8 91 471 95 376 125

10 54 525 259 266 89

12 18 543 380 163 54

14 0 543 471 72 24

16 543 525 18 6

18 543 543 0 0

Table 7: Derivation of 6 hours unit hydrograph from 2 hours unit hydrograph.

• The given 2 hr hydrograph and developed 6 hr hydrographs are shown inFigure 9. It may be noted that increasing the duration of unit hydrographhas decreased peak flow.



6 Hour UH from 2 Hour UH

0

20

40

60

80

100120

140

160

180

0 2 4 6 8 10 12

Time (Hours)

F l o w

( m ³ / s )

2 Hour Unit Hydrograph 6 Hour Unit Hydrograph

Figure 9: Two Hour and Six Hour Unit Hydrograph

C it H d h



Composite Hydrograph

The stream flow hydrograph may be due tomore than one storm. In that case we have toseparate the effect of each storm. This is

composite hydrograph separation. Here the analogy is used that ordinates of

composite hydrograph are proportional tointensity of rainfall.

Continued..



Example 5

Given is hydrograph resulting from two stormseach of 6 hours duration (Table 8). The rainfall

intensity of first storm is 0.8 cm and that of secondstorm is 0.6 cm each of 6 hr duration. Separatethe hydrograph of both storms and find 6 hr unithydrograph for each storm.

Continued..



Solution Refer to Table 9. The following steps illustrate the

procedure Total rainfall is 0.8 + 0.6 = 1.4 cm. The contribution of

second storm relative to the first storm is 0.6 / 0.8 =0.75. The ordinates of first hydrograph are, thereforeobtained by subtracting effect of second hydrographfrom composite hydrograph. Since second storm occurs6 hours later than the first one, the first ordinate of

composite hydrograph is due to contribution of firststorm. Therefore first ordinate of first hydrograph is 750- 0 = 750.

Continued..



The second ordinate is obtained by subtracting(0.6 / 0.8) x 750 = 562.50 from second ordinateof composite hydrograph i.e. 1400 - 562.50 =837.50. The third ordinate of first storm

hydrograph is obtained by subtracting 0.75 x837.50 = 628.12 from third ordinate ofcomposite hydrograph i.e. 1671.88 and so on.The calculations are shown in table 5.2

Continued..



The ordinates of second hydrograph areobtained simply by subtracting ordinates of firsthydrograph from composite hydrograph.

For unit hydrograph ordinates of eachhydrograph are divided by their respectiverainfall intensities

Continued..



Date Time(Hours)

CompositeHydrograph

(cumecs)

Date Time(Hours)

CompositeHydrograph

(cumecs)

Date Time(Hours)

CompositeHydrograph

(cumecs)

22-04-65 0 - 25-04-65 60 19,100.00 27-04-65 120 3,300.00

6 750.00 66 15,000.00 126 2,900.00

12 1,400.00 72 11,400.00 132 2,600.00

18 2,300.00 78 9,300.00 138 2,100.00

23-04-65 24 3,000.00 84 7,800.00 28-04-65 144 1,700.00

30 7,000.00 90 6,300.00 150 1,300.00

36 13,000.00 26-04-65 96 5,700.00 156 900.00

42 23,000.00 102 4,800.00 162 600.00

24-04-65 48 29,700.00 108 4,000.00 29-04-65 168 -

54 25,000.00 114 3,700.00

Table 8. Data of Composite Hydrograph

Continued..



Date Time CompositeHydrograph

First 6 HourHydrograph

Second 6 HourHydrograph

6 Hour UnitHydrograph (First

Storm)

6 Hour UnitHydrograph

(Second Storm)

(Hours) (cumecs) (0.8 cm) (0.6 cm)

22-04-65 0 - - - -

6 750.00 750.00 - 937.50 -

12 1,400.00 837.50 562.50 1,046.88 937.50

18 2,300.00 1,671.88 628.13 2,089.84 1,046.88

23-04-65 24 3,000.00 1,746.09 1,253.91 2,182.62 2,089.84

30 7,000.00 5,690.43 1,309.57 7,113.04 2,182.62

36 13,000.00 8,732.18 4,267.82 10,915.22 7,113.04

42 23,000.00 16,450.87 6,549.13 20,563.58 10,915.22

24-04-65 48 29,700.00 17,361.85 12,338.15 21,702.31 20,563.5854 25,000.00 11,978.61 13,021.39 14,973.27 21,702.31

60 19,100.00 10,116.04 8,983.96 12,645.05 14,973.27

66 15,000.00 7,412.97 7,587.03 9,266.21 12,645.05

Table 9: Separation of Component Hydrographs

Continued..



Date Time CompositeHydrograph

First 6 HourHydrograph

Second 6 HourHydrograph

6 Hour UnitHydrograph (First

Storm)

6 Hour UnitHydrograph

(SecondStorm)

(Hours) (cumecs) (0.8 cm) (0.6 cm)

25-04-65 72 11,400.00 5,840.27 5,559.73 7,300.34 9,266.21

78 9,300.00 4,919.80 4,380.20 6,149.74 7,300.34

84 7,800.00 4,110.15 3,689.85 5,137.69 6,149.74

90 6,300.00 3,217.38 3,082.62 4,021.73 5,137.69

26-04-65 96 5,700.00 3,286.96 2,413.04 4,108.70 4,021.73

102 4,800.00 2,334.78 2,465.22 2,918.47 4,108.70

108 4,000.00 2,248.92 1,751.08 2,811.14 2,918.47

114 3,700.00 2,013.31 1,686.69 2,516.64 2,811.14

27-04-65 120 3,300.00 1,790.02 1,509.98 2,237.52 2,516.64

126 2,900.00 1,557.49 1,342.51 1,946.86 2,237.52

132 2,600.00 1,431.88 1,168.12 1,789.85 1,946.86

138 2,100.00 1,026.09 1,073.91 1,282.61 1,789.85

Continued..



SEPARATION OF COMPOSITE HYDROGRAPH

-

5,000

10,000

15,000

20,000

25,000

30,000

35,000

0 20 40 60 80 100 120 140 160 180

Time (Hours)

D i s c h a r g e ( c

u m e c )

Composite Hydrograph First 6 hr Hydrograph Second 6 hr Hydrograph

Fig 10: Separation of Component Hydrographs from Composite Hydrograph



Example 6 Develop a complex hydrograph for the following data given

in Table 10, Take R1 / R2 = 1.2

Time 1 2 3 4 5 6 7 8 9 10 11

q1 0 200 400 1000 2800 1400 900 700 480 200 0

Table 10

Continued..



Solution Table 11 below gives calculations for determination of q2 & Q, in

which ordinates of q2 have been calculated by using a modifiedbut simple formula given in the table.

n TIME q1 Ordinates of q2=(R1 /R2)xq1(n-

1). Discharge Q of complex hydrograph

0 1 0 0

1 2 200 0 200

2 3 400 240 640

3 4 1000 480 1480

4 5 2800 1200 4000

5 6 1400 3360 4760

6 7 900 1680 2580

7 8 700 1080 1780

8 9 480 840 1320

9 10 200 576 776

10 11 0 240 240

11 12 0 0

T a b l e 1 1

Continued..



Composite hydrograph along with componenthydrograph is shown in Figure 11.

COMPLEX HYDROGRAPH

0

1000

2000

3000

4000

5000

6000

0 2 4 6 8 10 12 14

TIME

D I S C H A R G E

Figure 6.11

Continued..



Example 7 Repeat example 6 for the case when ordinates

of Q are given (as shown in Table 12 below)

for its separation into its components q1 & q2with R1/R2 =1.2

Continued..



n TIME Discharge Q ofcomplex

hydrograph

Ordinates of q1 = Qn -(R1/R2) x q1(n-1).

Ordinates ofq2 = (R1/R2)xq1(n-1).

0 1 0 0

1 2 200 200 02 3 640 400 240

3 4 1480 1000 480

4 5 4000 2800 1200

5 6 4760 1400 3360

6 7 2580 900 1680

7 8 1780 700 1080

8 9 1320 480 840

9 10 776 200 576

10 11 240 0 240

11 12 0 0

Solution: Table 12: Calculations for Separation of Complex hydrograph

Continued..



COMPONENTS OF A COMPLEX HYDROGRAPH

0

1000

2000

3000

4000

5000

6000

0 2 4 6 8 10 12 14

TIME

D I S C H A R G

E

Figure 12: (With Example 7)



Example 8 Develop a complex hydrograph for the following data given

in Table 13, TakeR1 / R2 = 1.2 and R1 / R3 = 1.5

Time 1 2 3 4 5 6 7 8 9 10 11

q1 0 200 400 1000 2800 1400 900 700 480 200 0

Table 13



Solution Table 14 below gives calculations for

determination of q2, q3 and Q, in which ordinatesof q2 and q3 have been calculated by using

modified but simple formulae given in the table.



n TIME q1 Ordinates ofq2=(R1/R2)xq1(n-1).

Ordinates ofq3=(R1/R3)xq1(n-2).

Discharge Q ofcomplex hydrograph

(1) (2) (3) (1) + (2) + (3)

0 1 0 0

1 2 200 0 200

2 3 400 240 0 640

3 4 1000 480 300 1780

4 5 2800 1200 600 4600

5 6 1400 3360 1500 6260

6 7 900 1680 4200 6780

7 8 700 1080 2100 3880

8 9 480 840 1350 2670

9 10 200 576 1050 1826

10 11 0 240 720 960

11 12 0 300 300

12 13 0 0

Table 14



Figure 13 shows complex hydrograph.

COMPLEX HYDROGRAPH

0

1000

2000

3000

4000

5000

6000

7000

8000

0 2 4 6 8 10 12 14

TIME

D I S C H A R G E



Example 9 Repeat example 8 for the case when ordinates

of Q are given (as shown in Table 15 below)for its separation into its components q1, q2 &q3 with R1/R2 =1.2 & R1/R3 =1.5.



N Time Discharge Qof complexhydrograph

Ordinates ofq1 = Qn - (R1/R2)xq1(n-1) -

(R1/R3)xq1(n-2).

Ordinates ofq2 = (R1/R2)x q1(n-1).

Ordinates ofq3=(R1/R3)x q1(n-2).

0 1 0 0

1 2 200 200 0

2 3 640 400 240 0

3 4 1780 1000 480 3004 5 4600 2800 1200 600

5 6 6260 1400 3360 1500

6 7 6780 900 1680 4200

7 8 3880 700 1080 2100

8 9 2670 480 840 1350

9 10 1826 200 576 1050

10 11 960 0 240 720

11 12 300 0 300

12 13 0 0

Table 15: Calculations for Separation of Complex hydrograph (see Fig. 14)



Synthetic Hydrograph



Synthetic Hydrograph

For un-gauged catchments, unit hydrographdeveloped from characteristics of similar catchmentsis called Synthetic hydrograph.

Synthetic hydrograph is obtained by mathematically

correlating catchment characteristics.

Different approaches are available for developing asynthetic hydrograph, some of which are:

Bernard’s method

McCarthy’s Method

Snyder’s Method

Taylor and Schartz Method

Continued..



Of these, Snyder’s Method is the mostcommonly used method, which is brieflydescribed here.

Snyder’s method Following are the steps involved in unit

hydrograph construction according to Snyder’s

method.

Continued..



Step (i) To find ‘Time to Peak’, Snyder proposed,

tp = C1 Ct (L Lc)0.3 - - - - - - - - - - - - - - - - - - - - -(5)

Where,

tp = Basin Lag in Hours. Basin Lag is the time between centre of

mass of unit rain and peak flow (See Figure 15).L = Distance from gauging station to catchment boundary.

Lc = Distance in kilometers between gauging station andcentroid of catchment. It is measured from gauging stationalong the main stream to the point nearest to centroid of

catchment.Ct = A coefficient determined from the gauged catchment in

same or similar region. Its value ranges from 1.35 to 1.65.

C1 is a constant whose value in MKS system of units is 0.75.

Continued..



Step (ii)The peak discharge is given by:

qp = C2 Cp - - - - - - - - - - - - - - - - - - - - - - - - (6)

tp

Where,

C2 is a constant having value of 2.75

Cp = A coefficient determined from the gaugedcatchment in same or similar region. Its value

ranges from 0.56 to 0.69. (See Figure 16)

Continued..



Synthetic Unit Hydrograph by Snyder's Method

0

5

10

15

20

25

30

35

40

45

0 10 20 30 40 50 60

Time , t


qp

W75

W50

0.5qp

0.75qp

Figure 16: Definition of q p

, W 50

and W 75 Continued..



The coefficients ‘Ct’ and ‘Cp’ are determined asexplained below:

Determination of ‘Ct’ and ‘Cp’

L is measured from map

Lc is measured from maptR - Effective duration measured from unithydrograph derived for similar catchment.

tPR - Basin lag measured from unit hydrograph

derived for similar catchment.qPR - Peak Discharge from unit hydrographderived for similar catchment.

Continued..



If,tPR = 5 tR then tR = tr

Else,tPR = tp and qPR = qP

By back substitution, ‘Ct’ and ‘Cp’ can be determined from Equations (5)and (6).

If,tPR ≠ 5.5 tR , then the standard basin lag istP = tPR + tr-tR - - - - - - - - - - - - - - - - - - - - - - - - - (7)

4Also, tp = 5.5 tr - - - - - - - - - - - - - - - - - - - - - - - - - - - - (8)

Equations (7) and (8) are solved simultaneously for ‘tr’ and ‘tP’. ‘Ct’ and‘Cp’ are then found from Equations (5) and (6) with conditions qPR = qP and tPR = tp. (Figure 17)

Continued..



Required Synthetic Unit Hydrograph

0

5

10

15

20

25

30

35

40

45

0 10 20 30 40 50 60

Time, t

D i s c h a r g e p e r u

n i t a r e a

tR

tPR

tb

qPR

Figure 17: Definition of tR, tPR and tb

Continued..



Step (iii) The Peak Discharge of required hydrograph is found using

equation 9.

qPR = qP tp - - - - - - - - - - - - - - - - - - (9)

tPR

Step (iv) The base time of required hydrograph is determined by

equation 10.

Base Time tb = C3 - - - - - - - - - - - - (10)

qPR

Where ‘C3’ is a constant having value of 5.56.

Continued..



Step (v) Width of Hydrograph ‘W’ is given by equation 11.

The coefficient ‘Cw’ defines value of peak discharge

where width of hydrograph is required.

For example, width at discharge which is 75% and50% of Peak discharge is to be calculated:

W = Cw qPR-1.08 - - - - - - - - - - - - - - - (11)

For qPR = 75%, Cw = 1.22For qPR = 50%, Cw = 2.14

Continued..



Example 10 Develop 3-hour and 6-hour unit hydrographs for a

catchment area measuring135.8 Sq. miles with following data:

Ct = 1.50, Cp = 0.60 L = 29.5 miles Lc = 15 miles

Also calculate total run-off in inches for thecatchment in each case.

Continued..



Solution Step-1

Basin Lag, tp = C1 x Ct x (L x Lc)0.30

= 1 x 1.50 x (29.5 x 15)0.30

= 9.33 hours

Standard duration of rainfall,

tr = tp /5.50

= 9.33/5.50= 1.70 hours

Continued..



Step-2 3-hour duration unit hydrograph tR = 3 hours > tr Basin Lag, tPR = tP + [tR – tr]/4

= 9.33 + [3-1.70]/4

= 9.66 hours

Peak discharge, qPR= 640 x Cp x A /tPR

= 640 x 0.60 x 135.8/9.66

= 5398.36 cfs

Continued..



Time base, TR = 3 x (1 + tPR /24)= 3 x (1 + 9.66/24)

= 4.21 days

= 101 hours

50% of qPR = 2694.5 cusecsW50 = 770 / [qPR / A ]1.08 = 770 / [5398.36 / 135.8 ]1.08

= 14.40 hours

75% of qPR = 4048.70 cusecs

W75 = 440 / [qPR / A ]1.08 = 440 / [5398.36 / 135.8 ]1.08

= 8.24 hours

Continued..



Plot unit hydrograph using above calculatedvalues. The graph is shown in Fig 18 below.It has been plotted using definitions given inFigs. 16 and 17 above.

From Fig. 18, tabulate the values ofdischarge against equal time interval of anylength say at 6 hours interval (17*6 = 102

close to 101). These values have beenshown in Table 16

Continued..



SYNTHETIC 3-HOUR UNIT HYDROGRAPH BY SNYDER'S METHOD

(EXAMPLE 6.10)

0

1000

2000

3000

4000

5000

6000

0 20 40 60 80 100 120

TIME (HOURS)

D I S C H A R G E

( c f

Figure 18

Continued..

Time (Hours) Q (cfs)



( ) Q ( )

0 0

6 2550.00

12 5200.00

18 3100.00

24 1300.00

30 950.00

36 530.00

42 300.00

48 200.00

54 150.00

60 110.00

66 100.00

72 80.00

78 60.00

84 30.00

90 20.00

96 10.00

101 (say 102) 0

Total 14690

Table 16



Total run-off = [14690 x 6 x 60 x 60]

[24 x 60 x 60]

= 3672.50 sfd

= 3672.50 / 26.90= 136.52 inches-mile

Run-off = 136.52 / 135.8

= 1.00 inch

Continued..



Step 3: 6-hour duration unit hydrograph tR = 6 hours > tr

Basin Lag, tPR = tP + [tR – tr]/4

= 9.33 + [6-1.70]/4= 10.41 hours

Peak discharge, qPR = 640 x Cp x A /tPR

= 640 x 0.60 x 135.8/10.41

= 5009.34 cfs

Continued..



Time base, TR = 3 x (1 + tPR /24)= 3 x (1 + 10.41/24)= 4.30 days= 103 hours

50% of qPR = 2504.67 cusecs

W50 = 770 / [qPR / A ]1.08= 770 / [5009.34 / 135.8 ]1.08= 15.64 hours

75% of qPR = 3757.00 cusecsW75 = 440 / [qPR / A ]1.08

= 440 / [5009.34 / 135.8 ]1.08= 8.94 hours

Continued..



Plot unit hydrograph using above calculatedvalues. The graph is shown in Fig 19 below.It has been plotted using definitions given inFigs. 16 and 17 above.

From Fig. 19, tabulate the values ofdischarge against equal time interval of anylength say at 6 hours interval (17*6 = 102 close to 101). These values have beenshown in Table 17

Continued..



SYNTHETIC 6-HOUR UNIT HYDROGRAPH BY SNYDER'S METHOD

(EXAMPLE 6.1)

0

1000

2000

3000

4000

5000

6000

0 20 40 60 80 100 120

TIME (HOURS)

D I S C H A R G E

( c f

Figure: 19

Continued..



Time (Hours) Q (cfs) Time (Hours) Q (cfs)

0 0 60 150.00

6 1880.00 66 110.00

12 4450.00 72 90.00

18 3800.00 78 80.00

24 1500.00 84 50.00

30 1050.00 90 30.00

36 600.00 96 20.00

42 350.00 101 (say 102) 0

48 250.00

Total 14610 54 200.00

Table 17

Continued..



Total run-off = [14610 x 6 x 60 x 60][24 x 60 x 60]

= 3652.50 sfd

= 3652.50 / 26.90

= 135.78 inches-mile

Run-off = 135.78 / 135.8

= 1.00 inch

Continued..

Questions



Define Hydrograph and explain its significance inHydrology. Discuss shape of hydrograph with respect to

catchment characteristics. Explain various methods for separating base flow

from direct runoff in hydrograph. Can a unithydrograph used directly to forecast DRO andpeak of a flood, if the intensity of rainfall is notconstant?

Define Unit Hydrograph and compare it with DROHydrograph. Criticize the assumptions involved inunit hydrograph theory.

Continued..



What is the practical significance of a unit hydrograph? Whatare the three propositions of the unit hydrograph theory?

Explain the step by step procedure for deriving unit hydrographfrom DRO hydrograph and vice versa?

What is S Curve? Explain its utility in hydrology.

How ordinates of unit hydrograph are related to rainfallintensity?

Why we need separation of composite hydrographs?

What is synthetic hydrograph? Explain the method employedfor deriving synthetic hydrograph. Comment on the merits and

demerits of these methods.

Exercise



Problem 1 Table P1 shows the ordinates at 24 hour intervals for a

hydrograph. Separate the base flow from the direct runoff.Compute the volume of DRO.

Time (days) Flow (m³/s) Time (days) Flow (m³/s)

1 2 8 3

2 28 9 2

3 20 10 2

4 11 11 2

5 7 12 1

6 5 13 1

7 3 14 1

Table P1 Hydrograph Data

Exercise



Problem 2 The hydrograph tabulated below in Table P2 resulted from

a 6-hour period of rainfall. Considering the base flow as 11m³/s, find the 6-hour unit hydrograph for the basin having148 sq. km drainage area.

Time

(hours) Flow

(m³/s) Time (hours) Flow (m³/s)

0 11 21 100

3 33 24 67

6 91 27 38

9 103 30 20

12 125 33 14

15 135 36 11

18 120

Table P2: Unit Hydrograph Data

Exercise



Problem 3 Given below in Table P3 is the data for the unit hydrograph

resulting from a 4-hours duration rainfall, for a certainbasin. Derive the S-curve ordinates. Hence derive theordinates for unit hydrographs of 2-hours and 6-hours

durations

Time

(hours) Flow

(m³/s) Time

(hours) Flow (m³/s)

0 0 8 109

2 117 10 64

4 197 12 21

6 145 14 0

Table P3:Unit Hydrograph Data

Continued..

Exercise



Problem 4 Table P4 shows the data for the hydrograph resulting from a 3-

hours duration rainfall for a catchment having area of 3,200 km².Derive the unit hydrograph.

Time Day 1 Flow (m³/s) Day 2 Flow (m³/s)

3 AM 17 130

9 AM 170 99

12 Noon 269 88

6 PM 198 68

9 PM 173 59

12 PM 150 54

Table P4: Hydrograph Data

Exercise



Problem 5 Three consecutive storms of 5 cm 7.5 cm and 4 cm of 6

hours durations each on a basin produced the stream flowsas given in Table P5. The basin area is 54 squarekilometers. Assuming a constant base flow of 7 m³/s andaverage storm loss of 0.4 cm/hour, derive a 6 hour unithydrograph.

Time Flow (m³/s) Time Flow (m³/s)

0 8 21 42

3 12 24 30

6 15 27 21

12 39 30 14

15 46 36 9

18 49 39 8

Table P5: Discharge Data

Exercise



Problem 6 The hydrograph of Table P6 resulted from three successive

6-hour period rainfall, having runoff estimated as 1.5, 3.0and 2.3 cm, respectively. The area of the basin is 148.5km². Assuming a constant base flow of 10 m³/s, derive theordinates of a 6-hour unit hydrograph for the basin.

Time (Hours) Flow (m³/s) Time (Hours) Flow (m³/s)

0 10 21 99

3 31 24 65

6 89 27 36

9 90 30 19

12 198 33 13

15 132 36 10

18 193


Exercise



Problem 7 The stream flows due to three successive storms of

3.7, 6 and 2.7 cm of 6-hours duration each on a basinare given in Table P7. The area of the basin is 51 km².Assuming a constant base flow of 8.5 m³/s and F indexof 0.25 cm/hour, derive a 6 hour unit hydrograph.

Time (Hours) Stream Flow (m³/s) Time (Hours) Stream Flow (m³/s)

0 8 21 42

3 12 24 31

6 15 27 21 9 27 30 14

12 39 33 10

15 46 36 9

18 49 39 8


Exercise



Problem 8 Two catchments A and B are considered meteorology

similar. Their catchment characteristics are given below inTable P8.

For catchment A, a 2-h unit hydrograph was developed and

was found to have a peak discharge of 50 m³/s. The time topeak from the beginning of the rainfall excess in this unithydrograph was 9.0 h. Using Snyder’s method, develop aunit hydrograph for catchment B.

Catchment A Catchment B L = 39 km L = 45 km

Lca = 15 km Lca = 25 km

A = 250 km² A = 400 km²

Table P8: Catchment Characteristics of Catchments A and B

Exercise



Problem 9 A basin has 400 sq. km. of area, L=35 km and Lca

= 10 km. Assuming Ct = 1.5 andCp = 0.70, develop a 3-hour synthetic unit

hydrograph for this basin using Snyder’s method.

Exercise



Problem 10 Table P10 shows the ordinates of the 9-hour unit

hydrograph for the entire catchment of a river up to a damsite

Time (Hours) Discharge (m³/s) Time (Hours) Discharge (m³/s)

0 0 36 118

9 69 45 74

18 1000 54 46

27 210


Exercise



The catchment characteristics are:A = 4480 km², L = 318 km, Lca = 198 km

Derive a 3 hour unit hydrograph of anothercatchment which is meteorologically and

hydrologically similar and has the followingcharacteristics.

A = 3780 km², L = 284 km, Lca=198km Use Snyder’s approach with necessary

modifications for the shape of the hydrograph

Exercise



Problem 11 Using Snyder’s method, derive the 6-hour unit

hydrograph by using the following data.

Ct= 2.65, C

p= 0.56, A = 3500 km²

L = 150 km, Lca = 75 km

complete ch#6 hydro graph

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