Estimation

of

Peak / Flood Discharge

Peak discharges occur in the rivers in the rainfall season.

Estimation methods:

1) Physical conditions of the past floods2) Flood discharge formulae3) Flood frequency studies4) Unit hydrograph

(1) Physical Conditions of the Past Floods

To know the maximum water level attained in the past 35 years.

Past flood marks on ancient monuments, etc., along river banks are located.

Old persons in the villages situated on the bank of the river

The cross-section pertaining to the highest flood marks can be drawn.

Obtain from the cross-section: Water flow area (A)Wetted perimeter (P)Hydraulic mean depth / hydraulic radius (R)

By leveling to u/s and d/s of the cross-section site, S can be determined.

Using above parameters velocity of flow (V) is computed by Manning’s orChezy's equation.

Using Continuity Equation, flood discharge can be calculated.

Procedure is repeated at several watermarks, to get consistent results.

(2) Flood Discharge Formulae

First Approach

The general form of the empirical formulae

Q = CAn

whereQ = flood discharge, A = catchment area, C = flood coefficient and n = flood index

Both C and n depend upon various factors, like

Size, Shape, Location and Topography of catchmentIntensity, Duration and Distribution pattern of the storm

(continued)

Depending upon the above factors flood discharge formula for each region is developed.

Following are few examples of regional based flood discharge formulae.

Dicken’s Formula: For various parts of India.Ryve’s Formula: Madras (India)Inglis formula: Former Bombay

presidency (India)Nawab Jang Bahadur formula: Hyderabad state (India)Fanning’s formula: American catchmentsCreager formula: American catchmentsMayer’s formula: American catchments

Second Approach

This includes formulae of the same form as the first approach, i.e.

Q = CAn

but here A refers to cross sectional area of the water way.

Example: Talbot formula.

Third Approach

This includes the formulae of the type:

A

SAciQ

wherei = expected average rainfallS = average slope of the watershed

Examples:Burkli-Ziegler formula andMcMath formula

Fourth Approach: Fuller’s formula

where

Peak 24-hr flood with a frequency of T years

A = Catchment area (km2)

Cf = Fuller’s constant = 0.18 – 1.88

Derived for U.S.A catchments.

TACQ fTplog8.018.0

pTQ

Fifth Approach: Rational Formula

It is characterized by:consideration of the entire drainage area as a single unit, estimation of flow at the most downstream point only, the assumption that rainfall is uniformly distributed over the drainage area.

The Rational Formula reads:

WhereQp= Peak runoff rate [m³/sec], C = Runoff coefficientI = Rainfall intensity [mm/hr], A = Drainage area [km²]

The Rational Formula follows the assumption that:the predicted peak discharge has the same probability of occurrence (return period) as the used rainfall intensity (I), the runoff coefficient (C) is constant during the rain storm, and the recession time is equal to the time of rise.

A I C 0.28pQ

Sixth Approach: Modified Rational Formula

In the modified version of the Rational Formula, a storage coefficient is included to account for a recession time larger than the time the hydrograph takes to rise. The Modified Rational Formula reads:

WhereCs= Storage coefficient

The peak flood discharge in a catchment is reached when all parts of the watershed are contributing to the outflow, i.e. at the time of concentration:

……………… Kirpich/Ramser formula

Wheretc= Time of concentration [min] L= Length of main river [m]S= Longitudinal or Distance weighted channel slope

A I C C 0.28 spQ

385.077.0 0195.0 SLtc

(3) Flood Frequency Studies

Probability concepts are used to study the probable variations in flow so that the design can be completed based on a calculated risk.

Flood Frequency Likelihood of flood being equaled or exceeded.

A 10% frequency means that the flood has 10 out of 100 chances of being equaled or exceeded.

Recurrence Interval (Tr) Number of years in which a flood can be expected (equaled or exceeded) once.

Gumbel's MethodThis method is useful for obtaining values of flood discharges for a high recurrence interval.

If Q1, Q2……… Qn comprise a series of extreme values of floods, the probability of occurrence of a value equal to or less than Q is given by

(1)

(2)

(3) Qf = value of flood discharge having highest frequency, and is given as:

(4)

Qav = average value of discharge

(5)

(6)

The recurrence interval in years (Tr) for the discharge Q is given by

(7)

zeP yez

)( fQQay

xavf QQ 45.0

N

QQQQQQ avnavavx

222

21 )(..........)()(

r

1T =

1-P

x

a28.1

Example # 1 [One Excel Sheet]

Example

With the data given in following table estimate the probability that the annual maximum discharge Q on the Guadalupe River will exceed 50,000 cfs at least once during the next three years.

Annual maximum discharges of the Guadalupe River near Victoria, Texas, 1935-1978, in cfs

Year 1930 1940 1950 1960 1970 0 55900 13300 23700 9190

1 58000 12300 55800 9740

2 56000 28400 10800 58500

3 7710 11600 4100 33100

4 12300 8560 5720 25200

5 38500 22000 4950 15000 30200

6 179000 17900 1730 9790 14100

7 17200 46000 25300 70000 54500

8 25400 6970 58300 44300 12700

9 4940 20600 10100 15200

Solution:From 1935 to 1978 discharge of 50,000 cfs exceeds 9 times as

shown in the table.

First exceedence occurred in 1936

Last exceedence occurred in 1977

Difference b/w first and last exceedences = 1977-1936 = 41 years

For 9 exceedences there will be 8 recurrence intervals.

Therefore, Average Recurrence Interval, T = 41/8 = 5.125 years

 

Probability of occurrence of an event in any observation, p = 1/T

Therefore, probability that discharge will equal or exceed 50,000 cfs, p ≈ 1/5.125 = 0.195

 

P(Qmax ≥ Q at least once in N years) = 1 - (1 - p)N

 

Therefore,

Probability that maximum discharge will equal or exceed 50,000 cfs at least once in 3 years = 1 - (1 - 0.195)3 = 0.48 = 48%

Exceeedence Year

Reccurence Interval (years)

1936

4

1940

1

1941

1

1942

16

1958

3

1961

6

1967

5

1972

5

1977

(4) Unit Hydrograph Method

Hydrograph It is a graph showing variations of discharge with time at a particular point of a stream.

Rainfall Excess or Effective RainfallIt is the rainfall that is neither retained on land surface nor filtered to the soil.

Unit Hydrograph

“It is a type of hydrograph that represents 1 cm or 1 inch of a runoff from rainfall of some unit duration”.

e.g. if a unit hydrograph results from a 3 hour unit rainfall duration it is known as 3-hour unit hydrograph.

In other words:“a hydrograph produced by surface runoff from a storm lasting for 3-hours and yielding a rainfall excess of 1 cm spread on the entire basin”.

Preparing a unit hydrograph from an isolated storm

First of all, the runoff rates (in cm/hr) resulted from the given storm is plotted against time (in hours), so as to obtain Direct Runoff Hydrograph (DRH).

The total area (D) contained within this hydrograph is calculated by a planimeter or by simpson׳s rule, etc. This will represent nothing but the volume of runoff in cm depth of water on the basin.

The ordinates of this hydrograph (ABC) are divided by D so as to obtain the ordinates of a unit hydrograph AB’C.

For example, suppose the area ABC is 3 cm depth of water, then the ordinates of unit hydrograph can be obtained by dividing the ordinates of storm hydrograph by 3.

The final step is to assign an effective storm duration from the study of the rainfall records. Periods of low rainfall at the beginning and end of storm are omitted if they do not contribute substantially to the RUNOFF.

In the fig. the duration of rainfall is 15 hours, but still it can be assumed to have an effective duration equal to 8 hours. In this way, the number of durations can be reduced, and the various storms of different durations can be assigned the same unit duration . Unit storms are therefore not the storms of same durations. But they are the storms of the like durations.

ProblemIn a typical 4 hr. storm producing 5 cm of run off from a basin, the flow in the stream are as follow:

Time Flow (hours) (cumecs)

0 0.0002 1.2004 4.0506 6.7508 5.67012 3.37516 1.35020 0.000

(a) Plot the unit hydrograph of runoff for this storm.

(b) Estimate, as accurately as possible, the peak flow and the time of its occurrence, in a flood created by an 8 hours storm, which produces 2.5 cm of effective rainfall during the 1st four hours & 3.75 cm of effective rainfall during the 2nd four hours. Assume the base flow to be negligible.

T

(hr)

Ordinate of strom

hydrograph

of 5 cm

Ordinate of

unit hydrograph of 1 cm

Ordinate of 1st 4 hr

hydrograph producing

2.5 cm rain

Ordinate of 2nd

4 hr

hydrograph producing 3.75

cm rain

Total 8 hr hydrograph ordinates in

Cumecs

(1) (2) (3) = (2) / 5 (4) = (3) * 2.5 (5) = (3) * 3.75 (6) = (4) + (5)

0 0 0 0 -- 0

2 1.22 0.24 0.61 -- 0.61

4 4.05 0.81 2.02 0 2.02

6 6.75 1.35 3.37 0.90 4.27

8 5.67 1.15 2.83 3.03 5.86

10 -- -- 2.26 5.06 7.32

12 3.375 0.68 1.69 4.31 6.00

16 1.35 0.27 0.67 2.55 3.32

20 0 0 0 1.01 1.01

24 -- -- -- 0 0

SolutionSolution