Hydrologic cycle,weather and temperature, Rain gauge, precipitation, discharge measurement,stream flow, hydrographs, flood and flood routing. unit hydrographs

1. 1
Lecture 12
FLOODS
Definition of Flood
The level at which the river overflows its banks and
inundates the adjoining area is called flood. This is
normally an unusually high stage in a river.
To estimate the magnitude of a flood peak four alternative
methods are available:
1.Rational method
2.Empirical method
3.Unit-hydrograph technique, and
4.Flood-frequency studies.

2. 2
Lecture 12 (contd.)
Uses of these previous four Methods
The use of the each particular methods depend on
 the desired objective
 the available data and
 the importance of the project.
The rational method is only applicable to small-size (< 50 km2 )
catchments
The unit-hydrograph method is normally restricted to
moderate-size catchments with areas less than 5000 km2.

3. 3
The frequency-distribution functions
applicable in hydrologic studies.
The frequency-distribution functions can be expressed by the
following equation known as the general equation of hydrologic
frequency analysis:
xT = x + K σ (9-1)
Where, xT = value of the variate X of a random hydrologic
series with a return period T,
x = mean of the variate, σ = standard deviation of the
variate.
K = frequency factor which depends upon the return period,
T = the assumed frequency distribution.
Lecture 12 (contd.)

4. 4
The commonly used frequency distribution
functions for the prediction of extreme flood values
are
1. Gumbel’s extreme –value distribution
2. log-Pearson Type III distribution and
3. log normal distribution
1. Gumbel extreme value-distribution
The extreme value distribution was introduced by
Gumbel(1941) and is commonly known as Gumbel’s
distribution method. This method is used for prediction of
flood peaks, maximum rainfalls, maximum wind speed, etc.
Lecture 12 (contd.)

5. 5
Gumbel defined
a flood as the largest of the 365 daily flows and the annual
series of flood flows constitute a series of the largest values
of flows.
According to his theory of extreme events, the probability of
occurrence of an event equal to larger than a value x0 is
P( x ≥ x0 ) = 1 – e-e-y (9-2)
where, y = dimensionless variable given by
y = α (x- a)
a = x – 0.45005 σx
α = 1.2825 / σx
Lecture 12 (contd.)

6. 6
y = {1.2825 ( x – x )}/ σx + 0.577 (9-3)
where x = mean and σx = standard deviation of the variate
X- it is the value of X for a given P that is required and as
such the Eq 9-3 is transposed as
yp = -ln {- ln ( 1 – P)} (9-4)
T = Return period = 1/P and designating
yT = the value of y, commonly called the reduced variate
for a given T
y1/T = - ln x[ - ln { (T-1) / T }]
yT = - ln x ln { T /(T-1) }] (9-5)
Lecture 12 (contd.)

7. 7
Now rearranging Eq 9-3, the value of the variate X with return
period T with Equation (9-1)
( xT = x + K σx )
Where, K = ( yT – 0.577) / 1.2825 (9–6)
This Equation (9-7) is of the same form as the general
equation of hydrologic-frequency analysis.
Lecture 12 (contd.)

8. 8
Gumbel’s Equation for Practical Use
xT = x + K σn-1 (9- 7)
where, σn-1 = standard deviation of the sample of size N
σn-1 = √ {∑(x – x )2 ÷ (N – 1)}
where, K = frequency factor expressed as
KT = (yT – yn) / Sn (9-8)
where, yT = reduced variate, a function of T and is given by
yT = - [ ln x ln {T/(T-1 )} ] (9-9)
Lecture 12 (contd.)

9. 9
Lecture 12 (contd.)
yT = - [ 0.834 + 2.303 {log log T / (T -1)} ] (9-10)
y = reduced mean, a function of sample size N and is given
in prescribed table for
N ∞ , yn 0.577
S n = reduced standard deviation, a function of sample size N
and is taken from prescribed table for
N ∞ , yn 1.2825

10. 10
Example
By using Gumbel’s method , Estimate the flood magnitude
in the Meghna river with a return period of 500 years for
the Flood-frequency computations for the river at
chandpur yielded the following results in the table:
Lecture 12 (contd.)
Return PeriodT(years) Peak Flood(m3/s)
50 40809
100 46300
Solution
Given :
Return Period for 50 and 100 years are mentioned in the table
40809 and 46300 m3/s respectively.
To be estimated 500 years flood magnitude in the Meghna river.

11. 11
Lecture 12 (contd.)
We know,
xT = x + K σn-1
For T = 100 years
x100 = x + K100 σn-1
For 50 years
x50 = x + K50 σn-1
(K100 - K50 ) σn-1 = x100 - x50
= 46,300 – 40,809 = 5491 m3/s
But as per Eq 9-9 , KT = (yT - yn ) / Sn

12. 12
where, Sn and yn are constants for given
data series.
(y100 – y50 ) * (σn-1/ Sn ) = 5491(m3/s) (9-1)
as per equation 9-10 : yT = - [ ln x ln {T/(T-1 )} ]
y100 = - ln. ln ( 100/99)= 4.60015
y50 = - ln. ln ( 50/49) = 3.90194
Putting the respective values in Equation (9-1)
σn-1/ Sn = 5491/( (y100 – y50 ) = 5491/( 4.60015 – 3.9094)
= 7864 m3/s
Lecture 12 (contd.)

13. 13
Lecture 12 (contd.)
For T = 500 years,
As per Equation 9- 10, yT = - [ 0.834 + 2.303 {log log T / (T -1)} ]
Here T = 500 years
y500 = - ln. ln { 500/( 500-1)} = 6.21361 m3/s
Putting the concerned values in the above Equation
(y500 – y100 ) * (σn-1/ Sn ) = x500 - x100 = x500 - 46,300 m3/s
(6.21361 – 4.60015) *7864 = x500 - 46,300
x500 = 58988 m3/s, say 59,000 m3/s Ans

14. 14
Estimation of Flood Peak
1) Rational Method
Consider a rainfall of uniform intensity and very long duration
occurring over a basin. The runoff rate gradually increases
from zero to a constant value as indicated in the following
figure(9-1). The runoff increases as more and more flow from
remote areas of the catchment reach the outlet. Designating
the time taken for a drop of water from the farthest part of
catchment to reach the outlet as tc = time of concentration, it
is obvious that if the rainfall continues beyond tc, the runoff will
be constant and at the peak value. The peak value of the
runoff is given by (basic equation of rational method) as
mentioned below:
Lecture 12 (contd.)

15. 15
RunoffandRainfall
Rainfall
End of Rainfall
Time
tc
Runoff
Peak value Recession
(volume of the two hatched portions are equal)
Figure 9- 1- Runoff hydrograph due to uniform rainfall
Qp = CAi ; for t ≥ tc
(9- 11)
where,
C =runoff / rainfall
coefficient
Qp = Peak flow
A = Catchment
Area and
i = rainfall intensity
Lecture 12 (contd.)
Ci
i

16. 16
The Eq 9-11 Commonly used equation for field
application as
Qp = (1/3.6 ) X C(itc,p) A (9-12)
where, Qp = peak discharge (m3/s)
C = Coefficient off runoff
itc,p = the mean intensity of precipitation (mm/h) or a
duration equal to tc and an exceedence probability
P
A = drainage area in km2
Lecture 12 (contd.)

17. 17
Time of concentration(tc)
The time taken for a drop of water from the
farthest part of catchment to reach the outlet is
called time of concentration and designated as tc
.
Lecture 12 (contd.)

18. 18
Lecture 12 (contd.)
Time of concentration may be estimated by the
two different empirical formulas :
(a) US Practice
For small drainage basins, the time of concentration is
approximately equal to the lag time of the peak flow. Thus
tc = tp = CtL (LLca / √S) n (9-13)
tc = time of concentration in hours

19. 19
where
CtL = basin constants –
values=1.715 mountainous drainage areas,
= 1.03 for foot hill drainage areas and
= 0.50 for valley drainage areas
n = basin constants = 0.38
Lca = distance along the main water course from the
gauging station to a point opposite the watershed
centroid in km.
L = basin length measured along the water course
from the basin divide to the gauging station in km.
Lecture 12 (contd.)

20. 20
Lecture 12 (contd.)
Ct= a regional constant representing watershed slope and storage.
The value of Ct depends upon the region under study and
wide variations with the value ranging from 0.30 to 0.6
Snyder adopted a standard duration tr hours of
effective rainfall given by
tr = tp / 5.5 (9-14)
Where,
tr = Standard duration of time in hours
tp = time of peak in hours

21. 21
Lecture 12 (contd.)
The peak discharge QPS (m3/s) of a Unit
hydrograph of standard duration tr h is given by
Synder as
QP = (2.78 CP A ) / tP (9-15)
where, A = catchment area in km2 and
CP = a regional constant. Values ranges from 0.56 — 0.69
This equation is based on the assumption that the peak discharge is
proportional to the average discharge (1 cm × catchment area) of
duration of rainfall excess.
i.e. CP = (1 cm × catchment area) / (duration of rainfall excess)

22. 22
(c) Kirpitch Equation(1940)
tc = 0.01947L0.77 S - o.385 (9 -16)
Where,
tc = time of concentration (minutes)
L = maximum length of travel of water (m) and
S = Slope of the catchment = H/L in which
H = difference in elevation between the most
remote point on the catchment and outlet
Lecture 12 (contd.)

23. 23
(2) Empirical Formulae
i) Dicken’s formula(1865)
QP = CD A ¾ (9-17)
where
QP = maximum discharge (m3/s)
A = catchment area(km2)
CD = Dicken’s constant
value between 6 to 30
Lecture 12 (contd.)

24. 24
Lecture 12 (contd.)
ii) Ryves Formula (1884)
QP = CR A ⅔ (9-18)
where,
QP = maximum discharge (m3/s)
A = catchment area(km2)
CR = Ryeve’s coefficient,
value = 6.8 for areas within 80 km from the east coast
= 8.5 for areas within 80-160 km from the east coast
= 10.2 for limited areas near hills

25. 25
Lecture 12 (contd.)
iii) Inglis Formula(1930)
QP = 124 A /( √ A + 10.4) (9-19)
where, A = catchment area.
This equation is used in Maharashtra, India

26. 26
Lecture 12 (contd.)
(3) Unit Hydrograph method
The unit –hydrograph technique can be used to predict the
peak flow-flood flow hydrograph if the rainfall producing
flood, infiltration characteristics of the catchment and
appropriate unit hydrograph are available. For design
purposes, extreme rainfall situation are used to obtain the
design storm, viz, the hydrograph of the rainfall excess
causing extreme floods.
The unit-hydrograph method is normally restricted to
moderate-size catchments with areas less than 5000
km2.

27. 27
Lecture 12 (contd.)
(4) Flood frequencies studies
The values of the annual maximum flood from a given catchment
area for large number of successive years constitute a hydrologic
data series called the annual series. The data then arranged in
decreasing order of magnitude and the probability P of each event
being equaled to or exceeded (plotting position) is calculated by
the plotting-position formula (Weibul Formula)
P = m/( N+1) (9-20)
where, m = order number of the event,
N = total number of events in the data.

28. 28
Lecture 12 (contd.)
The recurrence interval, T also called Return period
(i.e. event happening recurrently in a certain interval
is Return period or frequency) or frequency is
calculated as follows:
T = 1/P (9-21)
The relationship between T and the probability of
occurrence various events is the same as described
earlier. Thus, for example, the probability of
occurrence of the event r times in n successive years
is given by
Pr,n = nCr Pr qn-r = n! / (n-r) ! r! Pr qn-r (9-22)
where, q = 1 –P

29. 29
Lecture 12 (contd.)
Example 9-2
(a) An urban catchment has an area of 0.85 km2 .The slope of the
catchment is 0.006 and the maximum length of travel of water is
950 m. The maximum depth of rainfall with a 25 year return period
is shown in the table-1. If a culvert for drainage at the outlet of this
area is to be designed for a return period of 25 years, estimate the
required peak-flow rate, by assuming the runoff coefficient as 0.30.
Duration(min) 5 10 20 30 40 60
Depth of rainfall (mm) 17 26 40 50 57 62
(b) If in the urban area mentioned in (a), the land use of the area and the
corresponding runoff coefficients are shown in Table-2. Calculate the
equivalent runoff coefficient.
Land use Area(ha) Runoff coefficient
Roads 8 0.70
Lawn 17 0.10
Residential area 50 0.30
Industrial area 10 0.80
Table-1
Table-2

30. 30
Lecture 12 (contd.)
Solution
Given : catchment has an area = 0.85 km2
Slope of the catchment = 0.006
Return Period T = 25 years
Runoff coefficient = 0.30
Length of travel of water = 950 m.
Maximum Rainfall for different duration in the Table-1
To be estimated maximum peak discharge (Qp)
We know, Peak discharge (Qp) as per Equation(9-12)
Qp = (1/3.6 ) X C(itc,p) A

31. 31
Hence we again know, the time of concentration by the
Kirpitch formula
tc = 0.01947L0.77 S - o.385
Putting concerned values in the above Equation
= 0.0194 ×7 (950) 0.77 ×(0.006) - o.385
= 27.4 minutes = 27.4 /60 = 0.45 h
Lecture 12 (contd.)

32. 32
Lecture 12 (contd.)
Maximum depth of rainfall (by interpolation)
(x1 –x2) / (x2 – x) = (y1 –y2) / (y2 – y)
(20 – 30)/(30—27.4) = (40— 50)/(50— y) = 47.4 mm
Average intensity (tic,p )= Rainfall / time of concentration
tic,p = (47.4/ 0.45) = 103.8 mm/h
Peak flow,
Qp = (1/3.6 ) X C ×(itc,p) A
Qp = (1/3.6 ) X 0.3 ×103.8 ×0.85 = 7.35 m3/s

33. 33
Lecture 12 (contd.)
(b) Solution:
Given, Coefficient of urban area and the corresponding
runoff coefficients are in the table-2 .
To be estimated equivalent runoff coefficients= Ce
N
We know equivalent runoff coefficient, Ce =( ∑ Ci Ai ) /A
i=1
Putting respective values i8n the above equation-
Ce = {( 0.7×8) + ( 0.1×17) +( 0.3×50) + ( 0.8×10) } /
{( 8 + 17 + 50 + 10)}
Ce = 30.3 /85 = 0.36 Ans.