Watershed Hydrology Lecture Notes PDF

Lecture Notes Course: Watershed Hydrology (SWC-201) (Instructor : Dr. R. Suresh, Professor SWE) Department of Soil & Water Engineering College of Agricultural Engineering R.P.C.A.U., Pusa (Samastipur)-848 125  Lecture – 1...

Lecture Notes Course: Watershed Hydrology (SWC-201) (Instructor : Dr. R. Suresh, Professor SWE) Department of Soil & Water Engineering College of Agricultural Engineering R.P.C.A.U., Pusa (Samastipur)-848 125  Lecture – 1 Introduction Definition and Importance Hydrology, by its term meaning is the science of water. However it does not gives a complete view of hydrology in respect to water generating source for creating water resource on earth system, essential for mankind. After a great exercise and deal the scientists defined the hydrology as the science dealing the occurrence of rainfall, its distribution & circulation on and below the earth and physical and chemical reaction with earth materials. In hydrology, the land unit is the watershed, which also may be referred to as a basin or catchment. A watershed is defined as an area of land in which all of the incoming precipitation drains (i.e., “sheds”) to the same place – toward the same body of water or the same topographic low area (e.g., a sinkhole) – as a result of its topography. This means that a watershed's boundary is defined by its topographic high points. Watersheds are fairly simple to identify in mountainous or hilly terrain because their boundaries are defined by ridges (Figure 1). However, in flatland watersheds, such as in the Coastal Plain of the Southeast, identifying topographic high points can be very challenging because the highest and lowest elevations may differ only by a few centimetres. The study of hydrology is important to have the information is following aspects:  The variation of water production from catchments can be calculated and described by hydrology.  Engineering hydrology enables us to find out the relationship between a catchments’s surface water and groundwater resources  The expected flood flows over a spillway, at a highway at a Culvert, or in an urban storm drainage system can be known by this very subject.  It helps us to know the required reservoir capacity to assure adequate water for irrigation or municipal water supply in droughts condition.  It tells us what hydrologic hardware (e.g. rain gauges, stream gauges etc) and software (computer models) are needed for real-time flood forecasting  Used in connection with design and operations of hydraulic structure  Hydrology is an indispensable tool in planning and building hydraulic structures.  Hydrology is used for city water supply design which is based on catchments area, amount of rainfall, dry period, storage capacity, runoff evaporation and transpiration.  Dam construction, reservoir capacity, spillway capacity, sizes of water supply pipelines and affect of afforest on water supply schemes, all are designed on basis of hydrological equations.  Crop planning based on available water in the area. Divisions of Hydrology: Hydrology can generally be divided into two main branches Engineering Hydrology: Engineering hydrology deals with the planning, design and Operation of Engineering projects for the control and use of water Applied Hydrology: Applied hydrology is the study of hydrological cycle, precipitation, runoff, relationship between precipitation and runoff, hydrographs, Flood Routing Chemical Hydrology: Study of chemical characteristics of water. Eco-hydrology: Interaction between organisms and the hydrological cycle. Hydrogeology: Also referred to as geo-hydrology, is the study of the presence and movement of ground water. Hydro-informatics: is the adaptation of information technology to hydrology and water resource applications Hydrometeorology: It is the study of the transfer of water and energy between land and water body surfaces and the lower atmosphere. Isotope Hydrology: It is the study of isotropic signatures of water (origin and age of water). Surface Water Hydrology: It is the study of hydrologic processes that operate at or near earth’s surface. Ground Water Hydrology: It is the study of underground water. The Hydrologic Cycle It refers to the continuous circulation of water within the earth’s hydro-sphere. Water moves into and from the various sources on, over and below the earth, with the total mass of water remaining fairly constant. The water cycle is highly crucial to maintain the life on earth, as it replenishes the world’s freshwater resources and moderates extremes in climate. The physical processes involved in hydrologic cycle are  Evaporation  Condensation  Sublimation  Precipitation  Transpiration,  Interception,  Infiltration,  Percolation and  The runoff Sun is the source of energy to activate Evaporation - It involves the vaporization of water from the water sources due to heat energy of solar radiation. The evaporated water gets converted into cloud. Through which water gets fall on the earth system in terms of precipitation. In water transfer process about 90% of atmospheric water is contributed by evaporation. Condensation- It refers to the transformation of evaporated water vapours into liquid water droplets suspended in the air as clouds or fog. It is important process to convert the evaporated water into liquid state enabling formation of clouds with the aid of condensation nuclei. Sublimation- This is the process in which there is direct conversion of solid ice into water vapour. By this process water mass is also added to atmosphere for cycling. Precipitation- It is the fall of atmospheric water to the ground surface. Under this process the water becomes available for its distribution (surface and sub-surface) and circulation on the above and below the earth surface. It mostly takes place in the form of liquid (rainfall) and very little in solid form (snow, sleet, hail fog etc.) Transpiration- It is a process of water loss from plants' leaves through respiration. The water loss through transpiration and evaporation coupled together is referred to Evapotranspiration (ET). In hydrologic cycle about 10% water or moisture is added to the atmosphere by transpiration process. Interception- This is the process in which a part of precipitation is abstracted by the objects lying on the ground surface. The objects may be the crop, tree, natural vegetations and any other in live or dead conditions. Intercepted precipitated water is ultimately lost through evaporation process. Rate and quantity of water loss under this process varied with the type and characteristics of vegetation/objects and climatic condition, mainly. Infiltration- It is defined as the entry of water into the soil by crossing the imaginary boundary between soil and atmosphere and its rate called infiltration rate. Under this process the precipitated water moves into the soil media and ultimately joins to the water –table or deposited on impervious layer, if there occurs across water movement path. It is treated as the input process for ground water occurrence. Runoff- The flow of joined rain water in the stream is designated as the channel flow or the runoff. The characteristics associated to the climate and watershed affects the quantum of runoff at the outlet. Runoff is categorised into surface and sub-surface runoff. In which surface runoff is that part of the runoff which travels over the ground surface thought the channels/ streams /rivers to reach the basin outlet, and sub- surface or indirect runoff points to the flow of precipitated water below the soil surface leading to water- table. The view of hydrologic cycle is presented in Fig- Fig- View Hydrologic cycle Hydrologic Budget It consists of inflows, outflows, and storage, presented by the following equation: Inflow = Outflow +/- Changes in Storage Inflows contribute or add water to the different parts of the hydrologic system, outflows remove water from them, and storage is the retention of water by parts of the system. Since, water movement is cyclical; therefore, an inflow for one part of the system is an outflow to another. As example, for an aquifer the percolation of water into the ground is the inflow to the aquifer while discharge of groundwater from the aquifer to a stream is an outflow. Over time, if inflows to the aquifer are greater than its outflows, the amount of water stored in the aquifer will increase. Conversely, if the inflow to the aquifer is less than the outflow, the amount of water stored decreases. Assignment - Describe various atmospheric/weather variables affecting hydrologic cycle. Suggested Reference 1. Watershed Hydrology- R. Suresh 2. Soil & Water Conservation Engineering- R. Suresh 3. Hydrology and Soil Conservation – G. Das 4. Engineering Hydrology – K. Subramanya  Lecture- 2 Watershed- Concept and Laws Definition- Watershed is an isolated area with a well demarcated boundary line, draining the rainwater to a single outlet. Within the boundary a watershed contains various natural resources such as the soil, water and natural vegetations. Also, there is a network of stream system to drain the rain water. The stream network is also called drainage system of watershed. A watershed having the measured hydrological parameters such as rainfall, runoff and others is called gauged watershed. For hydrological study of watershed the hydrological input data is essential. There is a variation in one to other watersheds regarding shape, size, morphology and other aspects. This variation enforces to cause variations in rainfall- runoff – soil erosion /loss relationship. The stream network system makes the watershed dynamic in regard to hydrologic processes. View of watershed is shown in figure-1. Figure-1.0 View of Watershed Watershed Component The followings are the major components of the watershed, 1. Watershed boundary, 2. Stream network, 3. Watershed soils/land ,and 4. Land use system. Boundary- It defines the size and shape of watershed. Boundary of a watershed may be the any continuously elevated ridge lines covering a particular extent of land segment. Also the river line, roads, etc may be as the watershed boundary. Stream Network- The watershed within it boundary assembles several interconnected streams leading to the outlet. Stream network constitutes drainage system for the watershed. Drainage system is assessed by the term called drainage density, which is defined as the ratio of total length of steams to the total area of watershed. A watershed with high drainage density is highly drained and vice- versa. In addition, the stream frequency which is the ratio of total number of streams to the area of watershed also signifies the degree of watershed drainage. Soil- There is possibility of a large variation in occurrence of soil in respect to its type, geography/topography etc in the watershed. Availability of good and healthy soil makes the watershed more productive which force the population to a good level with their high standard of livings. In hilly watersheds; however, the availability of land for cultivation is very limited, normally located at the river bed and few at the slope faces, which may be under constraint of high slope. The soils at steep slopes cannot be directly used because of problems in use of tillage or cultural practices for crop growing. Normally, such steep lands are converted into terraces of different types depending on the soil types and rainfall of the area for cultivating the crops. Land use- Mainly, the land use system comprises, agricultural, orchard, forest land, grassland / pasture and waste land uses. The waste land use may be with vegetation or without vegetations. Land use system affects the hydrological behaviour of watershed. For example, a watershed dominated by the forest land use system involves very less runoff or rain water availability for harvesting point of view than another watershed which involves very less extent of forest lands. To a large extent the land use system maintains the watershed towards its management aspects. Watershed Morphology- It includes overall surface characteristics of watershed including the stream network comprising the stream ordering, stream length, stream slope, areal aspect , relief aspects etc mainly. Related to watershed morphology there are various laws defining the features of watershed; they are the law of steam order, law of stream length, law of stream slope, law of stream area etc. Stream order: In order to facilitate the study of watershed behaviour on morphological aspects the stream ordering is followed as an approach for categorizing the streams into different orders as per their sequence of their origin. This also provides a basis for dividing the entire area of watershed for grouping, stream wise. There has been devised a rule for proving order to a given stream of watershed. For stream ordering the watershed map containing the stream network is essential. Rule-1: The finger tips like streams are taken first for ordering. These are provided stream order –Ist. Rule-2: When two same order streams join together the resulting steam will be of next higher order. Say for example, if Stream A and B of Ist order join together the order of resulting stream C will be 2nd order. Rule-3: If a lower order stream joins to a higher order stream the order of resulting stream will be the same, i.e., remain the higher order as it is. (Fig-2.0 illustrates the procedure of stream ordering). Figure- 2.0. Stream ordering Law of Stream Number It states that the number of stream of each order forms an inverse geometric sequence with the stream order number, expressed as, 𝑁𝑢 = 𝑅𝑏𝑘−𝑢 Where, Nu = Number of steam of order u. The number of stream of any order u is fewer than for next lower order, but more in numbers than the next higher order Rb = Bifurcation ratio, defined as the ratio of number of stream segments of a given order u to the number of stream segment of next higher order, expressed as under, K = trunk order of stream segment in the watershed. 𝑁𝑢 𝑅𝑏 = 𝑁𝑢+1 Problem(1) Under morphological study of a watershed the following details have been collected: i. Total number of Ist order streams = 35 ii. Trunk order stream = 3. Determine bifurcation ratio of watershed. Solution – Using following formula for bifurcation ratio, 𝑁𝑢 = 𝑅𝑏𝐾−𝑈 in which, 𝑁𝑢 is the number of stream of order u (in present case it is taken as Ist order, which total number is given as 35); K is the trunk order stream (3) and u is the order of stream is given as 1. Substituting the values in above formula and solving, we have, 35 = 𝑅𝑏3−1 𝑅𝑏 = 𝟓. 𝟗𝟐 𝑨𝒏𝒔. Problem.(2) Using following data set predict the shape of three different watersheds based on bifurcation ratio. S. No Parameter Watershed A B C 1. Number of stream of order u (Nu) 25 20 14 2. Number of stream of order u+1 (Nu+1) 1 3 5 Solution- Using following formula for bifurcation ratio, 𝑁 𝑅𝑏 = 𝑁 𝑢 𝑢+1 Accordingly, the computed Rb is shown below, S. No Parameter Watershed A B C 1. Number of stream of order u (Nu) 25 20 14 2. Number of stream of order u+1 (Nu+1) 1 3 5 3. Bifurcation ratio (Rb) 25 6.66 2.80 The computed value of bifurcation ratio of watersheds A, B and C advocates following views about their shapes, S. Particular Watershed No. A B C 1. Bifurcation ratio 25 6.66 2.80 2. Watershed shape Elongated or leaf shape Semi- elongated shape Approaching semi – circular Law of Stream Length The law of stream length states that the mean length of stream segment of successive order basin approximates a direct geometric sequence represented by the following expression 𝐿̅𝑢 = 𝐿̅1. 𝑅𝑙𝑢−1 In which 𝐿̅𝑢 is the mean length of stream of order u; 𝐿̅1 is the mean length of stream of order 1; Rl is the length ratio and K is the trunk order of the stream. The length ratio is given as under, 𝐿̅𝑢 𝑅𝑙 = ̅ 𝐿𝑢−1 This equation is the non- linear equation, which can be linearize by taking of the equation. Law of Stream Area It states that the mean area of drainage basin of progressively higher order stream constitute a direct geometric sequence, expressed as under, 𝐴̅𝑢 = 𝐴̅1. 𝑅𝑎𝑢−1 In which 𝐴𝑢 is the mean area of drainage basin of stream of order u; 𝐴̅1 is the mean area of drainage ̅ basin of stream of order 1; Ra is the area ratio and K is the trunk order of the stream, is analogous to length ratio. This equation is the non- linear equation, which can be linearize by taking log of the equation. Law of Stream Slope It is inverse geometric series law. It states that the stream slope is progressively increased with increase in their order. Mathematically, it is expressed as under, 𝑆𝑢̅ = 𝑆1̅. 𝑅𝑠𝑘−𝑢 In which 𝑆𝑢 is the mean slope of stream of order u; 𝑆1̅ is the mean slope of stream of order 1; Rs is ̅ the slope ratio (constant) and K is the trunk order of the stream, is analogous to bifurcation ratio. This equation is the non- linear equation, which can be linearize by taking log of the equation. Drainage Density Drainage density of a watershed is the ratio of cumulative length of streams of all orders existing in the watershed to the total area of watershed projected on horizontal plane. It is expressed by the following expression, ∑𝑘𝑖=1. ∑𝑁 𝑖=1 𝐿𝑢 𝐷𝑑 = 𝐴𝑘 In which 𝐷𝑑 is the drainage density of a watershed (km per sqkm); 𝐴𝑘 is the drainage area of trunk order stream, which refers to the total area watershed. Drainage density varies among watersheds. Problem (3)- The order-wise stream lengths of two different watersheds are given as under. Particular Watershed-A Watershed-B Stream order I II III I II III IV Stream length, m 500 210 55 3500 2350 1900 750 Area of watershed, ha 100 100 Determine drainage density and narrate which watershed is dominated by overland flow, and why. Solution- Using following formula for drainage density: ∑𝐾 𝑖=1 𝐿𝑢 𝐷𝑑 = 𝑖 𝐴𝑢 In which, in case of Watershed- A: ∑𝐾 𝑖=1 𝐿𝑢𝑖 = (500 +210 + 55) =765m and 𝐴𝑢 = 100ℎ𝑎. Substituting value of the parameter in above formula and solving, we have, 765 𝐷𝑑 = 100 𝑥 10000 = 0.000765 𝑚 −1 Watershed- B: ∑𝐾 𝑖=1 𝐿𝑢𝑖 = (3500 +2350 +1900 + 750) =8500m and 𝐴𝑢 = 100ℎ𝑎. Substituting value of the parameter in above formula and solving, we have, 8500 𝐷𝑑 = 100 𝑥 10000 = 0.0085 𝑚 −1 On comparison of drainage density of both the watersheds, it is found that the watershed – A comprises very less value of drainage density than the watershed-B, which signifies the watershed –A to be under more influence of overland flow. Stream Frequency It is defined as the ratio of cumulative number of streams of all order existing in a watershed to the total area of watershed, expressed by the following formula, ∑𝑘 𝑖=1 𝑁𝑢 𝐹= 𝐴𝑘 Associated parameters have already been defined above. This morphological parameter is used for assessing the stream network of watershed. Problem (4). The order wise number of streams of a given watershed is mentioned below.  Stream order I II III IV Stream length (m) 450 175 30 1 Determine stream frequency, if area of watershed is 100ha. Solution- The formula for drainage density is given as under: ∑𝐾 𝑖=1 𝑁𝑢 𝐹= 𝐴𝑢 𝑖 in which, ∑𝐾 𝑖=1 𝑁𝑢𝑖 = (450 +175 + 30 +1) =656.0 and 𝐴𝑢 = 100ℎ𝑎. Substituting value of the parameter in above formula and solving, we have, 656 𝐹= = 100 𝑥 10000 = 𝟔. 𝟓𝟔𝒙 𝟏𝟎−𝟒 𝒎−𝟐 Ans. Problem (5) A watershed comprises its drainage density as 0.005m-1. Calculate length of overland flow of watershed, if its channel slope is 0.5% and average ground slope is 1.5%. Solution- The formula for length of overland flow, accounting the effect of channel slope and average ground slope, is given as under: 1 𝐿𝑔 = 𝜃 2𝐷𝑑 √1−( 𝑐 ) 𝜃 𝑔 in which 𝐷𝑑 is the drainage density of watershed (0.005m-1); 𝜃𝑐 is the slope of channel (0.005) and 𝜃𝑔 is the average ground slope of watershed is given as 0.015. Substituting these values in above formula and solving, we have, 1 𝐿𝑔 = 0.005 2 𝑥 0.005√1−( ) 0.015 = 𝟏𝟐𝟐. 𝟎𝒎 𝑨𝒏𝒔. Relief It is defined as the elevation difference between the reference points located in the damage basin. Relief Ratio It is the ratio of relief to the horizontal distance on which relief is measured. Maximum Basin Relief It is the elevation difference between basin outlet and the highest point located at the perimeter of the basin. Relative Relief It is the ratio of maximum basin relief to the basin perimeter, expressed as under, 𝐻 𝑅ℎ𝑝 = 𝑥100 𝑃 In which, 𝑅ℎ𝑝 is the relative relief (%); H is the maximum basin relief (m) and P is the basin perimeter (m) Problem (6) A watershed of 100ha size has been surveyed for determining its relief status. The measured elevations in different aspects are given as under: 1. Elevation of watershed out let= 12.5m 2. Elevation of highest point located at watershed boundary =7.75m 3. Elevation of highest point within boundary = 9.75m 4. Elevation of lowest point within watershed = 11.35m Determine followings: i. Maximum relief and ii. Maximum basin relief. Solution- Computations are presented as under: 1. Maximum relief- It is the elevation difference between highest and lowest points located within watershed. Accordingly, Rmax = 11.35-9.75= 1.60m Ans. 2. Maximum basin relief- It is the elevation difference between basin outlet and highest point located at watershed perimeter. Accordingly,  Rb max = 12.5- 7.75= 4.75m Ans. Problem (7)-. In a watershed the elevations of highest points (A) and lowest point (B) are 12.50 and 15.50m, respectively. Determine relief ratio, if horizontal distance between both the points is 1500m. Solution- Relief ratio is presented by the following relationship, 𝐻 𝑅ℎ = 𝐿 in which, H is the relief is elevation difference between highest and lowest points is 15.5- 12.5m=3.0m and L is the horizontal distance between both the points is given as 1500m. Accordingly, 3.0 𝑅ℎ = 1500 = 𝟎. 𝟎𝟎𝟐 𝑨𝒏𝒔. Problem (8). Determine the value of relative relief of a watershed, if maximum basin relief is 3.75m and perimeter of drainage basin is 1250m. Solution- Using following formula for relative relief of watershed, 𝐻 𝑅ℎ𝑝 = 𝑥 100 𝑝 in which, H is the maximum basin relief, given as 3.75m and p is the perimeter of watershed is 1250. Therefore, 3.75 𝑅ℎ𝑝 = 1250 𝑥 100 = 𝟎. 𝟑𝟎% 𝑨𝒏𝒔 Problem (9). The elevation of outlet and highest point located at basin perimeter is 15.0 and 11.45m, respectively. The length of basin perimeter is 2.55km. Determine the value of maximum basin relief and relative relief of watershed. Solution- Computations are presented as under: 1. Maximum basin relief- It is the elevation difference between basin outlet and highest point located at watershed perimeter. Accordingly, 15-11.45=3.55m Ans. 2. Relative relief- It is given by the following formula, 𝐻 𝑅ℎ𝑝 = 𝑥 100 𝑝 in which, H is the maximum basin relief, given as 3.55m and p is the perimeter of watershed is 2550m. Therefore, 3.55 𝑅ℎ𝑝 = 2550 𝑥 100 = 𝟎. 𝟏𝟒% 𝑨𝒏𝒔 Suggested References 5. Soil & Water Conservation Engineering- R. Suresh 6. Hydrology and Soil Conservation – G. Das  Lecture-3 Clouds- It is the source of precipitation. Precipitation is resulted mainly from two types of clouds. They are the nimbostratus and cumulonimbus type clouds. The nimbostratus clouds appear at mid height in dark gray colour result the rainfall throughout day, continuously. The cumulonimbus clouds are low lying in the sky. These appear as thick puffy columns climbing into the atmosphere. These are thunderstorm clouds, cause intense, short rain bursts, heavy snowfall, hail etc. Precipitation- The moisture emanating from the cloud and falling to the earth surface is called precipitation. Precipitation in the form of rainfall develops water resource potential of the region, on which various activities like crop cultivation; industrial needs, house hold needs, water requirement for electricity generation etc are met. Also, the precipitation is important  Precipitation replenishes the water to the earth.  Without precipitation the earth would behave like desert. The amount and duration of precipitation affect the water level and water quality as well.  Precipitation supplies freshwater to an estuary, which is important source of dissolved oxygen and nutrients.  Drought effects are lowered. Forms of Precipitation Precipitation is water released from clouds in the form of rain, freezing rain, sleet, snow, or hail. It is the primary connection in the water cycle that provides for the delivery of atmospheric water to the Earth. Most precipitation falls as rain. Broadly, precipitation takes place in liquid and solid forms. Liquid form of precipitation comprises the forms of rainfall and drizzle, mainly. Liquid precipitation is any precipitation that falls as a liquid and remains liquid after striking on object, i.e. the earth surface. Liquid form of Precipitation Rainfall- It is in liquid form (drops) falling from the clouds to the earth surface. Size of water droplets is about to 0.5 mm or little bit bigger. Rate of rainfall varies from time to time. A light rain ranges from 2.5mm/h, moderate rain from 2.5-7.5mm/h and heavy rain above 7.5mm/h. Rainfall is the most important component of hydrologic cycle which replenishes large percentage of fresh water on earth. Rain and drizzle are beneficial for plants. Drizzle- It is also in liquid form but its droplets size is less than 0.5mm diameter. Its intensity is lesser than 2.5mm/h. It contributes moisture to the lower atmosphere effective for cooling and generating warm air mass to create cloud in the sky. Drizzle usually falls from low stratus clouds and is frequently accompanied by fog. Acid rain- It occurs when rain becomes mixed with pollutants such as Sulphur Oxides and Nitrous Oxides. It kills plants and pollutes the water sources lying on the earth surface. Solid form of Precipitation Various forms are the snow, sleet, hail etc, Snow- It consists of white or translucent ice crystals. Originally these are highly complex, hexagonal, branched structures. Snow falls as a combination of individual ice-crystals, fragments of crystals, or clusters of crystals. Snow pellets – These are small hails, appear in the form of white, opaque, round/ conical. Average diameter varies from 0.08 to 0.2 inch. Have tendency to burst upon striking a hard surface. These occur almost exclusively in snow showers. Hail- These are opaque ball of hard ice, ranging in diameter from 1/8 inch or so to 5 inches or larger. Hails are basically the ice that falls from the sky, often in round shape. Hailstones are large chunks of ice that fall from large thunderstorms. Precipitation size and speed Have you ever watched a raindrop hit the ground during a large rainstorm and wondered how big the drop is and how fast it is falling? Or maybe you've wondered how small fog particles are and how they manage to float in the air. The table below shows the size, velocity of fall, and the density of particles (number of drops per square foot/square meter of air) for various types of precipitation, from fog to a cloudburst. Types Intensity Median diameter Numbers of drops (per (cm/h) (mm) sqm) Fog (0.013) 0.01 67,425,000 Mist 0.005 0.1 27,000 Drizzle 0.025 0.96 151 Light rain 0.10 1.24 280 Moderate rain 0.38 1.60 495 Heavy rain 1.52 2.05 495 Excessive rain 4.06 2.40 818 Cloudburst 10.2 2.85 1,220 (Source: Lull, H.W., 1959, Soil Compaction on Forest and Range Lands, U.S. Dept. of Agriculture, Forestry Service, Misc. Publication No.768) Precipitation Types Orographic precipitation- This type of precipitation is caused by air masses striling some natural topographic barriers like mountains. The greater amount of precipitation falls on the windward side. Few important features are given as under:  Lifting of warm air mass is due to orographic barriers causing formation of cloud and precipitation thereof.  Rainfall is steady rainfall.  Southern slope of the Himalayas is a good example of this type of precipitation.  Similarly, the winds coming from ocean strike the western slopes of coastal ranges causing heavy rains are another example of orographic precipitation.  The rainfall at u/s side is intense to that of the d/s side.  The d/s orographic precipitation is called Rain shadow. Cyclonic precipitation- Cyclone is atmospheric disturbance caused by air mass circulating clockwise in southern and anticlockwise in northern hemispheres. Basically, cyclone is violently rotating wind storm. The precipitation as result of cyclone is termed as cyclonic precipitation. Cyclone is very large mass of air ranging from 800 to 1600km in diameter and moving with the velocity of 50 kmph. Few important features are mentioned below:  The cyclonic precipitation occurs in the form of drizzle, intermediate rain or steady rain.  Precipitation caused by cold front is intense and of short duration.  Precipitation caused by warm front is more continuous Convective Precipitation- In this precipitation the lifting of warm air mass is due to convective effect, called convective uplift. The convective uplift takes place especially when air near to ground gets warm due to sun energy, and begins to rise upward in the sky. The process of rising of warm air mass and its cooling is governed by adiabatic cooling process. This leads to the formation of clouds and precipitations, sometimes. In this precipitation the precipitation occurs in the form of showers of high intensity and for short duration. Rainfall Measurement Rain gauge- The rain gauge is the instrument used for rainfall measurement. The measured rainfall is termed as the point rainfall. The point rainfall is used for determining the mean areal rainfall by using various computing methods. The mean aerial rainfall can be used for determining the volume of rainwater received over the surface area of watershed by multiplying the mean depth of rainfall and area of watershed/region. Types of Rain gauge Broadly, it is classified as 1. Non- recording type; and 2. Recording type rain gauge. Non- Recording Type Rain gauge (Simon type) - It is most common type of rain gauge, consists of 127mm diameter cylindrical vessel with a base width 210mm diameter for making stability. At top a funnel is provided with circular brass rim which is exactly 127mm to fit into vessel correctly. This funnel shank is inserted in the receiving bottle placed below. The height of receiving bottle is 75 to 100mm. The bottle receives the rainfall. Capacity of receiving bottle is to measure the rainfall depth is 100mm. During heavy rainfall, the rainfall amount is likely to exceed the bottle capacity. In this condition it is suggested to take the observations frequently, normally 3 to 4 times in a day. The water collected in the receiving bottle is measured by a graduated measuring cylinder. The measuring accuracy of graduated cylinder is being up to 0. 1mm. The timing of rainfall measurement is uniformity done, every day at 8:30Am IST. For accurate measurement the proper care, maintenance and inspection of rain gauge should be carried out during dry weather. Recording Type Rain gauge It records the information about start and end of rainfall event taking place. With the help of this information, one can determine the rainfall intensity and depth of the place under measurement. The following rain gauges are commonly used as recording type rain gauge, 1. Float type rain gauge, 2. Weight type rain gauge, and 3. Tipping bucket type rain gauge. Float Type Rain Gauge- It is also known as natural siphon type rain gauge. In India this rain gauge is adopted as the standard recording rain gauge. Working of this rain gauge is similar to the weighing type rain gauge. In this a funnel receives the rain water which is collected into a container equipped with a float at the bottom. The position of float rises as the water level rises in the container depending on rain water coming into. The movement of float is transmitted to a pen which traces a curve on the rain chart mounted on a clockwise rotating drum. When float rises to the top of container the siphon comes into action and drain the total water from the container. At this stage the pen traces a straight line. If rainfall is continued and water is coming into the container then further float rises up and pen traces the curve. This process is continued. If rainfall is stooped the pen traces a horizontal line on the chart. The obtained curve is the mass curve. View of this rain gauge is shown in Fig-. Weighing Type Rain Gauge- It is most common self-recording rain gauge, consists of a receiver (bucket) supported by a spring/ lever balance or some other weighing mechanism. The movement of bucket due to its increasing weight because of accumulation of rainwater is transmitted to a pen, which traces a curve on the rain chart wrapped on clock. The obtained rainfall record in terms of curve is mass curve, i.e. the plot of cumulative rainfall vs elapsed time. View of rain gage is shown in Fig- Tipping Bucket Type Rain Gauge- It is a 30cm size rain gauge, used as recording type rain gauge. US weather bureau uses this rain gauge for measuring the rainfall. Its construction includes a 30cm diameter sharp edged receiver. At its end a funnel is provided for directing the rainwater into the receiver. One pair of buckets is pivoted on a fulcrum below the funnel in such a way that when one bucket receives 0.25mm depth of rainfall, it tips and empties its rainfall into the container, and immediately the second bucket comes below the funnel (Fig..). The rainfall measurement is recorded in terms of number of tips made for a given rainfall event, which is indicated on a dial actuated by electrical circuit.. Assignment- Describe various types of rain gauges used for recording the rainfall data with a neat sketch. Suggested Reference 7. Watershed Hydrology- R. Suresh 8. Soil & Water Conservation Engineering- R. Suresh 9. Hydrology and Soil Conservation – G. Das 10. Engineering Hydrology – K. Subramanya  Lecture-4 Rain gauge Installation The surrounding exposure of rain gauge station affects the catching of rainfall by a raingauge. Therefore, before establishment of raingauge, the selection of a suitable place is very important, otherwise the rainfall measurement would not be accurate as should be. There are few important points to follow for installation of raingauge station in the area. 1. The ground surface must be level and firm. The places such as the building roof, sloppy surface, and terrace wall should be avoided. 2. In hilly areas the valley and hill top should not be selected for installation of the raingauge. 3. The site should be representative of the area or watershed. 4. Wind affected site should be avoided. 5. Site should be open from all the sides. 6. In forest area the raingauge should be instated at the distance twice the height of tallest tree from the forest plantation. Also the gauge’s upper point should make an angle ranging from 20 to 30 degree from the upper point of the raingauge. 7. The receiver‘s height from the ground surface should be around 75cm. 8. The position of raingauge must be vertical. Rain gauge Distribution The rain gauge distribution in watershed for measurement of rainfall, in accurate way is very important. If number of rain gauge differs than the required depending on the areal extent and topography, there is likely to get effects on the results various estimates. The topography and extent of area of watershed decide the number of rain gauge stations to be there. The number of rain gauge stations for highly undulating watershed comparatively more to that of the flat topography watershed. Rain gauge stations as per WMO and IMD is described below, As per WMO 1. Flat reasons of temperate mediterrian and tropical reasons Ideal -600 to 900 sq km per raingauge Acceptable - 900 to 3000 sq km/per raingauge 2. For mountainous reasons of temperate, mediterrian and tropical reasons Ideal - 100 to 250 sq km/per raingauge Acceptable - 250 to 1000 sq km/per raingauge 3. In arid and polar zones – 1500 to 10,000 sq km per raingauge As per IMD Region 1. Plain reasons - 520sqkm/raingauge 2. Reasons Region of average elevation - 260-360 sq km/per raingauge 3. Pre dominantly hilly reasons Region with heavy rainfall – 130 sq km/per raingauge Note: 10% of total raingauge required is taken as recording type of raingauge for installation. Problem (1)- In watershed the tipping bucket type rain gauge is installed for taking measurement. Determine the depth of rainfall, if rainfall duration is 2.30 hour. Solution- The following standards are involved with the tipping bucket type rain gauge: 1. For single tipping, the time involvement = 6 to 7 seconds 2. Capacity of one compartment is =0.25mm. 2.30𝑥3600 As per above, total number of tipping = 6 = 1380 Therefore, depth of rainfall=1380x0.25=345mm Ans. Problem (2)- In a watershed of mountainous region of temperate zone determine the number of ideal and acceptable rain gauge stations, if area of watershed is 1500sqkm. Solution- As per WMO, the distribution of rain gauge station in mountainous watershed of temperate regions is as per below: Ideal- 1 station/100 to 250sqkm; and acceptable- 250 to 1000sqkm. Accordingly, the number of rain gauge stations would be as below: 1500 Ideal- 200 = 7.5 = 8.0  (Assuming 1station/200sqkm) 1500 Acceptable- 750 =2.0 (Assuming 1 station/750sqkm) Raingauge Adequacy – It Is determined by using the following formula 𝐶 2 𝑁 = ( 𝜀𝑣 ) Where, N= adequate No. of raingauge stations CV= Coefficient of variations in percent 𝜀 = percent allowable error taken as 10% Problem (3)- In a watershed there are total 7 rain gauge stations, out of which one is siphon type rain gauge. On the basis of statistical analysis of long term rainfall data of said watershed the coefficient of variation is obtained as 35%. Determine the adequate number of rain gauge stations to be in the watershed. Take the value of allowable percent error in estimation as 10%. Solution- Adequacy of rain gauge stations is given by the following formula, 𝐶 √𝑁 = ( 𝜖𝑣 ) in which, N is the adequate number of rain gauge stations to be; 𝐶𝑣 is the coefficient of variation given as 35% and percent allowable error is 10%. Substituting these values in above formula; and after solving, we have, 35 √𝑁 = (10) N= 12.25 = 12.0 Normal Rainfall- It is an average of the rainfall values over a 30-year period. Rainfall may very often be either above or below the seasonal average, or "normal." Common Errors in Rainfall Measurement Few important errors in rainfall measurement by the raingauge are mentioned as under, 1. In non- recording type raingauge (Symons’s type) about 2% error is introduced due to displacement of water level by measuring scale. 2. Possibility of errors due to initial wetting of dried surface of the catch can or receiver. 3. The dents in catch can or receiver also introduces errors in measurement. 4. A high temperature cause evaporation loss also signifies a kind of errors in rainfall measurement. The errors may be up to 2%. 5. A high wind velocity deflects the rainfall to fall at the mouth of the rain gauge, introduces an errors in rainfall catching. The research revealed that at the wind velocity of 10mile per hour the catching of rainfall is declined to the tune of about 17% while at 30mile per hour it may be up to 60%. Missing Rainfall Data In normal course, sometime what happens, because of several reasons such as absence of observer, instrumental fault etc there is short breaks in the rainfall records. In this condition to fill the break the estimation of missing rainfall data is essentially required. The following methods are commonly used for computing the value of missing rainfall data 1. Arithmetic Mean Method 2. Normal Ratio Method Arithmetic Mean Method- This method follows following formula for determining the mean aerial rainfall, in which 'n' is the number of raingauge stations in nearby area, 'Pi' is rainfall depth at ith station and 'Px' is missing rainfall data. Solve example….. illustrates the computation procedure. Normal Ratio Method- This method is used when normal annual rainfall at any of the index station differs from the interpolation station by more than 10%. Missing rainfall data is predicted by weighing the rainfall of index stations by the ratios of their normal annual rainfall. Formula is given as under, For 3 number of defined index raingauge stations the above formula is expanded as in which Px is the missing rainfall at raingauge station 'x' of a given rainfall event, Pi is the precipitation for the same period and same rainfall event of "ith" raingauge station among group of index stations, Nx the normal annual rainfall (NAR) of station x and Ni the normal annual rainfall of 'ith' station. The solve example – illustrates the procedure. Problem ()- In a watershed four rain gage stations namely a, B, C and D are instatlled for recording rainfall data. The normal annual rainfall of these four stations is 75, 60, 70.5 and 87 cm, respectively. The rain gauge station A does not have the annual rainfall observation for one year during total length of record, because of disorder of the rain gauge. Calculate the missing value of rainfall data of rain gauge station A, if the annual rainfall recorded at other three stations for that particular year was 85, 67.5 and 75 cm, respectively at B, C and D, respectively. Solution- The variation in normal rainfall data is more than 20% at all the four rain gauge stations. In this condition, the normal ratio method for computing the missing value of annual rainfall of station A is suitable. Accordingly, the formula for computing the missing annual rainfall is given as under. 𝑁 𝑃 𝑃 𝑃 𝑃1 = 1 ( 2 + 3 + 4 ) 𝑚−1 𝑁2 𝑁3 𝑁4 in which, 𝑃2 = 85𝑐𝑚; 𝑃3 = 67.5𝑐𝑚; 𝑃4 = 75𝑐𝑚, 𝑎𝑛𝑑 𝑁1 = 75𝑐𝑚; 𝑁2 = 60𝑐𝑚; 𝑁3 = 70.5𝑐𝑚; 𝑁4 = 87𝑐𝑚 𝑎𝑛𝑑 𝑚 = 4. Substituting these values in above formula and solving , we have, 75 85 67.5 75 𝑃1 = ( + + ) 4−1 60 70.5 87 = 81𝑐𝑚 𝑨𝒏𝒔. Assignment- (1) Describe design of rain gauge network. (2) Describe WMD and IMD guidelines for distribution of rain gauge. Suggested Reference 11. Watershed Hydrology- R. Suresh 12. Soil & Water Conservation Engineering- R. Suresh 13. Hydrology and Soil Conservation – G. Das 14. Engineering Hydrology – K. Subramanya  Lecture-5 Mean Areal Rainfall Average rainfall is the representative of large area, which is computed with the help of rainfall data generated from well distributed raingauge network system of the watershed. The computing methods are elaborated as under, 1. Arithmetic or station average method 2. Thiessen Polygon Method 3. Isohyetal Method. Arithmetic Average Method This method computes arithmetic average of the rainfall by considering point rainfall observations of all the raingauge stations installed in the area. This method computes accurate value when rainfall is uniformly distributed in the entire area, as in this situation equal weightage of area is assigned to the point rainfall data. Formula is given as under, where 𝑃̅ is the mean rainfall is over an area, P is the point rainfall at individual station i, and n is the total number of stations. Solve problem (1) illustrates the computation procedure. Problem (1)- In a topographically homogeneous watershed total four number of non- recording and one recording type rain gauges have been installed for recording the rainfall measurements. The point rainfall of four non- recording type rain gauge stations have been observed to the tune of 250,175,225 and 270mm, respectively during a given rainfall event. Determine the mean areal rainfall of the watershed for the said rainfall event. Solution- The mean areal rainfall of the watershed can be computed by using the simple arithmetic mean method, given as under: 𝑃 +𝑃 +𝑃 +𝑃 𝑃𝑎 = 1 2 4 3 4 250+175+225+270 = 4 = 230𝑚𝑚 Ans. Thiessen Polygon Method This is a graphical method for computing MAP. It computes by weighing the relative area of each raingauge station equipped in the watershed. It follows the concept that the rainfall varies by its intensity and duration, spatially. Therefore, the rainfall recorded by each station should be weighed as per the influencing area (polygons). This method computes better for the areas having flat topography and size ranging from 500 to 5000 km2. Computing steps are described as under, 1. Plot the locations of raingauge stations on map of the area drawn to a scale. 2. Join each station by straight line. 3. Draw perpendicular bisectors of each line. These bisectors form polygons around each station. Area enclosed within polygon is the effective area for the station. For a raingauge station close to the boundary, the boundary lines forms its effective area. 4. Determine effective area of each raingauge station. For this the planimeter can be used. 5. Calculate MAP by using the following formula, in which, Pi is the rainfall depth of raingauge station i and A is the total area of watershed. Solve problem(2) illustrates the computation procedure. Problem (2)- Compute the mean areal rainfall of the watershed by using Theissen Polygone Method. The details are cited below. Rain gauge station A B C D E  Measured rainfall (cm) 10.5 11.56 9.57 10.50 11.63 Area of enclosed polygon 15.0 23.5 35.9 8.50 12.35 (sqkm) Solution- In Theissen Polygon Method the following formula is used for computing the value of mean areal rainfall depth. ∑𝑛𝑖=1 𝑃𝑖 𝐴𝑖 𝑃̅ = 𝐴 in which, 𝑃𝑖 is the rainfall depth for rain gauge station i and 𝐴𝑖 is the area of polygon enclosed by the rain gauge station i and A is the total area of watershed. Computation is shown below. Rain gauge Measured rainfall Enclosed area of Rainfall x enclosed area of station (cm) polygon (sqkm) polygon (cm.sqkm) Col.II x Col.IV I II III IV A 10.5 15.0 157.50 B 11.56 23.5 271.66 C 9.57 35.9 343.56 D 10.50 8.5 89.25 E 11.63 12.35 143.63 Total 95.25 1005.60 Therefore, mean areal rainfall= 1005.60/95.25=10.56cm Ans. Isohyetal Method This is also a graphical method, in which an isohyets map is prepared with the help of measured rainfall data of various raingauge stations located in the watershed. An isohyet map includes a network of isohyet lines. Each line represents a fixed value of rainfall depth. Computation of MAP under this method is done by using following steps, 1- Collect the map of area/watershed. The map should to the scal. 2- Draw isohyet map with the help of measured rainfall data of various raingauge stations installed in the watershed. 3- Find the area enclosed between each isohyet. 4- Multiply the area enclosed between each isohyet by the average precipitation, i..e 𝑃 +𝑃 𝐴 ( 1 2 2 ). 5- Find the sum of product of area enclosed and average of rainfall for all segments of Isohyet map. 6. Divide the sum of the values found in step- 5 by the total area of the watershed to get MAP of watershed. Computing formula is mentioned as under, 𝑃1 + 𝑃2 𝑃2 + 𝑃3 𝑃3 + 𝑃4 𝐴1 ( 𝑃̅ = 2 ) + 𝐴2 ( 2 ) + 𝐴3 ( 2 ) 𝐴 Solve problem (3) illustrates the computation procedure. Problem (3)- In a watershed total five rain gauge stations (A, B, C, D, and E) are installed for taking rainfall measurements. Calculate the mean areal rainfall depth using Isohyetal Method for a particular rainfall event. The details about measured rainfall and area enclosed by respective rain gauge station are given as under. Rain gauge station A B C D E F Measured depth of rainfall (cm) 5.35 4.75 6.45 5.00 4.55 3.50 Area enclosed by isohyets (sqkm) 75 125 65 70 100 70 Solution- In Isohyetal method of mean areal rainfall computation the following formula is used. 𝑃 +𝑃 𝑃 +𝑃 𝑃 +𝑃 ( 1 2)𝐴1+( 2 3)𝐴2+( 3 4)𝐴3+⋯ ̅𝑃 = 2 2 2 𝐴 Computation is shown as below. Rain gauge Measured Isohyet area Average of rainfall Pav x enclosed station rainfall (sqkm) of two consecutive isohyet area depth(cm) isohyets (cm) (sqkm.cm) I II III IV V  A 5.35 75.0 - - B 4.75 125.0 5.05 378.75 C 6.45 65.0 5.60 700.00 D 5.00 70.0 5.73 372.45 E 4.55 100.0 4.78 334.60 F 3.50 70.0 4.03 403.0 Total 505.00 25.19 2188.80 The mean areal rainfall is equal to 2188.8/505= 4.33cm Ans Suggested Reference 15. Watershed Hydrology- R. Suresh 16. Soil & Water Conservation Engineering- R. Suresh 17. Hydrology and Soil Conservation – G. Das 18. Engineering Hydrology – K. Subramanya  Lectures-6 Rainfall Analysis Mass Curve- It is the plot of accumulated rainfall against time, in chronological order (Fig-). The rainfall record generated by float type and weighing-bucket type gauges is in terms of mass curve. Mass curve acts as tool to determine the duration & magnitude and intensity of rainfall event. In case of the observation of non-recording rain gauges, the mass curve is prepared on the basis of knowledge of approximate beginning and end of rainfall event and by taking guidance from the mass curve of adjacent recording raingauge stations. Fig.1. View of mass curve of rainfall Double Mass Curve- It is used to check the consistency of many kinds of hydrologic data. Also, it can be used to adjust inconsistent rainfall data. The plotting between cumulative data of one variable versus the cumulative data of a related variable, if produces a straight line indicates consistency in record, otherwise, break in the double-mass curve shows inconsistency in record. The break point indicates the time from when there is development of inconsistency, which could be because of several reasons like change in original location/ exposure of the instrument/ device used for measurement. Consistency of Rainfall Data In the condition when a long time has been to a raingauge establishment, then there is possibility of change in the surroundings of the rain gauge in reference to the original condition. In result the rainfall to be measured by the raingauge gets change, and thus the rainfall data becomes inconsistence. The change in raingauge surroundings may be due to various reasons, such as, 1. Construction of new infrastructures like buildings apartments etc. 2. Plantation of orchards etc. 3. Introduction of instrumental errors. 4. Repetition of observational errors from a certain period. Consistency test provides the time from when the inconsistency is introduced in the data. It is tested with the help of double mass analysis method. The stepwise procedure is described as under, 1. Collect the rainfall data (normally annual) of the raingauge station which consistency is to be tested (say station X) and of the surrounding rain gauge stations. 2. Find the mean of the surrounding rain gauge data called base station data. 3. Calculate the cumulative rainfall of inconsistence raingauge station and of the base stations. 4. Plot the cumulative rainfall of base station on X -axis and corresponding rainfall of station X on Y-axis. It is shown in Fig- 5. Find the regime where rainfall data are under inconsistence nature. It can be demarcated by getting the change in slope of the plotted curve.  6. Rectify the inconsistence nature of the data, by multiplying a factor to them. The multiplying factor is the ratio of slope of straight line of consistence data to the slope of inconsistence data. Hyetograph-Hyetograph is the plot of rainfall intensity vs time presented as bar chart (Fig.). It is derived with the help of mass curve. Hyetograph represents the characteristics of a rainfall event and acts as a tool to develop a design storm for predicting extreme floods. Also, a hyetograph casts the information about total depth of rainfall occurred during rainfall event. Hyetograph is used hydrological analysis of catchment for (i) predicting flood; (ii) for estimating runoff and (iii) for deriving unit hydrograph. Fig. View of hyetograph of a storm Problem (1)- The rainfall intensity vs time data derived from the mass curve of a specific rainfall event is given as under: Rainfall time (hour) 0-1.5 1.5-3.0 3.0-4.5 4.5 – 6.0 6.0 – 7.5 7.5 – 9.0 Rainfall intensity 3.50 5.25 6.00 5.00 6.25 3.50 (cm/h) Determine the depth of rainfall of the rain event. Solution- The relationship between rainfall intensity and time is the hyetograph. Area of hyetograph represents the depth of rainfall. Calculation is shown in following table. Rainfall time Rainfall Time Depth of rainfall (cm) interval (hour) intensity (cm/h) (hour) Col. (2)x Col.(3) (1) (2) (3) (4) 0 -1.5 3.50 1.5 5.250 1.5- 3.0 5.25 1.5 7.875 3.0 – 4.5 6.00 1.5 9.000 4.5 – 6.0 5.00 1.5 7.500 6.0 – 7.5 6.25 1.5 9.375 7.5 – 9.0 3.50 1.5 5.250 Total depth of rainfall 44.250 cm Total rainfall depth is computed as 44.25cm. Ans. Problem (2)- Using the data of above problem; calculate the depth of effective rainfall if average loss of rainwater during rainfall duration is @2.50cm/h. Solution- Effective rainfall depth is determined by subtracting the loss of rain water during rainfall. The average loss of rainwater is termed as - index. In this problem the - index is given as 2.50cm/h. Calculation is presented in following table. Rainfall time interval (hour) Rainfall intensity Time (h) Depth of rainfall (cm) (cm/h) Col. (2)x Col.(3) (1) (2) (3) (4) 0 -1.5 3.50 1.5 5.250 1.5- 3.0 5.25 1.5 7.875 3.0 – 4.5 6.00 1.5 9.000  4.5 – 6.0 5.00 1.5 7.500 6.0 – 7.5 6.25 1.5 9.375 7.5 – 9.0 3.50 1.5 5.250 Total depth of rainfall 44.250 cm Effective rainfall depth= Total rainfall depth- Loss of rainwater during rainfall = (44.25- 9x2.5) cm = 21.75cm. Ans Depth-Area Relationship This relationship provides the view about variation in overall depth of rainfall with respect variation in aerial extent of watershed. In general, for a rainfall of specific duration the average depth decreases with increase in area, exponentially. Depth- area relationship is presented below, in which 𝑃̅ is the average depth of rainfall (cm) over an area A (km²), P0 is the highest rainfall depth (cm) at the storm centre and K and n are the constant for the region specific. Dhar and Bhattacharya (1975) have determined the value of K and n for different duration storms on the basis of 42 severe most storms in north India, shown in Table-. Duration K n 1 day 0.0008526 0.6614 2 days 0.0009877 0.6306 3 days 0.0017454 0.5961 In this formula to determine the exact or accurate value of P0 is not possible, because it is very unlikely that the established raingauge station coincides the storm centre. Considering this fact in view the analysis of rain events covering a large area the highest station rainfall is taken as the average depth over an area of 25 km². Problem (3)- Determine the average depth of rainfall in the watershed of 1000sqkm size by using Depth- Area relationship, if the highest rainfall at the centre of rain storm is 15cm. Take the constants K=8.256x10 -4 and n=6.614x10-1. Solution - The depth- area relationship is given by the following expression, ̅𝑃 = 𝑃0 𝑒 −𝐾𝐴𝑛 in which, ̅𝑃 is the average depth of rainfall (cm); 𝑃0 is the highest rainfall depth at the centre of storm (cm):A is the area of watershed; and K and n are the empirical constants. Substituting the values of different parameters in above formula, and solving, we have, ̅𝑃 = 15𝑒 −8.256x0.00010x10000.6614 = 6.21cm Ans. Rainfall Intensity – Return Period Relationship Return period or recurrence interval is the number of years in which an event can be expected once. The relationship between rainfall intensity and return period is very important to determine the rainfall intensity for different return periods or rainfall frequencies, which is mainly required for computation of runoff rate to be used for design of hydraulic structures. For example the soil conservation structures like drop structures, grassed waterways, farm ponds, reservoirs, dams etc are designed on the basis of runoff rates for different return periods. The rainfall duration increases when intensity decreases and vice-versa. The rainfall intensity increases when return period increases and vice-versa. The formula for rainfall intensity – return period is presented by the following expression. 𝐾𝑇𝑎 𝑖 = (𝑡+𝑏)𝑑 in which, i is the rainfall intensity (cm/h) for given return period (T, year) and duration of rainfall (h) and K ,a ,b and d are the regional constants. Solve problem (3) illustrates the computation procedure. Problem (4)- Calculate the rainfall intensity of 1000 sqkm size catchment for 25-yeras return period to be used for design of semi- permanent gully control structure. Take the duration of rainfall is 6.0 hours and constants, K=45.216; a= 0.50; b=4 and d=0.6870. Solution- The formula for rainfall intensity – return period is presented by the following expression. 𝐾𝑇𝑎 𝑖 = (𝑡+𝑏)𝑑 in which, i is the rainfall intensity for given return period and duration of rainfall (cm/h) and K ,a ,b and d are the empirical constants. Substituting the values of above parameters in above formula and solving, we have, 45216𝑥 250.50 𝑖= (6+4)0.6870 226.08 = 4.864 = 46.45𝑐𝑚/ℎ Ans. Rainfall Frequency It is also known as plotting position. The design of hydraulic structures, flood control structures, soil conservation structures, drains, culverts etc are based on probability of occurrence of extreme rainfall events. The relationship for plotting position/rainfall frequency is given by the Weibul’s equation presented as under: 𝑚 𝑝% = 𝑥100 𝑛+1 in which, 𝑝% is the plotting position; m is the rank number of event after arranging in descending order and n is the length of record. In terms of return period (T) the relationship as under, 1 𝑇= 𝑝% Solve example illustrates the computation procedure. Problem (5)- Determine the plotting position of highest rainfall of 50cm during 50 years period. Solution- The relationship for plotting position is given by the Weibul’s, given as under: 𝑚 𝑝% = 𝑛+1 𝑥100 In which, 𝑝% is the plotting position; m is the rank number of event after arranging in descending order and n is the length of record. Sustituing the values of different parameters (m=1 and n=50) in above formula, and solving, we have, 1 𝑝% = 50+1 𝑥100 = 1.96% Ans Problem (6) - At Pusa Farm (Samastipur), Bihar the highest rainfall depth to the tune of 25cm was recorded in the years 1987 and 1995 during 75 years period. Determine their plotting positions and return periods. Solution- In this case the highest rainfall 25cm is observed in the years 1987 and 1995. The rank number will be 1 for the 1987 as it has first place first and rank 2 will be for the year 1995 because this event was taken place after first event. Accordingly, plotting positions for years 1987 and 1995 will be as below. Using Weibul’s method, 𝑚 𝑝% = 𝑁+1 𝑥100 in which, 𝑝% is the plotting position(percent); m is the rank number and N is the length of record. Accordingly, (i) Plotting position of the rainfall depth 25cm for the year 1987- For this case, m=1 and N=75. Substituting these values in above formula and solving, we have, 1 𝑝% = 75+1 𝑥100 = 1.25% Ans. 1 Return period (T) is the inverse of plotting position. For this case it is 𝑇 = 1.25 =80 years. ⁄100 (i) Plotting position of the rainfall depth 25cm for the year 1996- For this case, m=2 and N=75. Substituting these values in above formula and solving, we have,  2 𝑝% = 75+1 𝑥100 = 2.50% Ans. 1 For this case return period (T) is = 2.5 =40 years. ⁄100 Problem (7)- Determine the length of rainfall record, if return period of highest rainfall of 75 cm is 25 years. Solution- Using Weibul’s method, 𝑚 𝑝% = 𝑁+1 𝑥100 1 1 In this formula, plotting position ( 𝑝% ) is equal to 𝑇 , i.e. 25 𝑥100 = 4.0%. Substituting the value of m =1 and plotting position ( 𝑝% ) as 4.0% and solving, we have, 1 4.0 = 𝑁+1 𝑥100 100 𝑁= −1 4 = 24 𝑦𝑒𝑎𝑟𝑠 Ans. Suggested Reference 19. Watershed Hydrology- R. Suresh 20. Soil & Water Conservation Engineering- R. Suresh 21. Hydrology and Soil Conservation – G. Das 22. Engineering Hydrology – K. Subramanya  Lecture-7 Rainfall Abstractions and Initial Loss Initial Loss of Rainwater In course of rainfall occurrence there is significant water loss from various sources such as interception, evaporation, transpiration, infiltration, depression storage. In result the overland flow and runoff yield against rainfall gets reduce. These loses are referred as initial loss. The prediction of initial rainwater loss is very important for determining the runoff and hydrograph derivation. Interception- It is the amount of rainwater loss due to abstractions from initially dry surfaces of the objects lying on the ground surface. The objects may be the live vegetations e.g. herbs, shrubs & trees and any dry surfaces like building etc. From a tree the interception is mainly from the canopy, is called canopy interception (Fig-) Intercepted rainwater is lost due to evaporation is called interception loss. The extent of interception loss depends on a host of factors such as types and characteristics of vegetations, rainfall, temperature, season of the year, wind velocity etc, mainly. Average interception loss from few forest cover and crops is presented in following table. S. No Vegetation Average Interception loss (of total rainfall) 1. Forest Coniferous forests 15-35% broad-leaved forests 9-25% natural grasses 14-19% 2. Crops Oats about 7% corn about 16% clover about 40% Interception losses generally occur during the first part of a precipitation event and the interception loss rate trends toward zero rather quickly (Figure 1). Interception losses are described by the following equation (Horton reprinted by Viessman 1996): 𝐿𝑖 = 𝑆 + 𝐾. 𝐸. 𝑡 in which 𝐿𝑖 is the total volume of water intercepted, S is the interception storage, K is the ratio of the surface area of the leaves to the area of the entire canopy, E is the rate of evaporation during rainfall and t is the time. This equation assumes that the rainfall is enough to satisfy the storage capacity of vegetation. Horton equation suggests that the total interception is dependent on the duration of rainfall as longer duration rainfall event allows more evaporation from the canopy. Brooks (2003) also suggested following formula for predicting interception loss, 𝐿𝑖 = 𝑃𝑔 − 𝑇ℎ − 𝑆𝑓 in which 𝐿𝑖 is the canopy interception loss, 𝑃𝑔 is the gross precipitation, 𝑇ℎ is the through fall and 𝑆𝑓 is the stem flow. The intensity of the storm also plays a role in canopy interception (Viessman 1996); however, there is debate as to whether intensity increases or decreases interception storage in canopy (Keim 2003). Interception is mainly at two levels depending on features of vegetation, given as under, 1. Primary interception, and 2. Secondary Interception Primary Interception takes place from the vegetations of uniform canopy like crops etc where secondary interception is from the vegetations having more that one level canopy such as found n the forest covers. In forest the tall tress constitutes primary level of interception. And the vegetative layer existing below the tall tree canopy is the secondary canopy, which intercepts the rainwater falling from the upper canopy, called secondary interception and loss of it is as secondary interception. Through fall – It is the process of falling of rainwater through the sp[aces of plant canopy. Through fall is affected by the factors such as plant leaf and stem density, type of precipitation, rainfall intensity and duration of the rainfall event. Through fall may be bys direct falling of rain water or dropping of intercepted water through tips of the leaf. The measurement of through fall can be carried out by putting a bucket below the tree canopy. Stem flow – In course of rainfall a part of rainwater is also absorbed by the branches and stem of the tree. If rainfall is continued, then after some time the absorbed water starts flow through the branches and joins to the soil surface through stem, is called stem flow. By this process a part of rain water is also absorbed by the surfaces of tree branches and stem, which is lost due to evaporation. The characteristics of tree branches and stem affect the rate of stem flow. A tree with rough surface of branches /stem involves high level of stem flow loss. Depression storage- Depressions are the small size low pockets lying on the ground surface. Such appearances are very common in undulating lands. Presence of depressions on the ground surface cause significant level of water retention in them, known as depression storage. It also constitutes a kind of rainwater loss because the retained water is lost due to infiltration and evaporation actions. Depression storage affects the rainfall- runoff relationship of the area. Linsley (1982) suggested following formula for predicting the volume of retained in depression storage at any time during a rainfall event 𝑉 = 𝑆𝑑 (1 − 𝑒 𝑘𝑃𝑒 ) In which V is the volume of water retained in depression storage, 𝑆𝑑 is the maximum storage capacity of, 𝑃𝑒 is the rainfall excess and k is a constant equal to 1/ 𝑆𝑑. As per various research studies the level of depression storage for different land covers are presented in following table, Land cover Depression storage Reference (inch) Impervious, 1% slope, flat roofs, 0.0625 – 0.125 Tholin and Kiefer(1960) parking lots, roads Impervious, 2.5% slope and sloped 0.05 Viessman (1996) roofs Turf grass 0.25 Tholin and Kiefer(1960) Open fields 0.40* Urban Drainage and Flood Control District (2008) Wooded Areas 0.40* Urban Drainage and Flood Control District (20080 *These values include interception losses by vegetation Infiltration It is defined as the entry of water into the soil by crossing the imaginary boundary between soil and atmosphere. Infiltration is treated as one of the most important factors making rainwater loss from the total. Runoff generating potential of area/ soil is very much affected by infiltration rate. A sandy soil belt involves very less potential for runoff yield to that of the heavy soil belt, because of the reason that the infiltration rate is quite high of sandy soil as compared to the heavy soil. The rate of water soaking by the soil is called as infiltration rate and its maximum rate is termed as infiltration capacity. Infiltration rate, Initial Infiltration rate and Basic Infiltration rate The velocity or speed at which water enters into the soil is called infiltration rate. It is usually measured by the depth (in mm) of the water layer that can enter the soil in one hour. For example an infiltration rate of 15 mm/h reveals that a 15 mm water depth standing on soil surface will infiltrate into the soil in one hour time. Initially, infiltration rate is very high and becomes slower as time proceeds. This is because of the reason that at beginning there is sufficient space or reservoirs are there for storing water in the soil media and likely to get reduce because of filling of water content in them with time advancement. In dry soil, the water infiltrates rapidly called initial infiltration rate. As more water replaces the air in the pores, the water from the soil surface infiltrates more slowly and eventually reaches a steady rate, called the basic infiltration rate. Basic infiltration rate of various soils are narrated as under, Soil type Basic infiltration rate (mm/h) sand less than 30 sandy loam 20 - 30 loam 10 - 20 clay loam 5 - 10 clay 1-5 Infiltration rate depends on host of factors such as soil texture (size of soil particles) and soil structure (arrangement of soil particle), soil depth, land topography, water- table position, vegetations and climatic factors, mainly. Field Test for Infiltration Measurement The most common method to measure the infiltration rate is by using the Double Ring Infiltrometer. The diameter of inner ring is 30 cm and of outer rings 60 cm. View of double ring infiltrometer is shown in Fig-. Experimental procedure is described as under. Procedure Different steps followed are numerated as under, Step (1) Insert the inner ring (30cm dia.) into the soil at least for 15 cm depth by using hammer.  Care should be taken that there should not be any kind of damage in the ring. Normally for this purpose a wooden plate is placed at the top of ring and hammering is done on that.. The side of ring must be in vertical position. After it, insert the measuring rod in the area of inner ring for taking observations about depth of water entered into the soil Step 2: Also, insert the outer ring (60cm dia.) ring into the soil for 15cm depth to create a protection bund or buffer area around inner ring. Step 3: Pouring water into both the rings for the depth of 70-100 mm. The pounding water should be done quickly. The water in outer area of inner ring or within the two rings prevents the lateral movement of water from the infiltrometer. Step 4: Record the time when test begins, and note the water level on the measuring rod. Initially the observations should be taken at lesser time intervals. Step 5: After lapse of 1-2 minutes, record the drop in water level in the inner ring area. It is noted from measuring rod. And Add water to bring the water level at original level. Also, maintain the water level outside the ring similar to the inside. Take the readings frequently (e.g. every 1-2 minutes) at the beginning, but extend the interval as the time goes on (e.g. every 20-30 minutes) and rate of water level drop also becomes slow. Step 6: Continue the test until the rate of water level drop becomes constant. It is suggested that at a given site at least two infiltration tests should be carried out to have a correct result. Water Year: It starts from Ist June to 31st May during the year. Suggested Reference 23. Watershed Hydrology- R. Suresh 24. Soil & Water Conservation Engineering- R. Suresh 25. Hydrology and Soil Conservation – G. Das 26. Engineering Hydrology – K. Subramanya  Lecture-8 Runoff and Its Computation Runoff Definition and Occurrence Runoff is defined as the flow of excess rainwater through a channel, gully, river or any fluvial path. The overland flow is the main input for generating runoff. On watershed scale, the rainwater after getting satisfied with the initial losses such as, abstractions, evapotranspiration, infiltration etc the remaining rainfall water is called excess rainfall or effective rainfall, is converted into a head of water on the ground surface, which attains motion due to land slope. In result the standing water starts moving down the slope. This phenomenon is called overland flow. As soon as the overland flow joins any flow path like channel etc, the runoff takes place. Since, this flow is through channel; therefore, runoff is also called channel flow. The length of overland flow is limited to a very short distance, normally maximum up to 150 m. Runoff Classification It is classified as 1. Direct runoff, and 2. Indirect runoff Direct Runoff - It is the surface runoff, takes place on the ground surface through the streams / channels etc. Since, it takes place very soon after start of rainfall event; therefore, it is called as the direct runoff. Direct runoff is also known as surface runoff. The interflow, in which the infiltrated rainwater joins to the stream flow in terms of influent flow, soon after start of rainfall, is also the part of surface or direct turnoff. The reason behind this is that the time gap between rainfall occurrence and interflow is very less, say as few minutes. The interflow always takes place above the main ground water- table. Indirect Runoff- This type of runoff takes place below the ground surface. In the course of occurrence of rainfall a part of rain water which is infiltrated into the soil media moves downward and joins to the water –table. The joined rainwater starts moving or flow along with ground water to the other places in forward direction, called indirect runoff. Since, this runoff takes place below the ground surface; therefore, it is also called sub-surface runoff. Sometimes, it is also known as the delayed runoff because of the reason that there is very large gap between occurrence of rainfall and formation of runoff say for example 1- year or more. Factors Affecting Runoff Conceptually, in runoff formation the rainfall is an input and watershed is the system on which rainfall take place, and runoff is the output. It means that the parameters associated to the rainfall (climate) and watershed affect the runoff yield. Broadly, the list of factors affecting the runoff are listed in following table- S. No Parameters Factors affecting 1. Climatic 1.1 Rainfall 1.1.1 Intensity 1.1.2 Duration 1.1.3 Direction Temperature 1.3 Wind velocity 1.4 Humidity  2. Watershed 2.1 Physiographic factors 2.1.1 Size 2.1.2 Shape 2.1.3 Slope 2.1.4 Form factor 2.1.5 Compactness factor 2.1.6 Drainage density 2.1.7 Stream frequency 2.1.8 Channel slope 2.1.9 Channel length 2.1.10 Channel size 2.2 Soil 2.2.1 Type 2.2.2 Depth 2.2.3 Slope 2.3 Land use Form Factor-It is defined as the ratio of average width to the axial length of the watershed. Axial length of watershed is the distance between outlet and the remotest point of the watershed. Width is determined by dividing the area of watershed by its axial length. Compactness factor- It is the ratio of watershed perimeter to the circumference of a circle which area is equal to the area of watershed. It is given by 𝑾𝒂𝒕𝒆𝒓𝒔𝒉𝒆𝒅 𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝑷 𝑪𝒐𝒎𝒑𝒂𝒄𝒕𝒏𝒆𝒔𝒔 𝒇𝒂𝒄𝒕𝒐𝒓 = Circumference of a circle which area is equal to the area of watershed = 𝟐𝝅𝑨 In which, P is the perimeter and A is the area of watershed. Problem (1)- Determine form factor of an elongated watershed, which axial length is 15000m and average width as 750m. Solution- The formula of form factor is given, below. 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟𝑠ℎ𝑒𝑑 𝐹𝑓 = 𝐴𝑥𝑖𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟𝑠ℎ𝑒𝑑 in which, average width and axial length of watershed is given as 15000 and 750m, respectively. Substituting these values in above equation and solving, we have, 750 𝐹𝑓 = 1350 = 𝟎. 𝟓𝟓 Ans. Problem (2)- Determine compactness factor of an elongated watershed, which perimeter is 2500m and area is 1.5sqkm. Solution- The formula of compactness factor is given, below. 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟𝑠ℎ𝑒𝑑 𝐶𝑓 = 𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑎𝑛𝑐𝑒 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒 𝑤ℎ𝑜𝑠𝑒 𝑎𝑟𝑒𝑎 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟𝑠ℎ𝑒𝑑 in which, perimeter of watershed is given as 2500m and circumference of circle is determined as below. 𝜋𝑟 2 = 1.5𝑥1000𝑥1000𝑚2 r = 691m Circumference of circle = 2 𝜋r = 2x3.14x 691 m = 4339.48m Therefore, 2500 = 4339.48 = 0.576 Ans. Stream Classification Manly, streams are classified as under, 1. Perennial stream, 2. Intermittent stream, and 3. Ephemeral stream Perennial stream carries runoff flow throughout the year. In off-season, i.e. summer season the flow of water is contributed by ground water. Resulted hydrograph is extended for the entire year duration. Intermittent streams do not have continuous flowing water year-round and are not relatively permanent. Water flowing water period is limited during wet season (winter-spring) but are normally dry during hot summer months. The water flow in Ephemeral streams is confined with the occurrence of rainfall. Comparatively, these streams have less flow than the intermittent stream. Typically these are shallow and have very less flowing periods. Effective Rainfall Hyetograph: ERH is the plot of rainfall intensity and time after deducting the Phi – index. It is plotted in the form of bar diagramme. The area of ERH is the depth of effective rainfall. The time duration of ERH is the duration ER. Direct Runoff: It is the runoff directly formed due to rainfall. In other words, the runoff excluding base flow is the direct runoff. It can be presented in terms of volume and rate both. The depth of effective rainfall multiplied by area of watershed is the direct runoff volume. The depth of effective rainfall is the area of ERH. Relationship between ER and DR: In terms of depth both are same. However, the volume of direct runoff is the product of ER and area of watershed. In other words, area of ERH multiplied with the area of watershed casts the volume of direct runoff. Runoff Computation There are host of method and empirical formulae for computing the runoff from a watershed, few important amongst them are listed as under, 1. Rational method, 2. SCS method 3. Cooks method 4. Infiltration Indices method 5. Hydrograph method 6. Empirical formulae Rational Method This method computes the peak runoff of small watershed. Peak runoff is required for design of hydraulic structures such as culverts, bridges, drop structures, and others. The rational method is appropriate for estimating peak discharge for small drainage areas of up to about 80 hectares with no significant flood storage. The method provides the designer with a peak discharge value, but does not provide a time series of flow nor flow volume. This method follows the hypothesis that, 1. Runoff is directly proportional to the area of watershed , and 2. Directly proportional to the rainfall intensity 3. Rainfall intensity must be for the duration equal to time of concentration of watershed. Accordingly, if the area of watershed is A (-) and rainfall intensity for the time equal to time of concentration of watershed is I then, the equation of peak runoff (𝑄𝑝 ) is given as under, 𝑄𝑝 = 𝐶𝐼𝐴 Time of Concentration- The time of concentration (Tc) of watershed is defined as the time required for movement of rainwater from remotest point to the outlet of watershed. If rainfall duration is equal to or greater than the TOC then from entire watershed area the excess rainwater or runoff is started to generate, which cumulatively comes to the watershed outlet. The cumulative runoff joining to the watershed outlet is at highest level, called peak runoff. Fig- shows the view of formation of peak runoff depending on rainfall duration and TOC. In contrast when rainfall duration does not coincide to the TOC the runoff yield at outlet is not at peak level. The following formula given by Kirpitch (1940) can be used for determining the TOC of watershed. 𝑇𝑐 = 0.02𝐿0.77. 𝑆 −0.385 In which 𝑇𝑐 is the time of concentration (minute); L is the longest length of water course (m) and s is the average slope of water course (m/m). The above formula is revised by Haan etal (1982) by including the component of overland flow. They reported that the Rational Method does not compute well when size of watershed is less than 5sqkm area. Such watersheds are dominated by overland flow rather channel flow or the runoff. In this condition after incorporating the effect of overland flow the revised formula for TOC is mentioned as under, 2𝐿0 0.467 𝑇𝑐 = 0.02𝐿0.77. 𝑆 −0.385 + ( 𝑛0.5 ) 𝑆0 In which, Lo is the length of overland flow (m) and n is the Manning’s roughness coefficient and s0 is the slope of overland flow path (m/m). The value of Manning’s roughness coefficients are given in following table, Table- Manning’s roughness coefficient (n) S.No Surface condition n 1. Smooth and impervious surface 0.02 2. Smooth and bare surface 0.10 3 Cultivated new crops 0.20 4 Pasture or average grassed surface 0.40 5. Forest area with dense grass cover 0.80 Rainfall Intensity- It is the ratio of rainfall depth and duration of rainfall event. As per this definition the computed rainfall intensity does not fit for rational method. The rainfall duration must be taken as the TOC of watershed. Accordingly, the rainfall intensity is presented as the ratio of rainfall depth to the TOC. Sometimes, for design of hydraulic structures such as the drop structure or grassed waterways etc the design runoff for a given return period or rainfall frequency is required. For such cases the formula for rainfall intensity – return period is given by the following expression. 𝐾𝑇𝑎 𝑖 = (𝑡+𝑏)𝑑 in which, i is the rainfall intensity (cm/h) for given return period (T, year) and t is the TOC (h) and K ,a ,b and d are the regional constants. Solve problem (3) of lecture-6 illustrates the computation procedure. Runoff Coefficient- It is the fraction of total rainfall converted into runoff. In other terms it is the ratio of Runoff depth and total rainfall depth. Its value is dimensionless varies from 0 to maximum 1 in which 0 is for soils having very high rate of infiltration, i.e. there is no excess rainwater available for generating runoff from the surface. The value of runoff coefficient for sandy soil may be approaching 1.0 at the beginning of rainfall occurrence when total rain water is likely to get infiltrated into the soil. And the value of runoff coefficient as 1.0 which is maximum may be for concrete or any hard formation in which the infiltration of rainwater is about to zero. However, the runoff coefficient for use in rational method is cited in Table- 1 Table-1. Runoff coefficient S. No Land use and topography Soil Type Sandy loam Clay and silt Tight clay loam 1.0 Cultivated land Flat 0.30 0.50 0.60 Rolling 0.40 0.60 0.70  Hilling 0.52 0.70 0.82 2.0 Pasture land Flat 0.10 0.30 0.40 Rolling 0.16 0.36 0.55 Hilling 0.22 0.42 0.60 3.0 Forest land Flat 0.10 0.30 0.40 Hilling 0.30 0.50 0.60 4.0 Populated land Flat 0.40 0.55 0.65 Rolling 0.50 0.65 0.80 Normally, there is a large variation in soil types , slope gradient, vegetations in the watershed. Because of this reasion the selction of single runoff coefficnt value is not accurate. In this condition the coideraion of weighted runoff coeffcient is most appropriate, which should be interms of soil types , types of vegations an slope gradient , if so. The formual for detrmining the weighted runoff coefcient is mentioned as under, 𝐶 𝐴 +𝐶 𝐴 +𝐶 𝐴 +⋯𝐶𝑛 𝐴𝑛 𝐶𝑤 = 1 1 2 2 𝐴 3 3 In which, 𝐶𝑤 is the weighted runoff coefficient, A is the total land area and C1 … Cn are runoff coefficient for the area A1 ….An. Assumptions and Limitations Assumptions counted under rational method are as follows,  Applicable when TOC of watershed is at least equal to or greater than the duration of peak rainfall intensity.  The calculated runoff is directly proportional to the rainfall intensity.  Rainfall intensity is uniform throughout the duration of the storm.  The frequency of occurrence for the peak discharge is the same as the frequency of the rainfall producing that event.  Rainfall is distributed uniformly over the drainage area.  The minimum duration to be used for computation of rainfall intensity is 10 minutes. If the time of concentration computed for the drainage area is less than 10 minutes, then 10 minutes should be adopted for rainfall intensity computations.  The rational method does not count for storage in the drainage area. Available storage is assumed to be filled. Problem (3)- For a watershed of varying land use systems, soils and topography, deter