Hydrology and Water Management 1

7/28/2019 Hydrology and Water Management 1

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Hydrology and water management (LEC 1A

& LEC 1B)

HydrologyHydrology deals with the origin,

occurrence, circulation,

distribution, the physical and chemical properties of water

and its interaction with living organisms.

Hydrology is an essential field of science since everything

from tiny organisms to individuals to societies to the whole

of civilization - depends so much on water.

Water management

The activity of planning, developing, distributing and

managing the optimum use of water resources.

Branches of Hydrology

Hydrology can generally be divided into two main branches:

o Engineering Hydrology

(Planning, design and Operation of Engineering projects for

the control and use of water)

o Applied Hydrology

(Hydrological cycle, precipitation, runoff, relationshipbetween precipitation and runoff, hydrographs, Flood Routing)

Hydrology can be divided into the following branches

* Chemical Hydrology : Study of chemical characteristics ofwater.


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* Ecohydrology : Interaction between organisms and thehydrological cycle.

*Hydrogeology : Also referred to as geohydrology, is the studyof the presence and movement of ground water.

* Hydroinformatics : is the adaptation of informationtechnology to hydrology and water resource applications.=

* Hydrometeorology : is the study of the transfer of water andenergy between land and water body surfaces and the lower

atmosphere.

* Isotope Hydrology : is the study of isotropic signatures ofwater (origin and age of water).

* Surface water Hydrology : is the study of hydrologicprocesses that operate at or near earths surface.

* Ground water Hydrology : is the study of underground water.=

* Drainage basin management : covers water storage in the formof reservoir and flood protection.

* Water quality : includes the chemistry of water in rivers andlakes, both of pollutants and natural solutes.

Applications of Hydrology

o Determining the water balance of a region.

o Determining the agricultural water balance.


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o Designing riparian restoration projects.

o Mitigation and predicting floods, landslides and droughtrisk.

o Flood forecasting and flood warnings.

o Designing irrigation schemes and managing agriculturalproductivity.

o Designing dams for water supply or hydroelectric powergeneration.

o Designing bridges.

o Designing sewers and urban drainage system.

o Predicting geomorphologic changes, such as, erosion orsedimentation.

o Assessing the impact of natural and anthropogenicenvironmental change.

o Assessing containment transport risk and establishingenvironmental policy guidelines

Hydrological cycle


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Hydrological cycle also known as Water cycle, describes the

continuous Movement of water on, above and below the surface

of the earth.


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The main processes involved in hydrological cycle are

* Evaporation

* Condensation

* Precipitation

* Interception

* Infiltration

* Percolation

* Transpiration

* Runoff

*Storage

The water cycle begins with the evaporation of water from

oceans and other water bodies. The resulting vapors are

transported by moving air and under proper conditions, the

vapor are condensed to form clouds, which in turn results in

precipitation.

The precipitation which falls upon land is dispersed in

several ways. The greater part is temporarily retained in the

soil near where it falls and is ultimately returned to the

atmosphere by evaporation and transpiration by plants. A

portion of the water flows over surface soil to stream

channels, while other penetrates into the ground to become

part of the ground water. Under the influence of gravity,

both surface and underground water move towards lower

elevations and may eventually discharge into the oceans.


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o This Hydrologic Cycle recycles the earths valuable watersupply. In other words, the water keeps getting reused over

and over.

o Just think, the next glass of water you drink could have beenused by a dinosaur in the Mesozoic Era one hundred million

years ago!

o Water in that glass could have been a liquid, a solid, and agas countless times over thanks to the water cycle.

Precipitation

The term precipitation as used in hydrology is meant for all

forms of moisture emanating from the clouds and all forms of

water like rain, snow, hail and sleet derived from

atmospheric vapors, falling to the ground.

Precipitation is one of the most important events of

hydrology. Floods and droughts are directly related to the

occurrence of precipitation. Water resources management,

water supply schemes, irrigation, hydrologic data for designof hydraulic structures and environmental effects of water

resources development projects are related to precipitation

in one way or the other. So it is important to study various

aspects of precipitation.

Forms of Precipitation

* Drizzle: These are the minute particles of water at start ofrain. These consist of water drops under 0.5 mm diameter and

its intensity is usually less than 1.0 mm/hr. Their speed is

very slow and we cannot even feel them. Therefore they cannot

flow over the surface but usually evaporate.


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* Rain: It is form of precipitation in which the size of dropsis more than 0.5 mm and less than 6.25 mm in diameter. It can

produce flow over the ground and can infiltrate and

percolate. Both the duration as well as rate of rainfall are

important. If the rainfall per unit time is greater than therate of infiltration, the rain water can flow over the

surface of earth.

* Glaze: It is the ice coating formed on drizzle or rain dropsas it comes in contact with the cold surfaces on the ground.

* Sleet: Sleet is frozen rain drops cooled to the ice stagewhile falling through air at subfreezing temperatures.

* Snow: Snow is precipitation in the form of ice crystalsresulting from sublimation i.e. change of water vapor

directly to ice.

* Snowflakes: A snowflake is made up of a number of icecrystals fused together.

* Hail: Hail is the type of precipitation in the form of ballsor lumps of ice over 5 mm diameter formed by alternate

freezing and melting as they are carried up and down by

highly turbulent air currents. The impact of these is also

more. A single hailstone weighing over a pound has been

observed.


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Factors influencing Precipitation formation

o A lifting mechanism to produce cooling of the air.

o A mechanism to produce condensation of water vapors andformation of cloud droplets.

o A mechanism to produce growth of cloud droplets to sizecapable of falling to the ground against the lifting force of

air.


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* Mechanism of cooling

* When air ascends from near the surface to upper

Levels in the atmosphere it cools.

* This is the only mechanism capable of producing the degreeand rate of cooling needed to account for heavy rainfall.

* Cooling lowers the capacity of a given volume of air to holda certain amount of water vapor.

* As a result super saturation occurs and the excess moistureover saturation condenses through the cooling process.

o Condensation of water vapor

* Condensation of water into cloud droplets takes place onhygroscopic nuclei which are small particles having an

affinity for water.

* The source of these condensation nuclei are the particles ofsea salt or products of combustion of certain sulfurous and

nitrous acid and carbon dioxide.

* There are always sufficient nuclei present in the atmosphere.


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o Growth of Droplet

* Growth of droplets is required if the liquid water present inthe cloud is to reach the ground. The two processes regardedas most effective for droplet growth are:

* Coalescence of droplets through collision due todifference in speed of motion between larger and smaller

droplets.

* Co-existence of ice crystals and water droplets.

* Co-existence effect generally happens in the temperaturerange from 100 to 20o F.

If in a layer of clouds there is a mixture of water droplets

and ice crystals, the saturation vapor pressure over ice is

lower than that over water. This leads to the evaporation of

water drops and condensation of much of this water on ice

crystals causing their growth and ultimate fall through the

clouds. This effect is known as Bergerons effect.

* The ice crystals will further grow as they fall and collidewith water droplets.

Classification of Precipitation Based on the Lifting

Mechanism


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The precipitation is often classified according to the factor

responsible for lifting of air to higher altitudes. Following

are the various types of precipitation based on this

classification.

* Convectional Precipitation

* Orographic Precipitation

* Cyclonic Precipitation

o Convectional Precipitation

* The main cause of Convectional precipitation is thermalconvection of the moisture laden air (rising of warmer,

lighter air in colder, denser surroundings).

* A major portion of the solar radiation is utilized in heatingthe earth. As the earth conducts heat slowly, the heat

accumulates at the surface of the earth and air which comes

in its contact is heated up and the lapse rate near the

surface of the earth increases rapidly. With the passage of

time as the sun gets higher and higher the lapse rate

increases further and air becomes unstable.

* Vertical currents are then set up which carry heat and themoisture laden air is picked up from the surface to higher

levels. Due to convection, the moist air in the lower levels

of the atmosphere rises up to the condensation level where

clouds develop and with further convection these clouds

finally grow resulting in a thunderstorm.

o Orographic Precipitation

* In the orographic precipitation, expansion and condensationoccurs because moisture laden air masses are lifted by

contact with orographic (mountain) barriers.


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* This type of precipitation is most pronounced on the windwardside of mountain range, generally heaviest precipitation

occurs where favorable orographic effects are present.

* Orographic precipitation also occurs in the inland areaswhere mountain ranges rise above the surrounding areas in the

path of the moisture laden air masses.

o Cyclonic precipitation

* Precipitation in plain regions is generally cyclonic incharacter.

* Cyclonic precipitation results from the lifting of airconverging into a low-pressure area or cyclone.

* Cyclonic precipitation can be frontal or nonfrontal.

* Frontal precipitation results from the lifting of warm air onone side over a colder denser air on the other side.

* Warm-front precipitation is formed in the warm air advancingupward over a cold air mass.

* Cold-front precipitation is formed in the warm air is forcedupward by an advancing mass of cold air.


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* In the Indo-Pak Subcontinent, the cyclonic storms form in theBay of Bengal in different months. During April, May and Junemost of these storms do not reach Pakistan. But some of them

affect Bangladesh and give very heavy rainfall there. During

the summer monsoon season, the cyclonic storms reach Pakistan

and are fed with moisture from the Arabian sea resulting in

heavy rainfall over the Northern areas of Pakistan. In

September, October and November these storms are very

destructive in Bangladesh. Such storms cause considerable

loss of life and property over the coastal districts.

Cyclonic storms also form in Arabian sea but their number is

far less.

Measurement of precipitation

o All forms of precipitation are measured on the basis of thevertical depth of water that would accumulate on a level

surface if the precipitation remained where it fell. The

amount of precipitation is measured in units of length

(millimeters/inches).

oThe precipitation is measured by rain gauges/precipitationgauges. There are two types of rain gauges.

* Non-recording rain gauge (standard rain gauge)

* Recording rain gauge


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* The main difference between these rain gauges is that withthe help of recording rain gauges we get the rain recorded

automatically with respect to time, so intensity of rain fall

is also known whereas an observer has to take readings from

non recording rain gauge for rain and he has to record thetime also, for calculation of intensity of rain fall.

* Non-Recording/Standard gauge

The standard gauge of U.S. Weather Bureau has a collector of 200

mm diameter and 600 mm height. Rain passes from a collector

into a cylindrical measuring tube inside the overflow can.

The measuring tube has a cross sectional area 1/10th of the

collector, so that 2.5 mm rain fall will fill the tube to 25

mm depth. A measuring stick is marked in such a way that

1/10th of a cm depth can be measured. In this way net rainfall

can be measured to the nearest 1 mm. The collector and tube

are removed when snow is expected. The snow collected in the

outer container or overflow can is melted, poured into the

measuring tube and then measured. This type of rain gauge is

one of the most commonly used rain gauges.

Standard rain gauge(Non Recording)


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o Sources of Error

* Some water is used to wet the surface of instrument.

* The rain recorded may be less than the actual rainfall due tothe direction of the rainfall as affected by wind.

* Dents in the collector and tube may also cause error.

* Some water is absorbed by the measuring stick.

* Losses due to evaporation can also take place.

* The volume of stick replaces some water which causes someerror.


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o Recording Rain gauge

Recording rain gauges can be divided into the following types:

* Float type

* Weighing type

* Tipping bucket type

o Float Type Rain Gauge

This type of rain gauge also has a receiver and a float chamber

along with some recording mechanism or arrangement. In thistype the rain is led into a float chamber containing a light,

hollow float. The vertical movement of the float as the level

of water rises is recorded on a chart with the help of a pen

connected to float. The chart is wrapped around a rotating

clock driven drum. To provide a continuous record for 24

hours the float chamber has either to be very large, or some

automatic means are provided for emptying the float chamber

quickly when it becomes full, the pen then returning to the

bottom of the chart. This is usually done with some sort of

siphoning arrangement. This arrangement activates when the

gauge records a certain fixed amount of rain (mostly 10 mm of

rainfall.). Snow cannot be measured by this type of rain

gauge.


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Float type rain gauge

o Weighing Type Rain Gauge

The weighing type rain gauge consists of a receiver, a bucket, aspring balance and some recording arrangement. The weighing

type gauge weighs the rain or snow which falls into a bucket

which is set on a lever balance. The weight of the bucket and

content is recorded on a chart by a clock driven drum. The

record is in the form of a graph, one axis of which is in

depth units and the other has time. The records show the

accumulation of precipitation. Weighing type gauges operate

from 1 to 2 months without stopping. But normally one chart

is enough only for 24 hours. This type of rain gauge has

advantage of measuring snow also. The receiver is removed

when snow is expected.

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o Tipping Bucket Type Rain Gauge

This type of gauge is equipped with a remote recorder locatedinside the office which is away from the actual site. The

gauge has two compartments pivoted in such a way that one

compartment receives rain at one time. A certain amount of

rain (usually 0.25 mm fills one compartment and over

balances it so that it tips, emptying into a reservoir and

bringing the second compartment of the bucket into place

beneath the funnel of receiver. As the bucket is tipped by

each 0.25 mm of rain it actuates an electrical circuit,

causing a pen to mark on a revolving drum. This type of gauge

is not suitable for measuring snow without heating the

collector. Plotting is similar to that of other recording

rain gauges.


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o Sources of Error

* Dents in the collector.

* Moistening of inside-surface of the funnel and the tube.

* Rain drops splashing from the collector.

* For very intense rain some water is still pouring into thealready filled bucket.

* Inclination of the gauge may result in catching less or morerain than the actual amount.

* Error in measurement due to wind.

o Remedial measures for Error in Precipitationmeasurement

* Removal of error due to dents obviously needs repair of theinstrument. For rain recorded with dents a correction should

be applied.

* Errors such as moistening of the inside surfaces of thegauge, splashing of rainwater from the collector and pouring

of water into the already filled bucket during an intenserain can only be corrected by some correction factor.

* Inclined instrument needs to be reinstalled. The correctionfactor however can be calculated from the angle of

inclination.


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* For wind protection certain wind shields are designed andused which are called Splash Guards. Proper setting of gauge

above ground level is necessary.

Example:

A rain gauge recorded 125 mm of precipitation. It was found

later that the gauge was inclined at an angle of 20 degree

with the vertical. Find the actual precipitation.

Solution:

P(measured) = 125 mm

Angle of inclination () = 20o with the vertical

P(actual) = P(measured)/cos() = 125/cos20o

= 133 mm

o Measurement of precipitation by Radar

This is a modern technique for measurement of rainfall rate. It

can also detect local movement of areas of precipitation. The

electromagnetic energy released and received back by radar is

a measure of rainfall intensity. The measurement isappreciably affected by trees and buildings. However extent

of rainfall can be estimated with reasonable accuracy. Use of

radar is useful where number of rain gauges installed in an

area is not sufficient.


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Rain Gauge Network

The number of rain gauges and their distribution affect the

nature of collected precipitation data. The larger the number

of rain gauges the more representative will be the data

collected. But on the other hand we have to observe other

factors also, like economy of the project, accessibility of

certain areas and topography of the area. So, one has to look

for some optimum solution. In this regard the World

Meteorological Organization (WMO) has made following

recommendations for minimum number of rain gauges in a

catchment:

oIn comparatively flat regions of temperate, Mediterranean and

Tropical Zones, the ideal is at least one station for 230

345 sq. miles. However one station for 345 1155 sq. miles

is also acceptable

o In mountainous regions of Temperate, Mediterranean andTropical Zones, the ideal is at least one station for 35 95

sq. miles. However one station for 95 385 sq. miles is also

acceptable.

o In arid and polar zones, one station for 575 3860 sq. milesis acceptable.


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Analysis of Precipitation data

o Estimation of missing precipitation data

o Consistency of precipitation data or Double Mass Index

1. Estimation of missing Precipitation data


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Some precipitation stations may have short breaks in the records

because of absence of the observer or because of instrumental

failures. It is often necessary to estimate this missing record.

o In the procedure used by the U.S. Weather Bureau, the missingprecipitation of a station is estimated from the observations

of precipitation at some other stations as close to and as

evenly spaced around the station with the missing record as

possible.

o The station whose data is missing is called interpolationstation and gauging stations whose data are used to calculate

the missing station data are called index stations.

There are two methods for estimation of missing data.

o Arithmetic mean method

oNormal ratio method

* Simple Arithmetic Mean method

According to the arithmetic mean method the missing

precipitation Px is given as:

Pi

ni

in=

=1

1

Pi

ni

in

=

= 1

1


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Px=

Where n is the number of nearby stations, Piis precipitation

at ith station and Px is missing precipitation.

In case of three stations 1, 2 and 3,

Px = (P1 + P2 + P3)/3

Naming stations as A, B and C instead of 1, 2 and 3

Px = (Pa + Pb + Pc)/3

Where Pa , Pb and Pc are defined above.

* Normal Ratio method

According to the normal ratio method the missing precipitation

is given as:

PNN i

ni

ii

x

n=

=1

1

PN

Ni

ni

ii

x

n=

=1

1


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Px =

Where Px is the missing precipitation for any storm at the

interpolation station x, Pi is the precipitation for the

same period for the same storm at the ith station of a

group of index stations, Nx the normal annual precipitation

value for the x station and Ni the normal annual

precipitation value for ith station.

o For example, for the symbols defined above for three indexstations in a catchment area.

][3

1

3

3

2

2

1

1 PN

N

PN

N

PN

N xxx

++


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Px=

o If the normal annual precipitation of the index stationslies within 10% of normal annual precipitation of

interpolation station then we apply arithmetic mean method to

determine the missing precipitation record otherwise the

normal ratio method is used for this purpose.

Consider that record is missing from a station X.

Now let,

N = Normal annual precipitation. (Mean of 30 years of annual

precipitation data)

P = Storm Precipitation.

Let Px be the missing precipitation for station X and Nx , the

normal annual precipitation of this station, Na, Nb and Nc are

normal annual precipitations of nearby three stations, A, B

and C respectively while Pa, Pb and Pc are the storm

precipitationof that period for these stations.

Now we have to compare Nx with Na , Nb and Nc separately. If

difference of Nx - Na, Nx - Nb, Nx - Nc is within 10% of Nx then

we use simple arithmetic mean method otherwise the normal

ratio method is used.

o Example : Find out the missing storm precipitation of stationC given in the following table:

Station A B C D E

][3

13

3

2

2

1

1

PN

NP

N

NP

N

N xxx++


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Storm precipitation (cm) 9.7 8.3 ---- 11.7 8.0

Normal Annual precipitation (cm) 100.3 109.5 93.5 125.7 117.


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o Solution:

In this example the storm precipitation and normal annual

precipitations at stations A, B, D and E are given and

missing precipitation at station C is to be calculated

whose normal annual precipitation is known. We will determine

first that whether arithmetic mean or normal ratio method isto be applied.

10% of Nc = 93.5x10/100 = 9.35

After the addition of 10% of Nc in Nc, we get 93.5 + 9.35=102.85

And by subtracting 10% we get a value of 84.15

So Na, Nb, Nd or Ne values are to be checked for the range 102.85

to 84.15.


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If any value of Na, Nb, Nd or Ne lies beyond this range, then

normal ratio method would be used. It is clear from data in

table above that Nb, Nd and Ne values are out of this range so

the normal ratio method is applicable here, according to

which:

PN

Ni

ni

ii

x

n=

=1

1

PN

Ni

ni

ii

x

n=

=1

1


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Px =

Pc= (1/4 )(93.5 x 9.7/100.3+ 93.5 x 8.3/109.5+ 93.5 x

11.7/125.7+ 93.5 x 8.0/117.5) = 7.8 cm

o Example

Precipitation station X was inoperative for part of a month

during which a storm occurred. The storm totals at three

surrounding stations A, B and C were respectively 10.7, 8.9

and 12.2 cm. The normal annual precipitation amounts at

stations X, A, B and C are respectively 97.8, 112, 93.5 and

119.9 cm. Estimate the storm precipitation for station X.

oSolution

Pa = 10.7 cm Na = 112 cm

Pb = 8.90 cm Nb = 93.5 cm

Pc = 12.2 cm Nc = 119.9 cm

Px = ? Nx = 97.8 cm

10% of Nx = 97.8 x 10/100 = 9.78 cm.

* Nx - Na = 97.8 - 112 = -14.2 cm More than + 10% of Nx (noneed of calculating Nx Nb and Nx Nc


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Px = (1/3)( 97.8x 10.7/112+ 97.8x 8.90 /93.5 + 97.8x 12.2 /

119.9)

Px = 9.5 cm

2. Consistency of Precipitation Data or DoubleMass Analysis

oIn using precipitation in the solution of hydrologicproblems, it is necessary to ascertain that time trends in

the data are due to meteorological changes. Quite frequently

these trends are the result of the changes in the gauge

location, changes in the intermediate surroundings such as

construction of buildings or growth of trees, etc. and

changes in the observation techniques.

o Due to such changes the data might not be consistent. Theconsistency of the record then is required to be determined

and the necessary adjustments be made. This can be achievedby the method called the double mass curve technique.

o The double mass curve is obtained by plotting the accumulatedprecipitation at the station in question along X-axis and the

average accumulated precipitation of a number of other nearby

stations which are situated under the same meteorological

conditions along Y-axis. If the curve has a constant slope,

the record of station X is consistent. However, if there

is any break in the slope of the curve, the record of the

station is inconsistent and has to be adjusted by theformula.

Pa = (Sa / So)x Po

Where Pa = Adjusted precipitation.

Po = Observed precipitation.


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Sa = Slope prior to the break in the curve

So = Slope after the break in the curve.

All values after break are to be adjusted.

o Example

Check consistency of the data given in table below and adjust

it if it is found to be inconsistent.

Year Annual

precipitation at x (mm)

Mean of annual

precipitation of20 surrounding

stations (mm)

Year Annual

precipitation at

x (mm)

Mean of annual

precipitation of20 surrounding

stations (mm)

1972 188 264 1954 223 360

1971 185 228 1953 173 234

1970 310 386 1952 282 333

1969 295 297 1951 218 236

1968 208 284 1950 246 251

1967 287 350 1949 284 284

1966 183 236 1948 493 361

1965 304 371 1947 320 282

1964 228 234 1946 274 252

1963 216 290 1945 322 274

1962 224 282 1944 437 302

1961 203 246 1943 389 350


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1960 284 264 1942 305 228

1959 295 332 1941 320 312

1958 206 231 1940 328 284

1957 269 234 1939 308 315

1956 241 231 1938 302 280

1955 284 312 1937 414 343

o Solution

A double mass curve is plotted by taking cumulative of

average precipitation of surrounding stations along x-axis

and accumulative precipitation of station X along

y-axis for which consistency of data is being investigated.

The double mass curve is shown in Figure

Year Cummulative Annual

precipitation at x (mm)

Cummulative ppt: of 20

srndg, stations (mm)

Corrected

Precipitation

Remarks

1972 188 264 188 No

coro1971 373 492 185

1970 683 878 310

1969 978 1175 295

1968 1186 1459 208

1967 1473 1809 287


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1966 1656 2045 183

1965 1960 2416 304

1964 2188 2650 228

1963 2404 2940 216

1962 2628 3222 224

1961 2831 3468 203

1960 3115 3732 284

1959 3410 4064 295

1958 3616 4295 206

1957 3885 4529 269

1956 4126 4760 241

1955 4410 5072 284

1954 4633 5432 223

1953 4806 5666 173

1952 5088 5999 282

1951 5306 6235 218

1950 5552 6486 246

1949 5836 6770 190.8 Precipit

n Sta

'X

0

1948 6329 7131 345

1947 6649 7413 224


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1946 6923 7665 192

1945 7245 7939 225.4

1944 7682 8241 306

1943 8071 8591 272.3

1942 8376 8819 213.5

1941 8696 9131 224

1940 9024 9415 229.6

1939 9332 9730 215.6

1938 9634 10010 211.4

1937 10048 10353 290


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Where

Pa = (Sa / So)x Po

Pa = Adjusted precipitation.

Po = Observed precipitation.

Sa = Slope prior to the break in the curve

So = Slope after the break in the curve.

o The correction for slope is applied to readings beyond breakin slope. The calculations are shown in table.

Slope of 1st line = Sa = 0.854

Slope of deviating line = So = 1.176

Correction to values (multiplying factor) = 0.854/1.176 =0.70

So up to 1950 no correction is required. Before 1950 all

readings are multiplied by slopes ratio of 0.7 to get

corrected precipitation. Note that data in latter interval

(1973-1950) is considered more authentic so kept in initial

reach of the graph.


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Estimation of Average Precipitation over a Basin

oTo find out runoff from a catchment and most of otherhydrologic analyses, it is important to know the average

precipitation of a certain part of catchment or for the whole

of the catchment area. To find out average precipitation of

watershed, records of precipitation from different rain gauge

stations is used. There are many factors which affect the

reliability of average precipitation of watershed determined

by using the data from individual stations in the watershed.

o For example : the total number of rain gauges and theirdistribution in the catchment (larger the number of raingauges, the reliable will be the calculated average

precipitation), the size and shape of area of catchment,

distribution of rainfall over the area and topography of the

area and the method used for calculating average

precipitation.

o There are three methods to find average precipitation over abasin. Accuracy of estimated average precipitation will

depend upon the choice of an appropriate method. These

methods are described below:

o Arithmetic Mean Method

o Thiessen Polygon Method

o Isohyetal Method


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o Arithmetic Mean Method

In this method the average precipitation over an area is the

arithmetic average of the gauge precipitation values. We take

data for only those stations which are within the boundary.This is the simplest method but is applicable only for flat

areas and not for hilly areas i.e. this method is used when:

o Basin area is flat.

o All stations are uniformly distributed (withinpractical limits) over the area.

o The rainfall is also nearly uniformly distributed overthe area.

According to this method

=

n

i

iPn 1

1

P (average) =

Or Pav = [P1+P2+P3++Pn]/n

Where Pi is precipitation at station i and there are n

number of gauges installed in the catchment area from where

the data has been collected.

=

n

i

iPn 1

1


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o Example

Six rain gauges were installed in a relatively flat area and

storm precipitation from these gauges was recorded as 3.7,

4.9, 6.8, 11.4, 7.6 and 12.7 cm respectively from gauges 1,2, 3, 4, 5, and 6. Find average precipitation over the

catchment.

o Solution

As the area is relatively flat so we apply the arithmetic

mean method. According to arithmetic mean method

P(average) = (3.7 + 4.9 + 6.8 + 11.4 + 7.6 + 12.7)/6 = 7.85

cm.


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o Thiessen Polygon Method

The fundamental principle followed in this method consists of

weighing the values at each station by a suitable proportion

of the basin area. In this method, a special weighing factoris considered.

The following steps are used to determine average

precipitation by Thiessen Polygon Method.

* Draw the given area according to a certain scale and locatethe stations where measuring devices are installed.

o Join all the stations to get a network of non-intersectingsystem of triangles.

o Draw perpendicular bisectors of all the lines joining thestations and get a suitable network of polygons, each

enclosing one station. It is assumed that precipitation over

the area enclosed by the polygon is uniform.

o Measure area of the each polygon.

o Calculate the average precipitation. For the whole basin bythe formula.

P (average) = (P1 A1 + P2 A2 + ...........+ Pn An)/A

Where,

P1 = Precipitation. at station enclosed by polygon of area A1

P2 = Precipitation. at station enclosed by polygon of area A2

and so on

Pn =Precipitation. at station enclosed by polygon of area An

And A represents the total area of the catchment.


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o Example : Following is shown map of a catchment having 6rainfall recording stations. Find the Average Precipitation

over the wholecatchment.

The precipitation and polygon area are given below.

Solution: The calculations are done in tabular form

Station Precipitation (mm) Polygon Area (km)


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Daggar 48 5,068.76

Besham 33 4,349.17

Shinkiari 25 1,399.25

Phulra 32 1,693.80

Tarbela 56 2,196.33

Oghi 30 2,234.29

=

=

=

=ni

i

i

Ai

1

ni

1

PiAi


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Mean Precipitation = =

666.02x106x10/16941.60x106

Mean Precipitation = 39.3 mm

* Example : From the data given in Table below, which wasobtained from Thiessen Polygon map of a catchment, find out

the average precipitation of the catchment.

* Solution : According to Thiessen Polygon Method


The calculations are shown in tabular form in Table.

Station Precipitation P (mm) Polygon Area A (km) P x A (x106 m)

Daggar 48 5,068.76 243.30

Besham 33 4,349.17 143.52

Shinkiari 25 1,399.25 34.98

Phulra 32 1,693.80 54.20

Tarbela 56 2,196.33 122.99

Oghi 30 2,234.29 67.03

=

=

=

=

ni

i

i

Ai

1

ni

1

PiAi


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Total 16,941.60 666.02

P (average) = 27890.20 2561=10.9 cm

* Example : There are 10 observation stations, 7 being insideand 3 in neighborhood of a catchment. Thiessen Polygons were

drawn for a storm data from these observation stations and

the data given in Table below was obtained. Find out the

average precipitation of the catchment.

Sr No Gauge

precipitat

ion (cm)

Area of

Thiessen

Polygonenclosing

the

station

(sq. km)

Sr No Gauge

precipitat

ion (cm)

Area of

Thiesse

Polygonenclosin

the

station

(sq. km)

1 10.2 416 4 9.4 520

2 8.1 260 5 15.2 390

3 12.7 650 6 7.6 325


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* Solution : According to Thiessen Polygon Method


P (average) = 5157.5 1200=4.3 cm

Isohyetal Method

The most accurate method of averaging precipitation over an area is the isohyetal

method.

For estimation of average precipitation of the catchment by isohyetal method the

following steps are used:

1. Draw the map of the area according to a certain scale.

2. Locate the points on map where precipitation measuring gauges are installed.

3. Write the amount of precipitation for stations.

4. Draw isohyets (Lines joining points of equal precipitation).

5. Measure area enclosed between every two isohyets or the area enclosed by an isohyet and

boundary of the catchment.

6. Find average precipitation by the formula.


Where,

P1= Mean precipitation of two isohyets 1 and 2

A1= Area between these two isohyets.

P2 = Mean precipitation of two isohyets 2 and 3


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A2 = the area b/w these two isohyets.

and, so on

Pn = Mean precipitation of isohyets n-1 and n

An = the area between these two isohyets.

It may be noted that the last and first areas mentioned above may be between an

isohyet and boundary of the catchment. In this case the precipitation at the boundary line

is required which may be extrapolated or interpolated.

Example: From the data given in table below, which was obtained from isohyetal map of

a catchment, find out the average precipitation of the catchment.

Isohyet

No.

Isohyetal

precipitation (cm)

Area enclosed between two

isohyets. (sq km)

1 2.5 390

2 5.0 520

3 7.5 650

4 10.0 390

5 10.0 390

6 7.5 442

7 5.0 546

8 2.5

Note that the isohyet No. 1 and 8 were out of the boundary of the catchment. The

area between isohyet No. 2 and the boundary was estimated to be 312 sq. km and that of

between isohyet No. 7 and boundary was 494 sq. km. Precipitation on these boundaries

was interpolated as 3.0 and 3.1 cm, respectively.


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Solution: In isohyetal method we have to calculate the average precipitation of every two

consecutive isohyets. This is given in Table below.

Isohye

t No.

Isohyetal

precipitation

(cm)

Average of

precipitation of two

consecutive isohyets

(cm)

Area enclosed

between two

isohyets

(sq km)

Volume

(x104

m)

(1) (2) (3) (4) (5) = (3) x (4)

Boundary 3 4 (for isohyet

and boundary)

312 (for

isohyet and

boundary)

1248.00

2 5.0 6.25 520 3250.00

3 7.5 8.75 650 5687.50

4 10.0 10.0 390 3900.00

5 10.0 8.75 390 3412.50

6 7.5 6.25 442 2762.50

7 5.0 4.05(for

isohyet and

boundary)

494 (for

isohyet and

boundary)

2000.70

Boundary 3.1

3198 22260.2


= 22260.2/3198= 6.96 cm

Example : In a catchment of area 1,000 sq km, there are 8 rain gauges, 5 inside the area

and 3 outside, in its surroundings. Isohyets were drawn from the data of these rain gauges

for a storm. From the isohyetal map the following information was obtained: areas

between 1 and 2 cm isohyets, 2 and 3 cm, 3 and 4 cm and 4 and 5 cm isohyets was 105,

230, 150 and 220 sq. km, respectively. The area between one end boundary which has


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0.75 cm rainfall and 1 cm isohyet was 120 sq. km and the other end boundary which has

precipitation of 5.5 cm and isohyet of 5 cm was 175 sq. km. Find average precipitation.

Solution

Isohyet No Isohyetalprecipitation

(cm)

Average ofprecipitation of two

consecutive isohyets

(cm)

Area enclosedbetween two

isohyets(sq km)

Volume(x10

4

m)

Boundar

y

0.75 0.875 (for

isohyet and

boundary)

120 (for

isohyet and

boundary)

105.00

1 1 1.5 105 157.50

2 2 2.5 230 575.003 3 3.5 150 525.00

4 4 4.5 220 990.00

5 5 5.25 (for

isohyet and

boundary)

175 918.75

Boundar

y

5.5

1000.00 3271.25


= 3271.25/1000 = 3.27 cm

Example : From the isohyetal map shown in Fig. below find out average precipitation.


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The isohyets are drawn on the topographic map by interpolating rainfall depths at given stations.

Once isohyets are drawn, the area enclosed between consecutive isohyets is determined either by

planimeter or other suitable more precise method

Isohyte value

(mm)

Av. Isohyte Value

(mm)

Area Between

Consecutive Isohytes

(km)

Volume (x106 m)

Boundary

and 2525.0 310.53 7.76

25 and 30 27.5 2220.71 61.07

30 and 35 32.5 2968.38 96.47

35 and 40 37.5 2231.86 83.69

40 and 45 42.5 2303.52 97.90

45 and 50 47.5 2731.90 129.77

50 and 55 52.5 2689.70 141.21

55 and

Boundary55 1484.99 81.67

Total 16,941.60 699.54

Mean Precipitation Depth = Volume/Area

= 699.54x106x10/16941.60x106 = 41.29 mm


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References:-

Hydrology for Engineers by Linsely, Kohler, Paulhus

Applied Hydrology by Dr. Abdur Razzaq Ghumman

Wikipedia

Documents

Hydrology and Water Management 1