Learning Packet 2 Hydrology PDF

i 2 | HYDROLOGY PREFACE The main purpose of this module is to provide the student with a clear and detailed presentation of the theory and application of hydrology. The course deals with the hydrologic cycle and the different processes such as p...

i 2 | HYDROLOGY PREFACE The main purpose of this module is to provide the student with a clear and detailed presentation of the theory and application of hydrology. The course deals with the hydrologic cycle and the different processes such as precipitation, evaporation, infiltration, overland flow, groundwater flow and surface runoff generation. To achieve this objective despite of this pandemic due to COVID-19, this work has been shaped by the comments and suggestions of the peer reviewer in the teaching profession, as well as the other faculty members who will ensure quality of the modules that will be distributed to the LGU.  1 2 | HYDROLOGY UNIT 3: PRECIPITATION 3.0 Intended Leaning Outcomes 1. Identify the different formation and types of precipitation. 2. Identify the rainfall measurements. 3. Solve problems related to precipitation. 3.1 Introduction The precipitation denotes all forms of water that reach the earth from the atmosphere. In this chapter, a brief introduction is given to familiarize the engineer with important aspects of rainfall, and in particular, with the collection and analysis of rainfall data. 3.2 Topics 3.2.1 Formation of Precipitation Precipitation can reach the ground surface in different forms such as: a) Drizzle is a very light, usually uniform, precipitation consisting of numerous minute droplets with diameters in excess of 0.1 mm but smaller than 0.5 mm. https://medium.com/poets-unlimited/drizzle-of-this-winter-329434bb796d b) Rain is precipitation consisting of water drops larger than 0.5 mm. It can be classified as light rain when the intensity is smaller than 2.5 mm h−1, moderate when it is between 2.5 - 7.5 mm h−1, and heavy when it exceeds 7.5 mm h−1. https://www.treehugger.com/why-does-rain-smell-4863931  2 2 | HYDROLOGY c) Snow is precipitation in the form mainly of branched hexagonal or star- like ice crystals, resulting from direct reverse sublimation of the atmospheric water vapor; snow particles can reach the ground as single crystals, but more often than not they do so after agglomerating as snowflakes. These flakes tend to be larger at temperatures close to freezing. The specific gravity of snow can vary over a wide range (Judson and Doesken, 2000), but as a rule of thumb for fresh snow it is often taken around 0.1. https://www.canberratimes.com.au/story/6740886/perisher-records-first-snowfall-of-the-year/ d) Sleet (North American usage) is precipitation consisting of fairly transparent pellets or grains of ice, formed as a result of the passage of raindrops through a layer of colder air near the ground. In British usage the word sleet refers to precipitation consisting of melting snow or a mixture of snow and rain. https://www.britannica.com/science/sleet e) Glaze or freezing rain is ice deposited by drizzle or rain on cold surfaces. https://www.metoffice.gov.uk/weather/learn-about/weather/types-of-weather/rain/freezing-rain  3 2 | HYDROLOGY f) Snow pellets (also called granular snow or graupel) are a form of precipitation consisting of white, opaque, small grains with diameters between roughly 0.5 and 5 mm. https://www.travelblog.org/Photos/7027298 g) Small hail is precipitation consisting of white, semitransparent or translucent grains with diameters ranging from about 2 to 5 mm. These grains are mostly round, and sometimes conical in shape, and they have a glazed appearance. Small hail falls usually accompanied by rain, when the temperature is above freezing. https://midwest64.zenfolio.com/stormsweatherphotos/h31B30EBA#h31b30eba h) Soft hail consists of round, opaque grains in the same size range as small hail, but they are softer in appearance and tend to disintegrate more easily. https://www.shutterstock.com/image-photo/close-dof-tiny-grains-soft-hail-1508143631  4 2 | HYDROLOGY i) Hail consists of balls or irregular chunks of ice with diameters between 5 and 50 mm, or even larger. These lumps of ice can be transparent or they can consist of concentric layers of clear and opaque ice; such layered structure is the result of the alternating rising and falling movements during the hail formation. Hail usually falls during violent and prolonged convective storms under above-freezing temperature conditions near the ground; it can cause severe damage. https://www.nationalgeographic.org/encyclopedia/hail/ j) Dew consists of moisture in the form of liquid drops on the ground surface and on the vegetation and other surface elements, as a result of direct condensation of atmospheric water vapor. It typically occurs at night on surfaces that have been cooled by outgoing long-wave radiation. http://www.differencebetween.net/science/nature/difference-between-dew-and-fog/ k) Hoar frost forms in the same way as dew, but the water vapor condenses directly into ice. These ice crystals can assume a wide variety of shapes. https://www.accuweather.com/en/weather-news/what-is-hoarfrost/7092  5 2 | HYDROLOGY 3.3.2 Different Types of Precipitation The precipitation may be due to: a) Thermal convection (convectional precipitation)—This type of precipitation is in the form of local whirling thunder storms and is typical of the tropics. The air close to the warm earth gets heated and rises due to its low density, cools adiabatically to form a cauliflower shaped cloud, which finally bursts into a thunder storm. When accompanied by destructive winds, they are called ‘tornados.’ b) Conflict between two air masses (frontal precipitation)—When two air masses due to contrasting temperatures and densities clash with each other, condensation and precipitation occur at the surface of contact. This surface of contact is called a ‘front’ or ‘frontal surface’. If a cold air mass drives out a warm air mass’ it is called a ‘cold front’ and if a warm air mass replaces the retreating cold air mass, it is called a ‘warm front’. On the other hand, if the two air masses are drawn simultaneously towards a low-pressure area, the front developed is stationary and is called a ‘stationary front’. Cold front causes intense precipitation on comparatively small areas, while the precipitation due to warm front is less intense but is spread over a comparatively larger area. Cold fronts move faster than warm fronts and usually overtake them, the frontal surfaces of cold and warm air sliding against each other. This phenomenon is called ‘occlusion’ and the resulting frontal surface is called an ‘occluded front’. c) Orographic lifting (orographic precipitation)—The mechanical lifting of moist air over mountain barriers, causes heavy precipitation on the windward side.  6 2 | HYDROLOGY d) Cyclonic (cyclonic precipitation)—This type of precipitation is due to lifting of moist air converging into a low-pressure belt, i.e., due to pressure differences created by the unequal heating of the earth’s surface. Here the winds blow spirally inward counterclockwise in the northern hemisphere and clockwise in the southern hemisphere. There are two main types of cyclones—tropical cyclone (also called hurricane or typhoon) of comparatively small diameter of 300-1500 km causing high wind velocity and heavy precipitation, and the extra-tropical cyclone of large diameter up to 3000 km causing wide spread frontal type precipitation. 3.2.3 Rain Fall Characteristics a) Rainfall Depth – the total height of rainfall in a certain area usually measured in mm. b) Rainfall Duration - The duration refers to the length of time rainfall occurs. A high intensity rainfall for a short duration may affect tender seedlings, but it will not likely have much effect on soil erosion and runoff. Rainfall of longer duration can significantly affect infiltration, runoff, and soil erosion processes. c) Rainfall Intensity - Rainfall intensity is defined as the ratio of the total amount of rain (rainfall depth) falling during a given period to the duration of the period It is expressed in depth units per unit time, usually as mm per hour (mm/h). d) Hyetograph - A hyetograph is a graphical representation of the distribution of rainfall intensity over time. 3.2.4 Point-Rainfall Measurement Point rainfall, also known as station rainfall refer to the rainfall data station. Depending upon the need, data can be listed as daily, weekly, monthly, seasonal, or annual values for various periods. Graphically this data is represented as plots of magnitude vs. chronological time in the form of a bar diagram. Such a plot is not, however, is not convenient for discerning a trend in the rainfall as there will be considerable variations in the rainfall values leading to rapid changes in the plot. The trend is often discerned by the method of moving averages, also known as moving means. Moving Average is a technique for smoothening out the high fluctuations of a time series and to enable the trend, if any, to be noticed. The basic principle is that a widow of time range m years is selected. Starting from the first set of m years of data, the average of the data m years is calculated and placed in the  7 2 | HYDROLOGY middle year of the range m. the window is next moved sequentially one time unit(year) at a time and the mean of the m terms in the window is determined at each window location. The value of m can be 3 or more years; usually an odd value. Generally, the larger the size of the range m, the greater is the smothering. There are many ways of averaging (and consequently the plotting position of the mean) and the method described above is called the Central Simple Moving Averages. Example: Annual rainfall values recorded at station M for the period 1950 to 1979. Represent this data a bar diagram with time in chronological order. (i) Identify those years in which the annual rainfall is (a) less than 20% of the mean, and (b) more than the mean. (ii) Plot the three-year moving mean of the annual rainfall time series. Year Annual Rainfall (mm) 1950 676 1951 578 1952 95 1953 462 1954 472 1955 699 1956 479 1957 431 1958 493 1959 503 1960 415 1961 531 1962 504 1963 828 1964 679 1965 1244 1966 999 1967 573 1968 596 1969 375 1970 635 1971 497 1972 386 1973 438 1974 568 1975 356 1976 685  8 2 | HYDROLOGY 1977 825 1978 426 1979 612 Solution: (i) Solve the Mean of the Rainfall by Average Rainfall/No. of Years. a.) Mean = Average Rainfall/No. of Years. 17060 = = 468.67 𝑚𝑚 30 b.) Less than 20% of the mean = Mean – 20% Mean = 468.67 − 20%(468.67) = 454.936 𝑚𝑚 Identify the answer for (i) by plotting a bar graph. (Use Excel to answer) (iii) Plot the three-year moving mean of the annual rainfall time series. 1 2 3 4 Year Annual Three consecutive Year total for 3-year Moving Rainfall moving mean Mean mm Preceding + Current + Succeeding Column 3 / 3 Annual Rainfall 1950 676 1951 578 676+578+95 = 1349 449.67  9 2 | HYDROLOGY 1952 95 578+95+462 = 1135 378.33 1953 462 95+462+472 = 1029 343.00 1954 472 462+472+699 = 1633 544.33 1955 699 472+699+479 = 1650 550.00 1956 479 699+479+431 = 1609 536.33 1957 431 479+431+493 = 1403 467.67 1958 493 431+493+503 = 1427 475.67 1959 503 493+503+415 = 1411 470.33 1960 415 503+415+531 = 1449 483.00 1961 531 415+531+504 = 1450 483.33 1962 504 531+504+828 = 1863 621.00 1963 828 504+828+679 = 2011 670.33 1964 679 828+679+1244 = 2751 917.00 1965 1244 679+1244+999 = 2922 974.00 1966 999 1244+999+573 = 2816 938.67 1967 573 999+573+596 = 2168 722.67 1968 596 573+596+375 = 1544 514.67 1969 375 596+375+635 = 1606 535.33 1970 635 375+635+497 = 1507 502.33 1971 497 635+497+386 = 1518 506.00 1972 386 497+386+438 = 1321 440.33 1973 438 386+438+568 = 1392 464.00 1974 568 438+568+356 = 1362 454.00 1975 356 568+356+685 = 1609 536.33 1976 685 356+685+825 = 1866 622.00 1977 825 685+825+426 = 1936 645.33 1978 426 825+426+612 = 1863 621.00 1979 612  10 2 | HYDROLOGY 3.2.5 Different Types of Rain Gauges The rain gauge is also known as Pluviometer, Ombrometer, hyetometer. No object should be nearer to the instrument than 30 m or twice the height of the obstruction. Types of Rain Gauges: 1. Non-Recording Type: It gives total depth of precipitation. The example of which is the Symon’s rain gauge (commonly used in India) it has a gauge diameter of 12.5 cm and the rim collector is at a height of 30 cm above the ground level. The Symon’s rain gauge can also be used to measure snow fall by removing the funnel and receiving bottle. 2. Recording Type Gauge: It gives rainfall depth with respect to time (i.e., intensity of rainfall, mass curve of rainfall). Examples of which are the tipping bucket, weighing bucket and natural siphon’s type. a) Tipping bucket rain gauge. This consists of a cylindrical receiver 30 cm diameter with a funnel inside. Just below the funnel a pair of tipping buckets is pivoted such that when one of the buckets receives a rainfall of 0.25 mm it tips and empties into a tank below, while the other bucket takes its position and the process is repeated. The tipping of the bucket actuates on electric circuit which causes a pen to move on a chart wrapped round a drum which revolves by a clock mechanism. This type cannot record snow. b) Weighing type rain gauge. In this type of rain-gauge, when a certain weight of rainfall is collected in a tank, which rests on a spring-lever balance, it makes a pen to move on a chart wrapped round a clock driven  11 2 | HYDROLOGY drum. The rotation of the drum sets the time scale while the vertical motion of the pen records the cumulative precipitation. c) Float type rain gauge. In this type, as the rain is collected in a float chamber, the float moves up which makes a pen to move on a chart wrapped round a clock driven drum. When the float chamber fills up, the water siphons out automatically through a siphon tube kept in an interconnected siphon chamber. The clockwork revolves the drum once in 24 hours. The clock mechanism needs rewinding once in a week when the chart wrapped round the drum is also replaced. This type of gauge is used by IMD. 3.2.6 Estimation of Missing Rainfall Data Given the annual precipitation values P1, P2, P3, … Pm at a neighboring M stations 1, 2, 3, …., M respectively, it is required to find the missing annual precipitation Px at a station X not included in the above M stations. Further, the normal annual precipitations N1, N2,..., Ni... at each of the above (M+1) stations including station X are known. If the normal annual precipitations at various stations are within about 10% of the normal annual precipitation at station X, then a simple arithmetic average procedure is followed to estimate Px. Thus, 1 𝑃𝑥 = [𝑃 + 𝑃2 + ⋯ + 𝑃𝑚 ] 𝑀 1  12 2 | HYDROLOGY If the normal precipitations vary considerably, then Px is estimated by weighing the precipitation at the various stations by the ratios of normal annual precipitations. This method, known as the normal ratio method, gives P x as 𝑁𝑥 𝑃1 𝑃2 𝑃𝑚 𝑃𝑥 = [ + +⋯+ ] 𝑀 𝑁1 𝑁2 𝑁𝑚 Example: The normal annual precipitation at stations A, B, C and D in a basin are 80.97, 67.59, 76.28 and 92.01 cm respectively. In the year 1975, the station D was inoperative and the stations A, B and C recorded an annual precipitation of 91.11, 72.23 and 79.89 cm respectively. Estimate the rainfall at station D in that year. Solution: As the normal rainfall values vary more than 10%, the normal ratio method is adopted. 92.01 91.11 72.23 79.89 𝑃𝐷 = [ + + ] = 99.41 𝑐𝑚 3 80.97 67.59 76.28 3.2.7 Conversion of Point Rainfall to Areal Rainfall Arithmetic average method—It is obtained by simply averaging arithmetically the amounts of rainfall at the individual rain-gauge stations in the area. ∑ 𝑷𝟏 𝑷𝒂𝒗𝒆 = 𝒏 𝑊ℎ𝑒𝑟𝑒: 𝑃𝑎𝑣𝑒 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑟𝑎𝑖𝑛𝑓𝑎𝑙𝑙 𝑜𝑣𝑒𝑟 𝑎𝑛 𝑎𝑟𝑒𝑎 ∑𝑃1 = 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑎𝑖𝑛𝑓𝑎𝑙𝑙 𝑎𝑚𝑜𝑢𝑛𝑡𝑠 𝑎𝑡 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑟𝑎𝑖𝑛 𝑔𝑎𝑢𝑔𝑒 𝑠𝑡𝑎𝑡𝑖𝑜𝑛 𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑎𝑖𝑛 − 𝑔𝑎𝑢𝑔𝑒 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 This method is fast and simple and yields good estimates in flat country if the gauges are uniformly distributed and the rainfall at different stations do not vary very widely from the mean. These limitations can be partially overcome if topographic influences and aerial representativity are considered in the selection of gauge sites. Thiessen polygon method—This method attempts to allow for non-uniform distribution of gauges by providing a weighting factor for each gauge. The stations are plotted on a base map and are connected by straight lines. Perpendicular bisectors are drawn to the straight lines, joining adjacent stations to form polygons, known as Thiessen polygons (Fig. 2.10). Each polygon area is assumed to be influenced by the rain gauge station inside it, i.e., if P 1, P2, P3,.... are the rainfalls at the individual stations, and A 1, A2, A3,.... are the areas of  13 2 | HYDROLOGY the polygons surrounding these stations, (influence areas) respectively, the average depth of rainfall for the entire basin is given by ∑ 𝑨𝟏 𝑷𝟏 𝑷𝒂𝒗𝒆 = ∑ 𝑨𝟏 𝑤ℎ𝑒𝑟𝑒 ∑𝐴1 = 𝐴 = 𝑡𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑛 The results obtained are usually more accurate than those obtained by simple arithmetic averaging. The gauges should be properly located over the catchment to get regular shaped polygons. However, one of the serious limitations of the Thiessen method is its non-flexibility since a new Thiessen diagram has to be constructed every time if there is a change in the rain gauge network. The Isohyetal Method—In this method, the point rainfalls are plotted on a suitable base map and the lines of equal rainfall (isohyets) are drawn giving consideration to orographic effects and storm morphology, Fig. 2.11. The average rainfall between the successive isohyets taken as the average of the two Isohyetal values are weighted with the area between the isohyets, added up and divided by the total area which gives the average depth of rainfall over the entire basin: ∑ 𝑨𝟏−𝟐𝑷𝟏−𝟐 𝑷𝒂𝒗𝒆 = 𝑨𝟏−𝟐 Where 𝐴1−2 = area between the two successive isohyets 𝑃1 and 𝑃2 𝑃1 + 𝑃2 𝑃1−2 = 2 ∑𝐴1−2 = 𝐴 = 𝑡𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑛 This method if analyzed properly gives the best results. 3.2.8 Double Mass Analysis The trend of the rainfall records at a station may slightly change after some years due to a change in the environment (or exposure) of a station either due to coming of a new building, fence, planting of trees or cutting of forest nearby, which affect the catch of the gauge due to change in the wind pattern or exposure. The consistency of records at the station in question (say, X) is tested by a double mass curve by plotting the cumulative annual (or seasonal) rainfall at station X against the concurrent cumulative values of mean annual (or  14 2 | HYDROLOGY seasonal) rainfall for a group of surrounding stations, for the number of years of record. From the plot, the year in which a change in regime (or environment) has occurred is indicated by the change in slope of the straight-line plot. The rainfall records of the station x are adjusted by multiplying the recorded values of rainfall by the ratio of slopes of the straight lines before and after change in environment. Example: The annual rainfall at station X and the average annual rainfall at 18 surrounding stations are given below. Check the consistency of the record at station X and determine the year in which a change in regime has occurred. State how you are going to adjust the records for the change in regime. Determine the average annual rainfall (a.r.r.) for the period 1952-1970 for the changed regime. ANNUAL RAINFALL (cm) YEAR Station X 18-stn. average 0 1952 30.5 22.8 1953 38.9 35 1954 43.7 30.2 1955 32.2 27.4 1956 27.4 25.2 1957 32 28.2 1958 49.3 36.1 1959 28.4 18.4 1960 24.6 25.1 1961 21.8 23.6 1962 28.2 33.3 1963 17.3 23.4 1964 22.3 36 1965 28.4 31.2 1966 24.1 23.1 1967 26.9 23.4 1968 20.6 23.1 1669 29.5 33.2 1970 28.4 26.4  15 2 | HYDROLOGY Solution: YEAR CUMULATIVE ANNUAL RAINFALL STATION X 18-STN. AVERAGE 0 0 0 1952 30.5 22.8 1953 69.4 57.8 1954 113.1 88 1955 145.3 115.4 1956 172.7 140.6 1957 204.7 168.8 1958 254 204.9 1959 282.4 223.3 1960 307 248.4 1961 328.8 272 1962 357 305.3 1963 374.3 328.7 1964 396.6 364.7 1965 425 395.9 1966 449.1 419 1967 476 442.4 1968 496.6 465.5 1669 526.1 498.7 1970 554.5 525.1 The above cumulative rainfalls are plotted as shown. It can be seen from the figure that there is a distinct change in slope in the year 1958, which indicates that a change in regime (exposure) has occurred in the year 1958. To make the records prior to 1958 comparable with those after change in regime has occurred, the earlier records have to be adjusted by multiplying by the ratio of slopes m2/m1 i.e., 0.9/1.25.  16 2 | HYDROLOGY Cumulative rainfall 1958-1970 = 554.5 – 204.7 = 349.8 cm Cumulative rainfall 1952-1957 adjusted for changed environment = 204.7 × (0.88/1.23) = 146.45 cm Cumulative rainfall 1952-1970 (for the current environment) = 496.25 cm a.a.r. adjusted for the current regime = 496.25 cm/19 years = 26.12 cm.  17 2 | HYDROLOGY 3.3 Assessment Name: _____________________________________________ Year & Section: ________ Instructor: __________________________________________Date: _________________ Direction: Answer the following questions written below. For items that requires solution, you are required to present a complete solution and box the final answer. Avoid erasures in answering the assessment. Criteria for Rating: Timeliness – 10% Organization – 10% Cleanliness - 10% Content - 70% 100% 1. Discuss the current practice and status of rainfall recording in the Philippines. (5 pts.) 2. Identify and describe the different forms of precipitation you have experience in your municipality. (5 pts.) 3. Identify and describe the different rain gauges used in the Philippines. (5 pts) 4. The normal annual precipitation at stations A, B, C and D in a basin are 80.88, 69.51, 73.26 and 91.24 cm respectively. In the year 1990, the station D was inoperative and the stations A, B and C recorded an annual precipitation of 92.22, 70.33 and 77.95 cm respectively. Estimate the rainfall at station D in that year. (5 pts) 5. The normal annual precipitation of five rain gauge stations P, Q, R, S, and T are respectively 125, 102, 76, 113, and 137 cm. During a particular storm the precipitation recorded by stations P, Q, R, and S are 13.2, 9.2, 6.8 and 10.2 cm respectively. The instrument at station T was inoperative during the storm. Estimate the rainfall at station T during the storm. (5 pts) 6. The annual rainfall values at a station P for a period of 20 years are as follows: Year Annual Rainfall (cm) Year Annual Rainfall (cm) 1975 120.0 1985 101.0 1976 84.0 1986 109.0 1977 68.0 1987 106.0 1978 92.0 1988 115.0 1979 102.0 1989 95.0 1980 92.0 1990 90.0  18 2 | HYDROLOGY 1981 95.0 1991 70.0 1982 88.0 1992 89.0 1983 76.0 1993 80.0 1984 84.0 1994 90.0 a.) Identify those years in which the annual rainfall is less than 25% of the mean (5 pts) more than the mean. (5 pts) b.) Plot the three-year moving mean of the annual rainfall time series. (5 pts) c.) Plot the five-year moving mean of the annual rainfall time series. (5 pts) 7. Find the mean precipitation for the area shown by Thiessen polygon method. The area is composed of a square plus an equilateral triangular plot of side 2 km. Rainfall readings are in cm at the various stations indicated. (10 points) 8. Thiessen polygons and Isohyetal map are given in Figures 5.1 and 5.2 below, respectively, for different drainage basins. Areas and precipitation values are tabulated. Compute the average precipitation over the basins. (10 pts.) Station Thiessen Polygon Area Precipitation (cm) (km2) A 170 9.3 B 164 10.5  19 2 | HYDROLOGY C 156 10.9 D 150 12.3 E 116 13.5 F 36 14.0 G 124 14.2 H 42 12.8 Area of basin = 958 km2 Zone Area (km2) I 56 II 192 III 420 IV 244 V 44 VI 58 Area of Basin = 1014 km2 Figure 5.1 Thiessen Polygon  20 2 | HYDROLOGY Figure 5.2 Isohyetal Map 3.4 References: K Subramanya (2008). Engineering Hydrology, Third Edition [E-book]. Tata McGraw- Hill Publishing Company Limited. https://kupdf.net/download/k-subrahmanya- engineering hydrology_5af43798e2b6f5d42e25cdd9_pdf H.M. Raghunath (2006). Hydrology Principles, Analysis and Design, Revised Second Edition [E-book]. New Age International (P) Ltd., Publishers. https://kupdf.net/download/hydrology-principles-analysis- design_59f21ecce2b6f5c576a9756f_pdf 3.5 Acknowledgement The images, tables, figures and information contained in this module were taken from the references cited above.  21 2 | HYDROLOGY UNIT 4: INFILTRATION 4.0 Intended Leaning Outcomes 1. Define Infiltration. 2. Identify Factors affecting infiltration and the different infiltration measurements. 3. Solve problems related to infiltration. 4. Create Infiltration Models by using Excel as a tool. 4.1 Introduction This chapter deals about the infiltration which is one of the process involve in the hydrological cycle. This is the process by which water on the ground surface enters the soil. 4.2 Topics 4.2.1 Definition of Infiltration Infiltration is the flow of water into the ground water through the soil surface. The distribution of soil moisture within the soil profile during the infiltration process is given in the illustration. When water is applied at the surface of the soil, four moisture zones in the soil, as indicated can be identified. 0 Moisture Content 1. Saturation Zone Depth 2. Transition Zone 3. Transmission Zone 4. Wetting Zone Wetting Front  22 2 | HYDROLOGY Zone 1: At the top, a thin layer of saturated zone is created. Zone 2: Beneath Zone 1, there is a transition zone. Zone 3: Next lower zone is the transmission zone where the downward motion of the moisture takes place. The moisture content in this zone is above field capacity but below saturation. Further, it is characterized by unsaturated flow and fairly uniform moisture content. Zone 4: The last zone is the wetting zone. The soil moisture on this zone will be at or near field capacity and the moisture content decreases with the depth. The boundary of the wetting zone is the wetting front where a sharp discontinuity exists between the newly wet soil and the original moisture content of the soil. Depending upon the amount of infiltration and physical properties of the soil, the wetting front can extend from a few centimeters to meters. The infiltration process can be easily understood through a simple analogy. Consider a small container covered with a wire gauze. If water poured into the container a part of it will go into the container and a part over flows. Further, the container can hold only a fixed quantity and when it is full no more flow into the container can take place. While this analogy is highly simplified, it underscores two important aspects; (i) the maximum rate at which the ground can absorb water, the infiltration capacity (ii) the volume of water that the ground can hold, field capacity. Since the infiltered water may contribute to the ground water discharge in addition to increasing the soil moisture, the process may be systematically modeled as in the figure a and b wherein two situations, viz. low intensity rainfall and high intensity rainfall are considered.  23 2 | HYDROLOGY 4.2.2 Factors Affecting infiltration and Infiltration Measurements 4.2.2.1 Factors Affecting Infiltration The factors affecting infiltration are: 1. Slope of the Land 2. Degree of Saturation 3. Porosity 4. Permeability 5. Packing 6. Sorting 7. Shape 8. Capillarity 9. Vegetation 10. Land use 4.2.2.2 Infiltration Measurements The methods of determining infiltration are: 1. Single Ring Infiltrometer 2. Double Ring Infiltrometer 3. Observation in pits and ponds 4. Placing a catch basin below a laboratory sample 5. Artificial Rain simulators 6. Hydrograph Analysis  24 2 | HYDROLOGY 4.2.2.2 1 Single Ring Infiltrometers The ring is a metal cylinder which is driven into the soil. The ring (about 30 cm in diameter and 60 cm in length) is inserted in the soil to the 30 cm depth as shown in Figure 10.1. Water is then ponded on top of the soil surface in the infiltrometer. The soil surface must be protected with a light burlap cloth to reduce surface sealing or surface destruction. The decrease in depth of water with time can be recorded with a point gauge or ruler taped to the inside of the infiltrometer. Only in soils of high infiltration rates is the effect of falling water level significantly different when compared to the magnitude of the matric potential below the wetting front. In practice, the effect of water depth can be ignored for most soils but it is a good procedure to standardize the water depth and use a Marriotte bottle for control (a method described later). The advantages of a single-ring infiltrometer are that installation and measurements are rapid with only 1 ring required, providing a measure of 1- dimensional infiltration within the ring. A disadvantage of this device is the required pounding of the ring into the soil (a difficult task) to a depth that can fracture a dry soil. Cracks, or spaces around the inside of the ring, allow water to seep into these spaces which increases the value of the estimated infiltration rate of the soil. Accurate infiltration can only progress to the depth of the ring insertion; below this point, the infiltrating water can spread out which may invalidate the assumption of I-dimensional vertical infiltration. The results are difficult to use for prediction of infiltration along furrows because the single ring infiltrometer is a point measurement device which does not consider spatial variability. https://www.wikiwand.com/en/Infiltrometer 4.2.2.2.2 Double Ring Infiltrometers A double ring infiltrometer is shown in Fig. 3.7. The two rings (22.5 to 90 cm diameter) are driven into the ground by a driving plate and hammer, to penetrate into the soil uniformly without  25 2 | HYDROLOGY tilt or undue disturbance of the soil surface to a depth of 15 cm. After driving is over, any disturbed soil adjacent to the sides tamped with a metal tamper. Point gauges are fixed in the center of the rings and in the annular space between the two rings. Water is poured into the rings to maintain the desired depth (2.5 to 15 cm with a minimum of 5 mm) and the water added to maintain the original constant depth at regular time intervals (after the commencement of the experiment) of 5, 10, 15, 20, 30, 40, 60 min, etc. up to a period of at least 6 hours is noted and the results are plotted as infiltration rate in cm/hr. versus time in minutes as shown in Fig. 3.8. The purpose of the outer tube is to eliminate to some extent the edge effect of the surrounding drier soil and to prevent the water within the inner space from spreading over a larger area after penetrating below the bottom of the ring. https://www.wikiwand.com/en/Infiltrometer 4.2.2.3 Observation in pits and ponds By noting the depression in the level of water in the pits and ponds and deducting the loss due to evaporation, an idea about the infiltration rates in such soils can be obtained. 4.2.2.3 Placing a catch basin below a laboratory sample By placing a catch basin called a lysimeter under a laboratory sample or at some depth below the land surface, the infiltrating water can be measured and the infiltration rate in the soil can be obtained. 4.2.2.4 Artificial Rain Simulators Artificial rain simulators on a small area of land of 0.1 to 50 m2, water is applied by artificial showers at a uniform rate. The  26 2 | HYDROLOGY resulting surface runoff is measured and the infiltration capacity of the soil is determined. 4.2.2.5 Hydrograph Analysis By knowing accurately, the varying intensities of rainfall during a storm and the continuous record of the resulting runoff, the infiltration capacity can be determined. 4.2.3 Modeling Infiltration Capacity The figure shows a typical variation of infiltration capacity fp with time. Cumulative infiltration capacity Fp (t) is defined as the accumulation of infiltration volume over a time period since the start of the process and is given by 𝑡 𝐹𝑝 = ∫ 𝑓𝑝 (𝑡)𝑑𝑡 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 0 Thus, the curve Fp (t) vs. time in the figure is the mass curve of infiltration. It may be noted that from the equation 1 it follows that 𝑑𝐹𝑝 (𝑡) 𝑓𝑝 (𝑡) = 𝑑𝑡 Many equations have been proposed to express the curves fp(t) or Fp (t) for use in hydrological analysis. In this section, three equations will be described. 4.2.3.1 HORTON’S EQUATION (1933) Horton expressed the decay of infiltration capacity with time as an exponential decay given by: 𝑓𝑝 = 𝑓𝑐 + (𝑓0 − 𝑓𝑐 )𝑒 −𝐾ℎ𝑡 𝑓𝑜𝑟 0 ≥ 𝑡 ≤ 𝑡𝑐  27 2 | HYDROLOGY Where 𝑓𝑝 = 𝑖𝑛𝑓𝑖𝑙𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑎𝑡 𝑎𝑛𝑦 𝑡𝑖𝑚𝑒 𝑡 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑠𝑡𝑎𝑟𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑎𝑖𝑛𝑓𝑎𝑙𝑙 𝑓0 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑖𝑛𝑓𝑖𝑙𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑡 𝑡 = 0 𝑓𝑐 = 𝑓𝑖𝑛𝑎𝑙 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒 𝑖𝑛𝑓𝑖𝑙𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑐𝑐𝑢𝑟𝑟𝑖𝑛𝑔 𝑎𝑡 𝑡 = 𝑡𝑐 , 𝑎𝑙𝑠𝑜, 𝑓𝑐 𝑖𝑠 𝑠𝑜𝑚𝑒𝑡𝑖𝑚𝑒𝑠 𝑘𝑛𝑜𝑤𝑛 𝑎𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑟𝑎𝑡𝑒𝑜𝑟 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑖𝑛𝑓𝑖𝑙𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝐾ℎ = 𝐻𝑜𝑟𝑡𝑜𝑛𝑠 ′𝑠 𝑑𝑒𝑐𝑎𝑦 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑤ℎ𝑖𝑐ℎ 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑢𝑝𝑜𝑛 𝑠𝑜𝑖𝑙 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐𝑠 𝑎𝑛𝑑 𝑣𝑒𝑔𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑣𝑒𝑟. 4.2.3.2 PHILIP’S EQUATION (1957) Philip’s two term model relates Fp (t) as: 𝐹𝑝 = 𝑠𝑡 1/2 + 𝐾𝑡 Where: 𝑠 = 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑜𝑖𝑙 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑎𝑛𝑑 𝑐𝑎𝑙𝑙𝑒𝑑 𝑎𝑠 𝑠𝑜𝑟𝑝𝑡𝑖𝑣𝑖𝑡𝑦 𝐾 = 𝐷𝑎𝑟𝑐𝑦 ′𝑠 ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦 Infiltration capacity is expressed as 1 −1/2 𝑓𝑝 = 𝑠𝑡 +𝐾 2 4.2.3.3 GREEN-AMPT EQUATION (1911) Green and Ampt proposed a model for infiltration capacity based on Darcy’s law as: ∆𝜃𝑆𝑐 𝑓𝑝 = 𝐾 (1 + ) 𝐹𝑝 Where: ∆𝜃 = difference between the initial and final moisture contents of the soil ∆𝜃 = (1 − 𝑆𝑒 )𝜃𝑒 𝑆𝑒 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛 Sc = capillary suction at the wetting front and K = Darcy’s hydraulic conductivity  28 2 | HYDROLOGY The equation can be considered as: 𝑛 𝑓𝑝 = 𝑚 + 𝐹𝑝 Where m and n are Green-Ampt parameters of infiltration model. 4.2.4 Ponding Time In the preceding sections several methods for computing the rate of infiltration into the soil were presented. All of these methods used the assumption that water is ponded to a small depth on the soil surface so all the water the soil can infiltrate is available at the surface. However, during a rainfall, water will pond on the surface only if the rainfall intensity is greater than the infiltration capacity of the soil. The  29 2 | HYDROLOGY ponding time tp is the elapsed time between the time rainfall begins and the time water begins to pond on the soil surface. If rainfall begins on dry soil, the vertical moisture profile in the soil may appear as in figure. Prior to the ponding time (t < tp), the rainfall intensity is less than the potential infiltration rate and the soil surface is unsaturated. Ponding begins when the rainfall intensity exceeds the potential infiltration rate. At this time (t = t p), the soil surface is saturated. As rainfall continues (t > tp), the saturated zone extends deeper into the soil and overland flow occurs from the ponded water. How can the infiltration equations developed previously be used to describe this situation? Figure 4.4 (A) Soil moisture profiles before, during, and after ponding occurs. Mein and Larson (1973) presented a method for determining the ponding time with infiltration into the soil described by the Green-Ampt equation for rainfall of intensity i starting instantaneously and continuing indefinitely. There are three principle involved: (1) prior to the time ponding occurs, all the rainfall is infiltrated; (2) the potential infiltration rate / is a function of the cumulative infiltration F; and (3) ponding occurs when the potential infiltration rate is less than or equal to the rainfall intensity. In the Green-Ampt equation, the infiltration rate fp and cumulative infiltration Fp are related by: ∆𝜃𝑆𝑐 𝑓𝑝 = 𝐾 (1 + ) 𝑒𝑞. 4.4 9(𝑎) 𝐹𝑝 where K is the hydraulic conductivity of the soil, S c is the wetting front capillary pressure head, and ∆𝜃 is the difference between the initial and final moisture content  30 2 | HYDROLOGY of the soil. As shown in Figure 4.4 B, the cumulative infiltration at the ponding time tp is given by Fp = i tp and the infiltration rate by fp = i substituting into Eq. 4.4 9(𝑎). ∆𝜃𝑆𝑐 𝑖 = 𝐾 (1 + ) 𝑒𝑞. 4.4 (𝑏) 𝑖𝑡𝑝 Solving: 𝐾𝑆𝑐 ∆𝜃 𝑡𝑝 = 𝑒𝑞. 4.4 (𝑐) 𝑖(𝑖 − 𝐾) gives the ponding time under constant rainfall intensity using the Green-Ampt infiltration equation. Figure 4.4 (B) Infiltration rate and cumulative infiltration for ponding under constant intensity rainfall. Example: Compute the ponding time and the depth of water infiltrated at ponding for a silt loam soil of 30 percent initial effective saturation, subject to rainfall intensities of a.)1 cm/h and b.) 5cm/h Solution: K = 0.65 cm/h (from Green-Ampt Table) For a silt loam soil  31 2 | HYDROLOGY ∆𝜃𝑆𝑐 = (1 − 0.3)(0.286)= 5.68 cm For i = 1 cm/h Substituting the given to the formula: 𝐾𝑆𝑐 ∆𝜃 0.65 × 5.68 𝑡𝑝 = = = 10.5 ℎ𝑟. 𝑖(𝑖 − 𝐾) 5.0(5.0 − 0.65) And Fp = i tp 𝐹𝑝 = 1.0 × 10.5 = 10.5 𝑐𝑚 For i = 5 cm/h Substituting the given to the formula: 𝐾𝑆𝑐 ∆𝜃 0.65 × 5.68 𝑡𝑝 = = = 0.17 ℎ𝑟. 𝑖(𝑖 − 𝐾) 5.0(5.0 − 0.65) And Fp = i tp 𝐹𝑝 = 5.0 × 0.17 = 0.85 𝑐𝑚 In each case the infiltration rate fp equals the rainfall intensity i at ponding.  32 2 | HYDROLOGY  33 2 | HYDROLOGY 4.2.5 Fitting Infiltration Models to infiltration data using excel Example: Infiltration capacity data obtained in a flooding type infiltration test is given below. Time since 5 10 15 25 45 60 75 90 110 130 start (minutes) Cumulative Infiltration 1.75 3.00 3.95 5.50 7.25 8.30 9.30 10.20 11.28 12.36 Depth (cm) Determine the parameters in (a.) Hortons’s Infiltration capacity equation (b) Philip’s Equation and (c.) Green-Ampt Equation that best fit this data. (a.) Hortons’s Infiltration capacity equation A line graph that is represented by ln(𝑓𝑝 − 𝑓𝑐 ) 𝑣𝑠. 𝑡𝑖𝑚𝑒 (Note: You may read the references for additional information. Step 1: Identify all the important parameters to be input. Then, make a table for the values. For Horton here are the important parameters: Cum. Depth Incremental Depth Infiltration Rate Time Ln (fp - fc) Time (Fp) in the interval (fp) min. cm cm (cm/h) hrs. 0 0 5 1.75 1.75 21.00 2.877 0.083 10 3.00 1.25 15.00 2.465 0.167 15 3.95 0.95 11.40 2.099 0.250 25 5.50 1.55 9.30 1.802 0.417 45 7.25 1.75 5.25 0.698 0.750 60 8.30 1.05 4.20 -0.041 1.000 75 9.30 1.00 4.00 -0.274 1.250 90 10.20 0.90 3.60 -1.022 1.500 110 11.28 1.08 3.24 1.833 130 12.36 1.08 3.24 2.167 Note: Infiltration rate fp = Incremental Depth in the interval (cm) / (current time-previous time) (hrs.)  34 2 | HYDROLOGY Step 2. You are already done with the data. Remember, that x=time and the y = ln (f p – fc). Plot your data. Identify your x and y and Highlight the values  35 2 | HYDROLOGY  36 2 | HYDROLOGY You can add title, x and y axis title and design to your graph. Add the trendline to your graph.  37 2 | HYDROLOGY To add the slope of the best fit line, go to more options. Check the encircled options to display the slope and the R squared Step 3: Now use this data to solve for the last missing parameter which is the f0. Remember, that x = time (t) and the y = ln (fp – fc). Therefore, y = -2.6751x + 2.8868 ln (f – fc) = -2.6751 t + 2.8868  38 2 | HYDROLOGY -Kh = slope of the best fit line = -2.6751, thus Kh = 2.6751 h-1 ln (f0 – fc) = intercept = 2.8868, thus f0 – fc = 17.94, f0 =21.18 Additional information: (You may want to read more the references to understand more the topic) These values are attained by manipulating the given formula of Horton: 𝑓𝑝 = 𝑓𝑐 + (𝑓0 − 𝑓𝑐 )𝑒 −𝐾ℎ𝑡 (𝑓𝑝 − 𝑓𝑐 ) = (𝑓0 − 𝑓𝑐 )𝑒 −𝐾ℎ𝑡 applying ln to both sides 𝑙𝑛(𝑓𝑝 − 𝑓𝑐 ) = 𝑙𝑛[(𝑓0 − 𝑓𝑐 )𝑒 −𝐾ℎ𝑡 ] by properties of logarithm this is equivalent to 𝑙𝑛(𝑓𝑝 − 𝑓𝑐 ) = 𝑙𝑛(𝑓0 − 𝑓𝑐 ) + ln (𝑒 −𝐾ℎ𝑡 ) by properties of logarithm this is also equivalent to 𝑙𝑛(𝑓𝑝 − 𝑓𝑐 ) = 𝑙𝑛(𝑓0 − 𝑓𝑐 ) + −𝐾ℎ 𝑡 ln 𝑒 𝒍𝒏(𝒇𝒑 − 𝒇𝒄 ) = 𝒍𝒏(𝒇𝟎 − 𝒇𝒄 ) + −𝑲𝒉 𝒕 This is the final form used to solve for the parameters of Horton’s Equation NOW IT’S YOUR TURN! (ASSIGNMENT) DETERMINE THE PARAMETERS OF PHILIP’S EQUATION AND GREEN- AMPT USING THE EXCEL. SHOW YOUR SOLUTION. (10 pts. Each) HINT: PHILIP’S EQUATION IS PLOT THROUGH fp vs. t -0.5 GREEN-AMPT EQUATION IS PLOT THROUGH fp vs. 1/Fp. SOLVE FOR PARAMETERS “m” AND “n”.  39 2 | HYDROLOGY 4.3 Assessment Name: _______________________________________ Year & Section: ________ Instructor: ___________________________________ Date: ___ / ____ / _______ Direction: Answer the following question below. Always show solution for problem solving. Criteria for Rating: Timeliness – 10% Organization – 10% Cleanliness - 10% Content - 70% 100% 1.) In the given factors affecting the infiltration, explain how each factor affects the soil infiltration. At least 1-2 sentence only for each explanation. (2 pts. each) 2.) Distinguish between infiltration capacity and infiltration rate. (5 pts.) 3.) Results of an infiltrometer test on a soil are given below. Determine the Horton’s infiltration capacity equation for this soil. (10 pts.) Time since 0.25 0.5 0.75 1.00 1.25 1.500 1.75 2.00 start (hrs.) Infiltration Capacity 5.6 3.20 2.10 1.50 1.20 1.10 1.0 1.0 in cm/h 4.) The infiltration capacity of a catchment is represented by Horton’s Equation as: 𝑓𝑝 = 0.5 + 1.2𝑒 −0.5𝑡 Where fp is cm/hr an t is in hours. Assuming the infiltration to take place at capacity rates in a storm in a 4 hours duration, estimate the average rate of infiltration for the duration of the storm. (10 pts.) 5. Results of an infiltrometer test on a soil is as follows:  40 2 | HYDROLOGY Time since 5 10 15 20 30 40 60 120 150 start (min.) Infiltration Capacity 1.00 1.80 2.50 3.10 4.20 5.10 6.60 11.0 12.90 in cm/h Determine the parameters of Green-Ampt equation and Philip’s Equation. (10 pts each) 4.4 References: K Subramanya (2008). Engineering Hydrology, Third Edition [E-book]. Tata McGraw-Hill Publishing Company Limited. https://kupdf.net/download/k-subrahmanya- engineering hydrology_5af43798e2b6f5d42e25cdd9_pdf H.M. Raghunath (2006). Hydrology Principles, Analysis and Design, Revised Second Edition [E-book]. New Age International (P) Ltd., Publishers.https://kupdf.net/download/hydrology-principles-analysis- design_59f21ecce2b6f5c576a9756f_pdf Ven Te Chow, David R. Maidment, Larry W. Mays (1988). Applied Hydrology. McGraw- Hill Inc. http://ponce.sdsu.edu/Applied_Hydrology_Chow_1988.pdf 4.5 Acknowledgement The images, tables, figures and information contained in this module were taken from the references cited above.