Chapter No. 01 And 02 Introduction To Hydrology. Precipitation.pptx

Chapter 01 and 02 1) introduction to hydrology and the hydrological cycle. 2) Precipitation “hydrology and water management fall 2017”

By. Engr.Rahat Ullah

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outline • Hydrology and water management definitions • Branches of Hydrology • Applications of hydrology • The hydrological cycle • Precipitation and forms of precipitation • Factors affecting the formation of precipitation • Classification of precipitation based on lifting mechanism • Measurement of precipitation • Analysis of precipitation data • Estimation of average precipitation over a basin By. Engr.Rahat Ullah

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Hydrology • The word hydrology means science of water which deals with the characteristics of the earth’s water in all its aspects such as occurrence, circulation, distribution, physical and chemical properties, and impact on environment and living things. 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.

By. Engr.Rahat Ullah

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Water management • Water resource management is the activity of planning, developing, distributing and managing the optimum use of water resources.

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Branches of Hydrology Hydrology can generally be divided into two main branches: 1) Engineering Hydrology (Planning, design and Operation of Engineering projects for the control and use of water)

2) Applied Hydrology (Hydrological cycle, precipitation, runoff, relationship between precipitation and runoff, hydrographs, Flood Routing)

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Branches of Hydrology (cont.) Hydrology can be divided into the following branches • Chemical hydrology is the study of the chemical characteristics of water. • Ecohydrology is the study of interactions between organisms and the hydrologic cycle. • Hydrogeology is the study of the presence and movement of ground water. • Hydroinformatics is the adaptation of information technology to hydrology and water resources applications. • Hydrometeorology is the study of the transfer of water and energy between land and water body surfaces and the lower atmosphere. • Isotope hydrology is the study of the isotopic signatures of water. • Surface hydrology is the study of hydrologic processes that operate at or near Earth's surface. • Drainage basin management covers water-storage, in the form of reservoirs, and flood-protection. • Water quality includes the chemistry of water in rivers and lakes, both of pollutants and natural solutes. • Surface water Hydrology : is the study of hydrologic processes that operate at or near earth’s surface. By. Engr.Rahat Ullah

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Application of Hydrology • Determining the water balance of a region. • Determining the agricultural water balance. • Mitigation and predicting floods, landslides and drought risk. • Designing irrigation schemes and managing agricultural productivity. • Designing dams for water supply or hydroelectric power generation.

• Designing bridges. • Designing sewers and urban drainage system. • Predicting geomorphologic changes, such as, erosion or sedimentation. • Assessing the impact of natural and anthropogenic environmental change.

• Assessing containment transport risk and establishing environmental policy guidelines By. Engr.Rahat Ullah

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Hydrological cycle • Hydrological cycle also known as Water cycle or H₂O cycle, describes the continuous Movement of water on, above and below the surface of the earth.

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The hydrological cycle

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Hydrological cycle (cont.) The main processes involved in hydrological cycle are • Evaporation • Condensation • Precipitation • Interception • Infiltration • Percolation • Transpiration • Runoff • storage By. Engr.Rahat Ullah

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The main processes…  Evaporation • Water is transferred from the surface to the atmosphere through evaporation, the process by which water changes from a liquid to a gas. The sun’s heat provides energy to evaporate water from the earth’s surface. Land, lakes, rivers and oceans send up a steady stream of water vapour and plants also lose water to the air (transpiration). • Approximately 80% of all evaporation is from the oceans, with the remaining 20% coming from inland water and vegetation.  Condensation • Condensation is the process in which water vapor changes into liquid water. Condensation occurs as air with water vapor in it cools. Clouds are evidence of condensation. Clouds are formed when water vapor cools and condenses into tiny liquid water droplets. • The transported water vapor eventually condenses, forming tiny droplets in clouds. By. Engr.Rahat Ullah

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Continued..  Precipitation The primary mechanism for transporting water from the atmosphere to the surface of the earth is precipitation. When the clouds meet cool air over land, precipitation, in the form of rain, sleet or snow, is triggered and water returns to the land (or sea). A proportion of atmospheric precipitation evaporates.  Interception Interception refers to precipitation that does not reach the soil, but is instead intercepted by the leaves and branches of plants and the forest floor. Because of evaporation, interception of liquid water generally leads to loss of that precipitation for the drainage basin. By. Engr.Rahat Ullah

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Continued..  Infiltration The infiltration of water from precipitation into the soil is an important topic. In some circumstances a dry soil may not absorb rainfall as readily as a soil that is already wet. Infiltration can sometimes be measured by an infiltrometer.  Percolation water that enters the soil can move either vertically or laterally through the soil. Significant lateral movement of water through soil is called through flow or interflow. Downward movement of water through the soil is called percolation.  Transpiration Some water clings to soil particles and is drawn into the roots of growing plants. It is transported to leaves where it is lost to the atmosphere as vapor. This process is called transpiration. By. Engr.Rahat Ullah

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Continued..  Run-off Groundwater flows through rock and soil layers of the earth until it discharges as a spring or seep into a stream, lake, or ocean. The groundwater contribution to a stream is called baseflow, while the total flow in a stream is called runoff.  Storage There are three basic locations of water storage that occur in the earth ‘s water cycle. Water is stored in the atmosphere; water is stored on the surface of the earth, and water stored in the ground.

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Hydrological cycle (cont.) • 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. By. Engr.Rahat Ullah

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Hydrological cycle (cont.) • This Hydrologic Cycle recycles the earth’s valuable water supply. In other words, the water keeps getting reused over and over. • Just think, the next glass of water you drink could have been used by a dinosaur in the Mesozoic Era one hundred million years ago! • Water in that glass could have been a liquid, a solid, and a gas countless times over thanks to the water cycle.

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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 design of hydraulic structures and environmental effects of water resources development projects are related to precipitation in one way or the other.

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Forms of Precipitation  Drizzle :

• These are the minute particles of water at start of rain. • 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. • They cannot flow over the surface but usually evaporate.

 Rain : • It is form of precipitation in which the size of drops is 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 the rate of infiltration, the rain water can flow over the surface of earth. By. Engr.Rahat Ullah

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Forms of Precipitation (cont.)  Glaze : • It is the ice coating formed on drizzle or rain drops as it comes in contact with the cold surfaces on the ground.  Sleet : • Sleet is frozen rain drops cooled to the ice stage while falling through air at subfreezing temperatures.

 Snow : • Snow is precipitation in the form of ice crystals resulting from sublimation i.e. change of water vapor directly to ice.

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…continued  Snowflakes : • A snowflake is made up of a number of ice crystals fused together.  Hail : • Hail is the type of precipitation in the form of balls or lumps of ice over 5 mm diameter. • They are 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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Source:http://kvgktrailblazers.weebly.com/forms-of-precipitation.html By. Engr.Rahat Ullah

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Source:http://www.slideshare.net/mahasabri/precipitation-and-its-forms-hydrology

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Factors influencing Precipitation formation • A lifting mechanism to produce cooling of the air. • A mechanism to produce condensation of water vapors and formation of cloud droplets. • A mechanism to produce growth of cloud droplets to size capable of falling to the ground against the lifting force of air.

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Factors influencing Precipitation formation (cont.)  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 degree and rate of cooling needed to account for heavy rainfall. Cooling lowers the capacity of a given volume of air to hold a certain amount of water vapor. As a result super saturation occurs and the excess moisture over saturation condenses through the cooling process.

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Factors influencing Precipitation formation (cont.)  Condensation of water vapor • Condensation of water into cloud droplets takes place on hygroscopic nuclei which are small particles having an affinity for water. • The source of these condensation nuclei are the particles of sea salt or products of combustion of certain sulfurous and nitrous acid and carbon dioxide. • There is always sufficient nuclei present in the atmosphere.

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Factors influencing Precipitation formation (cont.)  Growth of Droplet • Growth of droplets is required if the liquid water present in the cloud is to reach the ground. The two processes regarded as most effective for droplet growth are: i. Coalescence of droplets through collision due to difference in speed of motion between larger and smaller droplets. ii.Co-existence of ice crystals and water droplets. • Co-existence effect generally happens in the temperature range from 100 to 20o F. By. Engr.Rahat Ullah

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Factors influencing Precipitation formation (cont.) • 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 Bergeron’s effect. • The ice crystals will further grow as they fall and collide with water droplets. By. Engr.Rahat Ullah

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Classification of Precipitation w.r.t Lifting Mechanism 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.

i. Convectional Precipitation ii. Orographic Precipitation iii. Cyclonic Precipitation

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Classification of Precipitation Based on the Lifting Mechanism (cont.)

 Convectional Precipitation • The main cause of Convectional precipitation is thermal convection of the moisture laden air (rising of warmer, lighter air in colder, denser surroundings). • A major portion of the solar radiation is utilized in heating the 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 the moisture 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. By. Engr.Rahat Ullah

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Classification of Precipitation Based on the Lifting Mechanism (cont.)  Orographic Precipitation •In the orographic precipitation, expansion and condensation occurs because moisture laden air masses are lifted by contact with orographic (mountain) barriers. •This type of precipitation is most pronounced on the windward side of mountain range, generally heaviest precipitation occurs where favorable orographic effects are present. •Orographic precipitation also occurs in the inland areas where mountain ranges rise above the surrounding areas in the path of the moisture laden air masses.

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Classification of Precipitation Based on the Lifting Mechanism (cont.)  Cyclonic precipitation • Precipitation in plain regions is generally cyclonic in character. • Cyclonic precipitation results from the lifting of air converging into a lowpressure area or cyclone. • Cyclonic precipitation can be frontal or non frontal. • Frontal precipitation results from the lifting of warm air on one side over a colder denser air on the other side. • Warm-front precipitation is formed if the warm air advancing upward over a cold air mass. • Cold-front precipitation is formed if the warm air is forced upward by an advancing mass of cold air. By. Engr.Rahat Ullah

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Classification of Precipitation Based on the Lifting Mechanism (cont.) • In the Indo-Pak Subcontinent, the cyclonic storms form in the Bay of Bengal in different months. • During April, May and June most 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. By. Engr.Rahat Ullah

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Classification of Precipitation Based on the Lifting Mechanism (Cont.)

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Measurement of precipitation • All forms of precipitation are measured on the basis of the vertical 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).

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Measurement of precipitation (cont.) • The precipitation is measured by rain gauges/precipitation gauges. There are two types of rain gauges.  Non-recording rain gauge (standard rain gauge)  Recording rain gauge

• The main difference between these rain gauges is that with the 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 the time also, for calculation of intensity of rain fall. By. Engr.Rahat Ullah

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Measurement of precipitation (cont.)  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. By. Engr.Rahat Ullah

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Measurement of precipitation (cont.)

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Measurement of precipitation (cont.) • Sources of Error • Some water is used to wet the surface of instrument. • The rain recorded may be less than the actual rainfall due to the 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 some error.

By. Engr.Rahat Ullah

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Measurement of precipitation (cont.)  Recording Rain gauge Recording rain gauges can be divided into the following types:

• Float type • Weighing type

• Tipping bucket type By. Engr.Rahat Ullah

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Measurement of precipitation (cont.)  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 this type 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 can not be measured by this type of rain gauge. By. Engr.Rahat Ullah

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Measurement of precipitation (cont.)

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Measurement of precipitation (cont.)  Weighing Type Rain Gauge • The weighing type rain gauge consists of a receiver, a bucket, a spring 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. By. Engr.Rahat Ullah

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Measurement of precipitation (cont.)

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Measurement of precipitation (cont.)  Tipping Bucket Type Rain Gauge • This type of gauge is equipped with a remote recorder located inside 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. By. Engr.Rahat Ullah

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Measurement of precipitation (cont.)

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Measurement of precipitation (cont.) • 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 the already filled bucket. Inclination of the gauge may result in catching less or more rain than the actual amount. • Error in measurement due to wind.

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Measurement of precipitation (cont.) • Remedial measures for Error in Precipitation measurement • Removal of error due to dents obviously needs repair of the instrument. For rain recorded with dents a correction should be applied. • Errors such as moistening of the inside surfaces of the gauge, splashing of rainwater from the collector and pouring of water into the already filled bucket during an intense rain can only be corrected by some correction factor. • Inclined instrument needs to be reinstalled. The correction factor however can be calculated from the angle of inclination. • For wind protection certain wind shields are designed and used which are called Splash Guards. Proper setting of gauge above ground level is necessary. By. Engr.Rahat Ullah

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Measurement of precipitation (cont.) 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 (θ) = 20o with the vertical P(actual) = P(measured)/cos(θ) = 125/cos20o = 133 mm By. Engr.Rahat Ullah

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Measurement of precipitation (cont.)  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 is appreciably 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. By. Engr.Rahat Ullah

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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: By. Engr.Rahat Ullah

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Rain Gauge Network (cont.) • In 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 • In mountainous regions of Temperate, Mediterranean and Tropical Zones, the ideal is at least one station for 35 – 95 sq. miles. However one station for 95 – 385 sq. miles is also acceptable. • In arid and polar zones, one station for 575 – 3860 sq. miles is acceptable.

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Analysis of Precipitation data • Estimation of missing precipitation data • Consistency of precipitation data or Double Mass analysis

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Estimation of missing Precipitation data (cont.) • 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. • In the procedure used by the U.S. Weather Bureau, the missing precipitation 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. • The station whose data is missing is called interpolation station and gauging stations whose data are used to calculate the missing station data are called index stations. By. Engr.Rahat Ullah

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Estimation of missing Precipitation data (cont.) • There are two methods for estimation of missing data. • Arithmetic mean method • Normal ratio method

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Estimation of missing Precipitation data (cont.)  Simple Arithmetic Mean method • According to the arithmetic mean method the missing precipitation ‘Px’ is given as: Px =

1 i n  Pi n i 1

• Where ‘n’ is the number of nearby stations, ‘Pi’ is precipitation at ith station and ‘Px’ is missing precipitation.

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Estimation of missing precipitation data • 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.

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Estimation of missing Precipitation data (cont.)  Normal Ratio method • According to the normal ratio method the missing precipitation is given as: Px =

1 in N x  Pi n i 1 N i

• 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. By. Engr.Rahat Ullah

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Estimation of missing Precipitation data (cont.) • For example, for the symbols defined above for three index stations in a catchment area. Px=

1 Nx [ P1  N x P 2  N x P3] 3 N1 N2 N3

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Estimation of missing Precipitation data (cont.) • If the normal annual precipitation of the index stations lies 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.

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Estimation of missing Precipitation data (cont.) 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 precipitation of 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. By. Engr.Rahat Ullah

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Estimation of missing Precipitation data (cont.) • Example : Find out the missing storm precipitation of station ‘C’ given in the following table: Station

A

B

C

D

E

Storm precipitation (cm)

9.7

8.3

----

11.7

8.0

Normal Annual precipitation (cm)

100.3 109.5

93.5

125.7

117.5

• 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 is to be applied. By. Engr.Rahat Ullah

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Estimation of missing Precipitation data (cont.) 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. 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: By. Engr.Rahat Ullah

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Estimation of missing Precipitation data (cont.) in Px =

1 Nx  Pi n i 1 N i

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

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Estimation of missing Precipitation data (cont.) • 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’.

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Estimation of missing Precipitation data (cont.) • Solution 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 (no need of calculating Nx – Nb and Nx – Nc Px = (1/3)( 97.8x 10.7/112+ 97.8x 8.90 /93.5 + 97.8x 12.2 /119.9) Px = 9.5 cm By. Engr.Rahat Ullah

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Consistency of Precipitation Data or Double Mass Analysis • In using precipitation in the solution of hydrologic problems, 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. • Due to such changes the data might not be consistent. The consistency of the record is then, required to be determined and the necessary adjustments to be made. This can be achieved by the method called the double mass curve technique. By. Engr.Rahat Ullah

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Consistency of Precipitation Data or Double Mass Analysis (cont.) • The double mass curve is obtained by plotting the accumulated precipitation at the station in question along Y-axis and the average accumulated precipitation of a number of other nearby stations which are situated under the same meteorological conditions along X-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 the formula. Pa = (Sa / So)x Po Where Pa = Adjusted precipitation. Po = Observed precipitation. 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. By. Engr.Rahat Ullah

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Consistency of Precipitation Data or Double Mass Analysis (cont.) • Example Check consistency of the data given in table below and adjust it if it is found to be inconsistent.

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Consistency of Precipitation Data or Double Mass Analysis (cont.) Year

Annual precipitation at x (mm)

Mean of annual precipitation of 20 surrounding stations (mm)

1972 1971 1970 1969 1968 1967 1966 1965 1964 1963 1962 1961 1960 1959 1958 1957 1956 1955

188 185 310 295 208 287 183 304 228 216 224 203 284 295 206 269 241 284

264 228 386 297 284 350 236 371 234 290 282 246 264 332 231 234 231 312

Year

1954 1953 1952 1951 1950 1949 1948 1947 1946 1945 1944 1943 1942 1941 1940 1939 1938 By. Engr.Rahat Ullah 1937

Annual precipitation at x (mm)

Mean of annual precipitation of 20 surrounding stations (mm)

223 173 282 218 246 284 493 320 274 322 437 389 305 320 328 308 302 414

360 234 333 236 251 284 361 282 252 274 302 350 228 312 284 315 280 343

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Consistency of Precipitation Data or Double Mass Analysis (cont.) • 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

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Cummulative precipitation of 20 surrounding stations (mm) 264 492 878 1175 1459 1809 2045 2416 2650 2940 3222 3468 3732 4064 4295 4529 4760 5072 5432 5666 5999 6235 6486 6770 7131 7413 7665 7939 8241 8591 8819 9131 9415 9730 10010 10353

By. Engr.Rahat Ullah

Corrected Precipitation 188 185 310 295 208 287 183 304 228 216 224 203 284 295 206 269 241 284 223 173 282 218 246 198.8 345 224 192 225.4 306 272.3 213.5 224 229.6 215.6 211.4 290

Remarks

No correction

1972 1971 1970 1969 1968 1967 1966 1965 1964 1963 1962 1961 1960 1959 1958 1957 1956 1955 1954 1953 1952 1951 1950 1949 1948 1947 1946 1945 1944 1943 1942 1941 1940 1939 1938 1937

Cummulative Annual precipitation at x (mm) 188 373 683 978 1186 1473 1656 1960 2188 2404 2628 2831 3115 3410 3616 3885 4126 4410 4633 4806 5088 5306 5552 5836 6329 6649 6923 7245 7682 8071 8376 8696 9024 9332 9634 10048

Precipitation of Station 'X' x 0.7

Year

72

Where Pa = Pa = Po = Sa = So =

(Sa / So)x Po Adjusted precipitation. Observed precipitation. Slope prior to the break in the curve Slope after the break in the curve. By. Engr.Rahat Ullah

73

Consistency of Precipitation Data or Double Mass Analysis (cont.) • The correction for slope is applied to readings beyond break in 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. By. Engr.Rahat Ullah

74

Estimation of Average Precipitation over a Basin • To find out runoff from a catchment and most of other hydrologic 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. • For example : the total number of rain gauges and their distribution in the catchment (larger the number of rain gauges, 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. By. Engr.Rahat Ullah

75

Estimation of Average Precipitation over a Basin (cont.) • There are three methods to find average precipitation over a basin. Accuracy of estimated average precipitation will depend upon the choice of an appropriate method. These methods are described below: • Arithmetic Mean Method • Thiessen’s Polygon Method • Isohyetal Method.

By. Engr.Rahat Ullah

76

Estimation of Average Precipitation over a Basin (cont.)  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: • Basin area is flat. • All stations are uniformly distributed (within practical limits) over the area. • The rainfall is also nearly uniformly distributed over the area. According to this method P (average) =

1 n    Pi  n  i 1

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.

By. Engr.Rahat Ullah

77

Estimation of Average Precipitation over a Basin (cont.) • 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. • 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. By. Engr.Rahat Ullah

78

Estimation of Average Precipitation over a Basin (cont.)  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 factor is considered. The following steps are used to determine average precipitation by Thiessen Polygon Method. 1. Draw the given area according to a certain scale and locate the stations where measuring devices are installed. 2. Join all the stations to get a network of non-intersecting system of triangles. 3. Draw perpendicular bisectors of all the lines joining the stations 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. 4. Measure area of the each polygon. 5. Calculate the average precipitation. For the whole basin by the formula. By. Engr.Rahat Ullah

79

Estimation of Average Precipitation over a Basin (cont.) 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. By. Engr.Rahat Ullah

80

Estimation of Average Precipitation over a Basin (cont.) • Example : Following is shown map of a catchment having 6 rainfall recording stations. Find the Average Precipitation over the whole catchment.

By. Engr.Rahat Ullah

81

Estimation of Average Precipitation over a Basin (cont.) The precipitation and polygon area are given below. Precipitation (mm)

Polygon Area (km²)

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

Station

By. Engr.Rahat Ullah

82

Estimation of Average Precipitation over a Basin (cont.) • Solution : The calculations are done in tabular form Station Daggar Besham Shinkiar i Phulra Tarbela Oghi

Precipitation P (mm) 48 33 25

Polygon Area A (km²) 5,068.76 4,349.17 1,399.25

P x A (x106 m³)

32 56 30 Total

1,693.80 2,196.33 2,234.29 16,941.60

54.20 122.99 67.03 666.02

243.30 143.52 34.98

in

Mean Precipitation =

 PiAi i 1 i n

= 666.02x106x10³/16941.60x106

 Ai i 1

Mean Precipitation = 39.3 mm By. Engr.Rahat Ullah

83

Estimation of Average Precipitation over a Basin (cont.) • Example : From the data given in Table below, which was obtained from Thiessen Polygon map of a catchment, find out the average precipitation of the catchment.

Sr No

Gauge precipitation (cm)

Area of Thiessen Polygon enclosing the station (sq. km)

Sr No

Gauge precipitation (cm)

Area of Thiessen Polygon enclosing 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

By. Engr.Rahat Ullah

84

Estimation of Average Precipitation over a Basin (cont.) • Solution : According to Thiessen Polygon Method P (average) = (P1 A1 + P2 A2 + ...........+ Pn An)/A The calculations are shown in tabular form in Table. Volume = Pi x Ai (x104 m³)

(1) = Pi

Area of Thiessen Polygon enclosing the station (sq. km) (2) = Ai

10.2 8.1 12.7 9.4 15.2 7.6 Total

416 260 650 520 390 325 2561

4243.20 2106.00 8255.00 4888.00 5928.00 2470.00 27890.20

Gauge precipitation (cm)

(3) = (1)x(2)

P (average) = 27890.20 ÷ 2561=10.9 cm By. Engr.Rahat Ullah

85

Estimation of Average Precipitation over a Basin (cont.) • Example : There are 10 observation stations, 7 being inside and 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 1 2 3 4 5 6 7

Gauge precipitation (cm) 5 3 4 3.5 4.7 6 4

Area of Thiessen Polygon enclosing the station (sq. km) 100 160 200 215 250 175 100

By. Engr.Rahat Ullah

86

Estimation of Average Precipitation over a Basin (cont.) • Solution : According to Thiessen Polygon Method P (average) = (P1 A1 + P2 A2 + ...........+ Pn An)/A Gauge precipitation (cm) (1) = Pi 5

Area of Thiessen Polygon enclosing the station (sq. km) (2) = Ai 100

Volume = PixAi (x104 m³) (3) = (1)x(2) 500

3

160

480

4

200

800

3.5

215

752.50

4.7

250

1175

6

175

1050

4

100

400

Total

1200

5157.50

P (average) = 5157.5 ÷ 1200 = 4.3 cm By. Engr.Rahat Ullah

87

Estimation of Average Precipitation over a Basin (cont.)  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. 2. 3. 4. 5.

Draw the map of the area according to a certain scale. Locate the points on map where precipitation measuring gauges are installed. Write the amount of precipitation for stations. Draw isohyets (Lines joining points of equal precipitation). Measure area enclosed between every two isohyets or the area enclosed by an isohyet and boundary of the catchment. By. Engr.Rahat Ullah

88

Estimation of Average Precipitation over a Basin (cont.) Find average precipitation by the formula. P (average) = (P1 A1 + P2 A2 + ...........+ Pn An)/A 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 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. By. Engr.Rahat Ullah

89

Estimation of Average Precipitation over a Basin (cont.) • 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. 1 2 3 4 5 6 7 8

Isohyetal precipitation (cm) 2.5 5.0 7.5 10.0 10.0 7.5 5.0 2.5

Area enclosed between two isohyets. (sq km) 390 520 650 390 390 442 546

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. By. Engr.Rahat Ullah

90

Estimation of Average Precipitation over a Basin (cont.) • Solution : In isohyetal method we have to calculate the average precipitation of every two consecutive isohyets. This is given in Table below. Isohyet No.

Isohyetal precipitation (cm)

(1) Boundary

(2) 3

2 3 4 5 6 7

5.0 7.5 10.0 10.0 7.5 5.0

Boundary

3.1

Average of precipitation of two consecutive isohyets (cm) (3) 4 (for isohyet and boundary) 6.25 8.75 10.0 8.75 6.25 4.05(for isohyet and boundary)

Area enclosed between two isohyets (sq km)

Volume (x104 m³)

(4) 312 (for isohyet and boundary) 520 650 390 390 442 494 (for isohyet and boundary)

(5) = (3) x (4) 1248.00

∑

3198

22260.2

3250.00 5687.50 3900.00 3412.50 2762.50 2000.70

• P (average) = (P1 A1 + P2 A2 + ...........+ Pn An)/A = 22260.2/3198 = 6.96 cm By. Engr.Rahat Ullah

91

Estimation of Average Precipitation over a Basin (cont.) • 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 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.

By. Engr.Rahat Ullah

92

Estimation of Average Precipitation over a Basin (cont.) • Solution

Isohyet No

Isohyetal precipitation (cm)

Boundary

0.75

1

Average of precipitation of Area enclosed between two consecutive isohyets (cm) two isohyets (sq km)

Volume (x104 m³)

1

0.875 (for isohyet and boundary) 1.5

120 (for isohyet and boundary) 105

2

2

2.5

230

575.00

3

3

3.5

150

525.00

4

4

220

990.00

5

5

4.5 5.25 (for isohyet and boundary)

175

918.75

Boundary

5.5 ∑

1000.00

3271.25

P (average) = (P1 A1 + P2 A2 + ...........+ Pn An)/A = 3271.25/1000 = 3.27 cm By. Engr.Rahat Ullah

105.00 157.50

93

Estimation of Average Precipitation over a Basin (cont.) • Example : From the isohyetal map shown in Fig. below find out average precipitation.

By. Engr.Rahat Ullah

94

Estimation of Average Precipitation over a Basin (cont.) • Solution :

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. Av. Isohyte Value (mm)

Area Between Consecutive Isohytes (km²)

Volume (x106 m³)

Boundary and 25

25.0

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

1484.99

81.67

Total

16,941.60

699.54

Isohyte value (mm)

55 and Boundary

• Mean Precipitation Depth = Volume/Area = 699.54x106x10³/16941.60x106 = 41.29 mm By. Engr.Rahat Ullah

95

References cited • Hydrology for Engineers by Linsely, Kohler, Paulhus • Applied Hydrology by Dr. Abdur Razzaq Ghumman • Hydrology Principles, analysis, design by HM Raghunath Special thanks to Engr. Asad Ali, Lecturer CECOS University.

By. Engr.Rahat Ullah

96