Hydro Graph Analysis

8/2/2019 Hydro Graph Analysis

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CE 205 Engineering HydrologyLecture Note by Dr. Uditha Ratnayake

Hydrograph Analysis

Hydrology of rivers is required by engineers for estimation of water, design of dams, diversions, and flood controlthrough reservoirs or dykes, etc. Information is gathered through a network of stream gauges. Hydrograph analysis

deals with the study of runoff records at a stream gauge. Hydrograph analysis is often combined with rainfall

analysis to investigate how a watershed responds to rainfall. In many cases, hydrometric information is not

available. This is especially true for small watersheds. In such situations, rainfall information must be combinedwith rainfall-runoff models

Streamflow measurement

Flow rate is measured in units of cms (cubic meter per second) or cfs (cubic feet per second). Direct measurement of

flow rates requires knowledge of the complete cross sectional velocity profile, which varies with flow rate. While it

is tedious to measure flow rate directly, it is straightforward to measure river stage, for example by a gauge.Therefore, flow rates are measured only a few times, enough to establish a rating curve that describes the

relationship between flow rate and stage. Regular measurement of stage is then combined with the rating curve to

produce time series of streamflow.

Components of the hydrograph

The hydrograph describes flow as a function of time usually known as a time series of flow. The interest may lie inthe hydrograph of a long period of several years or only few selected rainfall events of few hours or days. The latter

situation frequently occurs in the development of a rainfall-runoff relationship for a watershed.

It is customary to consider two components of the hydrograph:

1. Direct runoff the flow that results directly from the rainfall event. Usually after considering the associatedlosses from the gross rainfall. The volumes of effective rainfall and the volume of direct

runoff should be equal.

2. Base flow flow that is unrelated to the rainfall event.

The rainfall-runoff relationship describes the time distribution of direct runoff as a function of excess rainfall (gross

rainfall minus losses). Therefore, in developing the rainfall-runoff relationship for a watershed based on observed

hyetographs and hydrographs, one must first subtract the baseflow from the hydrograph. Even after long periodswithout rain, water still flows in many streams and rivers. This flow is the result of seepage from groundwater

aquifers into the stream channel. In larger rivers, baseflow can be significant. In periods without rain, the baseflow

in a stream will slowly decline as a result of the draw down of the groundwater aquifers. This phenomenon is called

baseflow recession. It is often assumed that baseflow declines exponentially. Baseflow separation involves dividing

the hydrograph into a direct runoff component and a baseflow component.

Unit Hydrograph

Rainfall-runoff modeling is the essence of much engineering hydrology. Because flow data are rarely available,

design event are usually determined by a combination of rainfall information and rainfall-runoff relationships. When

it comes to derive a rainfall-runoff relationship the concept of the runoff caused by a unit rainfall or in other words

the unit hydrograph plays a major role.

The unit hydrograph of time T is based on a hypothetical case of 1 unit (1 mm) of rain falling uniformly over the

whole catchment during a time interval T. The unit hydrograph gives the runoff response of the catchment to that

rain.The basic assumptions of the unit hydrograph are:

The time base of the hydrograph remains the same irrespective of the rain intensity The unit hydrograph is linear (proportionality and superposition applies) The unit hydrograph is time invariant

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Unit Hydrograph DerivationTo derive the unit hydrograph from a simple rainstorm divide the direct discharge values of time-discharge curve by

the height of effective rainfall to get the unit hydrograph.

For example if the total effective rain volume is 5.4 mm then divide all direct discharge values of time-

discharge curve by 5.4 to get the unit hydrograph. A detailed example follows.

Suppose that there are M pulses of excess rainfall and N pulses of direct runoff is observed in a storm. Then N-M+1values will be needed to define the unit hydrograph. The discrete convolution equation, given below, it allows the

computation of direct runoff Qn given excess rainfall Pm and the unit hydrographUn-m+1.

Q P Un m nm

n M

= +

=

11

m

The reverse process, called deconvolution, is needed to derive a unit hydrograph given data on Pm and Qn. Suppose

that there are M pulses of excess rainfall and N pulse of direct runoff in the storm considered; then N equation can

be written for Qn, with n = 1,2,.,N. The equations will consists N M + 1 unknown values of the unit hydrograph.

Few of the equations will be redundant, because there are more equations (N) than unknowns (N M + 1).

The following table shows the set of equations for discrete time convolution

Q1 = P1U1Q2 = P2U1 + P1U2Q3 = P3U1 + P1U2 + P1U3

..QM = PMU1 + PM-1U2 + + + P1UM

QM+1 = 0 + PMU2 + + P2UM + P1UM+1.

QN 1 = 0 + 0 + + + 0 + 0 + + + PMUN-M +1 PM-1UN-M+1QN = 0 + 0 + + + 0 + 0 + + + 0 PMUN-M+1

ExampleAn observed hydrograph is given below with the corresponding excess rainfall. The time interval is 6 hours

between readings. Observed hydrograph for this event is shown in the figure.

Hours ExcessRainfall (mm)

ObservedDirect

Discharge(m3/s)

1 10 10

2 30 70

3 20 200

4 460

5 1780

6 3880

7 3160

8 1120

9 620

10 34011 150

12 70

13 20

0

500

1000

1500

2000

2500

3000

3500

4000

4500

0 5 10 15

Time Interval (1/2hr)

ObservedDischarge(cms)

1st define the number of equations. There are 3 pulses of rainfall so M = 3. There are 13 pulses of observed direct

runoff so N = 13. The total number of unit hydrograph ordinates are N M +1 = 13 3 + 1 = 11 ordinates. So we

have to solve 11 linear equations as follows.

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Q1 = P1U1 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0

Q2 = P2U1 + P1U2 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0

Q3 = P3U1 + P2U2 + P1U3 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0

Q4 = 0 + P3U2 + P2U3 + P1U4 + 0 + 0 + 0 + 0 + 0 + 0 + 0

Q5 = 0 + 0 + P3U3 + P2U4 + P1U5 + 0 + 0 + 0 + 0 + 0 + 0

Q6 = 0 + 0 + 0 + P3U4 + P2U5 + P1U6 + 0 + 0 + 0 + 0 + 0

Q7 = 0 + 0 + 0 + 0 + P3U5 + P2U6 + P1U7 + 0 + 0 + 0 + 0Q8 = 0 + 0 + 0 + 0 + 0 + P3U6 + P2U7 + P1U8 + 0 + 0 + 0

Q9 = 0 + 0 + 0 + 0 + 0 + 0 + P3U7 + P3U8 + P1U9 + + 0 + 0

Q10 = 0 + 0 + 0 + 0 + 0 + 0 + 0 + P3U8 + P2U9 + P1U10 + 0

Q11 = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + P3U9 + P2U10 + P1U11

Explanation of the TableWhat the unit hydrograph says is that each pulse of runoff (1-13 in this example) is generated by some linear

combination of the excess rainfall. For example, the very first pulse of rainfall (0.5 inches) caused the very first

pulse of runoff (5 ft3/s). Because it is the only rainfall that occurred during that time interval, it alone is responsiblefor the runoff that is occurring. Similarly, The second pulse of rainfall is caused by rainfall pulse 1 and 2 (0.5 and

1.2 inches) because they alone occurred during that time interval. So the unit hydrograph is merely a solution to a

set of linear equations that determine the contributions of rainfall over time to the direct runoff hydrograph. The unit

hydrograph becomes normalizedduring the deconvolution process to represent the flow that would occur fromone unit of rainfall occurring during the 1sttime interval.

Now plug in the numbers.

10 = 10U1 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0

70 = 30U1 + 10U2 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0

200 = 20U1 + 30U2 + 10U3 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0

460 = 0 + 20U2 + 30U3 + 10U4 + 0 + 0 + 0 + 0 + 0 + 0 + 0

1780 = 0 + 0 + 20U3 + 30U4 + 10U5 + 0 + 0 + 0 + 0 + 0 + 0

3880 = 0 + 0 + 0 + 20U4 + 30U5 + 10U6 + 0 + 0 + 0 + 0 + 0

3160 = 0 + 0 + 0 + 0 + 20U5 + 30U6 + 10U7 + 0 + 0 + 0 + 0

1120 = 0 + 0 + 0 + 0 + 0 + 20U6 + 30U7 + 10U8 + 0 + 0 + 0

620 = 0 + 0 + 0 + 0 + 0 + 0 + 20U7 + 30U8 + 10U9 + + 0 + 0

340 = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 20U8 + 30U9 + 10U10 + 0150 = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 20U9 + 30U10 + 10U11

We now have 11 linear equations with 11 unknowns. Each unknown is an ordinate of the unit hydrograph. We

have to solve these in a step-wise fashion starting with U1. Lets work them out.

EQ 1. 10 = 10U1 U1 = 10/10 = 1 cms direct runoff / mm excess rainfall

EQ 2. 70 = 30U1 + 10U2 U2= (70 30 U1)/10 = (70 30(1))/10 = 4

EQ 3. 200 = 20U1 + 30U2 + 10U3 U3 = (200 20U1 30U2)/10 = (200 20(1) 30(4))/10 = 6

EQ 4. 460 = 20U2 + 30U3 + 10U4 U4 = (460 - 20U2 - 30U3)/10 = (460 20(4) 30(6))/10 = 20

EQ 5. 1780 = 20U3 + 30U4 + 10U5 U5= (1780 -20U3 - 30U4)/10 = (1780 20(6) 30(20))/10 = 106

EQ 6. 3880 = 20U4 + 30U5 + 10U6 U6= (3880 - 20U4 - 30U5)/10 = (3880 - 20(20) 30(106)/10 = 30EQ 7. 3160 = 20U5 + 30U6 + 10U7 U7 = (3160 - 20U5 - 30U6)/10 = (3160 20(106) 30(30)/10 = 14

EQ 8. 1120 = 20U6 + 30U7 + 10U8 U8 = (1120 - 20U6 - 30U7)/10 = (1120 20(30) 30(14)/10 = 10

EQ 9. 620 = 20U7 + 30U8 + 10U9 U9 = (620 - 20U7 - 30U8)/10 = (620 20(14) 30(10)/10 = 4

EQ 10 340 = 20U8 + 30U9 + 10U10 U10= (340 - 20U8 - 30U9)/10 = (340 20(10) 30(4)/10 = 2

EQ 11 150 = 20U9 + 30U10 + 10U11 U11= (150 - 20U9 - 30U10)/10 = (150 20(4) 30(2) / 10 = 1

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The resulting unit hydrograph is shown below in tabular and graphical form. This hydrograph represents the flow

that would result from 1 mm of rainfall occurring during the 1st time interval (i.e. first half an hour).

TimeInterval

(1/2 hr)

UnitHydrograph

(cms)

1 1

2 4

3 6

4 20

5 106

6 30

7 14

8 10

9 4

10 2

11 1

0

20

40

60

80

100

120

0 2 4 6 8 10 1

Time Interval (1/2 hr)

Unithydrographordinate

(cms)

2

Using the Unit Hydrograph to derive a direct runoff hydrograph

The unit hydrograph can be used to determine the direct runoff hydrograph for any rainfall amount with any time

distribution. When it is needed to derive the time-discharge curve due to multiple storms when storm duration is nT(n is integer) use proportionality and principle of superposition to get the total hydrograph. The process is called

convolution. Lets use the following rainfall distribution to calculate the direct runoff hydrograph.

Derive the hydrograph for an excess rainfall of 10mm, 5mm in the first and second half hours. Use the above unit

hydrograph. The table for calculation looks like this.

Unit Hydrograph Ordinates Direct Runoff

TimeInterval Rainfall U1 U2 U3 U4 U5 U6 U7 U8 U9 U10 U11

1 P1 P1U1 0 0 0 0 0 0 0 0 0 0 P1U1

2 P2 P2U1 P1U2 0 0 0 0 0 0 0 0 0 P2U1 + P1U2

3 0 P2U2 P1U3 0 0 0 0 0 0 0 0 P2U2 + P1U3

4 0 0 P2U3 P1U4 0 0 0 0 0 0 0 P2U3 + P1U4

5 0 0 0 P2U4 P1U5 0 0 0 0 0 0 P2U4 + P1U5

6 0 0 0 0 P2U5 P1U6 0 0 0 0 0 P2U5 + P1U6

7 0 0 0 0 0 P2U6 P1U7 0 0 0 0 P2U6 + P1U7

8 0 0 0 0 0 0 P2U7 P1U8 0 0 0 P2U7 + P1U8

9 0 0 0 0 0 0 0 P2U8 P1U9 0 0 P2U8 + P1U9

10 0 0 0 0 0 0 0 0 P2U9 P1U10 0 P2U9 + P1U10

11 0 0 0 0 0 0 0 0 0 P2U10 P1U11 P2U10 + P1U11

0 0 0 0 0 0 0 0 0 0 P2U11 P2U11

The table with numbers plugged in is shown below. The sum of the columns equals the direct runoff hydrograph.

The resulting graph shows the unit hydrograph and the storm event or the runoff hydrograph resulting from the total

excess rainfall of 15mm occurred within one hour. The area under the unit hydrograph should equal 1.0mm ofrunoff and the area under the direct runoff hydrograph should equal 15 mm of runoff.

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Unit HydrographTime

Interval(1/2 hr)

Rainfall

(mm) 1 4 6 20 106 30 14 10 4 2 1

Direct

Runoff(cms)

1 10 10 0 0 0 0 0 0 0 0 0 0 10

2 5 5 40 0 0 0 0 0 0 0 0 0 453 0 20 60 0 0 0 0 0 0 0 0 80

4 0 0 30 200 0 0 0 0 0 0 0 230

5 0 0 0 100 1060 0 0 0 0 0 0 1160

6 0 0 0 0 530 300 0 0 0 0 0 830

7 0 0 0 0 0 150 140 0 0 0 0 290

8 0 0 0 0 0 0 70 100 0 0 0 170

9 0 0 0 0 0 0 0 50 40 0 0 90

10 0 0 0 0 0 0 0 0 20 20 0 40

11 0 0 0 0 0 0 0 0 0 10 10 20

0 0 0 0 0 0 0 0 0 0 5 5

0

200

400

600

800

1000

1200

1400

0 2 4 6 8 10 12 1

Time Interval (1/2 hr)

Runoff(cms)

4

Unit Hydrograph Runoff Hydrograph

Synthetic Unit Hydrograph

If there is no data for the specific catchment to derive the unit hydrograph, it is sometimes possible to construct a

synthetic unit hydrograph. This is usually based on empirical functions, which correlate unit hydrograph with somebasic morphologic data of the catchment such as area, slope, and land cover together with knowledge of catchments

in the region. Two models of synthetic unit hydrograph will be covered in the course. They are Snyders method

andSCS-dimensionless unit hydrograph.

The key properties of a unit hydrograph that will affect design flows are the peak flow rate, the time to peak, and the

duration of runoff. In many cases, the exact shape of the unit hydrograph is relatively unimportant as long as the

above three properties are reasonably correct. Synthetic unit hydrograph attempt to estimate these three keyproperties based on information of watershed characteristics.

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Snyders unit hydrographTo develop a unit hydrographs based on Snyders method, five inputs are required. They are A (watershed area), L

(length of main stream from outlet to divide), Lc (length to centroid of basin), Ct and Cp (model coefficients). These

model coefficients can be determined from gauged watersheds in the region and transferred to the ungauged designsite. Empirically, Ct values ranges from 0.3 to 6.0. Cp values are in the range of 0.31 to 0.93.

Procedure:1. Estimate basin lag. Basin lag is the time lag from the centroid of excess rainfall hyetograph to the peak

runoff. Snyders method estimates basin lag as

( ) 3.0catp LLCt = tp (hr), L and Lca (km).

2. Estimate peak discharge

ppp tACQ /78.2= Qp (cms), A (square km), tp (hr).

3. Estimate time base of unit hydrograph, which is the time of direct runoff.

8/3 pb tT += Tb (days), tp (hr).

This formula is intended for large watersheds. For small watersheds, the formula will give excessively large

time bases. Hence for small and moderate watersheds, the time base (in hours) should be calculated as 3-5

times the basin lag (use Tb=4tp [hrs] in lack of better knowledge).4. Determine the duration D to which the unit hydrograph corresponds

5.5/ptD = D (hr), tp (hr)5. Adjust the unit hydrograph to desired duration. In many cases, one is interested in a unit hydrograph with a

specific duration. For example, if the design hyetograph is given in time steps of 1 hour, it is desirable to

have a unit hydrograph with duration of 1 hour. The duration can be changed using the S-curve method;however, in Snyders method the following adjustment is recommended: First, adjust lag time as follows

( )DDtt pp += '' 25.0Dis the desired rainfall duration

tp is the corresponding basin lag

6. Calculate the time of rise. Basin lag, tp, is the time from the centroid of excess rainfall to the peak of theunit hydrograph. Hence, the time of rise is calculated as follows.

pR tDT += 2/

7. To assist in sketching the unit hydrograph, calculate unit hydrograph width at 50% and 75% of Qp.

( ) 08.150 /87.5 = AQW p 75.1/5075 WW = W75 and W50 (hrs), Qp (cms), A (km

2).

The endpoints of the intervals defined by W75 and W50 should be placed so that 1/3 appears before TR and 2/3 after.

The above information provides sufficient detail to allow a sketching of the unit hydrograph. Adjustments should be

made such that the volume of runoff (area under the curve) corresponds to 1 inch (or cm) of runoff.

Example

The following characteristics are given for a watershed. Develop a 2-hr unit hydrograph for the basinArea: A = 400 km2

Watershed length L = 45 km

Length to center Lc = 25 km

Coefficient Ct = 1.257Coefficient Cp = 0.576

Solution:

Step 1: Calculate basin lag (time to peak): ( ) ( ) hrsLLCt ctp 34.102545257.13.03.0 ===

Step 2: Calculated duration of excess rainfall hrs88.15.5

10.34

5.5

pt

D ===

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Step 3: Adjust lag time to correspond to 2hr excess rain.( )

( ) hrs

DDtt pp

37.1088.1225.034.10

25.0 ''

=+=

+=

Step 4: Calculate the base time of the unit hydrograph hrsdayst

tp

b 10329.48

37.103

83 ==+=+=

Step 5: Calculate peak discharge cmst

AC

Q p

p

p 77.6137.10

400576.078.278.2=

==

Which occurs at time T hrs37.1137.1012/ =+=+= pR tD

Step 6: Compute W50 and W75

( ) ( ) hrsAQp 44400/77.6187.5/87.5 08.108.150 === W W hrs2575.1/4475 ==

The unit hydrograph ordinates for a flow of 0.75Qp =

46.3 cms should be plotted at times TR 25/3

=3.04hrs and TR + 25(2/3) = 28.04hrs. The unithydrograph ordinates for a flow of 0.50Qp = 30.88

cms has a problem as it results in a negative value.

Therefore, it is plotted at times say 2/3(3.04) =2hrs

and 44+2=46hrs. This happens because the basin issmall.

Snyder's Unit Hydrograph

0

10

20

30

40

50

60

70

0 20 40 60 80 100 120

Time (hrs)

Discharge(cms)

Seven unit hydrograph points (time, discharge) are

now available: (0,0), (2, 30.88), (3.04, 46.3), (11.37,

61.77), (28.04, 46.3), (46, 30.88), and (103, 0). Thesynthetic unit hydrograph can be sketched. Some

adjustment may be needed to ensure that the volume

of runoff corresponds to 1 cm of net precipitation.

S-curve Method: S-curve is the hydrograph produced by a continuous series of effective rainfall at a constant rate.

Infinite number of unit hydrographs of this rainfall rate spaced at its duration is summed up to obtain the S-curve. Ifanother S-curve lagged by a given time duration T is subtracted from the original S-curve a hydrograph due to a T hr

rainfall can be obtained. Converting the runoff volume to unity a unit hydrograph of T hr duration can be obtained.

SCS Dimensionless Hydrograph

While peak discharge rates are adequate for many engineering design problems, they are inadequate for design

problems where watershed or channel storage is significant.

The SCS (US Soil Conservation Service) dimensionless hydrograph is an idealized shape that approximates the flow

from an intense storm from a small watershed. The dimensionless hydrograph arbitrarily has units of 100 units of

flow for the peak and 100 units of time for the duration of flow. The area under a dimensionless hydrograph has

2,620 square units of runoff. The SCS hydrograph has 19 constant ordinates that represent percentages of flow andtime. They can be seen on the figure below. To develop the design hydrograph for a watershed, the peak flow and

the runoff volume must be known for the desired return period storm. The design hydrograph is developed from the

dimensionless hydrograph by using approximate conversion factors. This allows us to determine the hydrographfrom different sized storms by scaling the hydrograph in both space and time.

Conversion FactorsThis dimensionless hydrograph is scaled to create design hydrographs of various storm events that have different

peak runoffs and durations. There are three scaling factors used in the dimensionless hydrograph. The first factor is

u and it is the ratio of the total runoff volume to the area under the dimensionless hydrograph. Recall that the area

under the dimensionless hydrograph has 2,620 units of runoff. So each single unit has a value ofu = Q /2620 where

Q is the total storm runoff volume. (Generally the peak flow and runoff volume are found from the Curve Number

Method, but other methods can also be used)

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SCS Dimensionless Hydrograph

The second factor is w and is the ratio of the peak runoff for the design storm to the peak flow of 100 on thedimensionless hydrograph. Each unit of flow on the dimensionless hydrograph has a value of w = q/100 where q is

the peak runoff(Generally found with the SCS peak flow equation).

The third factor is kand is it the value that each unit of time on the dimensionless hydrograph represents in the

design hydrograph. On the design hydrograph 1/100 of the peak flow times 1/100 of the duration of the runoff must

equal 1/2620 of the flood volume just as it does on the dimensionless hydrograph. Since w is equal to 1/100 of thedesign peak flow, k must be equal to 1/100 of the design duration, and u is 1/2620 of the design flood volume

therefore,

w*k = u and k = u/w

When runoff rate is measured in m3/s, runoff volume is measured in hectares-meters, and time is measured in

minutes. So,k = u(ha-m) * 10,000 (m2/ha) = 167 u

W(m3/s) * 60 (sec/min) w

The coordinates of the design hydrograph are obtained by multiplying the flow and time ordinates of the

dimensionless hydrograph by w and k respectively.

TheSCS Equivalent Triangular Hydrograph is also shown in the above figure. This triangular hydrograph is

defined by the ordinates (0,0), (20,100) and (53.33, 0).

ExampleSuppose a watershed has a peak runoff of 8.2 m3/s and a flood volume 2.9 ha-m derive the design hydrograph.

Step 1: Calculate u as: u = Q/2620 = 2.9 / 2620 = 0.00110687 ha-m / unit

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Step 2: Calculate w as: w = q / 100 = 8.2 / 100 = 0.082 m3/s / unit

Step 3: Calculate k as: k = 167 u/w = 167 * 0.00110687 / 0.082 = 2.2542 min/unit

Step 4: Multiply ordinates

k = 2.2542 w = 0.082

Design Hydrograph CoordinatesPoint Time Ordinate Flow Ordinate k * t w * q

a 0 0 0 0

b 2 3 4.5084 0.246

c 6 19 13.5252 1.558

d 8 31 18.0336 2.542

e 12 66 27.0504 5.412

f 14 82 31.5588 6.724

g 16 93 36.0672 7.626

h 18 99 40.5756 8.118

I 20 100 45.084 8.2

j 22 99 49.5924 8.118

k 24 93 54.1008 7.626

l 26 86 58.6092 7.052

m 30 68 67.626 5.576

n 34 46 76.6428 3.772

o 38 33 85.6596 2.706

p 44 21 99.1848 1.722

q 52 11 117.2184 0.902

r 64 4 144.2688 0.328

s 100 0 225.42 0

Resulting design hydrograph.

0

0.51

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

6.5

7

7.5

88.5

0 25 50 75 100 125 150 175 200 225

Time (minutes)

Flow

Rate(cms)

The result is a graph of flow rate versus time for a watershed with a peak flow rate of 8.2 m3/s with total runoffvolume equaling 2.9 ha-m. The length of time is 225.4 minutes.

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Instantaneous Unit Hydrograph

A main disadvantage of the unit hydrograph is that it is dependant on the duration of the excess rainfall. Given oneunit hydrograph it is difficult to arrive at a unit hydrograph of a different duration. To overcome this difficulty the

concept of Instantaneous Unit Hydrograph is proposed. Limiting the duration of a unit hydrograph to zero an

Instantaneous Unit Hydrograph is obtained. Instantaneous Unit Hydrograph is the hydrograph resulting from an

instantaneous rainfall of one unit uniformly over the basin.

Derivation of instantaneous unit hydrographConsider two S-curves Sa derived from D hr unit hydrograph. The average intensity of excess rainfall of Sa is i=1/D

cm/hr. Let there be another S-curve Sb which is leading Sa by a time interval t. Subtracting Sb from Sa and devidingit by itwill provide the thr unit hydrograph. Limiting t 0 will bring up the instantaneous unit hydrograph.

Thus, mathematically the instantaneous unit hydrograph can be expressed as follows.

( )dt

dS

iti

SS

t

tu ba1

0

lim =

=

Therefore, it is seen that the slope of the S-curve defines the instantaneous unit hydrograph. The most commonlyapplied models for the derivation of the unit hydrograph are the Clarks model and the Nashs model.

Applications of instantaneous unit hydrographAn effective rainfall distributed according to a function of duration t0 results in the runoff hydrograph Q(t) and this

hydrograph can be calculated using the equation given below. How the instantaneous unit hydrograph is used in the

calculation is shown in the following figure.

( ) ( ) ( ) dItutQt

=0

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Rational Method to Calculate Peak Discharge

The most widely used uncalibrated equation is the Rational Method. Mathematically, the rational method relates thepeak discharge (q, m3/sec) to the drainage area (A, ha), the rainfall intensity (i, mm/hr), and the runoff coefficient

(C).

SI Units q = 0.0028CiA

Where q = design peak runoff rate in m3/s

C = runoff coefficienti = rainfall intensity in mm/h for the design return period and for a duration

equal to the time of concentration of the watershed.

To use the rational method there are a few assumptions.

Rainfall intensity and duration is uniform over the area of study Storm duration must be equal to or greater than the time of concentration of the watershed.

Rational Method - Runoff CoefficientsThe rational method uses runoff coefficients in the same fashion as the SCS curve number method for estimating

runoff volume. They have been determined over the years and primarily focus on urban watershed applications.

Below are several tables for different land conditions.

from Schwab (1981)

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from McCuen (1998)

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HydrologicRouting

A hydrograph is function of discharge with time. In fact it describes the passage of a wave along the river. As thiswave moves down the river its shape gets distorted due to various factors such as channel storage, resistance and

lateral addition or withdrawal etc. The process is known as routing and it can be separated in to two categories as

follows.

1. Reservoir Routing2. Channel Routing.

Routing methods can be classified in to (1) Hydrologic routing and (2) Hydraulic routing. Hydraulic routing uses the

St. Venant equation and it tries to preserve the hydraulic properties and provides more accurate results while thehydrologic routing employs the continuity equation. Hydrologic routing is much simpler and is used mainly in

calculation of related design parameters.

Most common application of hydrologic routing is to obtain the design parameters of dams and spillways. In thiscase the outflow hydrograph is needed to determine the length of spillway and flood control storage. In urban

hydrology, the design of detention basins is dependent on an accurate routing of flow. For rivers the interest could

be related to the design of an early warning system. In this case we want to predict outflow from a river reach based

on a known inflow.

Basic principle of continuity can be stated as follows.

I - Q = dS/dtWhere,

I= inflow, Q = outflow, S = storage

In finite difference form the continuity equation becomes StQtI =

Using suffixes 1 and 2 to denote the conditions at the beginning and end of time interval , the above equation can

be written as follows.

t

122121

22SSt

QQt

II=

+

+

For a reservoir both initial Sand Q are known. Mostly, they are functions of water level (H), which is a known data

for a reservoir. If we know the initial condition, we can then solve for Q2 stepwise in time and we need to use

numerical or graphical methods.

Storage Routing

The continuity equation given above is rearranged as follows to use in this method of storage routing.

+=

+

+

222

22

11

21 tQStQ

StII

For a given time step the left-hand side of the equation is known and the right-hand side contain the unknowns. But,the available data of the reservoir can establish the relationships between reservoir stage (water level) and storage,

and between reservoir stage and outflow. This means that the right hand side quantity can be plotted against the

reservoir stage. Through this plot the value of right hand side of the equation can graphically point out the

corresponding reservoir stage, which in turn will show the outflow. Knowing these quantities the next time step canbe solved.

14


15/17


The graphical solution looks as the figure given below. The plot contains the set of curves, (1) reservoir stage verses

, (2) reservoir stage versesS

2

tQS

, (3) reservoir stage verses

+

2

tQS

, and (4) reservoir stage verses .

First step is to locate the water level corresponding to the initial storage and thereby the relevant point on the curve

of stage vs.

Q

2

tQS

as shown in figure. Then, starting from this point, the quantity t

II

+

2

21

is marked

horizontally. From the end point a vertical line is drawn until it meets the curve stage vs.

+

2

tQS . Moving

horizontally the storage, discharge at the end of the time step and the starting point of the next time step on the curve

of stage vs.

2

tQS

can be located.

Channel Routing

One way of looking at the channel routing is to see it as a system of interconnected storages with each storage has aninflow hydrograph and an outflow hydrograph. Each storage is considered to be consists of a prism storage and a

wedge storage as shown in the figure below.

In a river, storage is not uniquely defined by the outflow. This can be understood if one considers the case of an

approaching wave as opposed to a wave just leaving the reach. Equal storage volumes can then occur at totally

different outflows. Therefore, channel routing require a method to deal with this situation and most popular methodis the Muskingum method. It assumes the following linear relationship.

S = K [xI+ (1-x) Q]Where,

K = parameter (of dimension time) which is approximately the traveling time for a wave through the

storage unit or the reach

x = parameter (no dimensions) in the interval 0


16/17


Estimation of K and xIn order to find the parameters Sis plotted versus [xI+ (1-x) Q] for measured events, assuming different values for

x. When the data plot almost as a straight line; x, and K are evaluated. When an incorrect value for x is chosen the

plotted points will trace a looping curve. This procedure is illustrated in the following example.

Example

Following inflow and outflow hydrographs were observed in a river reach. Estimate the values of K and x applicable

to this reach.

Time (h) 0 6 12 18 24 30 36 42 48 54 60 66

Inflow (m3/s) 5 20 50 50 32 22 15 10 7 5 5 5

Outflow m3/s) 5 6 12 29 38 35 29 23 17 13 9 7

The variation of storage during this time interval is estimated first in [m3/s.h] and then it was compared with the

quantity K [xI+ (1-x) Q] for x = 0.35, 0.30 and 0.25 which has the units [m3/s]. Then the gradient of the plot of Sversus [xI+ (1-x) Q] will give K in hours. Following table and the figure shows this procedure.

Continuity equation can be written as; ( ) ( ) St

QQt

II =

+

+22

2121

Rearranging the above equation: ( ) ( )[ ] St

QIQI =

+2

2211

[xI+ (1-x) Q] (m3/s)Time

(h)

I

(m3/s)

Q

(m3/s)

I-Q

(m3/s)

Mean

I-Q

(m3/s)

S(m3/s.h)

S =

S(m3/s.h)

x=0.35 x=0.30 x=0.25

0 5 5 0 0 5.0 5.0 5.0

7.0 42

6 20 6 14 42 10.9 10.2 9.5

26.0 156

12 50 12 38 198 25.3 23.4 21.5

29.5 177

18 50 29 21 375 36.4 35.3 34.3

7.5 45

24 32 38 -6 420 35.9 36.2 36.5-9.5 -57

30 22 35 -13 363 30.5 31.1 31.8

-13.5 -81

36 15 29 -14 282 24.1 24.8 25.5

-13.5 -81

42 10 23 -13 201 18.5 19.1 19.8

-11.5 -69

48 7 17 -10 132 13.5 14.0 14.5

-9.0 -54

54 5 13 -8 78 10.2 10.6 11.0

-6.0 -36

60 5 9 -4 42 7.6 7.8 8.0

-3.0 -18

66 5 7 2 24 6.3 6.4 6.5

16


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0

10

20

30

40

0 100 200 300 400 500

[xI + (1-x) Q] (m3/s)

S(m3/s.h)

0

10

20

30

40

0 100 200 300 400 500

[xI + (1-x) Q] (m3/s)

S(m3/s.h)

x = 0.35 x = 0.30

0

10

20

30

40

0 100 200 300 400 500

[xI + (1-x) Q] (m3/s)

S(m3/s.h)

x = 0.25

This plot for x = 0.25 show a more straight line

relationship compared to x = 0.35 or x = 0.30.

Therefore, x = 0.25 is selected and the value of K isestimated from the gradient of the line as K = (400/30)

= 13.3hrs.

Procedure for channel routing

With known x, K a given inflow hydrograph can be routed through the reach using the following method.

Change in the storage according to Muskingum equation is;

( ) ( )( )[ ]121212 1 QQxIIxKSS +=

Using the above equation and the continuity equation Q2 is evaluated as follows.

1211202 QCICICQ ++=Where,

tKxK

tKxC

+

+=

5.0

5.00

tKxK

tKxC

+

+=

5.0

5.01

tKxK

tKxKC

+

=

5.0

5.02

For best results the routing interval ( t) should be chosen such that K > t> xK.

17

Hydro Graph Analysis

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