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Prof. Rajesh BhagatAsst. Professor
Civil Engineering Department
Yeshwantrao Chavan College Of Engineering
Nagpur
B. E. (Civil Engg.) M. Tech. (Enviro. Engg.)
GCOE, Amravati VNIT, Nagpur
Mobile No.:- 8483003474 / 8483002277
Email ID:- rajeysh7bhagat@gmail.com
Website:- www.rajeysh7bhagat.wordpress.com
ENGINEERING HYDROLOGY
UNIT-IV
Statistical Methods: Statistics in hydrological analysis, probability
and probability distribution.
Floods: Causes and effects, Factors affecting peak flows and its
estimation, Flood routing and Flood forecasting. Frequency analysis.
Economic planning for flood control.
Importance of Statistics and Probability in Hydrology:
1) If the outcome of a process can be precisely predicted is known as deterministic
process.
2) But in some cases, there is uncertainty or unpredictability regarding outcome.
3) Average weather condition are same from year to year, but yearly rainfall at a
given place is not same every year.
4) In all such a cases, process is treated as a probabilistic process and outcome
governed by some probability law.
Flood
1) Any flow which is relatively high and which overtops the natural or artificial
banks in any reach of a river may be called a flood.
2) Flood is the most worldwide natural disaster.
3) It occurs in every country and wherever there is rainfall or coastal hazards.
Different types of flood:-
Local flood:- caused due to high intensity of local rainfall which generate higher
surface run-off. Urbanization, Surface sealing and loss of infiltration capacity
increases the surface runoff causes local flood.
River floods: - River flood occur when the river volume exceeds due to excess of
run-off beyond the river capacity. The river levels rise slowly due to high intensity
rainfall for long lasting duration.
Flash floods:-Flash floods occur as a result of the rapid accumulation and release
of runoff waters from upstream mountainous areas, which can be caused by very
heavy rainfall, bursting of clouds, landslides, sudden break-up of an ice jam or
failure of flood control works.
Coastal floods:- High tides and storm caused by tropical depressions and cyclones
can cause coastal floods in urban areas, low-lying land near the sea in general.
Tidal effects can keep the river levels high for long periods of time and sustain
flooding.
Major parameters factors responsible for the occurrences of flood:
1) Climate:- Due to global warming or climate change many subsystems of the
global water cycle are effected causing an increase of flood magnitude as well as
flood frequency.
2) Nature of collecting basin:- A collecting basin is an area where surface water
from rain converges to a single point, it is usually exit of basin, where water
join another water body, such as a river. Larger the area of basin more will be the
water receiving capacity.
3) Nature of the streams: - Water flowing through a channel, such as a stream, has
the ability to transport sediment. Some percolates deep into the ground and
replenishes the groundwater supply.
4) Vegetative cover:- Trees reduce top soil erosion, prevent harmful land
pollutants contained in the soil from getting into our waterways, slow down
water run-off, and ensure that our groundwater supplies are continually being
replenished.
5) Rainfall: - It is difficult to predict and measure the precise changes in the
hydrological cycle because of the processes such as evaporation, transport, and
precipitation.
CAUSES OF FLOOD
1) Intense rainfall
2) Melting of snow
CAUSES OF FLOODING
1) Culvert, drain and gully entry blockage (partial or total)
Dumped rubbish.
Floating debris, including leaves and branches.
Sediment deposition
2) Poorly maintained urban watercourses (silted, overgrown with vegetation, or
blocked with rubbish.)
3) Bridge, culvert or tunnel waterway capacity exceeded.
4) Failure of drainage pump or jammed sluice gate.
5) High water level in receiving river that flows back up the drain or sewer.
FLOOD:-
1) Design of culvert, bridges, drainage work and irrigation diversion works, needs a
reliable estimates of the flood.
2) The maximum flood that any structure can safely pass is called the design flood.
3) As the magnitude of the design flood increase, the capital cost of the structure also
increases but the probability of annual damages will decrease.
4) Standard Project Flood (SPF) is the estimate of the flood likely to occur from the
most severe combination of the meteorological & hydrological conditions, which
are reasonably characteristics of the drainage basin being considered, but excluding
extremely rare combination.
5) Maximum Probable Flood (MPF) differs from the SPF in that it includes the
extremely rare and catastrophic floods and is usually confined to spillway design of
very high dams. The SPF is usually around 80% of the MPF for the basin.
DESIGN FLOOD:-
1) It is the flood adopted for the design of hydraulic structure like spillways, bridges,
flood banks, etc.
2) It may be MPF or SPF or flood of any desired recurrence interval.
3) The methods used in the estimation of the design flood can be grouped as under:
1) Physical indication of past flood
2) Envelop curves
3) Empirical flood formulae
4) Rational method
5) Unit hydrograph application
6) Frequency analysis or statistical method
Rational Method for Estimation of Design Flood:-
1) The time of concentration is the maximum time required by the surface runoff to
reach the basin outlet.
2) When storms continues beyond concentration time every part of the catchment
would be contributing to the runoff at outlet and therefore it represents condition of
peak runoff.
3) It is given by Q = C x A x I
Where, C is runoff coefficient or impermeability coefficient = (runoff/rainfall)
A = catchment area
I = intensity of rainfall.
If different portions of catchment have different runoff coefficient then C is
calculated by
Ceq = (C1A1 + C2A2 + C3A3……) / (A1 + A2 + A3……)
Rational Method for Estimation of Design Flood:-
1) Q = C x A x I
2) The rational formula is found to be suitable for a peak flow prediction in small
catchment up to 50 km2 in area.
3) Application in urban drainage designs and in the design of small culverts and
bridges.
Rainfall Intensity, I tc,p = (KTx) / (tc + a)m
Probability exceedance P : (return period T = 1/P)
K, a, x and m are constant.
Kirpich Equation, tc = 0.01947 L0.77 S-0.358
tc = Time of concentration in minutes
L = Max. Length of traveller of water in meter
S = Slope of catchment = ∆H / L
∆H = Difference in elevation
12
What is a return period?
1) The probability that events such as floods, wind storms & earthquake will
occur is often expressed as a return period.
2) The inverse of probability (generally expressed in %), it gives the estimated
time interval between events of a similar size or intensity.
3) For example, the return period of a flood might be 100 years; otherwise
expressed as its probability of occurring being 1/100, or 1% in any one year.
This does not mean that if a flood with such a return period occurs, then the
next will occur in about one hundred years' time - instead, it means that, in
any given year, there is a 1% chance that it will happen, regardless of when
the last similar event was. Or, put differently, it is 10 times less likely to occur
than a flood with a return period of 10 years (or a probability of 10%).
13
What is a return period?
1) It represents the avg. number of years within which a given event will be
equaled or exceeded..
2) It is calculated by using Weibuls formula, T = (n + 1) / m
Where T is Return period, n is number of years in record & m is the order or
rank ( of highest observed flood)
1) After either the partial or annual series is complied, the items are arranged in
descending order of magnitude and assigned an order or rank, starting with 1
for the highest observed flood.
1) 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 years
return period is as below:-
if culvert for drainage at the outlet of this area is to designed for a return period of 25 years,
estimates the required peak flow rate by assuming the runoff coefficient as 0.3.
Sol:- Time of concentration by Kirpich Equation, tc = 0.01947 L0.77 S-0.358
tc = 0.01947 x (9500.77 ) x (0.006)-0.358 = 27.4 minutes
Max. Depth of rainfall for 24.7 minute duration (by interpolation):
(50 - 40) / (30 - 20) = (50 - x) / (30 - 27.4)
x = 47.4 mm = max. Depth of rainfall for 27.4 minutes.
Avg intensity of rainfall = I tc,p = (47.4 / 27.4 ) 60 = 103.8 mm/hr
Q = C x A x I = (0.3 x 103.8 x 0.85 x 1000) / 60 x 60 = 7.35 m3 / s
Duration, Minute 5 10 20 30 40 60
Depth of rainfall, mm 17 26 40 50 57 62
2) If in the urban area, the land use of the area and the corresponding runoff
coefficient are as given below calculate the equivalent runoff coefficient.
Sol: Equivalent runoff coefficient
Ceq = (C1A1 + C2A2 + C3A3 + C4A4) / (A1 + A2 + A3 +A4)
Ceq = ((0.7x8) + (0.1x17) + (0.3x50) + (0.8x10)) / (8 + 17 + 50 +10)
= 0.36
Land Use Area (ha) Runoff coefficient
Roads 8 0.7
Lawn 17 0.1
Residential area 50 0.3
Industrial area 10 0.8
3) A small watershed consists of 1.5 km2 of cultivate area (C=0.2), 2.5 km2 under
forest (C=0.1) and 1.0 km2 under grass cover (C=0.35). There is a fall of 22 m in a
watercourse length of 1.8 km. The intensity frequency-duration relation for the
area may be taken as I = (80 x Tr0.2) / (tc + 13)0.46
Where I is in cm/hr, T is is in minutes. Estimate the peak rate of runoff for a 25
year frequency.
Sol: length of water course = 1.8 km = 1800 m & Fall in the elevation = 22 m
S = 22 / 1800 = 0.0122
tc = 0.0195 x (L0.77 ) x (S)-0.358
tc = 0.0195 x (18000.77 ) x (0.0122)-0.358 = 34 minutes
I = (80 x Tr0.2) / (tc + 13)0.46
I = (80 x 250.2) / (34 + 13)0.46 = 25.9 cm/hr
Ceq = (C1A1 + C2A2 + C3A3) / (A1 + A2 + A3)
Ceq = ((0.2x1.5) + (0.1x2.5) + (0.35x1)) / (1.5 + 2.5 + 1)
= 0.18
3) A small watershed consists of 1.5 km2 of cultivate area (C=0.2), 2.5 km2 under
forest (C=0.1) and 1.0 km2 under grass cover (C=0.35). There is a fall of 22 m in a
watercourse length of 1.8 km. The intensity frequency-duration relation for the
area may be taken as I = (80 x Tr0.2) / (tc + 13)0.46
Where I is in cm/hr, T is is in minutes. Estimate the peak rate of runoff for a 25
year frequency.
Sol:
S = 0.0122
Tc = 34 minutes
I = 25.9 cm/hr = 0.259 m / hr
Ceq = 0.18
Q = C x A x I = (0.18 x 5 x 1000000 x 0.259) / 60 x 60 = 64.75 m3 / s
4) A bridge is proposed to constant across a river having catchment area of 2027
hectors. The catchment has a slope of 0.007 and length of travel of water is
20270m. Estimates 30 years flood, the intensity frequency duration relationship is
given by
I = (9876 x Tr0.2) / (tc + 45)0.98
Where I is in mm/hr, Tr is in years & tc is in minutes. Assume runoff coefficient as
0.35.
Sol:- Time of concentration by Kirpich Equation, tc = 0.01947 L0.77 S-0.358
tc = 0.01947 x (202700.77 ) x (0.007)-0.358 = 272.45 minutes
Max. intensity of rainfall = I tc,p = ((9876 x 300.2 )) / ((272.45 + 45)0.98)
= 68.92 mm/hr = 6.892 cm/hr
A = 2027 Ha = 20.27 Km2
Q = C x A x I
Peak Discharge, Q = 2.78 x C x A x I = (2.78 x 0.35 x 6.892 x 20.27)
= 135.93 cumec
Estimation of Design Flood For A Particular Return Period:
1) The return period is calculated by using Weibull’s formula:
Tr = ( n + 1) / m
Tr = return period in years
n = number of years of record
m = order number
2) Return period represents the avg no. Of years within which a given event will be equalled
or exceeded. Probability of Exceedance, p = 1 / Tr
3) If the probability of occurrence of an event is p, then the probability of its no-occurrence
is q = 1- p
4) Probability of an event not occurring at all in ‘n’ successive years would be equal to qn
qn = (1- p)n
Probability of an event occurring at least once in ‘n’ successive years, = 1 – qn = 1- (1 - p)n
This probability is called risk.
A flood of a certain magnitude has a return period of 20 years. What is the probability
that a flood of this magnitude may occurs in next 15 years.
Return period represents the avg no. Of years within which a given event will be equalled or
exceeded. Probability of Exceedance, p = 1 / Tr
p = 1 / 20 = 0.05
Probability of an event not occurring at all in ‘n’ successive years would be equal to qn
qn = (1- p)n = (1 – 0.05)15 = 0.46
Probability of an event occurring at least once in ‘n’ successive years,
Risk = 1 – qn = 1- (1 - p)n
Risk = 1 – 0.46 = 0.54
The flood data analysis at river site yielded mean of 12000 m3/s & standard deviation of
650 m3/s for what discharge the structure should be designed so as to provide 95%
assurance that the structure will not fail in next 50 years.
Percentage of assurance = 95 %
Percentage risk = 5%
Risk = 1 – qn = 1- (1 - p)n
Probability of Exceedance, p = 1 / Tr
0.05 = 1- (1 - 1 / Tr)50
Tr = 975.286 years
YT = - ln ln (T / (T - 1)) = - ln ln (975.286 / (975.286 - 1)) =6.88
XT = X + K . σn-1 = 12000 + K . 650
K = Frequency factor = (YT – yn) / Sn = = (6.88 – 0.577) / 1.28225
XT = X + K . σn-1 = 12000 + K . 650
XT = 15195 m3/s
Estimation of Design Flood For A Particular Return Period:
1) The return period is calculated by using Weibull’s formula:
Tr = ( n + 1) / m
Tr = return period in years
n = number of years of record
m = order number
2) Return period represents the avg no. Of years within which a given event will be equalled
or exceeded. Probability of Exceedance = 1 / Tr = P
3) If a graph plotted between flood magnitude and its return period in simple plane co-
ordinates, the plot is called probability or empirical distribution. Extrapolation for large
return period may yield more result hence following method are used:
A. Gumble’s method
B. Log Pearson Type III Method
Gumble’s Method:
XT = X + K . σn-1
XT = value of flood magnitude of return period T years
X = mean value of past flood record = ( ∑X) / n = X
K = Frequency factor = (YT – yn) / Sn
YT = - ln ln (T / (T - 1))
No. Of years of records, n > 50 we use yn = 0.577 & Sn = 1.2825
σn-1 = Standard deviation = √ ( ( ∑ ( Xi – X )2 ) / ( n – 1 ) )
Gumble’s Probability Distribution Method for Prediction of Flood Peak:
1) This method of estimating the flood peak was given by Gumbel and is known as
Gumbel’s Distribution method.
2) This is a flood frequency method which is most widely used in hydrological analysis &
meteorological studies.
Procedure for Estimating Peak Flood:
1) From the given discharge data & for sample size n, X & Sx or σn are to be computed.
Sx OR σn-1 = Standard deviation = √ ( ( ∑ ( Xi – X )2 ) / ( n – 1 ) )
2) yn & σn are to be computed for given sample size, n.
(assumed if not given 0.577 & 1.28225)
1) For given Tr or T, YT is to be calculated by YT = - ln ln (T / (T - 1))
2) Frequency factor (K or KT) is calculated by K = (YT – yn) / Sn
3) Now the magnitude of flood XT can be obtained from XT = X + K . σn-1
Flood frequency computation for the river Chambal at Gandhisagar Dam by using
Gumble’s Method, yielded the following results:
Estimates the flood magnitude in this river with a return period of 500 years.
Sol: XT = X + K . σn-1
X100 = X + K100 . σn-1 & X50 = X + K50 . σn-1
X100 – X50 = (K100 - K50 ) σn-1
46300 – 40809 = (K100 - K50 ) σn-1
5491 = (K100 - K50 ) σn-1
Return Period, T (Years) Peak Flood, m3/s
50 40809
100 46300
Flood frequency computation for the river Chambal at Gandhisagar Dam by using
Gumble’s Method, yielded the following results:
Estimates the flood magnitude in this river with a return period of 500 years.
Sol: XT = X + K . σn-1
X100 = X + K100 . σn-1 & X50 = X + K50 . σn-1
(K100 - K50 ) σn-1 = X100 – X50 = 46300 – 40809 = 5491
KT = (YT – Yn) / Sn
K100 = (Y100 – Yn) / Sn & K50 = (Y50 – Yn) / Sn
(Y100 - Y50 ) (σn-1 / Sn )= 5491
No. Of years of records, n > 50 we use Yn = 0.577 & Sn = 1.2825
YT = - ln ln (T / (T - 1))
Y100 = - ln ln (100 / (100 - 1)) & Y50 = - ln ln (50 / (50 - 1))
Y100 = 4.60015 & Y50 = 3.90194
Return Period, T (Years) Peak Flood, m3/s
50 40809
100 46300
(K100 - K50 ) σn-1 = 5491
KT = (YT – Yn) / Sn
K100 = (Y100 – Yn) / Sn & K50 = (Y50 – Yn) / Sn
(Y100 - Y50 ) (σn-1 / Sn )= 5491
(σn-1 / Sn )= 5491 / (Y100 - Y50 )
(σn-1 / Sn )= 5491 / (4.60015 - 3.90194) = 7864
For T = 500 years
YT = - ln ln (T / (T - 1))
Y500 = - ln ln (500/ (500 – 1)) = 6.21361
(Y500 - Y100 ) (σn-1 / Sn ) = X500 - 46300
(6.21361 – 4.60015 ) (7864) = X500 - 46300
X500 = 58988 say 59000 m3/s
Flood Routing:-
1) It is the technique of determining the flood hydrograph at a section of a river by utilising
the data of flood flow at one or more upstream sections.
2) It is also used in analysis of a flood forecasting, flood protection, reservoir design,
spillway design etc.
3) The broad categories of flood routing are reservoir routing and channel routing.
4) In reservoir routing the effect of reservoir storage on a flood hydrograph is analysed.
This is used in design, location and sizing of the capacity of reservoir.
5) In channel routing the effect of storage of a specified channel reach on the flood
hydrograph is studied. It is used in flood forecasting and flood protection.
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