Irrigation and Drainage Engineering PDF

Irrigation and Drainage Engineering Lesson 1 Water Resources of the Philippines and its Demand in Various Sectors Definition Irrigation - Irrigation is defined as the science of artificial application of water to the land, in accordance with the ‘crop water requirements’ throughout t...

Irrigation and Drainage Engineering Lesson 1 Water Resources of the Philippines and its Demand in Various Sectors Definition Irrigation - Irrigation is defined as the science of artificial application of water to the land, in accordance with the ‘crop water requirements’ throughout the ‘crop period’ for full-fledged nourishment of the crops (Garg, 1996). Irrigation Engineering - the analysis and design of systems that optimally supply the right amount of water to the soil at the right time to meet the needs of the plant system. The enterprise may be growing plants for food, landscape irrigation, or other purposes. Irrigation water is supplied to supplement the water available from rainfall, soil moisture storage and capillary rise. However, in many cases, it is not possible to meet the full crop water requirement throughout the season due to limited water availability. In such cases, deficit irrigation is provided in the form of life saving or supplemental irrigation. Besides meeting the crop water requirement, irrigation is also provided for field preparation, climate control (crop cooling and frost control), and leaching of excessive salts. Advantages Increases agricultural productivity and allows for multiple cropping during a year Provide jobs. Reduces risk of crop failures. Higher productivity results in steady supply of food at lower prices (supply demand principle of economics) Improves socioeconomic conditions of farmers Disadvantages 1. Excessive irrigation may cause decrease in crop yield 2. Excessive irrigation may cause leaching of pesticide, insecticide, nitrogen and nitrates to groundwater and may also transport them to surface water systems. 3. In poorly drained soils water logging and salinity may occur. 4. In poorly maintained canals- excessive seepage may cause water logging. 5. Excessive groundwater pumping may cause decrease in groundwater levels which may damage aquifer structure and increase the risk of land subsidence Purpose of Irrigation Some of the main purposes of irrigation are enlisted below: 1. To supply essential moisture for plant growth 2. Transportation of fertilizers (Fertigation) 3. To leach or dilute salts in soil 4. To help in field preparation, dust control etc. 5. Other benefits of irrigation include cooling of the soil and atmosphere to create more favourable environment for crop growth and frost control Sources of Water 1. Natural Resources Rain, snow, hail and sleet are precipitated upon the surface of the earth as meteorological water and may be considered as the original source of all the water supplied. Surface and groundwater are main sources of irrigation water. Three aspects should be considered in appraising water resources which are the quantity, the quality, and the reliability of availability of water. Rainwater, rivers, lakes, streams, ponds and springs are natural sources of water. Dams, wells, tube wells, hand-pumps, canals, etc. are man- made sources of water. Irrigation water sources can be broadly classified into two main groups, namely, 1. Surface water sources a. Lakes c. Tanks e. Storage reservoir b. Ponds d. Rivers/streams  2. Groundwater sources a. Open wells c. Artesian wells e. Infiltration b. Tube wells d. Springs Irrigation water supply can be either obtained from surface water sources or groundwater sources or both. Both of these depend upon the precipitation 1. Surface water Water present on the surface of the earth in the form of oceans, rivers, lakes, ponds and streams is called surface water. Surface water accumulates mainly by direct runoff from precipitation i.e., rain or snow melting. The amount of available surface water depends largely upon rainfall. Surface water sources consists river, lake, and reservoir supplies. Dams or reservoirs are constructed to create artificial storage of water. Canals or open channels can be constructed to convey surface water from the rivers or reservoirs to the farm fields where it may directly applied to the field or stored in farm irrigation structures like ponds or tanks. The water is also conveyed through pipes by gravity or pumping. Thus, sources of surface water are i) Rivers and streams ii) Reservoirs iii) Tanks, ponds and lakes. a. River A river is a natural water course, usually of freshwater, flowing towards an ocean, a lake, a sea, or another river. In a few cases, a river simply flows into the ground or dries up completely before reaching another body of water. b. Reservoir A reservoir is a natural or an artificial lake, storage pond or impoundment from a dam which is used to store water. Reservoirs may be constructed across the rivers or may be excavated in the ground. c. Lake A lake is an inland water body of considerable size. Lakes can serve as the source or termination point for rivers or smaller streams. Lakes are distinct from lagoons as they are not part of the ocean. Lakes are larger and deeper than ponds d. Ponds A pond is a body of standing water, either natural or man-made, that is usually smaller than a lake. Generally, they contain shallow water with marsh and aquatic plants and animals. e. Tank Tanks are large excavations in which water is stored. They form an important source of water in many places. 2. Ground Water A part of the water which infiltrates into the soil after any rainfall event percolates to the groundwater table. Ground waters, generally, characterized by higher concentrations of dissolved solids, lower levels of colour, higher hardness (as compared with surface water), dissolved gasses and freedom from microbial contamination. Wells are generally used to extract groundwater. The extraction of groundwater is mainly by: 1. Dug well with or without straining walls 2. Dug cum bore wells 3. Cavity Bore 4. Radial collector wells 5. Infiltration galleries 6. Tube wells& bore wells. Groundwater that flows naturally from the ground is called a spring. Activity Present Status, Development and Utilization of Water resources in the Philippines 1. What is/are the major source/s of water in the Philippines? (include quantity of each source) 2. What are the major river basins in the Philippines? 3. What is the present status of groundwater in the Philippines? 4. What is/are the major users of surface water and groundwater in the country? 5. What is the total usable water resource of the country? 6. Determine the water demand by sector (irrigation, drinking, industry, energy and others)  Lesson 2 Irrigation in the Philippines Pre-spanish and Spanish period (17th -19th ¬century) Banaue rice terraces The zanjeras (Spanish for turn out) American Colonization (20th century) 1908 – government became involved in irrigation activities through division of BPW 1912 – irrigation act 2152, authorizes Irrigation Division to appropriate public waters, investigate, construct, operate and maintain irrigation systems; collect ISF 1916- Act 2652 authorized the granting of loans to private irrigation systems that were owned and operated by organized associations of landowners that had service areas of not more than 25 ha. 1924 - The Irrigation Division was downgraded to a mere section of the Design Division of the BPW. Japanese Regime (1937-1945) No significant irrigation development has occurred. Most of the irrigation systems during this period turned in poor condition Philippine Independence 1945 - The Irrigation Division was reactivated. 1963 - The Philippine Congress created the National Irrigation Administration through Republic Act 3601. The NIA is the government agency tasked to undertake planning, development, implementation, and maintenance of all irrigation projects/systems in the country (Mejia, 1999). Its mission is to develop water resources for irrigation and corollary physical and technical services in line with the development program of the national government. 1974 - The Martial Law administration of Pres. F.E. Marcos promulgated Presidential Decree (PD) 552 authorizing communal irrigation systems to receive assistance from NIA. However, IAs must pay back to NIA 10 percent of the direct cost of construction or rehabilitation of irrigation systems. PD 552 also implied that communal IAs must be viable and capable of collecting payments from farmers. 1975 - NIA entered into tie-up with the Farm Systems Development Corporation (FSDC). Under this partnership, NIA would continue to construct communal irrigation systems and FSDC would organize the corresponding IAs. But the partnership was short-lived. After only a few years, it was terminated because of failures to coordinate the technical and institutional aspects of irrigation development. 1976 - NIA launched the Participatory Pilot Projects in Laur, Nueva Ecija. The projects were aimed at testing the effectiveness of the community organizing approach in eliciting active involvement of IAs in the physical rehabilitation of the two communal irrigation systems (Bagting-Siclong and Pinagbaryuhan CIS). 1979 - The NIA participatory program was expanded to two additional communal irrigation systems (Aslong and Taisan) in Camarines Sur. 1980 - All communal irrigation system projects adopted the participatory approach. 1991 - The Local Government Code (LGC) was promulgated, directing all national agencies of government to devolve their functions to the local government units (LGUs). LGU was authorized to use their own funds for infrastructure projects to develop CIS, SWIP, and the like. 1997 - The Agriculture and Fisheries Modernization Act (AFMA) or RA 8435 was promulgated, directing NIA to hand over the management of communal irrigation systems to LGUs. It aims to strengthen the agriculture and fisheries sectors through modernization, greater participation of smallholder farmers and fisherfolk, food security and self-sufficiency, private sector participation and people empowerment. successful implementation of two large reservoir-type projects which primarily provide year-round irrigation: 1. Upper Pampanga River Integrated Irrigation System (UPRIIS) Project in Nueva Ecija 2. Magat River Multipurpose Project in Isabela  Comparison Between The National and Communal Irrigation Systems Communal For Comparison National Irrigation System Irrigation System Area (ha) > 1,000 < 1,000 NIA with farmers' Implementation /construction NIA participation NIA and Irrigators Irrigators Operation and maintenance Associations Associations Farmers pay irrigation Farmers pay Water charges service fee per amortization hectare/season/crop Operation and Capital cost Purpose of water charges maintenance recovery Small Water Impounding Projects (SWIPs) Small Water Impounding Project (SWIP) is a water harvesting and storage structure consisting of an earth embankment spillway, outlet works and canal facilities. It is designed for soil and water conservation and flood control by holding as much water as possible during the rainy season. The reservoir with its stored water is an important supplemental source of water for agriculture and is also used for fisheries. Small Farm Reservoir (SFR) The small farm reservoir (SFR) is a small water impounding earth dam structure to collect rainfall and runoff, designed for use in a single farm, and typically has an area of about 300-2,000 square meters. The embankment height above ground level is less than 4 meters. It can be easily constructed with a bulldozer or manual labor. Irrigation is done with PVC siphon pipes or pumps. SFR is used in rainfed-growing areas to provide supplemental irrigation to a wet season crop and partial irrigation to a dry season crop. Aside from irrigation and aquaculture, water in the reservoir could also be used for small scale livestock watering, wallowing areas for animals, e.g. ducks and picnic ground. Source: wocat and bswm TYPES OF SYSTEMS 1. According to Source a) Reservoir-type system - is being served by a storage reservoir used to store water that could be made available to the system on demand b) Diversion-type or run-of-the-river – gets water directly from a stream the flow of which is diverted to the service area of the system by a diversion dam across the stream; available water in the stream has to be used as it appears or else it will no longer be available for future use c) Pump-type – while the first two are gravity types, this type of system needs a pump to lift water from the source either underground or surface water 2. According to Management  a) National systems – are under the management of government agency which provides funds, constructs, operates and maintains the systems b) Communal systems – being managed by group of farmers; types of communal systems in the Philippines according to how they were built: b.1 indigenous systems – group of farmers built the system themselves without the assistance from the government b.2 assisted systems – constructed with the government’s assistance but system management is left entirely to the farmers after construction b.3 turned-over systems – constructed by the government and used to be under its management but later was turned over to the farmers’ associations for them to own and manage Activity 1. Look for the different irrigation projects in the Philippines and categorize it according to types (students will be assigned per region). 2. Locate the location of these projects using the map. 3. What are the environmental impacts of irrigation projects? Lesson 3 Methods of Water Measurements in Open Channels Units of Water Measurement Irrigation water is conveyed either through open channels or pipes and knowing the quantity of water available is essential for irrigation water management. Sometimes one will want to know only the volume of water used; while, at other times one will want to know the rate of flow. Conversion factors simplify changing from one unit of measurement to another. Water may be measured in two condition viz. (i) at rest and (ii) in motion. At rest means volume of water is measured and different units used for volume measurement are litre, cubic metre, hectare-centimetre, hectare- metre etc. Water is measured in motion means rate of flow is measured and different units used for this are litre per second, cubic metre per second, etc. 1. Litre: The volume equal to one cubic decimetre or 1/1000 cubic metre. 2. Cubic metre: A volume equal to that of a cube 1 metre long, 1 metre wide and 1 metre deep. 3. Hectare – centimetre: A volume necessary to cover an area of 1 hectare (10,000 sq.m) up to a depth of 1 centimetre (1 hectare – centimetre = 100 cu. m = 100,000 litres) 4. Hectare –metre: A volume necessary to cover 1 hectare (10,000 sq.m) up to a depth of 1 metre (1 hectare –meter = 10,000 cu. m = 10 M litres) 5. Litre per second: A continuous flow amounting to 1 litre passing through a point each second. 6. Cubic metre per second: A flow of water equivalent to a stream 1 metre wide and 1 metre deep, flowing at a velocity of 1 metre per second. Methods of Water Measurements There are several methods used for the measurements of irrigation water on the farm. They can be grouped into four categories as, 1. Volumetric or volume methods of water measurement 2. Area – Velocity Method 3. Measuring Structures (Orifices, Weirs and Flumes) 4. Tracer methods. 1. Volume Methods of Water Measurement This method is suitable for measuring small irrigation stream. In this case, water is collected in a container of known volume and the time taken to fill the container is recorded. The rate of flow is measured by the formula ( ) 2. Area-velocity Method The rate of flow of water passing a point in open channel is determined by multiplying the cross sectional area of the flow section at right angles to the direction of flow by the average velocity of water. The cross sectional area is determined by measuring the depths at various locations. The depth can be measured by different methods like sounding rods or sounding weights or echo-depth sounder for accurate measurement. For discharge calculation the entire cross section is divided into several subsections and the average velocity at each of these sub-sections is determined by current meters or floats. The accuracy of discharge measurement increases with the increase in the number of segments. Some guidelines for choosing the number of sections are: a) The discharge in the segment should not be more than 10% of total discharge. b) The difference in velocities between two adjacent sections should not be more than 20%. c) The segment width should not 1/15th to 1/20th of total width. Stream section for area-velocity method Calculation of Discharge The total discharge is calculated using the method of mid sections. It has been considered that the section is divided into N-1 sections. ∑ 3. Float Method It is inexpensive and simple. This method measures surface velocity. Mean velocity is obtained using a correction factor. The basic idea is to measure the time that it takes for the object to float a specified distance downstream. Because surface velocities are typically higher than mean or average velocities, Where, k is a coefficient that generally ranges from 0.8 for rough beds to 0.9 for smooth beds (0.85 is a commonly used value). Step 1- Choose a suitable straight reach with minimum turbulence (ideally at least 3 channel widths long). Step 2 - Mark the start and end point of your reach. Step 3 - If possible, travel time should exceed 20 seconds. Step 4 - Drop your object into the stream upstream of your upstream marker. Step 5 - Start the watch when the object crosses the upstream marker and stop the watch when it crosses the downstream marker. Step 6 -You should repeat the measurement at least 3 times and use the average velocity in further calculations.  Float Method 4. Current Meters In the area velocity method current meters are generally used to measure the velocity of flow at the different sections. The current meter consists of a small revolving wheel or vane that is turned by the movement of water. It may be suspended by a cable for measurements in deep streams or attached to a rod in shallow streams. The propeller is rotated by the flowing water and speed of propeller is proportional to the average velocity of flow. Corresponding to the number of revolutions, the velocity can obtained from calibration graphs or tables. Current Meter Procedure for velocity measurement using the current meter: Stretch a tape across the channel cross- section. Divide the distance across the channel to at least 25 divisions. Use closer intervals for the deeper parts of the channel. 1. Stretch a tape across the channel cross-section. Divide the distance across the channel to at least 25 divisions. Use closer intervals for the deeper parts of the channel. 2. Start at the water’s edge and call out the distance first, then the depth and then the velocity. Stand downstream from the current meter in a position such that the velocity is least affected by the meter. Hold the rod in a vertical position with the meter directly into the water. 3. To take a reading, the meter must be completely under water, facing the current, and free of interference. The meter may be adjusted slightly up or downstream to avoid boulders, snags and other obstructions. The note taker will call out the calculated interval, which the meter operator may decide to change (e.g., taking readings at a closer intervals in deep, high-velocity parts of the channel). Record the actual distance called out by the meter operator as the center line for the subsection.  Take one or two velocity measurements at each subsection.  If depth (d) is less than 60 cm, measure velocity once for each subsection at 0.6 times the total depth (d) measured from the water surface.  If depth (d) is greater than 60 cm, measure velocity twice, at 0.2 and 0.8 times the total depth. The average of these two readings is the velocity for the subsection. 4. Allow a minimum of 40 seconds for each reading. The operator calls out the distance, then the depth, and then the velocity. The note taker repeats it back as it is recorded, as a check.  5. Calculate discharge in the field. If any section has more than 5% of the total flow, subdivide that section and make more measurements. Current meters are designed in a manner such that the rotation speed of the blades varies linearly with the stream velocity. This can be expressed by the following equation: Where, v = stream velocity at measuring site in m/s Ns = revolutions per second of the meter a, b = constants of the meter. To determine the constants, which are different for each instrument, the current meter has to be calibrated before use. This is done by towing the instrument in a tank at a known velocity and recording the number of revolutions Ns. This procedure is repeated for a range of velocities. It has to be kept in mind that for shallow streams the measurement can be taken at a depth= 0.6 of the total depth, whereas for deeper streams two measurements are needed at 0.2 and 0.8 of total depth and then averaged to get the actual velocity. Example Data pertaining to a stream-gauging operation at a gauging site are given below. The rating equation of the current meter is v = 0.63Ns + 0.08m/s. where Ns = revolutions per second. Calculate the discharge in the stream. Distance from the left water edge (m) 0 5.0 8.0 11.0 14.0 17.0 20.0 24.0 Depth (m) 0 1.8 3.4 4.6 3.7 2.6 1.5 0 Revolutions of a current meter kept 0 42 55 93 87 48 28 0 at 0.6 depth Duration of observation (s) 0 120 120 125 135 110 100 0 Solution: Since the velocity is measured at 0.6 depth the measured velocity is the average velocity at that vertical The calculation of discharge is shown below: Segmental Distance from the Average Depth y Ns = rev. Velocity discharge left water edge (m) width (m) (m) /sec (m/s) (m3/s) 0 0 0 0.0000 5 4.225 1.8 0.350 0.3005 2.2853 8 3 3.4 0.458 0.3688 3.7613 11 3 4.6 0.744 0.5487 7.5723 14 3 3.7 0.644 0.4860 5.3946 17 3 2.6 0.436 0.3549 2.7683 20 4.225 1.5 0.280 0.2564 1.6249 24 0 0 0.0000 Sum = 23.4067 ∴ Discharge in the stream = 23.40 m3/s Ans.  5. Tracer Method In the tracer-dilution methods, a tracer solution is injected into the stream at one point and the tracer is measured at a point downstream to the first point. Knowing the rate and concentration of tracer in the injected solution and the concentration in the downstream section, the stream discharge can be computed. Either constant rate injection method or sudden injection method may be used for determining the discharge of a stream by tracer dilution. Constant rate injection method: In this method the tracer solution is injected at a constant rate into the stream till a constant concentration of the tracer in the stream flow at the downstream sampling cross section is achieved. Fig. below shows constant rate injection system. If the tracer is injected for a sufficiently long period, sampling of the stream at the downstream sampling cross section will produce a concentration-time curve similar to that shown in the Figure below. Concentration-time curve at downstream sampling site for constant-rate injection The stream discharge is computed from the equation for the conservation of mass, which follows: Where, q is the rate of flow of the injected tracer solution, Q is the discharge of the stream, Cb is the background concentration of the stream, C1 is the concentration of the tracer solution injected into the stream, an C2 is the measured concentration of the plateau of the concentration-time curve Example A 12g/L solution of a tracer was discharged into a stream at a constant rate of 15cm3/s. The background concentration of the dye in the stream water was found to be 2 parts per billion. At a downstream location sufficiently far away, the dye was found to reach an equilibrium concentration 7 parts per billion. Estimate the stream discharge. Solution: Using the equation, Given: q = 15 cm3 /s = 15 x 10-6 m3/s C1 = 0.012, C2 = 7 x 10-9, Cb = 2 x 10-9 Putting the above values in the equation, [ ] ⁄ Discharge through the stream = ⁄ Answer.  Lesson 4 Weirs Introduction Effective use of water for irrigation requires that flow rates and volumes be measured and expressed quantitatively. Measurement of flow rates in open channels is difficult because of non-uniform channel dimensions and variations in velocities across the channel. A weir is a calibrated instrument used to measure the flow in an open channel, or the discharge of a well or a canal outlet at the source. Terms 1. Weir Pond: Portion of the channel immediately upstream from the weir. 2. Weir Crest: The edge over which the water flows is the weir crest. 3. Broad-crested weir: A weir having a horizontal or nearly horizontal crest sufficiently long in the direction of flow. When the crest is "broad", the streamlines become parallel to the crest invert and the pressure distribution above the crest is hydrostatic. 4. Sharp Crested Weir: A weir having thin- edged crest such that the over flowing sheet of water has the minimum surface contact with the crest. A sharp-crested weir allows the water to fall cleanly away from the weir, e.g., V notch, Cipolleti weir etc. Figure below shows sharp crested weir. 5. Head: The depth water flowing over the weir crest measured at some point in the weir pond. 6. End Contraction: The horizontal distance from the ends of the weir crest to the sides of the weir pond. 7. Weir Scale or Gauge: The scale fastened on the sides of the weir or on a stake in the weir pond to measure the head on the weir crest. 8. Nappe: The sheet of water which overflows a weir is called a nappe. Advantages of Weirs a) Capable of accurately measuring a wide range of flows b) Can be both portable and adjustable c) Easy to construct d) Tends to provide more accurate discharge rating than flumes and orifices Profile of a Sharp-Crested weir Disadvantages a) Relatively large head required, particularly in free flow condition. b) The upstream pool must be maintained clean of sediment and kept free of weeds and trash. Otherwise, measurement accuracy will be compromised. Classification of Weirs Weirs are classified based on the shape of their opening or notch. The edge of the opening can be either sharp or broad-crested. Sharp-Crested Weir These are generally used for water measurement on the farm. They are generally of three types depending upon the shape of notch. These are Rectangular Weir Cipoletti Weir or Trapezoidal Weir V Notch Weirs or Triangular Weir  Sharp crested weirs (a) rectangular, (b) Cipoletti or trapezoidal and (c) V-notch or triangular Broad-crested Weirs A weir that has a horizontal or nearly horizontal crest sufficiently long in the direction of the flow so that the nappe will be supported and hydrostatic pressures will be fully developed for at least a short distance. Broad crested weir is shown in the figure below. Broad crested weir Weirs may also be classified as suppressed and contracted. Suppressed Weir A rectangular weir whose notch or opening sides are coincident with the sides of the approach channel, also rectangular, which extend unchanged downstream from the weir. It is the lateral flow contraction that is “suppressed”. Figure below shows suppressed and contracted rectangular weir. Suppressed and Contracted rectangular Weir Contracted Weir The sides and crest of a weir are far away from the sides and bottom of the approach channel. The nappe will fully contract laterally at the ends and vertically at the crest of the weir. Also called an “unsuppressed” weir. Rectangular Weir Rectangular weir takes its name from the shape of its notch. The discharge through a weir or notch is directly related to the water depth (H), (Fig. 6.5) and H is known as the head. This head is affected by the condition of the crest, the contraction, the velocity of approaching stream and the elevation of the water surface downstream from the weir. Rectangular weirs can be suppressed, partially contracted, or fully contracted. Derivation of Equation Definition sketch for rectangular weir Consider the above figure, Cross sectional area (a), a = L * dy Velocity (V), V=√ Total discharge (Q), ∫ √ Cd = coefficient of discharge (value bet 0.6 – 0.65) = √ [ ] = √ √ Let C = √ = Discharge coefficient of weir So, if Q = L/s; L and H in cm, then For one sided contraction, ( ) For both side contraction, ( ) Example 1. Water flows through a contracted rectangular weir 120 cm long to a depth of 30 cm, it then flows along a rectangular channel 150 cm wide and over a second weir which has length equal to the width of the channel. Determine the depth of water over the second weir. Solution: The first weir is contracted, i.e. both end contracted. Given, Length of the weir (L) = 120 cm Depth of water over the weir (H) = 30 cm So, Discharge of flow, ( ) ( )( )( ) In second weir, length of the weir (L) =150 cm Discharge through first weir and second weir is same. Let assume, depth of water over second weir is = H cm Now, ( )( ) Answer Cipoletti Weirs The Cipolletti weir (Figure below) is trapezoidal in shape. The slope of the sides, inclined outwardly from the crest, should be one horizontal to four vertical. The purpose of the slope, on the sides, is to obtain an increased discharge through the triangular portions of the weir, which, otherwise would have been decreased due to end contractions in the case of rectangular weirs. The selected length of notch (L) should be at least 3H and preferably 4H or longer. Cipoletti weirs are considered fully contracted. Cipoletti Weirs Let us split up the trapezoidal weir into a rectangular weir and a triangular notch. Now discharge over a rectangular weir, and discharge over a triangular notch so the total discharge, Q = Q1 + Q2 Q= Since the main idea of Cippoletti was to avoid the factor of end contraction, and as such the formula for the discharge, Or Example A Cipoletti weir has a breadth of 60 cm at its crest. The head of water flowing over the crest is 30 cm. Determine its discharge. Solution: Given, Crest width (L) = 60 cm Head of flow over the crest (H) = 30 cm So, discharge through the weir (Q), ( )( ) Q = 183.37 L/s V-notch Weir In this case, the notch is “V” in shape. Depth of water above the bottom of the V is called head (H). The Vnotch design causes small changes in discharge hence causing a large change in depth and thus allowing more accurate measurement than with a rectangular weir. Head (H) should be measured at a distance of at least 4H upstream of the weir.  Triangular notch weir Derivation of Equation Consider the above figure, Cross-sectional area, dA = xdy Velocity of flow, √ √ By similar triangle, and ( ) On substituting and integrating both sides of the equation, ∫ ( )√ ( ) √ ( )∫ ( √ √ ) √ ( )* + √ ( )[ ] √ ( ) [ ] √ ( ) Let √ ( ), then If , then, √ ( ); since ( ) √ , For ,H in cm and Q in L/s Example Determine discharge of 90o V-notch having 30 cm head of flow. Solution: Discharge through V-notch, ( ) Broad Crested Weir If the height of water above the weir crest is not greater than two times of the width of the crest of weir, the weir is called a Broad Crested weir. Broad Crested Weir Consider a broad crested weir as shown in figure. Let A and B be the upstream and downstream ends of the weir. Let, H = Head of water on the upstream side of the weir h = Head of water on the downstream side of the weir v = Velocity of the water on the downstream side of the weir Cd = Coefficient of discharge and L = Length of the weir Applying Bernoulli's equation at A and B, √ ( ) Discharge over the weir, √ √ The discharge will be maximum, when ( ) is maximum. Therefore differentiating the equation ( ) and equating to zero, ( ) Substituting h in the equation √ √ ( ) ( ) Example Determine the maximum discharge over a broad-crested weir 60 meters long having 0.6 m height of water above its crest. Take coefficient of discharge as 0.595. Also determine the new discharge over the weir, considering the velocity of approach. The channel at the upstream side of the weir has a cross-sectional area of 45 sq meters. Given, L = 60 m H = 0.6 m Cd = 0.595 A = 45 m2 ( )( )( ) Operation & Limitations Properly constructed and installed weirs provide most accurate flow measurement. However improper setting and operation may result in large errors in the discharge measurements. To ensure reliable results in measurement, the following precautions are necessary in the use of weirs. 1. The weirs should be set at lower end of a long pool sufficiently wide and deep having smooth flow at velocities less than 15cm/sec. 2. Baffles may be put in weir pond to reduce velocity. 3. The weir wall must be vertical. 4. The center line of the weir should be parallel to the direction of flow. 5. The crest of weir should be level so that water passing over it will be of the same depth at all points along the crest. 6. Notch should be of regular shape and its edge must be rigid and straight. 7. The weir crest should be above the bottom of the approach channel. 8. The crest of weir should be placed high enough so that water will fall freely below weir. 9. The depth of water flow over the rectangular weir should not less than about 5 cm and not more than about 2/3 crest width. 10. The scale or gauge used for measuring the head should be located at a distance of about four times the approximate head. Zero of scale should be exactly at the same level as the crest level of the weir. Limitations of Weirs: 1. Weirs are not always suitable for measuring flow. Sufficient head is required for operating any type of weir. 2. They are not accurate unless proper conditions are maintained. 3. They require a considerable loss of head which is mostly not available in channels on flat gradients. 4. Weirs are not suitable for water carrying silt. 5. Weirs are not easily combined with turnout structures.  Lesson 5 Flumes Parshall Flume Parshall flumes are devices for the measurement of flow of water in open channels when depth of flow is less i.e., head drop is very small, the volume of flow is less and channel bed slope is less. The flume consists a converging section with a level floor and walls converges towards the throat section, a throat section with a downward sloping floor and parallel walls, and a diverging section with an upward sloping floor and diverging walls towards the outlet. The size of flume is determined by the width of its throat. The size ranges from 7.5 cm to several metres in throat width. Figure 5.1 Parshall flume Parshall flumes are available in various sizes. Care must be taken while constructing the flumes exactly in accordance with structural dimensions. On the basis of the throat width, Parshall flumes have been classified into three main groups. a) Very small - 25.4 mm to 76.2 mm. b) Small 152.40 mm to 2438.4 mm. c) Large 3048 mm to 15240 mm. d) Standard dimensions of Parshall flumes with discharge values are presented in Table 5.1 and 5.2, respectively. Discharge through the flume can occur under either free or submerged flow condition. Flow is submerged when the down-stream water elevation retards the rate of discharge. To determine discharge through the flume under free flow condition, head is measured at upstream section (Ha).However, downstream head (Hb) is also measured for submerged flow condition. Free flow condition prevails if the submergence ratio (Hb/Ha) remains within 0.5, 0.6 and 0.7 for width of throat varying from 2.5 to 7.5 cm, 1.5 to 22.5 cm and 3.0 to 24.0 cm, respectively. Table 5.1.Dimensions and capacities of Parshall flume of various sizes (Letter, refer Fig. 5.1) Throat A B C D E F G K N X Y width Free-flow capacity cm cm cm cm cm cm cm cm cm cm cm cm Minimum, L/s Maximum, L/s 7.5 31 46 18 26 46 15 30.5 2.5 5.7 2.5 3.8 0.85 28.4 30. 15 41.4 61 39 39.7 61 61 7.6 12 5.1 7.6 1.4 110.8 5 30. 23 58.8 86 38 57.5 76 45.5 7.6 12 5.1 7.6 2.5 253 5 30 91.5 134 61 84.5 92 61 91.5 7.6 23 5.1 7.6 3.13 456.6 Table 5.2.Free flow discharge values for Parshall Flume Discharge, L/s Head Throat width cm 7.5 cm 15 cm 23 cm 30 cm 3 0.8 1.4 2.6 3.1 4 1.2 2.3 4 4.5 5 1.7 3.3 5.5 7 6 2.3 4.4 7.2 9.6 7 2.7 5.4 8.5 11.4 8 3.4 7.2 11.1 14.4 9 4.3 8.5 13.5 17.7 10 5 10.2 15.9 21.1 11 5.8 11.6 18.1 23.8 12 6.7 13.5 21.1 27.5 13 7.5 15 23.3 31 14 8.5 17.3 26.7 35 15 9.4 19.2 29.5 38.7 16 10.4 21.2 32.5 42.7 17 11.4 23.2 35.6 46.6 18 12.4 25.3 39 51.2 19 13.6 27.8 42.5 55 20 14.3 30 45.8 59.7 21 15.8 32.7 49.3 64.7 22 17.1 35.2 53.3 69.8 23 18.2 37.7 56.8 74 24 19.4 40.1 60.5 79 25 20.7 42.7 64.5 84.1 26 22 45.7 69.3 89 27 23.3 48.1 72.4 94.3 28 24.8 51.5 76.7 100 29 26 54 80.7 105.1 30 27.5 57.3 85.2 111 The discharge is measured by the following: Where, Q is the discharge; C, n are the flume coefficients which vary with size of the flume and H is the measuring head. Table 5.3 gives a set of standard values for the C,n for different dimensions (these co-efficient are in fps units so the calculated discharge would be in ft3/s and head has to be in ft,) Table 5.3. Value of C and n for different throat widths Throat width Coefficient (C) Exponent (n) 1 in 0.338 1.55 2 in 0.676 1.55 3 in 0.992 1.55 6 in 2.06 1.58 9 in 3.07 1.53 1 ft 3.95 1.55 2 ft 8.00 1.55 3 ft 12.00 1.57 4 ft 16.00 1.58 5 ft 20.00 1.59 6 ft 24.00 1.59 7 ft 28.00 1.60 8 ft 32.00 1.61 10 ft 39.38 1.60  12 ft 46.75 1.60 15 ft 57.81 1.60 20 ft 76.25 1.60 25 ft 94.69 1.60 30 ft 113.13 1.60 40 ft 150.00 1.60 50 ft 186.88 1.60 In the above figure, Ha is upstream head and Hb is downstream head Advantages a) This instrument is effective when the total head drop is small b) Its operation is independent of approaching velocity c) Being a self-cleaning device, it is not affected by sand and silt deposition Cut-throat Flume The geometry of the throat-less flumes with broken plane transition was first developed in Punjab by Harvey in 1912. In the cut throat flume however, the flume discharge and the modular limit are related to the piezometric heads at two points in the converging section (ha) and in the downstream expansion (hb). One of the advantages of a cut- throat measuring flumes is that there are only two walls on each side, resulting in an economic installation. The cross-section can be rectangular, trapezoidal or triangular, depending only on the availability of appropriate calibration. A U- shaped section can also be used for critical depth flumes. Cut-throat flume Design specifications of a cut-throat flume L = Total length of the flume, L1= Converging section = L/3 L2=Diverging section= 2L/3, La=Distance to piezometer tap a = 2L/9 Lb=Distance to piezometer tap b= 5L/9 B=width of the converging and diverging section= W + L/4.5 The cut-throat flume can operate either as a free flow or a submerged flow structure. Under free flow conditions, critical depth occurs in the vicinity of minimum width, w, which is called the flume throat or the flume neck. The attainment of critical depth makes it possible to determine the flow rate, knowing only an upstream depth, ha. The relationship between flow rate Q and upstream depth of flow ha, in a cut-throat flume under free flow conditions is given by the following experimental relationship: Where; Q = flow rate; C1 =free flow coefficient which is the value of Q when ha is 1 foot.; n exponent  Lesson 6 Orifices Free Flow Orifice Orifices may be used to measure rates of flow when the size and shape of the orifice and head acting upon them are known. Orifices used in measurement of irrigation water are commonly circular or rectangular in shape and are generally placed in vertical surfaces, perpendicular to the direction of channel flow. The section where contraction of the jet is maximal is known as the vena contracta. The vena contracta of a circular orifice is about half the diameter of the orifice itself. Vena contracta is the point in a fluid stream where the diameter of the stream is the least, and fluid is at its maximum. Free Discharging through orifice Derivation of Equation Velocity of flow through orifice √ Where, h = head Discharge through orifice, √ Where, Cd = discharge coefficient Co-efficient of Velocity (C v): It is defined as the ratio of the actual velocity of a jet of liquid at vena-contracta and the theoretical velocity of the jet. It is mathematically expressed as: √ Where, V = actual velocity; h = head Co-efficient of Contraction (Cc): It is defined as the ratio of the area of the jet at vena-contaracta to the orifice. It is mathematically expressed as: Co-efficient of Discharge (Cd): It is defined as the ratio of the actual discharge from an orifice to the theoretical discharge from the orifice. It is mathematically expressed as: Submerged Orifice 1. Fully Submerged Orifice Fully submerged orifice In fully sub-merged orifice, the outlet side is fully sub-merged under the liquid and it discharges a jet of liquid into the liquid of the same kind. It is also called totally drowned orifice. Discharge through fully sub-merged orifice is calculated as ( ) √ Where, H1= height of water above the top of the orifice on the upstream side H2= height of water above the bottom of the orifice H= difference in water level b= width of orifice 2. Partially Submerged Orifice Partially submerged orifice In this case, its outlet side is partially sub-merged under liquid. The discharge through partially sub-merged orifice is calculated as: ( ) √ √ [ ] The first term represents flow through drowned (submerged portion of orifice), whereas the second term represents discharge through free portion. Example The head of water over an orifice of diameter 100 mm is 10 m. The water coming out from orifice is collected in a circular tank of diameter 1.5 m. The rise of water level in the tank is 1.0 m in 25 seconds. Also the co-ordinates of a point on the jet, measured from vena-contracta are 4.3 m horizontal and 0.5 m vertical. Find the coefficients, Cd, Cv and Cc. Solution: Given, Head H= 10 m. Diameter of orifice d= 100 mm = 0.1 m So, area of orifice a= π/4 X (0.1) 2 = 0.007853 m2 Diameter of measuring tank, D= 1.5 m So, area = π/4 X (1.5)2 = 1.767 m2 Rise of water, h= 1 m Time t = 25 sec Horizontal distance x= 4.3 m Vertical distance y= 0.5 m Now, theoretical velocity, √ √ Theoretical discharge, ⁄ Actual discharge, ⁄ So, ⁄ ⁄ √ √ Example Find the discharge through a fully sub-merged orifice of width 2 m if the difference of the water levels on the both sides of the orifice is 50 m. The height of water from top and bottom of the orifice are 2.5 m and 2.75 m respectively. Take Cd= 0.6. Solution: Given, Width of the orifice b = 2 m Difference in water level H= 50 cm = 0.5 m Height of water from top of orifice H1= 2.5 m Height of water from bottom of orifice H1= 2.75 m Now, discharge through fully submerged orifice is, ( ) √ ( ) √ ⁄ Lesson 7 Water Flow Measurements in Pipes In the previous lesson, we have studied methods for water flow measurements in open channel. However, irrigation water also conveyed through pipes and therefore we will now study methods of flow measurements in pipes. Table 7.1 Difference between Pipe Flow and Open Channel Flow Open Channel Flow Pipe Flow  Defines as a passage in which liquid flows with  A pipe is a closed conduit which is used for its upper surface exposed to atmosphere. The carrying fluids under pressure. The flow in a pipe flow is due to gravity. Flow conditions are is termed as pipe flow only when the fluid greatly influenced by slope of the channel. completely fills the cross section & there is no free surface of fluid.  Hydraulic grade line coincides with the water surface  Hydraulic grade line does not coincide with the   The maximum velocity occurs at a little distance water surface. below the water surface.  The maximum velocity occurring at the pipe  The shape of the velocity profile is dependent on centre. the channel roughness.  Velocity distribution is symmetrical about the  For laminar flow, Reynolds number and for pipe axis. turbulent flow,. For laminar flow, Reynolds number and for  Flow cross section is unknown because flow turbulent flow, depth is unknown. Flow cross section is known and fixed.  Flow depth is deduced simultaneously from Velocity is deduced from continuity equation. solving both continuity and momentum equation. Figure 7.1 Pipe flow (left) and open channel flow (right) From the above Figure above, we can see that the in the pipe flow there is a pressure equal to a head y whereas in the open channel the surface is at atmospheric pressure denotes the head loss from section 1 to section 2.In case of open channel the conditions are much more varied than pipe flow in terms of surface geometry, surface roughness, depth and velocity of flow an uniformity of flow. Venturimeter Definition: Aventurimeter is a device used to measure the rate of flow of a fluid through a pipe and is often fixed permanently at different sections of the pipeline to know the discharge there. Description: Venturimeter consists of three parts: 1. A short converging part 2. A throat 3. A diverging part Due to the constriction there is an increase in the flow velocity and hence an increase in the kinetic energy. In the venturimeter (Figure 7.2) the fluid is accelerated through a converging cone of angle 15-20° and the pressure difference between the upstream side of the cone and the throat is measured and provides the signal for the rate of flow. Figure 7.2 Venturimeter and its operations The fluid slows down in a cone with smaller angle (5-7°) where most of the kinetic energy is converted back to pressure energy. Because of the cone and the gradual reduction in the area there is no "vena contracta". The flow area is at minimum at the throat. Principles of Operation It works on the Bernoulli’s principle. From Bernoulli’s principle the increase in kinetic energy gives rise to a reduction in pressure. Rate of discharge from the constriction can be calculated knowing the pressure reduction across the constriction, area of cross-section, density of fluid and the coefficient of discharge. Coefficient of discharge is the ratio of actual flow to the calculated flow and it takes into account the stream contraction and frictional effects. Energy and Head (review) The energy possessed by a flowing fluid consists of the kinetic energy and the potential energy. Potential energy may in turn to be subdivided into energy due to position or elevation above a given datum, and energy due to pressure in the fluid. The amount of energy per pound or Newton of fluid is called the head. Kinetic Energy The ability of the fluid mass to do work by virtue of its velocity Kinetic or velocity head For circular pipe of Diameter D flowing full; ( ) Elevation Energy (Potential Energy) The energy possessed by the fluid by virtue of its position or elevation with respect to a datum plane Elevation energy = Wz = Mgz Elevation head Pressure Energy (Potential Energy) The energy on the fluid because of the pressure at the top of it. Pressure energy Pressure head Where, p = fluid pressure; = spec. weight Total Energy, E The total energy or head in a fluid flow is the sum of kinetic and potential energies. For measuring discharge we should apply Bernoulli’s equation at point 1 and at point 2 (Fig.7.2).The following treatment is limited to incompressible fluids. Friction is neglected, the meter is assumed to be horizontal and there is no pump. If and are the average velocities at point 1 and point 2 respectively and ρ is the density of fluid, then Bernoulli’s equation can be written as Where, P1=P2 pressure at section 1 and 2 Since Z1=Z2, Now applying the equation of continuity at both points, we have Where d1, d2 and are the diameters at point 1(pipe) and at point 2(throat) respectively. Now putting the value of in equation (2) and if We have ( ) √ ( ) To account for the friction losses a coefficient of discharge, is introduced in the above equation and the final equation becomes: ( ) √ ( ) depends upon the type of flow, type of fluid and dimensions of venture tube and pipe. Given below are the formulae to calculate the value of head difference in terms of the liquid flowing through the venturi from the head difference observed in the manometric liquid: 1. When the manometric liquid is heavier than the liquid flowing through the pipe: [ ] 2. When the manometric liquid is lighter than the liquid flowing in the pipe: [ ] where, h = head difference in terms of the liquid flowing in the pipe x = head difference in the manometer Sh = specific gravity of the liquid flowing in the pipe So = specific gravity of the manometric liquid Merit: - Widely used particularly for large volume liquid and gas flows. Demerits: - Highly expensive - Occupies considerable space - Cannot be altered for measuring pressure beyond a maximum velocity. Example: A horizontal venturimeter with inlet and throat diameters 36 cm and 18 cm respectively is used to measure the flow of water. The reading of the differential manometer connected to the inlet and the throat is 15 cm of mercury. Determine the rate of flow. Take Cd = 0.98. Solution: Given Diameter at inlet, d1 = 36 cm Diameter at throat, d2 = 18 cm Using equation, ( ) √ ( ) We know that, ( ) Calculating h the equation, [ ] [ ] then, ( ) Putting the values in equation: ( ) √ ( ) √ ( ) ⁄ ⁄ Pitot Tube Definition It a device to measure the fluid flow velocity at any point in a pipe or channel. Description In its simplest form the pitot tube consists of a glass tube bent at right angles. Pitot tube Principle of Operation The velocity calculation is done by measuring the stagnation pressure and then applying the Bernoulli’s theorem, the basic working principle being the conversion of kinetic energy to pressure energy at the point where velocity becomes zero. Applying Bernoulli’s theorem at points 1 and 2 shown Consider a particle at point 1 to moving with a velocity of v. As the particle approaches point 2, its velocity is gradually retarded to 0 at point 2. Writing the energy equation between 1 and 2 neglecting friction: The symbols have same meaning as in case of venturimeter. = and =0 Also, and √ (Theoretical velocity) √ (actual velocity) Where: h = height to which the liquid rises above the pipe = the coefficient of pitot tube Merits: - Simple in construction having no moving parts. - Easy to install. - Requires no external power source. - Easy measurement and velocity. Demerits: - Can’t be used for turbulent flow, i.e. only used for laminar flow. - Less accurate in measurement of velocity due to assumption of ideal fluid. Example A pitot static tube placed in the centre of a 325 mm pipeline has one orifice pointing upstream and other perpendicular to it. The mean velocity in the pipe is 0.85 of the central velocity. Find the discharge if the pressure difference between the two orifices is 50 mm of water. Take the coefficient of pitot tube as: Cv = 0.98. Given, Diameter of pipe = 325mm = 0.325m Difference in pressure head = 50mm = 0.05m of water Cv = 0.98 Calculating the central velocity using equation √  √ Mean velocity = 0.85 x 0.97 = 0.825 m/s Discharge = mean velocity x area ( )  Lesson 8 Soil-Water-Atmosphere-Plants Interaction System design and selection involves three steps. The first step is to characterize crop water requirements, hydrology, and soil characteristics. The second step is the hydraulic design of potential alternatives. The third step is an economic and environmental analysis. To adequately evaluate alternatives, it is important to look at all relevant inputs to the system (water, energy, labor) and at the effect of the system on the environment. Soil Water Introduction Soil-water-plant relationship relates to the properties of soil and plant that affect the movement, retention and use of water. Due to inadequate and/or uneven distribution of rainfall during the cropping season, it becomes necessary to apply additional water to the soil for plant use in the form of irrigation. Therefore, proper understanding of the soil- water-plant relationship is a prerequisite for the sound design of any efficient irrigation system. 8.1 Soil Figure 8.1 Volume composition & sectional view of soil A soil matrix consists of solid, liquid and gaseous phases (Fig 8.1). The solid phase is the soil matrix comprising mineral, organic matter and various chemical compounds. The liquid phase contains all the dissolved substances. Liquid phase also referred by the soil moisture or soil water or soil solution. The gaseous portion of the soil consists of soil air and it occupies those spaces between the soils particles which are not filled with water (Fig. 8.1). In a completely dry soil, all of the pore spaces (i.e., space between soil particles) are filled with air, and in a completely wet soil all of the pores are filled with water. However, in most of the field situations the pore spaces are filled with both air and water. Finally, soil water and air vary in composition, both in time and space. 8.2 Soil Texture It refers to the relative proportion of sand, silt and clay in a given soil. Soil particles are classified based on particle diameter. There are three major categories of soil particle size: sand, silt, and clay. The USDA particle size classification scheme is in Table 8.1. Coarse sand has approximately the same diameter as pencil lead (Fig. 8.2) while the largest clay particle diameters are three orders of magnitude smaller. Based on the percentages of sand, silt, and clay, soils are classified by ―textural class names‖ such as sandy loam, clay, and silt loam. The USDA soil textural triangle is used to classify soil textures (Fig. 8.3). Soil textures can also be classified based on the way that the soil responds in the hand (the feel and appearance method) Sand—Sand is loose and single-grained. The individual grains can be readily seen and felt. Squeezed in the hand when dry, sand falls apart when pressure is released. Squeezed when moist, it forms a cast, but crumbles when touched. Sandy loam—A sandy loam is soil containing a high percentage of sand, but having enough silt and clay to make it somewhat coherent. The individual sand grains can be readily seen and felt. Squeezed when dry, a sandy loam forms a cast that falls apart readily. If squeezed when moist, a cast can be formed that bears careful handling without breaking. Loam—A loam is soil having a relatively even mixture of different grades of sand, silt, and clay. It is friable with a somewhat gritty feel, but is fairly smooth and slightly plastic. Squeezed when dry, it forms a cast that bears careful handling, and the cast formed by squeezing the moist soil can be handled freely without breaking. Silt loam—A silt loam is soil having a moderate amount of fine sand with a small amount of clay. Over half of the particles are silt size particles. When dry, a silt loam appears cloddy, but the lumps can be readily broken. When pulverized, it feels soft and floury. When wet, the soil runs together readily and puddles. Either dry or moist, silt loam forms a cast that can be handled freely without breaking. When moist and squeezed between thumb and finger, it does not ribbon, but has a broken appearance. Clay loam—A clay loam is moderately fine-textured soil that generally breaks into clods or lumps that are hard when dry. When the moist soil is pinched between the thumb and finger, it forms a thin ribbon that breaks readily, barely sustaining its own weight. The moist soil is plastic and forms a cast that bears much handling. When kneaded in the hand, clay loam does not crumble readily, but works into a heavy compact mass. Clay—A clay is fine-textured soil that usually forms very hard lumps or clods when dry and is very sticky and plastic when wet. When moist soil is pinched between thumb and finger, it forms a long flexible ribbon. Some clays are very high in colloids are friable and lack plasticity. Three laboratory methods are commonly used to measure soil particle size distribution: laser particle size analysis, hydrometer and sieve. Once the soil texture is known, the Soil Water Characteristics Calculator (download from http://hydrolab.arsusda.gov/soilwater/Index.htm )calculates soil hydraulic properties. The Calculator also includes compaction, which incorporates the soil structure concepts. Table 8.1 Particle size classification by USDA Material Diameter Stones >250 mm Cobbles 80–250 mm Course gravel 12.5–80 mm Fine gravel 2.0–12.5 mm Very coarse sand 1.0–2.0 mm Coarse sand 0.5–1.0 mm Medium sand 0.25–0.5 mm Fine sand 0.1–0.25 mm Very fine sand 0.05–0.1 mm Silt 0.002–0.05 mm Clay Less than 0.002 mm Fig. 8.2 Soil particle sizes (Courtesy of Don Post. The University of Arizona)  Fig. 3.3 USDA soil textural triangle (Credit NRCS, National Agronomy Manual. Part 504) 8.3 Soil Structure Soil structure is the arrangement and organization of soil particles into natural units of aggregation. These units are separated from one another by weakness planes that persist through cycles of wetting and drying and cycles of freezing and thawing. Structure influences air and water movement, root development, and nutrient supply. Soil structure, (Source: Rao, et al., 2010)  Soil structural types. (Source: Rao, et al., 2010) Soil structure strongly influences infiltration rate. Single grain soils are sands with no cementing agents attaching particles to each other. Granular soils are aggregates of small particles. Both of these soils have rapid infiltration rates. Blocky and prismatic soils are larger aggregates and have moderate infiltration rates. Although platy structures are aggregates, they have poor vertical infiltration. Massive soils are generally grossly swelling soils (montmorrilonite clay). These soils have extremely low infiltration rates. Cultural practices can either improve (increase porosity and hydraulic conductivity) or degrade soil texture (increase density). For example, driving a tractor in a wet field can decrease soil structure and form hardened layers with low infiltration rates. Soil-Water Relationships As we discussed earlier the basic components of the soil consists of solid mineral particles, organic matter, the voids among the particles and water and air occupying the voids. The figure below shows a schematic representation of the soil in relative proportions both in masses and volumes. The physical properties of the soil, including its ability to store water, are highly related to the fraction of the bulk soil volume that is filled with water and air. For plant growth and development to be normal, a balance of water and air in the pore space must be attained. If water is limited, plant growth may be inhibited by water stress. If air (aeration) is limited, usually by too much water, then growth may be limited by insufficient aeration. The relationship between the three phases of soil can be described in a number of mathematical relationships. These relationships can be used to calculate one soil property from another. Where; Va = Volume of air Vw = Volume of water Vs = Volume of solids Vv = Volume of voids (Va+Vw) Vt = Total Volume (Va+Vw+Vs) Ma = Mass of air (negligible) Mw = Mass of water Ms = Mass of solids Mt = Total mass (Ma+Mw+Ms) Three phases of soil Particle density ( ) It is the ratio of a given mass (or weight) of soil solids to that of its volume and it is given by; Sometimes it is referred to as true density. It is usually expressed in terms of g/cm3and varies between the narrow limits of 2.6 to 2.75 g/ cm3. Average particle density of mineral soils is 2.65 g/ cm3. Particle density is a constant for a soil with a given texture and is independent of size and arrangement of the soil particles. Dry Bulk Density ( ) It is the density of the undisturbed (bulk) soil sample which is the ratio of dry mass of the soil to its total volume. It is given by; This is expressed as gm/cm3.Drybulk density can be calculated by collecting a known volume of soil to get the soil volume (Vt), and drying the associated soil to get the mass of dry soil (Ms). Total (Wet) Bulk Density ( ) It is the mass of moist soil per unit volume and is represented as: Bulk density has a pronounced effect on the soil properties like permeability of soil for water and air, and penetration of plant roots through the soil. Compression or compaction of soil particles can increases bulk density but it reduces the soil porosity and in turn the soil water storage capacity. The bulk density values for different soil textural classes are given in Table below. Bulk density values of various soil types (USDA - SCS) Bulk density Soil texture (g/cm3) 1. Sandy 1.60–1.70 2. Loamy sand 1.60–1.70 3. Sandy loam 1.55–1.65 4. Fine sandy loam 1.50–1.60 5. Loamy soil 1.45–1.55 6. Silty loam 1.40–1.50 7. Silty clay loam 1.35–1.45 8. Sandy clay loam 1.40–1.50 9. Clay loam 1.30–1.50 10.Clay soil 1.25–1.35 Example 8.1 Calculate the dry bulk density from the following data Fresh weight of soil = 2505g; Weight of water = 740g; Height of core = 10cm; Diameter of the core = 12 cm Solution: Volume of the core, ( ) Dry Bulk density, ⁄ Porosity ( ) Porosity is the void space (air + water) in a given volume of soil that is occupied by air and water. The total porosity is calculated as follows: ( ) ( ) ( ) ( ) Generally total porosity varies from 30% to 60% for agricultural soils. Coarse textured soils are normally less porous (35%– 50%) than the fine textured soils (40% – 60%). However, the mean size of individual pores is greater (>0.06mm in diameter) in the coarse textured soils than the fine textured soils. From irrigation water management point of view, knowledge of porosity in a given volume of soil is very important, because it is an index of moisture storage capacity and the aeration conditions. These are two most important factors that influence the plant growth. Example 8.2 Calculate the porosity from the following data: Bulk density = 1.31 g/ cm3and particle density = 2.64 g /cm3 Solution: ( ) ( ) ( ) ( ) Void Ratio, ( ) It is the ratio of the pore space to the volume of solids and is given by, ( ) Soil Water Content Soil water content can be calculated by gravimetric method or volumetric method Gravimetric soil water content, ( ) The mass water content or soil moisture content (θm) is the ratio of the mass of water in a sample to the dry soil mass, expressed as either a decimal fraction or as percentage. It is often referred to as ‗gravimetric water content‘. The mass water content is found by Where; = gravimetric water content, g/g = mass of water, g = mass of soil after drying, g = mass of soil before drying, g It is determined by weighing the soil sample collected from field, drying the sample for at least 24 hours at 105 , and then weighing the dry soil. Difference in mass of the wet and dry sample represents the mass of water in the soil sample (Mw). The mass of the sample after drying represents the mass of dry soil (Ms) Volumetric water content, ( ) The volumetric water content ( ) represents the volume of water contained in total volume of undisturbed soil. The volumetric water content is defined as Determination of volumetric water content requires the volume of the undisturbed soil sample which is sometimes difficult to measure. However, it can also be determined from mass water content and specific gravity (ratio of bulk density of soil to density of water) as follows Where, = density of water which is 1 g/cm3 When comparing water amounts per unit of land area, it is frequently more convenient to speak in equivalent depths of water rather than water content. The relationship between volumetric water content and the equivalent depth of water in a soil layer is: Where, d = equivalent depth of water in a soil layer, and L = depth increment of the soil layer Example 8.3 A soil sample is collected just before irrigation, and another soil sample is collected two days after irrigation. The soil sample collected before irrigation weighs 1.73 kg, and the soil collected just after irrigation weights 1.94 kg. Both soils are placed in an oven and dried at 100 for 24 hours. After drying, the soil collected just before irrigation weighs 1.49 kg and the soil collected just after irrigation weighs 1.52 kg. What are the volumetric water contents just before and just after irrigation? Solution: Gravimetric water content just before irrigation Gravimetric water content just after irrigation Example 8.4 Based from example 8.3, if the bulk density is 1.3 g/cm3, find the volumetric water contents and porosity of the two samples. Solution: Volumetric water content just before irrigation Volumetric water content just after irrigation = 0.208 =20.8 = 0.36 =36 Porosity of the soil sample, The bulk density is 1.3 g/cm3 and soil particle density is 2.65 g/cm3. ( ) ( ) ( ) ( ) This means that 51% of the soil volume is filled with air or water Example 8.5 A field soil sample prior to being disturbed has a volume of 82 cm 3. The sample weighed 125 grams. After drying at 105 for 24 hours, the dry soil sample weighs 100 grams. What is the mass water content? What is the volumetric water content? What depth of water must be applied to increase the volumetric water content of the top 1m of soil to 0.40? Given: Find: , ,d Ms = 100g Mw = 125g – 100g = 25g = 82 Solution: Gravimetric moisture content; Volumetric moisture content; but, The current depth of water in 1 m of soil is, d= x L = 0.3025 x 1m = 0.3025m d = 0.4 x 1m = 0.4m (depth of water when ) Thus, Depth of water to be added = 0.4m – 0.3025m = 0.0975m Kinds of Soil Water Water present in the soil is referred to as the soil moisture. It is divided into three categories viz., gravitational water, capillary water and hygroscopic water (Fig. below). Physical classification of soil water Gravitational Water Water held between 0.0 to 0.33 bars (0 to −33 kPa) soil moisture tension that moves downward freely under the influence of gravity to the water table is termed as gravitational water (Fig. above). It is also referred to as free water. Gravitational water is of no use to plants as it drains out due to gravity. It reduces aeration in the soil and hence, its removal from soil is necessary for optimum plant growth. Capillary Water Capillary water is the water held in the capillary pores (micro pores). Capillary water is retained on the soil particles by surface forces, adhesion i.e., attraction of water molecules for soil particles, cohesion i.e., attraction between water molecules and surface tension phenomena. Adhesion is a process of the attraction of solid surface for water molecules and forms a very thin film of water at solid-liquid interface. On the other hand, cohesion is attraction of water molecules for each other. Capillary water is held between tension of about 0.33 bars (−33 kPa or 1/3 atmosphere, moisture content at field capacity) to 31 bars (−3100 kPa or 31 atmosphere, hygroscopic coefficient) as shown in the Fig. below. However, the water within the capillary range is not equally available i.e., it is readily available starting from 0.33 bars up to a certain point often referred to as critical soil moisture and thereafter up to 15 bars (−1500 kPa) it is available in lesser amounts. Further below, the water is held very tightly in thin films and is practically not available for plant use between 15 bars and 31 bars tension. Hygroscopic Water It is the water held tightly to the surface of soil particles by adsorption forces. Hygroscopic water is held tightly in thin films of 4 – 5 milli microns thickness on the surface of soil colloidal particles at 31 bars tension (−3100 kPa) and above (Fig. below). It is essentially non-liquid and moves primarily in the vapour form. This water is unavailable to the plants as huge pressure force would be needed to extract it. Diagrammatic representation of different types of water  Lesson 9 Soil Water Constants Soil Moisture Constants From previous discussion, it is clear that a part of capillary water is useful for plant uptake and thus we need to replenish this part of soil water during irrigation. In order to manage irrigation, we need to define soil water constants that are used as reference points for practical irrigation water management. These constants are briefly explained below: 1. Saturation Capacity, (SC) Saturation capacity of soil refers to the condition when all the macro and micropores are filled with water and the soil is at maximum water retention capacity (Fig. 9.1).The metric suction at this condition is almost zero and it is equal to free water surface. High water content is not desirable because it restricts oxygen diffusion from the atmosphere into the soil because most of the soil pores are occupied by water: oxygen diffusion is approximately 1,000 times greater in a gas than it is in water. This lack of oxygen restricts oxygen uptake by plant roots during respiration. Figure 9.1. Soil condition at Saturation, Field Capacity and Permanent wilting point 2. Field Capacity, (FC) The field capacity is the amount of water held in soil after excess water has been gravity drained and the rate of downward movement has relatively stable, which usually takes place within 1 – 3 days after a rain or irrigation. At field capacity, the soil moisture tension depending on the soil texture ranges from 0.10 to 0.33bars. Field capacity is the upper limit of available soil moisture. The field capacity is greatly influenced by soil texture, finer the soil particles higher the water retention due to very large surface are and vice versa. It can be seen from Table 9.1 that moisture content at field capacity of clay soil is much higher (40%) as compared to that of coarse sand (10%). Field capacity of soil can be determined by ponding water over the area of 2 to 5m2 for two to three days, with surface evaporation prevented by spreading polyethylene sheet on thick straw mulch over the soil surface. After three days soil samples from different depth will give the field capacity. As a rule of thumb, 1 day of drainage will generally be adequate for sandy soils,2 days for silt loam soils, and 3 days for silty clay loam soils. Table 9.1. Soil Water Characteristics for various soil Textue 3. Permanent Wilting Point, (PWP) Permanent wilting point is considered as lower limit of available soil moisture. At this stage, water is held tightly by the soil particles that the plant roots can no longer obtain enough water to satisfy the transpiration requirements; and remain wilted unless the moisture replenished. The soil moisture tension at permanent wilting point is about 15 bars. 4. Available Soil Moisture, (AM) Available soil moisture is the moisture between field capacity (0.33 bars) and permanent wilting point (15 bars) which is referred as readily available water (TAW) for plant growth. The water present above the field capacity and below the permanent wilting point is not available to the plant. The available soil moisture is expressed as depth of water per unit of soil and is calculated according the following formula: ( ) Where; TAW = Total available water (cm) FC = Volumetric moisture content at field capacity (fraction) PWP = Volumetric moisture content at Permanent wilting point (fraction) DRZ = depth of root zone (cm) Although plants are theoretically able to obtain water from the soil whenever water contents exceed PWP, the actual rate at which they transpire decreases as stomata close in response to declining soil water contents. Fig. 9.2 indicate that soil moisture stress coefficient (Ks) remains almost unity for soil water content reductions between FC and θt indicate that water is more readily available. However, as the moisture content decrease below θt soil moisture stress coefficient decreases below unity and causes decreases in the crop transpiration (ET) at the same rate. Therefore, irrigations are generally scheduled to maintain soil water contents above θt. The water content between FC and θt is called Readily Available Water (RAW). Figure 9.2. Soil moisture stress as a function of TAW Depletion is the depth of water removed from the soil (after the soil reaches field capacity by gravity drainage) by evapotranspiration. Percent depletion is percent of the TAW that is removed by evapotranspiration. Percent depletion is 0 % at field capacity and 100 % at permanent wilting point. Depletion and percent depletion are calculated as follows. ( ) ( ) ( ) ( ) The percent depletion with no yield reduction for specific crops is called the management allowed depletion (MAD). The MAD value for many crops changes with growth stage and is based on the crop sensitivity to water stress at that stage. The MAD value for a drought tolerant crop is approximately 0.5 and the MAD value for a drought sensitive crop is approximately 0.2 The readily available water (RAW) is the depth of water available to the plant between irrigation events. ( )( ) Example 9.1 The root depth of an orange tree is 1.5 m. The soil has 56 % solid particles, and 44 % voids. After drainage, the 50 % of the void volume contains water and 50 % holds air. At the wilting point, the void volume is 25 % water. What is the total available water and the depth of the total available water? Solution: Given: DRZ = 1.5m; porosity (n) = 44%; After drainage (means at FC), 50% of porosity (water); 50% of porosity (air) At PWP, 25% of porosity (water) Solution: TAW = FC –PWP but, FC = (0.44) (0.5) = 0.22 or 22% PWP = (0.44) (0.25) = 0.11 or 11% TAW = 0.22 – 0.11 = 0.11 DTAW = (TAW) (DRZ) = (0.11)(1.5) = 0.165m or 16.5cm NOTE: Only about half of TAW is available to the plant (since MAD = 50% for drought tolerant crop) so only about 8 cm of water would be available between irrigation events. This means that if the evapotranspiration rate is 1 cm/day, then the field would need to be irrigated every 8 days. Example 9.2 A soil sample is collected just before irrigation, and another soil sample is collected two days after irrigation. The soil sample collected before irrigation weighs 1.73 kg, and the soil collected just after irrigation weights 1.94 kg. Both soils are placed in an oven and dried at 100 for 24 hours. After drying, the soil collected just before irrigation weighs 1.49 kg and the soil collected just after irrigation weighs 1.52 kg. The bulk density of the soil is 1.3 g/cm3. If the soil has a permanent wilting point equal to 20% (volumetric water content). Calculate the percent depletion in the soil sample taken just before irrigation. If the root zone is 1 m deep, then what is the depth of depletion? Given: Just before irrigation; Just after irrigation; Soil Fresh weight = 1.73kg Soil fresh weight = 1.94kg Soil oven dry weight = 1.49kg Soil oven dry weight = 1.52kg = 1.3 g/cm3 PWP = 20% DRZ = 1m Required: % depletion just before irrigation; depth of depletion Solution: ( ) but, FC = = ( )=( )( )=( )( ) = 0.36 = = ( )=( )( )=( )( ) = 0.209 ( ) Depth = (FC - ) x DRZ = (0.36 – 0.209) (1m) = 0.151m or 15.1cm Example 9.3 Use the soil describe in example 9.2. A crop has a MAD value of 0.4 and a root zone of 1 m. What is RAW? The depth of evapotranspiration is 1 cm/day. How often would the field need to be irrigated?  Solution: RAW = TAW x MAD but, TAW = (FC-PWP) x DRZ = (0.36 – 0.20) x (1m) = 0.16m or 16cm RAW = (16cm) (0.4) = 6.4cm Thus, if the evapotranspiration rate was 1 cm/day, then, Hence, irrigation should be applied every after 6 days. NOTE: In order to find the preferred volumetric water content at which the next irrigation should take place, subtract AWC (MAD) from the field capacity. Example 9.4 From example 9.3. Find the preferred water content at which irrigation should be take place for a plant with 40 % MAD. Solution: Water content at next irrigation = FC – (AWC x MAD) = FC – (FC – PWP)(MAD) = (0.36) – (0.36 – 0.20)(0.40) = 0.30 or 30% NOTE: 30 % is a higher water content than the soil water content taken just before irrigation in Example 9.2. Thus, a plant with a 0.4 MAD would have been overly stressed before irrigation. 5. Soil Water Potential The driving force for water flow in the soil-plant-atmosphere-continuum (SPAC) is the difference in water potential. Water flows from high to low potential. Soil water potential (Ψt) is an indicator or measure of mount of work needed to be done to displace a unit quantity of water from a given reference point. The water potential (ψ) can be expressed as the potential energy per unit mass, volume or weight. Units of Water Potential Mass basis: Joules/kg Volume basis: Pascal Weight basis: Meters or mm The three major components of soil water potential are gravitational potential (Ψg), matric potential (Ψm), and osmotic potential (Ψo). The soil water potential (Ψt) then is Ψt = Ψp(m) + Ψz +Ψo Where, Ψp(m) = Pressure or (matric) potential Ψz = gravitational potential Ψo = osmotic potential 5.1. Gravitational Potential (Ψz) Gravitational potential energy at a point Z above a reference point can be expressed as Where, = density of water V = Volume of water, cm3 g = Acceleration due to gravity M = Mass of water Gravitational potential energy per unit mass of water is the gravitational constant multiplied by the distance of the reference position and can be expressed as  If the unit quantity of weight is used then the gravitational potential can be related to a distance. In other words, it is a vertical distance from an arbitrary reference elevation to the point of question. Example: Given: Two points is a soil. Each point is located a specified distance above a reference elevation. Point A is 150mm above the reference and point B is 100 mm below reference. Find: The Ψz at each point; Difference in Ψz (Δ Ψz) between the two points. Solution: Ψz (A) = 150mm; Ψz (B) = - 100mm Δ Ψz = Ψz (A) - Ψz (B) = 150mm – (- 100mm) = 250mm 5.2. Pressure Potential (Ψp) Pressure potential, under field conditions, applies mostly to saturated soil. For weight basis, it is the vertical distance from a point in question to tree water surface. The hydrostatic pressure of water with reference to atmospheric pressure can be expressed as: ( ) The potential energy of the water isPdv, where dv is the infinitesimal volume of water. The pressure potential on volume, weight and mass basis can be expressed as: Ψp(volume) per unit volume = Ψp(mass) per unit mass = Ψp(weight) per unit weight = 5.3. Metric Potential (Ψm) It is the negative pressure potential related to the adsorptive forces of the soil matrix. Considering an infinitesimal volume dv of water, with pressure deficit, P, the metric potential will be Pdv. The pressure potential on volume, weight and mass basis can be expressed as Metric potential per volume = Metric potential per unit mass = Metric potential per unit weight = On weight basis Ψm is the vertical distance between a point in the soil and the water level of manometer connected to