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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
Preface
1
Water is vital to life and development in all parts of the world. Water resources
plays a key role in economic growth, the management of water resources is an
item of high priority in developmental activities. The basic inputs in the
evaluation of water resources are from hydrological parameters and the subject
of hydrology forms the core in the evaluation and development of water
resources. In the civil engineering curriculum, this subject occupies an important
position.
UNIT-I
1) Introduction: Engineering hydrology, hydrological cycle, hydrological
equation, Importance of temperature, wind and humidity in hydrological
studies.
2) Precipitation: Definition, types and forms of precipitation, factors affecting
precipitation, measurement of precipitation (using rainguages & analytical
methods), optimum number of rain gauge stations, radar measurement of
rainfall, mass curves, data missing records. Intensity-Duration-Frequency and
Depth-Area-Duration analysis.
3
UNIT-II
1) Infiltration: Definition, mechanism, factors affecting Infiltration &
Infiltration indices.
2) Evaporation: Definition, mechanism, factors affecting evaporation,
estimation of evaporation (instrumental & analytical) & evaporation control.
3) Transpiration: Definition, mechanism, factors affecting transpiration, its
measurement and control.
4) Evapotranspiration & Its measurement.
5) Interception and its measurement.
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1) Ojha, C.S.P., Berndtsson, R., & Bhunya, P., Engineering Hydrology, Oxford
University Press.
2) Raghunath H.M., Hydrology, New Age International Publishers.
3) Reddy R., Hydrology, Tata McGraw-Hill New Delhi.
4) Linsley, R.K.,Kohler, M.A. and Paulhas, Hydrology for Engineers, Tata McGraw-
Hill Publishing Company Limited.
5) Todd, D.K., Ground Water Hydrology, John Wiley & Sons.
6) Subramnaya, K., Engineering Hydrology, Tata McGraw-Hill Publishing
Company Limited.
7) Sharma R.K. & Sharma T.K., Hydrology & Water Resources Engineering, Dhanpat
Rai Publications.
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Text Books & Reference Books
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Hydrology means the science of water. It is the science that deals with theoccurrence, movement or circulation and distribution of water of the earthand earth’s atmosphere.
Allied Sciences:-
Physics, Chemistry, Biology, Mathematics, Statistics, Probability & Geology.
Hydraulics & Fluid Mechanics
Soil Engineering
Meteorology & Climatology
Agronomy & Forestry
Potamology (streams), Limnology (lake), & Cryology (snow)
Oceanology
1. Estimation of water resources
2. The study of process such as precipitation, runoff, evapotranspiration andtheir interaction
3. The study of problems such as floods and droughts, and the strategies tocombat them.
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Applications or Importance of Hydrology:-
Design of hydraulic structures
Irrigation & hydropower development
Flood control
Water supply scheme
Navigation
Erosion & sediment control
Stream flow & flood forecasting
Pollution control
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The Hydrological Cycle:-
1) The entire process of circulation & redistribution of water by the atmosphere
& the earth surfaces which has no beginning on end is called hydrological
cycle.
2) Precipitation takes place from the vapour storage of atmosphere in the form of
rain, hail, snow, dew, frost, etc.
3) Part of it intercepted by trees & vegetation & remaining reaches to ground.
4) Some part evaporates, while other part is available for infiltration & excess
water becomes surface & subsurface runoff, which ultimately reaches to the
streams.
5) Part of infiltrated water is used for soil moisture storage, the maximum of
which is utilized by plants & reaches to the atmosphere through leaves by
transpiration.
6) Other part of infiltrated water slowly percolates in deeper zone & known as
ground water flow or base flow.
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The Hydrological Cycle:-
Thus the total stream flow is sum of surface runoff and ground water runoff.
A small portion of flow or storage evaporates & the remaining portion joins the
ocean where from it ultimately evaporates into atmosphere.
Thus the hydrological cycle is formed with its various phases like precipitation,
interception, surface runoff, infiltration, percolation, storage, stream flow,
transpiration & evaporation.
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TERMS IN WATER CYCLE:
1) Precipitation
2) Evaporation
3) Transpiration
4) Interception
5) Surface runoff
6) Groundwater
7) Percolation
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Temperature, Humidity & Wind in Hydrological Cycle:-
1) Hydrological cycle depends on meteorological factor such as wind,
temperature & humidity.
2) To form precipitation, humidity should be in atmosphere.
3) Weather condition must be good for condensation of water vapour..
4) Wind speed facilitates the movement of cloud while its turbulence retains the
water droplet in suspension.
5) The velocity of the air in contact with the water surface is more, the saturated
film of air containing the water vapour will move easily and diffusion &
dispersion of vapour will be easier causing more evaporation.
6) Proper temperature increases the saturation vapour pressure & evaporation
increase.
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WATER BUDGET EQUATION:
1) The quantities of water going through various individual paths of thehydrological cycle can be described by the continuity principle known aswater budget equation or hydrologic equation.
2) The equation which represents this balance of the moisture quantity isknown as hydrological equation. ( law of conservation)
3) It is important to note that the total water resources of the earth areconstant and the sun is the source of energy for the hydrologic cycle.
Mass Inflow – Mass Outflow = Change in Mass Storage
Vi – Vo = ∆S
P – R – G – E – T = ∆S
P: precipitation, R: surface runoff, G: net groundwater flow out of the
catchment, E: evaporation, T: transpiration, ∆S: change in storage
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Catchment Area or Watershed:-
1) The area of land draining into a stream or a water course at a givenlocation is known as catchment area.
2) Drainage area, drainage basin, drainage area, catchment basin, river basin,water basin, watershed, etc.
3) The catchment act as funnel by collecting all the water within the areacovered by catchment & channeling to single point.
4) It is normal to assume the groundwater divide to coincide with the surfacedivide.
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Topographic Water Divide or Watershed Divide:-
1) Each catchment area is separated topographically from adjacent catchment
area by a geographical barrier such as ridge, hill or mountain.
2) The line which divides the surface runoff between two adjacent river basins is
called the topographic water divide or the watershed divide.
3) If a permeable soil covers an impermeable substrate, the topographical
division of watershed will not always correspond to the line that is effectively
delimiting the groundwater.
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PRECIPITATION:
Water falling from the atmosphere to the earth.
1) Rain
2) Snowfall
3) Sleet
4) Hail
5) Drizzle
6) Glaze or Freezing Rain
Requires lifting of air mass so that it cools and condenses.
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MECHANISMS FOR AIR LIFTING:
1. Frontal lifting
2. Orographic lifting
3. Convective lifting
• Air mass : A large body of air with similar temperature and moisture
characteristics over its horizontal extent.
• Front: Boundary between contrasting air masses.
• Cold front: Leading edge of the cold air when it is advancing towards
warm air.
• Warm front: leading edge of the warm air when advancing towards cold
air.
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FRONTAL MECHANISM (Cyclonic)
Boundary between air masses with different properties is called a front
Cold front occurs when cold air advances towards warm air & warm air get
lifted upward (gives Rain)
Warm front occurs when warm air overrides cold air (Summer Cloud)
Cold front (produces cumulus cloud) Cold front (produces stratus cloud)
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Atmosphere is divided into four sphere:-
1. Troposphere
2. Stratosphere
3. Mesosphere
4. Thermosphere or Ionosphere
Atmospheric Layers:-
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OROGRAPHIC LIFTING
• Orographic uplift occurs when air is forced to rise because of the physical
presence of elevated land.
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CONVECTIVE LIFTING
• Convective precipitation occurs when the air near the ground is heated by
the earth’s warm surface. This warm air rises, cools and creates
precipitation.
Hot earth
surface
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Forms of Precipitation:
1) Rain
2) Snowfall
3) Sleet
4) Hail
5) Drizzle
6) Glaze or Freezing Rain
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RAIN: (Avg. rainfall for entire country (India) is 119cm)
1) Rain develops when growing cloud droplets become too heavy to remain
in the cloud and as a result, fall toward the surface as rain.
2) Precipitation in form of water drops of size greater than 0.5mm & less
than 6mm.
3) Rain can also begin as ice crystals that collect each other to form large
snowflakes. As the falling snow passes through the freezing level into
warmer air, the flakes melt.
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SNOWFALL:
1) Snow is formed when ice crystals form from water vapor that is in the
clouds.
2) The fall of larger snowflakes from the clouds on the ground surface is
called snowfall.
3) It occurs when the freezing level is so close to the ground surface(<300m)
4) Snowfall is measured by equivalent depth of water.
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Sleet:-
1) Sleet refers to mixture of rain & snow.
2) It is frozen raindrops & it begins as rain or snow and falls through a deep layer of
cold air that contains temperatures below freezing that exist near the surface.
3) Rain that falls through this extremely cold layer has time to freeze into small
pieces of ice.
4) It is transparent or semitransparent ice having diameter 5mm or less.
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Hail:-
1) Its solid precipitation consist of large pellets or sphere of ice.
2) It is formed when updrafts carry raindrops upwards into extremely cold areas of
the atmosphere
3) It can vary in size, from the size of a small stone to that of a baseball (5 to
50mm).
4) Hailstorm are very destructive(destroy agricultural crop, & claim animal &
human life).
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Drizzle:- (Drops float in air)
1) The fall of numerous uniform minute droplets of water having diameter of less
than 0.5 mm is called drizzle.
2) Drizzles fall continuously but the total amount of water received on ground
surface significantly low.
3) Intensity is usually less than 0.1 mm/Hr.
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Glaze or Freezing Rain:-
1) It’s a form of precipitation which fall as rain and freezes when come in
contact with cold ground at around 00c.
2) Water drops freeze to form an ice coating but not a ice is also called as
freezing rain.
3) The water is “super cooled”.
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Measurement of Precipitation:-
1) Total amount of precipitation on a given area expressed as the depth of water
(if accumulated over horizontal projection of area).
2) Any form of precipitation, if falling as snow or ice is to be accounted for in
its melted form.
3) Since it is not physical possible to catch all the rainfall over a drainage basin,
it is only sampled by rainguages.
4) Rainguages catch in a perfect exposure & represents the precipitation falling
on their respective surrounding areas.
5) Pluviometer, Ombrometer, Hyetometer & Udometer are also sometimes
used to designate a rainguage.
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Types of Rain Guages:-
A. Non-recording Gauges:- (Symon’s Rainguage)
B. Recording Gauges:-
1) Tipping Bucket Type,
2) Weighing Bucket Type
3) Float Gauge
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For sitting a rain gauge the following considerations are important:
1) Site should be true representation of the area for which station is to give the
rainfall.
2) The ground must be level (open) and the instrument must kept on a
horizontal surface. ( not on sloping ground, terrace, etc)
3) The gauge must be set as near the ground to reduce wind effects but it must
be sufficiently high to prevent splashing, flooding, etc.
4) The instrument must be surrounded by an open fenced area of at least 5.5m
* 5.5m.
5) No object should be nearer to the instrument than 30m or twice the height
of obstruction.
6) Site should be easily accessible at all times.
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Non- Recording Gauges:-
1) It gives total depth of precipitation.
2) Commonly used non-recording gauge in India is Symon’s rain gauge.
3) It is installed in open area on a concrete foundation.
4) It should never be on terrace or under a tree.
5) The gauge may be fenced with a gate, to prevent animals & unauthorized
person from entering the premises.
6) Measurement are made at fixed time, every day normally at 8:30AM.
7) In case of heavy rainfall measurements are made as often as possible.
8) Last reading must be taken at 8:30AM & that will b rainfall of the day.
9) If the rainfall in a day is greater than 2.5cm then day is called rainy day.
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RECORDING TYPE
1) Recording gauges measure both the amount of rainfall and the intensity
i.e. rainfall depth w.r.t. time.
2) It produces a continuouss plot of rainfall against time & valuable short
duration data on intensity & duration of rainfall for hydrological analysis
of storm.
3) Commonly used recording raingauges are:
1. Tipping Bucket Type
2. Weighing Bucket Type
3. Float Type or Natural Syphon Type.
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Tipping Bucket Gauge:
1) The catch from the funnel falls onto one of pair of small buckets.
2) These buckets are so balanced that when 0.25mm of rainfall collects in
one bucket, it tips & brings the other one in position.
3) When rain that enters the equipment from the receiver reaches 0.25 mm, it
tips and empties its contents to the reservoir below.
4) The tipping actuates an electrically driven pen to trace a record on clock
work driven chart. (paper wrapped round a rotating cylinder)
5) The record from tipping bucket gives data on the rainfall intensity.
6) Main advantage is it gives electronic pulse output that can be recorded at
remote distance from the rain gauge.
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38TIPPING BUCKET TYPE:-
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WEIGHING BUCKET TYPE:-
1) The weighing type is suitable for measuring all types of precipitation (snow,
sleet, rain, etc.)
2) The catch from the funnel empties into a bucket mounted on a weighing
scale.
3) The weight of rainfall is recorded on the weighing scale.
4) Recorded data transmitted & plotted on a clockwork driven chart.
5) This instrument gives a plot of the accumulated rainfall against the elapsed
time.
6) Mass curve of rainfall (accumulated precipitation against time)
7) Slope of mass curve at any time is equal to intensity of rainfall at that time.
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WEIGHING BUCKET TYPE
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WEIGHING BUCKET TYPE
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Natural Syphon Type:-
1) This type of recording rainguage is also known as Float Type Gauge.
2) Rainfall collected by a funnel shaped collector is led into a float chamber
causing a float to rise.
3) As the float rises, a pen attached to the float through a lever system records
the elevation of the float on a rotating drum driven by a clockwork
mechanism.
4) A syphon arrangement empties the float chamber when the float has reached
a pre-set maximum level which resets the pen to its zero level.
5) This type of rainguage is adopted as the standard recording type rainguage in
India.
6) This type of rainguage gives a plot of the mass curve of rainfall.
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Other Precipitation Measurement Methods:-
1) Radar Measurement of Rainfall:-
Rainfall over large area can be determined through radar with good
degree of accuracy.
Radar emits a regular succession of pulse of electromagnetic radiation in
a narrow beam.
Doppler type radar measures the velocity & raindrops distribution.
2) Observation by Satellites:-
Global Precipitation Measurement (GPM) is a joint mission between
JAXA and NASA as well as other international space agencies to make
frequent (every 2–3 hours) observations of Earth’s precipitation.
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Rain Gauge Network:-
1) For proper assessment of water resources, a good network of rain gauges is a
must.
2) Rain gauge density is expressed as area covered per gauge.
3) The ratio of total area of the catchment to the total number of gauges in the
catchment is rainguage density.
4) For better accuracy, catchment area per gauge should be small.
5) As per IS: 4987-1968, the recommended rain gauge density is as:
1) In plains: 1 station per 520 km2 ( 2 per 1000 km2 )
2) Moderately elevated area: 1 station in 260 to 390 km2 ( 4 per 1000 km2 )
3) Hilly areas: 1 station in 130 km2 ( 8 per 1000 km2 )
6) At least 10% of the rain gauge stations should be equipped with automatic
rain gauges.
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Adequacy of Rain Gauge Station:-
1) Quantity of being sufficient or minimum number of station required to
represent a reliable picture of the intensity & duration of rainfall in the area is
called adequate number of station
2) Number of rain gauge station in a given catchment must be sufficient so that
the error in precipitation measurement is not more than acceptable value.
• Optimum number of stations:
2
vCN
Optimum number of stations
Coefficient of variation of the rainfall
values at the existing ‘m’ stations
Allowable degree of error in
the estimate of the mean
rainfall
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• If there are m stations in the catchment each recording rainfall values
P1,P2,…Pm in a known time, the coefficient of variation Cv is calculated as:
2
11
1
m
i
m
P P
m
1100 mvC
P
1
1 m
iP Pm
Coefficient of variation
Standard deviation
Mean precipitation
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Que. 2 There are four rain gauge stations existing in the catchment area of a
river. The average annual rainfall values at these stations are 750, 600, 420
& 500 mm respectively. Determine the optimum number of a rain gauges
in the catchment if it is desired to limit the error in the mean value of a
rainfall in the catchment to 10%. How many more gauges are required to
be installed?
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Que. 2 There are four rain gauge stations existing in the catchment area of a
river. The average annual rainfall values at these stations are 750, 600, 420
& 500 mm respectively. Determine the optimum number of a rain gauges
in the catchment if it is desired to limit the error in the mean value of a
rainfall in the catchment to 10%. How many more gauges are required to
be installed?
Sol:- m = 4
Mean Precipitation, Pˉ = (750 + 600 + 420 + 500) / 4 = 567.5 mm
σm-1 = √(((750- 567.5)2 + (600- 567.5)2 +(420- 567.5)2 + (500- 567.5)2 ) / (m-1))
σm-1 = 142.21 = standard deviation
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Que. 2 There are four rain gauge stations existing in the catchment area of a
river. The average annual rainfall values at these stations are 750, 600, 420
& 500 mm respectively. Determine the optimum number of a rain gauges
in the catchment if it is desired to limit the error in the mean value of a
rainfall in the catchment to 10%. How many more gauges are required to
be installed?
Sol:- m = 4
Pˉ = (750 + 600 + 420 + 500) / 4 = 567.5 mm
σm-1 = √(((750- 567.5)2 + (600- 567.5)2 +(420- 567.5)2 + (500- 567.5)2 ) / (m-1))
σm-1 = 142.21 = standard deviation
Coefficient of variation , Cv = (100 x σm-1 ) / Pˉ = (100 x 142.21) / 567.5 = 25.06
Optimum Number, N = (Cv / ɛ)2 = (25.06 / 10)2 = 6.28 say 7
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Que. 3 The annual rainfall depths recorded at existing 5 rain gauge stations (A,
B, C, D & E) in a drainage area are 88, 104, 138, 78 & 56 cm respectively.
What is the percentage accuracy of the existing rain gauge network in
estimation of the rainfall over the basin? If it is desired to estimate the
rainfall over the basin with not less than 90 % accuracy, is there a need to
install additional rain gauges? If so how many?
Sol:-
Limiting Percentage error = (Cv / √m) = 33.059 / √5 = 14.784 %
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Que. 3 The annual rainfall depths recorded at existing 5 rain gauge stations (A,
B, C, D & E) in a drainage area are 88, 104, 138, 78 & 56 cm respectively.
What is the percentage accuracy of the existing rain gauge network in
estimation of the rainfall over the basin? If it is desired to estimate the
rainfall over the basin with not less than 90 % accuracy, is there a need to
install additional rain gauges? If so how many?
Sol:- m = 5
Pˉ = (88 + 104 + 138 + 78 + 56) / 5 = 92.8 mm
σm-1 = √(((88- 92.8)2 + (104- 92.8)2 +(138- 92.8)2 + (78- 92.8)2 + (78- 92.8)2 ) / (5-1))
σm-1 = = standard deviation
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Que. 3 The annual rainfall depths recorded at existing 5 rain gauge stations (A,
B, C, D & E) in a drainage area are 88, 104, 138, 78 & 56 cm respectively.
What is the percentage accuracy of the existing rain gauge network in
estimation of the rainfall over the basin? If it is desired to estimate the
rainfall over the basin with not less than 90 % accuracy, is there a need to
install additional rain gauges? If so how many?
Sol:- m = 5
Pˉ = (88 + 104 + 138 + 78 + 56) / 5 = 92.8 mm
σm-1 = √(((88- 92.8)2 + (104- 92.8)2 +(138- 92.8)2 + (78- 92.8)2 + (78- 92.8)2 ) / (m-1))
σm-1 = = standard deviation
Coefficient of variation , Cv = (100 x σm-1 ) / Pˉ = 33.059
Limiting Percentage error = (Cv / √m) = 33.059 / √5 = 14.784 %
Percentage of accuracy = 100 – 14.784 = 85.216 %
Optimum Number, N = (Cv / ɛ)2 = (33.059 / 10)2 = 10.9289 say 11
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Que. 4 The annual rainfall depths recorded at existing 5 rain gauge stations (A,
B, C, D & E) in a drainage area are 87, 102, 106, 72 & 68 cm respectively.
What is the percentage accuracy of the existing rain gauge network in
estimation of the rainfall over the basin? If it is desired to estimate the
rainfall over the basin with not less than 93 % accuracy, is there a need to
install additional rain gauges? If so how many?
Sol:-
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Que. 4 The annual rainfall depths recorded at existing 5 rain gauge stations (A,
B, C, D & E) in a drainage area are 87, 102, 106, 72 & 68 cm respectively.
What is the percentage accuracy of the existing rain gauge network in
estimation of the rainfall over the basin? If it is desired to estimate the
rainfall over the basin with not less than 93 % accuracy, is there a need to
install additional rain gauges? If so how many?
Sol:- m = 5
Pˉ = (87 + 102 + 106 + 72 + 68) / 5 = 87 mm
σm-1 = √(((87- 87)2 + (102- 87)2 +(106- 87)2 + (72- 87)2 + (68- 87)2 ) / (m-1))
σm-1 = = standard deviation
Coefficient of variation , Cv = (100 x σm-1 ) / Pˉ = 19.675
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Que. 4 The annual rainfall depths recorded at existing 5 rain gauge stations (A,
B, C, D & E) in a drainage area are 87, 102, 106, 72 & 68 cm respectively.
What is the percentage accuracy of the existing rain gauge network in
estimation of the rainfall over the basin? If it is desired to estimate the
rainfall over the basin with not less than 93 % accuracy, is there a need to
install additional rain gauges? If so how many?
Sol:- m = 5
Pˉ = (87 + 102 + 106 + 72 + 68) / 5 = 87 mm
σm-1 = √(((87- 87)2 + (102- 87)2 +(106- 87)2 + (72- 87)2 + (68- 87)2 ) / (m-1))
σm-1 = = standard deviation
Coefficient of variation , Cv = (100 x σm-1 ) / Pˉ = 19.675
Limiting Percentage error = (Cv / √m) = 19.675 / √5 = 8.799 %
Percentage of accuracy = 100 – 8.799 = 91.20 %
Need of additional rain gauges with accuracy greater than 93 % ( 100 – 93 = 7%)
Optimum Number, N = (Cv / ɛ)2 = (8.799 / 7)2 = 7.9 say 8
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Normal Precipitation:-
1) The normal rainfall is the average value of rainfall at a particular date, month
or year over a specified 30-year period.
2) Like normal rainfall of 5th March or normal rainfall of January or yearly
rainfall.
3) 30 year normal rainfall are recomputed every decade to account for change in
environment and land use, because these factors may affect amount of
rainfall on that area.
4) Normal rainfall is used to find out the missing data of rain gauges.
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Preparation of Data:-
1) Before use of rainfall data for hydrological analysis of storm, water resources,
floods, etc. It is necessary to check the data for continuity and consistency.
2) Continuity: means availability of continuous record of previous rainfall.
3) Consistency: means that rainfall data of previous year should be consistent
with the present environmental condition. (like if there is forest in a particular
area which did not exist 15 years ago then previous records will not be
consistent with current record)
4) Continuity: missing data can be calculated by Arithmetic Mean Method &
Normal Ratio Method.
5) Consistency: consistency can be achieved & verified by Double mass curve
method or technique.
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Estimation of Missing Data:-
1) Sometimes a station has break in record due to absence of observer or
failure of the instrument.
2) It is necessary to estimate that missing data.
3) To estimate the data, three or more stations close to this station are
selected.
4) Following are the different methods to calculate the missing data:
1) Arithmetic Mean Method.
2) Normal Ratio Method.
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Arithmetic Mean Method:-
1) If the normal precipitation at each of these station is within 10% of that for the
station with missing data, then simple arithmetical mean of the precipitation of
those stations will give the value of missing station.
Where, Px, P1, P2,…,Pn = Rainfall at different station.
m = number of known rainfall station.
1 2 3
1...x mP P P P P
m
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Normal Ratio Method:-
If the normal precipitation at any of these selected stations is above 10% of
that for station with missing data then normal ratio method is used.
Where, Px, P1, P2,…,Pn = Rainfall at different station.
Nx, N1, N2…, Nn = normal rainfall
m = number of known rainfall station.
31 2
1 2 3
...x mx
m
N P PP PP
m N N N N
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Que. 5 the normal annual rainfall at station P, Q, R, S & T are 79.98, 68.59,
72.34, 82.87 & 93.09 cm respectively. In the year 1989 the station T was
inoperative & station P, Q, R, & S recorded annual precipitation of 89.89,
73.73, 79.29 & 84.63cm respectively. Estimates the rainfall at station T in
that year. Name the method which is to be used.
Sol:-
68
68
Que. 5 the normal annual rainfall at station P, Q, R, S & T are 79.98, 68.59,
72.34, 82.87 & 93.09 cm respectively. In the year 1989 the station T was
inoperative & station P, Q, R, & S recorded annual precipitation of 89.89,
73.73, 79.29 & 84.63cm respectively. Estimates the rainfall at station T in
that year. Name the method which is to be used.
Sol:- 93.09 x 1.1 = 102.399 & 93.09 x 0.9 = 83.781
Normal annual rainfall at missing station is not within 10% of the normal rainfall at
adjoining station. Hence, Normal ratio method is to be used to find out missing data.
Stations P Q R S T
Normal annual Average rainfall, cm 79.89 68.59 72.34 82.87 93.09
Annual Rainfall in 1989, cm 89.89 73.73 79.29 84.63 PT
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Que. 5 the normal annual rainfall at station P, Q, R, S & T are 79.98, 68.59,
72.34, 82.87 & 93.09 cm respectively. In the year 1989 the station T was
inoperative & station P, Q, R, & S recorded annual precipitation of 89.89,
73.73, 79.29 & 84.63cm respectively. Estimates the rainfall at station T in
that year. Name the method which is to be used.
Sol:- 93.09 x 1.1 = 102.399 & 93.09 x 0.9 = 83.781
Normal annual rainfall at missing station is not within 10% of the normal rainfall at
adjoining station. Hence, Normal ratio method is to be used to find out missing data.
PT = (NT/m) ((PP/NP) + (PQ/NQ) + (PR/NR) + (PS/NS))
Stations P Q R S T
Normal annual Average rainfall, cm 79.89 68.59 72.34 82.87 93.09
Annual Rainfall in 1989, cm 89.89 73.73 79.29 84.63 PT
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Que. 5 the normal annual rainfall at station P, Q, R, S & T are 79.98, 68.59,
72.34, 82.87 & 93.09 cm respectively. In the year 1989 the station T was
inoperative & station P, Q, R, & S recorded annual precipitation of 89.89,
73.73, 79.29 & 84.63cm respectively. Estimates the rainfall at station T in
that year. Name the method which is to be used.
Sol:- 93.09 x 1.1 = 102.399 & 93.09 x 0.9 = 83.781
Normal annual rainfall at missing station is not within 10% of the normal rainfall at
adjoining station. Hence, Normal ratio method is to be used to find out missing data.
PT = (NT/m) ((PP/NP) + (PQ/NQ) + (PR/NR) + (PS/NS))
PT = (93.09/4) x ((89.89/79.98) + (73.73/68.59) + (79.29/72.34) + (84.63/82.87))
PT = 100.45 cm
Stations P Q R S T
Normal annual Average rainfall, cm 79.89 68.59 72.34 82.87 93.09
Annual Rainfall in 1989, cm 89.89 73.73 79.29 84.63 PT
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Inconsistency of Records:-
1) Some of the common causes of inconsistency of the records are:
1) Shifting of station at new location.
2) Neighborhood of the station undergoing a marked changes.
3) Replacement of old station by new one.
4) Change of observer or change of observation method.
5) Change in the ecosystem.
6) Occurrence of observational error from certain date.
2) Inconsistency of record is corrected by using double mass curve.
3) Thus on correction previous records becomes consistent with present
condition.
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Consistency Verification by Double Mass Curve Method:-
1) To draw this curve, a group of station (say 10) is taken as a base
station in the neighborhood of the problematic station X.
2) The accumulated rainfall of station X (∑ Px ) & accumulated values
of average of group of base station (∑ PAV ) are calculated starting
from the latest record.
3) ∑ Px on Y-axis & ∑ PAV on X-axis.
4) In the plot, if a break in the slope is observed, it indicates a change in
precipitation of station X.
5) The values of precipitation at X beyond the break point are corrected
based on the slope of both the lines.
6) A change in a slope is normally taken as significantly only where it
persists for more than 5 years.
7) Correction factor = corrected slope / original slope = c / s
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75
75
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DOUBLE MASS CURVE
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Presentation of Rainfall Data:-
1) Rainfall data is presented in the form of:
1) Mass Curve
2) Hyetograph
3) Moving Average
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1) A graph showing the cumulative depth of rainfall against time is known as mass curve.
2) Hyetograph can be obtained from mass curve.
3) Time plotted in chronological order.
4) Mass curve of rainfall are very useful in extracting the information on the duration and
magnitude of a storm.
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Hyetograph:-
1) A graph showing the variation of rainfall intensity with time is called a
hyetograph.
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Hyetograph:-
1) A hyetograph is a plot of the average intensity of rainfall against time interval.
2) The hyetograph is derived from the mass curve and usually represented as a
bar chart.
3) The area under a hyetograph represents the total precipitation received in the
period.
4) Hyetograph is more preferred than mass curve since it is convenient to
determine the area of hyetograph.
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Moving Average:-
1) If we plot point rainfall with the time in chronological order the fluctuations
will be large in the time series of rainfall.
2) From this it will be difficult to determine the trend of the rainfall.
3) Thus a moving plot is made which smoothen out the fluctuation in time series
& help in determining trend of rainfall.
4) To find out moving average for say 3 years, average of rainfall of 1st, 2nd &
3rd years is plotted against 2nd year. Average of 2nd, 3rd & 4th years is plotted
against 3rd year & so on.
5) Similarly, for say 5 years moving average, average of rainfall of 1st, 2nd , 3rd,
4th & 5th years is plotted against 3rd year. Average of 2nd, 3rd , 4th , 5th & 6th
years is plotted against 4th year & so on.
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Calculation of Average Depth of Precipitation Over A Catchment:-
1) The precipitation over a catchment is actually measured as point values at a finite
number of precipitation stations. (Raingauge Station)
2) However hydrological analysis requires knowledge of rainfall over catchment
area and not of point rainfall.
3) Convert the point rainfall values at various stations into an average value over
catchment, several methods are available:
A. Arithmetic Mean Method
B. Thiessen Polygon Method
C. Isohyetal Method
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Arithmetic Mean Method:-
1) Gives equal weight or importance to all the rain gauges.
2) Quick , easy & yields fair result if rain gauges are distributed uniformly.
3) Rarely used bcoz least accurate method in normal condition.
4) Doesn’t account the rain gauges outside the catchment.
P1, P2,….,Pn = rainfall recorded at rain gauges.
P = Average rainfall in the catchment area.
N = number of raingauges in the catchment (n).
1 2 3
1
.... .... 1 Ni n
i
i
P P P P PP P
N N
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Thiessen Polygon Method:- (Weighted Mean Method)
1) Rainfall varies in intensity & duration from place to place, hence rainfall recorded
by each raingauge station should be weighted according to area.
2) Consider the representative area for each rain gauge.
3) Also consider the rain gauge outside catchment area.
4) Take care of non-uniform distribution of raingauges.
5) Suitable for plain area & elevation difference is not taken in consideration.
6) More accurate than arithmetic mean method.
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Steps for Thiessen Polygon Method:-
1) The basin area is plotted & location of the rain gauge stations are indicated.
2) Joining the adjacent rainguage station by straight lines to form triangle.
3) Bisecting the edges of the triangle to form Polygon.
4) Calculate area enclosed around each rain gauge station bounded by the polygons
edges to find out the area of influence corresponding to the rain gauge.
5) Polygon needs to calculated only once for a given distribution of rainguages
89
1 1 2 2 3 3 4 4 5 5 6 6
1 2 3 4 5 6
P A P A P A P A P A P AP
A A A A A A
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Isohyetal Method:-
1) Isohyetal is a line joining points of equal rainfall magnitude.
2) The area between two adjacent isohyets is determined by a planimeter.
3) This is most accurate method.
4) Topographic influence are taken into account.
5) New isohyets have to be made for each rainfall.
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Steps for Isohyetal Method:-
1) Prepare isohyetal map of the area from the rainfall values recorded at different
stations.
2) Measure the areas enclosed between successive isohyets by using planimeter
or graph paper.
3) Multiply each of these areas by average rainfall intensity between two
successive isohyets.
4) Compute average rainfall.
Where, P1, P2,…., Pn = the rainfalls recorded at rain gauge station
a1, a2,……an = area between successive isohyets.
2 3 11 21 2 1.......
2 2 2n n
n
P P P PP Pa a a
PA
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1) Greater than 5000 sq.km = Isohyetal method.
2) Hilly area = Isohyetal Method.
3) 500 – 5000 sq.km = Thiessen Polygon.
4) Plain Area = Thiessen Polygon.
5) Limited number of rain gauge station as compare to basin = Thiessen
polygon otherwise Isohyetal.
6) Less than 500 sq. Km = Arithmetic Mean.
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Que. 6. The influence area and the average annual rainfall observed of 5 rain
guage stations in a catchment of 30 km2 is as follows:
Find out the average annual rainfall of the catchment.
Sol:-
Stations A B C D E
Influence area in % 25 30 10 20 15
Annual average rainfall in cm 100 110 120 95 90
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Que. 6. The influence area and the average annual rainfall observed of 5 rain
guage stations in a catchment of 30 km2 is as follows:
Find out the average annual rainfall of the catchment.
Sol:-Total Area = 30 km2
Using Thiessen Polygon Method,
Stations A B C D E
Influence area in % 25 30 10 20 15
Annual average rainfall in cm 100 110 120 95 90
Stations Influence area, % Influence area, km2 Annual average rainfall
ABCDE
2530102015
7.59364.5
1001101209590
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Que. 6. The influence area and the average annual rainfall observed of 5 rain
guage stations in a catchment of 30 km2 is as follows:
Find out the average annual rainfall of the catchment.
Sol:-Total Area = 30 km2
Using Thiessen Polygon Method,
PAV = ((P1A1) + (P2A2) + (P3A3) + (P4A4) + (P5A5)) / (A1+A2+A3+A4+A5)
PAV = ((7.5 x 100) + (9 x 110) + (6 x 120) + (3 x 95) + (4.5 x 90)) / (30)
PAV = 102.50 cm
Stations A B C D E
Influence area in % 25 30 10 20 15
Annual average rainfall in cm 100 110 120 95 90
Stations Influence area, % Influence area, km2 Annual average rainfall
ABCDE
2530102015
7.59364.5
1001101209590
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Que. 7. Find the mean precipitation for a rectangular area ABCD having AB =
10 km and BC = 5 km. The precipitation recorded at four corner are A
(10cm), B (12cm), C(15cm) & D(11cm) by arithmetic mean method &
Thiessen polygon method.
Sol:-
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Que. 7. Find the mean precipitation for a rectangular area ABCD having AB =
10 km and BC = 5 km. The precipitation recorded at four corner are A
(10cm), B (12cm), C(15cm) & D(11cm) by arithmetic mean method &
Thiessen polygon method.
Sol:-
Arithmetic Mean Method,
Pm = (P1 + P2 + P3 + P4) /4
Pm = (10 + 12 + 15 + 11) /4
Pm = 12 cm
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Que. 7. Find the mean precipitation for a rectangular area ABCD having AB =
10 km and BC = 5 km. The precipitation recorded at four corner are A
(10cm), B (12cm), C(15cm) & D(11cm) by arithmetic mean method &
Thiessen polygon method.
Sol:-
Arithmetic Mean Method,
Pm = (P1 + P2 + P3 + P4) /4
Pm = (10 + 12 + 15 + 11) /4
Pm = 12 cm
Using Thiessen Polygon Method,
PAV = ((P1A) + (P2B) + (P3C) + (P4D)) / (A1+A2+A3+A4)
PAV = ((10 x 12.5) + (12 x 12.5) + (15 x 12.5) + (11 x 12.5)) / (50)
PAV = 12 cm
Stations Rainfall, cm Figure Area, km2
ABCD
10121511
AEOGGOFBFCHOOHDE
12.512.512.512.5
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EXAMPLE
100
10
0
EXAMPLE
101
10
1
EXAMPLE
102
10
2
EXAMPLE
103
10
3
EXAMPLE
104
105
10
5
INTENSITY-DURATION-FREQUENCY
10
6
SUMMARY
1) Explain in brief the importance of temperature, humidity and wind in hydrological studies.
2) What are various factors affecting precipitation?
3) Explain hydrological cycle with the help of neat sketch.
4) Enlist the various forms of precipitation and explain any one of it.
5) What is hydrological equation? Describe briefly the components of this equation.
6) Explain in brief the different types of precipitation according to the factors responsible for lifting of air
mass.
7) Define rain gauge. what are the different types of rain gauges? describe with neat sketch the principle of
working of “Tipping Bucket Type “ recording rain gauge.
8) How do you measure rainfall using radar?
9) Explain Intensity-Duration-Frequency and Depth-Area-Duration analysis.
10) Explain data missing records, with the help of an example.
11) The annual rainfall measured over a catchments having an existing rain gauge network of six rain
gauge stations are as follows:- Compute the % accuracy of the following rain gauge network
Station Annual rainfall in (mm)
A 826
B 1029
C 1803
D 1103
E 988
F 1367
.
12) The isohyetal weights of the stations in and around a river basin are
0.30,0.04,0.12,0.08,0.25,0.05,0.10,0.06 respectively. Station 3,5,7 lieoutside the basin while the remaining
are inside. The rainfalls recorded at these stationsare 123,234,321,195,252,274,281,246 respectively.
Determine the average depth of rainfall over the catchment by arithmetic and isohyetal mean methods.
TYPICAL QUESTIONS
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