estimation of peak flood discharge

2. HYDROLOGY
Second Term, Final Year
B.E. Civil Engineering
Presented by.
Engineer Ghulam Shabir Solangi

3. Estimation
of
Peak / Flood Discharge

4. Peak discharges occur in the rivers in the rainfall season.
Estimation methods:
1) Physical conditions of the past floods
2) Flood discharge formulae
3) Flood frequency studies
4) Unit hydrograph

5. (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 or
Chezy's equation.
Using Continuity Equation, flood discharge can be calculated.
Procedure is repeated at several watermarks, to get consistent results.

6. (2) Flood Discharge Formulae

7. First Approach
The general form of the empirical formulae
Q = CAn
where
Q = 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 catchment
Intensity, Duration and Distribution pattern of the storm
(continued)

8. 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 catchments
Creager formula: American catchments
Mayer’s formula: American catchments

9. 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.

10. Third Approach
This includes the formulae of the type:
A
S
AciQ =
where
i = expected average rainfall
S = average slope of the watershed
Examples:
Burkli-Ziegler formula and
McMath formula

11. 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 fTp
log8.018.0
+=
=pT
Q

12. 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:
Where
Qp= Peak runoff rate [m³/sec], C = Runoff coefficient
I = 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.
AIC0.28=pQ

14. 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:
Where
Cs= 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
Where
tc= Time of concentration [min]
L= Length of main river [m]
S= Longitudinal or Distance weighted channel slope
AICC0.28 s=pQ
385.077.0
0195.0 −
= SLtc

15. (3) Flood Frequency Studies

16. 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.

17. Gumbel's Method
This 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)
z
eP −
=
y
ez −
=
)( fQQay −=
xavf QQ σ45.0−=
N
QQQQQQ avnavav
x
∑ −++−+−
=
22
2
2
1 )(..........)()(
σ
r
1
T =
1-P
x
a
σ
28.1
=

19. 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

20. 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

21. Example # 2 [E-5]

22. (4) Unit Hydrograph Method

23. Hydrograph
It is a graph showing variations of discharge with time at a particular
point of a stream.
Rainfall Excess or Effective Rainfall
It is the rainfall that is neither retained on land surface nor filtered to the
soil.

24. 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”.

25. 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 simpsons 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.

27.  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.

28. Problem
In 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 0.000
2 1.200
4 4.050
6 6.750
8 5.670
12 3.375
16 1.350
20 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.

29. 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