LP3 Hydrology PDF

i 3 | 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 3 | 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 3 | HYDROLOGY UNIT 5: EVAPORATION 5.0 Intended Leaning Outcomes 1. Define Evaporation. 2. Identify the factors affecting evaporation and the different measurement for the different factors. 3. Identify the different methods/procedures for estimating evaporation from open water. 5.1 Introduction Evaporation is one of the major process in the hydrological, thus, it is important that we understand the physics of evaporation, the factors affecting the evaporation, the measurements for different factors affecting this process, and methods or procedures used to estimate evaporation from open source. This chapter is chiefly intended to explain the evaporation. 5.2 Topics 5.2.1 Physics of Evaporation Evaporation is the process in which a liquid change to gaseous state at the free surface below the boiling point through the transfer of heat energy. Consider a body of water in a pond. The molecules of water in constant motion with a wide range of instantaneous velocities. An addition of heat causes this range and average speed to increase. When some molecules possess sufficient kinetic energy, they may crossover the water surface. Similarly, the atmosphere in the immediate neighborhood of the water surface contains water molecules within the water vapor in motion and some of them may penetrate the water surface. The net scape of water molecules from the liquid state to gaseous state constitutes evaporation. Evaporation is a cooling process in the latent heat of vaporization (at about 585 cal/g of evaporated water) must be provided by the water body. 5.2.2 Factors Affecting Evaporation The rate of evaporation is dependent on the following factors given below: 1. Vapor Pressure The rate of evaporation is proportional to the difference between the saturation vapor pressure at the temperature, ew and actual vapor pressure in the air, ea. Thus 𝐸𝐿 = 𝐶 (𝑒𝑤 − 𝑒𝑎 )  2 3 | HYDROLOGY Where EL = rate of evaporation (mm/day) and C = a constant; ew and ea are in mm of mercury. This equation is known as “Dalton’s law of Evaporation” named after John Dalton (1802) who first recognized this law. Evaporation continues till 𝑒𝑤 = 𝑒𝑎. If 𝑒𝑤 > 𝑒𝑎 condensation takes place. 2. Temperature Other factors remaining the same, the rate of evaporation increases with an increase in the water temperature. Regarding the air temperature, although there is a general increase in the evaporation rate with increasing temperature, a high correlation between evaporation rate and air temperature does not exist. Thus, for the same mean monthly temperature it is possible to have evaporation to different degrees in a lake in different months. 3. Wind Wind aids in removing the evaporated water vapor from zone of evaporation and consequently creates a grater scope for evaporation. However, if the wind velocity is large enough to remove all the evaporated water vapor, any further increase in the wind velocity does not influence the evaporation. Thus, the rate of evaporation increases with the wind speed up to a critical speed beyond which any further increase in the wind speed has no influence on the evaporation rate. This critical wind speed value is a function of the size of the water surface. For large water bodies high speed turbulent winds are needed to cause a maximum rate of evaporation. 4. Atmospheric Pressure Other factors remaining the same, a decrease in the barometric pressure, as in high altitude, increases evaporation. 5. Soluble Salts When a solute dissolve in water, the water vapor pressure of the solution is less than of pure water and hence causes reduction in the rate of evaporation. The percent of reduction in evaporation approximately corresponds to the percentage increase in the specific gravity. Thus, for example, under identical conditions evaporation from a sea water is about 2-3% than that from fresh water. 6. Heat Storage in Water Bodies Deep water has more heat storage than shallow ones. A deep lake may store radiation energy received in summer and release it in winter causing less evaporation in summer and more evaporation in winter compared to a shallow lake exposed to a similar situation. However, the effect of heat storage is essentially to change the seasonal evaporation rates and the annual evaporation rate is seldom affected.  3 3 | HYDROLOGY 5.2.3 Measurements of Different factors for Evaporation 1. Lysimeter A lysimeter is a measuring device which can be used to measure the amount of actual evapotranspiration which is released by plants, usually crops or trees. By recording the amount of precipitation that an area receives and the amount lost through the soil, the amount of water lost to evapotranspiration can be calculated. In general, a lysimeter consists of the soil-filled inner container and retaining walls or an outer container, as well as special devices for measuring percolation and changes in the soil- moisture content. There is no universal international standard lysimeter for measuring evapotranspiration. The surface area of lysimeters in use varies from 0.05 to some 100 m2 and their depth varies from 0.1 to 5 m. According to their method of operation, lysimeters can be classified into non- weighable and weighable instruments. Each of these devices has its special merits and drawbacks, and the choice of any type of lysimeter depends on the problem to be studied. Monolithic weighable lysimeters are a tool for water balance studies and solute transport determination. Lysimeters are of two types: 1. Weighing 2. Non-weighing Non-weighable (percolation-type) lysimeters can be used only for long-term measurements, unless the soil-moisture content can be measured by some independent and reliable technique. Large-area percolation-type lysimeters are used for water budget and evapotranspiration studies of tall, deep rooting vegetation cover, such as mature trees. Small, simple types of lysimeters in areas with bare soil or grass and crop cover could provide useful results for practical purposes under humid conditions. This type of lysimeter can easily be installed and maintained at a low cost and is, therefore, suitable for network operations. Weighable lysimeters, unless of a simple micro lysimeter-type for soil evaporation, are much more expensive, but their advantage is that they secure reliable and precise estimates of short-term values of evapotranspiration, provided that the necessary design, operation and siting precautions have been taken. Several weighing techniques using mechanical or hydraulic principles have been developed. The simpler, small lysimeters are usually lifted out of their sockets and transferred to mechanical scales by means of mobile cranes. The container of a lysimeter can be mounted on a permanently installed mechanical scale for continuous recording. The design of the weighing and recording system can be considerably simplified by using load cells with strain gauges of variable electrical resistance. The  4 3 | HYDROLOGY hydraulic weighing systems use the principle of fluid displacement resulting from the changing buoyancy of a floating container (so called floating lysimeter), or the principle of fluid pressure changes in hydraulic load cells. The large weighable and recording lysimeters are recommended for precision measurements in research centers and for standardization and parameterization of other methods of evapotranspiration measurement and the modelling of evapotranspiration. Small weighable types of lysimeters are quite useful and suitable for network operation. Micro lysimeters for soil evaporation are a relatively new phenomenon. 2. Pan Evaporation Pan evaporation is a measurement that combines or integrates the effects of several climate elements: temperature, humidity, rain fall, drought dispersion, solar radiation, and wind. Evaporation is greatest on hot, windy, dry, sunny days; and is greatly reduced when clouds block the sun and when air is cool, calm, and humid Pan evaporation measurements enable farmers and ranchers to understand how much water their crops will need. There are many types of evaporation pans used by farmers. However, the universal pan is the United States Weather Bureau (USWB) Class A pan evaporimeter. It is important to use the same dimensions as this universal pan, mainly because the effect of wind and temperature on evaporation will vary with the surface area and the depth of water in the pan. Evaporation and irrigation replacements cannot be compared between sites if non-standard pans are used. Construction: There are three parts to an evaporimeter. All parts can be made very cheaply with common materials. Alternatively, a complete unit can be purchased at considerably greater cost. The following is a description of how to construct the three components of the evaporimeter.  5 3 | HYDROLOGY Evaporation Pan: The evaporation pan must be made to the standard specifications of an internal diameter of 1207 mm and height of 254 mm using 20 gauges galvanized iron. The standard material is galvanized iron as alternatives will have different thermal and reflectance properties, therefore altering the evaporation rate. It is best to have the pan made by either a galvanized tank manufacturer or an engineering firm. Before the pan is sited in the field it should be checked for leaks. Fixed Pointer: The fixed pointer that sits inside the pan can be made from standard irrigation fittings and a piece of stainless-steel rod. There are three parts to the fixed pointer: 1. The base, a 100 mm PVC flange 2. The pointer support, a 230 mm long piece of 100 mm PVC pipe. Four equally spaced 9 mm holes 70 mm from the base are drilled to allow the water height around the fixed pointer to quickly adjust to the water height in the pan. A single 15 mm long 5 mm wide elongated hole is also drilled 70 mm from the base of the PVC pipe. 3. The pointer, a 170 mm long piece of 5 mm stainless steel rod bent at a right angle 60 mm from one end. From the shorter end a thread is tapped for about 15 mm and a point is ground on the other end of the rod. After fitting the PVC pipe into the flange, the stainless-steel rod is inserted into the elongated hole with nuts located on the inside and outside of the PVC pipe. To initially set the stainless-steel rod in the correct position, the fixed pointer is placed in the pan and the pan is filled with water to a depth of 190 mm. The rod is then slid up or down in the 5 mm elongated hole so that the point of the rod just breaks the surface of the water. Measuring Cylinder: To measure evaporation the pan must be refilled with a known volume of water. The surface area of the pan is 1.14 square meters, so for every mm of evaporation 1.14 liters of water must be added to the pan. A transparent plastic 2 liters measuring jug with vertical sides is an excellent measuring cylinder if it is scaled properly. It is important that the jug actually holds more water than 2 liters so the sides of the jug must extend  6 3 | HYDROLOGY past the 2 liters mark. The jug is filled with 2.28 liters of water and the water level marked. This can conveniently be done by weighing the jug and adding 2.28 kilograms of water. For most jugs this will just about overflow, which is perfect. A jug of water filled to the marker will be equivalent to 2 mm of evaporation. To scale the jug when less than 2 mm of water is required to fill the pan, the distance from the top marker to the bottom of the jug is measured and divided by 20. The numbers 0 to 2.0 in increments of 0.1 are then written with a permanent marking pen from the top marker to the bottom of the jug. These numbers are equivalent to the same number of mm of evaporation from the pan. Measurement: With evaporation the water level in the pan will fall. To measure the amount of evaporation, water is added to the pan with the measuring jug filled to the top mark. Water is added until the pointer just breaks the surface of the water. The PVC pipe supporting the pointer will help by reducing wave motion. It is important to keep track of the number of jugs used to refill the pan and the reading on the last jug when the pan water level is just broken by the pointer. The total amount of water added equals the amount of evaporation. It is also essential to measure rainfall in conjunction with evaporation. Both measurements enable evaporation to be calculated on rainy days. After heavy rain the pan may have to be emptied to bring the water level down to the pointer. After rainfall on a hot summer day, less water may have to be removed than actually fell as rain. For example, after a 25 mm rainfall there might only be 12 mm of water removed from the pan with the measuring jug to bring the water level back to the pointer. The difference between the rainfall (25 mm) and the water removed from the pan (12 mm) is the evaporation. In this example it is 13 mm. If the rain does not fill the pan above the pointer, the rainfall must still be added onto the measured evaporation to give the actual evaporation. For example, if there was 7 mm of rainfall and 6 mm of water was added to the pan with the measuring jug then the evaporation would be 13 mm. Evaporation measurements should be routinely done every day at 9.00 am and clearly recorded. If measurements are not done routinely then the volume of water in the pan will decrease and take less time to heat up during the day and cool at night. This will induce an error which will become greater as the volume of water in the pan decreases. Evaporation measurements are very simple and take less than 5 minutes. 5.2.4 Available Methods/Procedures for Estimating Evaporation from Open Water The analytical methods for determination of lake evaporation can be broadly classified into three categories as:  7 3 | HYDROLOGY 1. Water Budget Method The Water Budget Method is the simplest of the three analytical methods and is also the least reliable. It involves writing the hydrological continuity equation for the lake and determining the evaporation from a knowledge or estimation of other variables. Thus, considering the daily average values for a lake, the continuity equation is written as: 𝑃 + 𝑉𝑖𝑠 + 𝑉𝑖𝑔 = 𝑉𝑜𝑠 + 𝑉𝑜𝑔 + 𝐸𝐿 + ∆𝑆 + 𝑇𝐿 Where: 𝑃 = 𝑑𝑎𝑖𝑙𝑦 𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑉𝑖𝑠 = 𝑑𝑎𝑖𝑙𝑦 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑖𝑛𝑓𝑙𝑜𝑤 𝑖𝑛𝑡𝑜 𝑡ℎ𝑒 𝑙𝑎𝑘𝑒 𝑉𝑖𝑔 = 𝑑𝑎𝑖𝑙𝑦 𝑔𝑟𝑜𝑢𝑛𝑑𝑤𝑎𝑡𝑒𝑟 𝑖𝑛𝑓𝑙𝑜𝑤 𝑉𝑜𝑠 = 𝑑𝑎𝑖𝑙𝑦 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝑖𝑛𝑡𝑜 𝑡ℎ𝑒 𝑙𝑎𝑘𝑒 𝑉𝑜𝑔 = 𝑑𝑎𝑖𝑙𝑦 𝑠𝑒𝑒𝑝𝑎𝑔𝑒 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐸𝐿 = 𝑑𝑎𝑖𝑙𝑦 𝑙𝑎𝑘𝑒 𝑒𝑣𝑎𝑝𝑜𝑟𝑎𝑡𝑖𝑜𝑛 ∆𝑆 = 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑙𝑎𝑘𝑒 𝑠𝑡𝑜𝑟𝑎𝑔𝑒 𝑖𝑛 𝑎 𝑑𝑎𝑦 𝑇𝐿 = 𝑑𝑎𝑖𝑙𝑦 𝑡𝑟𝑎𝑛𝑠𝑝𝑖𝑟𝑎𝑡𝑖𝑜𝑛 𝐿𝑜𝑠𝑠 All quantities are in units of volume (m3) or depth (mm) over a reference area. It can also be written as: 𝑬𝑳 = 𝑷 + (𝑽𝒊𝒔 − 𝑽𝒐𝒔 ) + (𝑽𝒊𝒈 + 𝑽𝒐𝒈 ) − ∆𝑺 − 𝑻𝑳 In these terms, P, Vis, Vos and ∆𝑆 can be measured. However, it is not possible to measure Vig, Vog, and TL and therefore these quantities can only be estimated. Transpiration losses can be considered to be insignificant in some reservoirs. 2. Energy-Budget Method The energy-budget method is an application of the law of Conservation of energy. The energy available for evaporation is determined by considering the incoming energy, outgoing energy and energy stored in the water body over a known interval. Considering the water body as in Fig. 3.4, the energy balance to the evaporation surface in a period of one day is give by 𝐻ₙ = 𝐻𝑑 + 𝐻𝑒 + 𝐻𝑔 + 𝐻𝑠 + 𝐻𝑖 (3.8) where 𝐻ₙ = 𝑛𝑒𝑡 ℎ𝑒𝑎𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑟𝑒𝑐𝑒𝑖𝑣𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑤𝑎𝑡𝑒𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒  8 3 | HYDROLOGY = 𝐻𝑐 (𝑙 − 𝑟) − 𝐻ℎ in which 𝐻𝑐 (𝑙 − 𝑟) = incoming solar radiation into a surface of reflection coefficient (albedo) r. 𝐻ℎ = back radiation (long wave) from water body 𝐻𝑎 = sensible heat transfer from water surface to air 𝐻𝑒 = heat energy used up in evaporation = 𝜌 𝐿𝐸𝐿 , where 𝜌 = density of water, 𝐿 = latent heat of evaporation and 𝐸𝐿 = evaporation in mm 𝐻𝑔 = heat flux into the ground 𝐻𝑠 = heat stored in water body 𝐻𝑖 = net heat conducted out of the system by water flow (advected energy) All the energy terms are in calories per square mm per day. If the time periods are short, the terms 𝐻𝑠 and 𝐻𝑖 can be neglected as negligibly small. All the terms except 𝐻𝑎 can either be measured or evaluated indirectly. The sensible heat term 𝐻𝑎 which cannot be readily measured is estimated using Bowen's ratio 𝛽 given by the expression 𝐻 𝑇𝑤 −𝑇𝑎 𝛽 = 𝜌𝐿𝐸𝑎 = 6.1 𝑥 10−4 𝑥 𝑃𝑎 (3.9) 𝐿 𝑒𝑤 −𝑒𝑎 where Pa = atmospheric pressure in mm of mercury, e = saturated vapour pressure in mm of mercury, e = actual vapour pressure of air in mm of mercury, T = temperature of water surface in C and T = temperature of air in C. From Eqs (3.8) and (3.9) E can be evaluated as 𝐻𝑛 +𝐻𝑔 +𝐻𝑠 +𝐻𝑖 𝐸𝐿 = 𝜌𝐿(1+𝛽) (3.10) Estimation of evaporation in a lake by the energy balance method has been found to give satisfactory results, with errors of the order of 5% when applied to periods less than a week. Further details of the energy-budget method are available in Refs 2, 3 and 5. 3. Mass-Transfer Method This method is based on theories of turbulent mass transfer in boundary layer to calculate the mass water vapour transfer from the surface to the surrounding atmosphere. However, the details of the method are beyond the scope of this book  9 3 | HYDROLOGY and can be found in published literature. With the use of quantities measured by sophisticated (and expensive) instrumentation, this method can give satisfactory results.  10 3 | HYDROLOGY 5.3 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.) Explain briefly the evaporation process. (5 pts.) 2.) Discuss the factors that affect the evaporation from a water body. (5 pts.) 3.) Identify and discuss the different evaporation pans used to measure evaporation. (5 pts.) 4.) A class A pan was set up adjacent to a lake. The depth of water in the pan at the beginning of certain week was 195 mm. In that week there was a rainfall of 45 mm and 15 mm of water was removed from the pan to keep the water level within the specified depth range. If the depth of water in the pan at the end of the week was 190 mm calculate the pan evaporation. Using the suitable pan coefficient estimate the lake evaporation in that week. (10 pts.) 5.) A reservoir had an average surface area of 20 km2 during June 1982. In that month the mean rate of inflow = 10 m3/s, outflow = 15 m3/s, monthly rainfall is 10 cm and change in storage = 16 million m3. Assuming the seepage losses to be 1.8 cm, estimate the evaporation in that month.  11 3 | HYDROLOGY 5.4 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 Lesson 12 Evaporation. Watershed Hydrology. http://ecoursesonline.iasri.res.in/mod/page/view.php?id=125268 5.5 Acknowledgement The images, tables, figures and information contained in this module were taken from the references cited above.  12 3 | HYDROLOGY UNIT 6: Basic Subsurface Flow 6.0 Intended Leaning Outcomes 1. Explain the Law of Darcy, confined and unconfined aquifers. 2. Solve problems related to Ground Water Confined/Unconfined Aquifers and Law of Darcy 6.1 Introduction Part of the infiltrated effective rainfall circulates more or less horizontally in the superior soil layer and appears at the surface through drain channels. This flow is called subsurface flow (in the past it was called hypodermic flow). The presence of a relatively impermeable shallow layer favors this flow. Subsurface flows in water bearing formations have a drainage capacity slower than superficial flows, but faster than groundwater flows. The essential condition for the appearance of the subsurface flows is: the hydraulic lateral conductivity of the environment has to be superior to the vertical conductivity. The subsurface flow in unsaturated regimes can be the base flow in the area with large slopes, and it is dominant in humid regions with vegetal covering and well-drained soils. 6.2 Topics 6.2.1 Law of Darcy, Confined and Unconfined aquifers 6.2.1.1 Darcy’s Law Flow of ground water except through coarse gravels and rockfills is laminar and the velocity of flow is given by Darcy’s law (1856), which states that ‘the velocity of flow in a porous medium is proportional to the hydraulic gradient’, Fig. 7.2, i.e., 𝑽 = 𝒌𝒊 ∆𝒉 𝒊= 𝑳 𝑸 = 𝑨𝑽 = 𝑨𝒌𝒊 𝑊ℎ𝑒𝑟𝑒: 𝑨 = 𝑾𝒃, 𝑻 = 𝑲𝒃 𝑸 = 𝑾𝒃𝑲𝒊  13 3 | HYDROLOGY 𝑸 = 𝑻 𝒊𝒘 Where: 𝑉 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑜𝑤 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑡ℎ𝑒 𝑎𝑞𝑢𝑖𝑓𝑒𝑟 𝐾 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑝𝑒𝑟𝑚𝑒𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑎𝑞𝑢𝑖𝑓𝑒𝑟 𝑠𝑜𝑖𝑙 ∆ℎ 𝑖 = ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = , ∆ℎ = ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑡 𝑖𝑛 𝑎 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑓𝑙𝑜𝑤 𝑝𝑎𝑡ℎ 𝐿 𝐿 𝐴 = 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑞𝑢𝑖𝑓𝑒𝑟 (= 𝑤𝑏) 𝑤 = 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑎𝑞𝑢𝑖𝑓𝑒𝑟 𝑏 = 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑎𝑞𝑢𝑖𝑓𝑒𝑟 𝑇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑞𝑢𝑖𝑓𝑒𝑟 𝑄 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑓𝑙𝑜𝑤 𝑜𝑓 𝑔𝑟𝑜𝑢𝑛𝑑 𝑤𝑎𝑡𝑒𝑟 (𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑟 𝑦𝑖𝑒𝑙𝑑) Darcy’s law is valid for laminar flow, i.e., the Reynolds number (R e) varies from 1 to 10, though most commonly it is less than 1 𝜌𝑉𝑑 𝑅𝑒 = ≤ 1.0 𝜇 where 𝜌 = 𝑚𝑎𝑠𝑠 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 µ = 𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑑 = 𝑚𝑒𝑎𝑛 𝑔𝑟𝑎𝑖𝑛 𝑠𝑖𝑧𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑞𝑢𝑖𝑓𝑒𝑟 𝑠𝑜𝑖𝑙  14 3 | HYDROLOGY In aquifers containing large diameter solution openings, coarse gravels, rockfills and also in the immediate vicinity of a gravel packed well, flow is no longer laminar due to high gradients and exhibit non-linear relationship between the velocity and hydraulic gradient. For example, in a gravel-packed well (mean size of gravel ≈ 5 mm) Re ≈ 45 and the flow would be transitional at a distance of about 5 to 10 times the well radius. 6.2.1.2 Types of Aquifers and Formations A water bearing geologic formation or stratum capable of transmitting water through its pores at a rate sufficient for economic extraction by wells is called ‘aquifer’. Formations that serve as good aquifers are: unconsolidated gravels, sands, alluvium lake sediments, glacial deposits sand stones limestones with cavities (caverns) formed by the action of acid waters (solution openings in limestones and dolomites) granites and marble with fissures and cracks, weathered gneisses and schists heavily shettered quartzites vescicular basalts slates (better than shales owing to their jointed conditions) A geologic formation, which can absorb water but cannot transmit significant amounts is called an ‘aquiclude’. Examples are clays, shales, etc. A geologic formation with no interconnected pores and hence can neither absorb nor transmit water is called an ‘aquifuge’. Examples are basalts, granites, etc. A geologic formation of rather impervious nature, which transmits water at a slow rate compared to an aquifer (insufficient for pumping from wells) is called an ‘aquitard’. Examples are clay lenses interbedded with sand. Specific yield. While porosity (n) is a measure of the water bearing capacity of the formation, all this water cannot be drained by gravity or by pumping from wells as a portion of water is held in the void spaces by molecular and surface tension forces. The volume of water, expressed as a percentage of the total volume of the saturated aquifer, that will drain by gravity when the water table (Ground Water Table (GWT) drops due to pumping or drainage, is called the ‘specific yield (Sy)’ and that percentage volume of water, which will not drain by gravity is called ‘specific retention (S r)’ and corresponds to ‘field capacity’ i.e., water holding capacity of soil (for use by plants and is an important factor for irrigation of crops). Thus,  15 3 | HYDROLOGY 𝑝𝑜𝑟𝑜𝑠𝑖𝑡𝑦 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑦𝑖𝑒𝑙𝑑 + 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑟𝑒𝑡𝑒𝑛𝑡𝑖𝑜𝑛 𝒏 = 𝑺𝒚 + 𝑺𝒓 Specific yield depends upon grain size, shape and distribution of pores and compaction of the formation. The values of specific yields for alluvial aquifers are in the range of 10–20% and for uniform sands about 30%. 6.2.2 Confined and Unconfined Aquifers If there is homogeneous porous formation extending from the ground surface up to an impervious bed underneath (Fig. 7.1), rainwater percolating down in the soil saturates the formation and builds up the ground water table (GWT). This aquifer under water table conditions is called an unconfined aquifer (water-table aquifer) and well drilled into this aquifer is called a water table well. On the other hand, if a porous formation underneath is sandwiched between two impervious strata (aquicludes) and is recharged by a natural source (by rain water when the formation outcrops at the ground surface—recharge area, or outcrops into a river-bed or bank) at a higher elevation so that the water is under pressure in the aquifer (like pipe flow), i.e., artesian condition. Such an aquifer is called an artesian aquifer or confined aquifer. If a well is drilled into an artesian aquifer, the water level rises in the well to its initial level at the recharge source called the piezometric surface. If the piezometric surface is above the ground level at the location of the well, the well is called ‘flowing artesian well’ since the water flows out of the well like a spring, and if the piezometric surface is below the ground level at the well location, the well is called a non-flowing artesian well. In practice, a well can be drilled through 2-3 artesian aquifers (if multiple artesian aquifers exist at different depths below ground level). Sometimes a small band of impervious strata lying above the main ground water table (GWT) holds part of the water percolating from above. Such small water bodies of  16 3 | HYDROLOGY local nature can be exhausted quickly and are deceptive. The water level in them is called ‘perched water table’. Storage coefficient. The volume of water given out by a unit prism of aquifer (i.e., a column of aquifer standing on a unit horizontal area) when the piezometric surface (confined aquifers) or the water table (unconfined aquifers) drops by unit depth is called the storage coefficient of the aquifer (S) and is dimensionless (fraction). It is the same as the volume of water taken into storage by a unit prism of the aquifer when the piezometric surface or water table rises by unit depth. In the case of water table (unconfined) aquifer, the storage coefficient is the same of specific yield (Sy). Since the water is under pressure in an artesian aquifer, the storage coefficient of an artesian aquifer is attributable to the compressibility of the aquifer skeleton and expansibility of the pore water (as it comes out of the aquifer to atmospheric pressure when the well is pumped) and is given by the relation. 𝟏 𝟏 𝑺 = 𝜸𝒘 𝒏𝒃 ( + ) 𝑲𝒘 𝒏𝑬𝒔 Where: 𝑆 = 𝑠𝑡𝑜𝑟𝑎𝑔𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 (𝑑𝑒𝑐𝑖𝑚𝑎𝑙) γw = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑛 = 𝑝𝑜𝑟𝑜𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑜𝑖𝑙 (𝑑𝑒𝑐𝑖𝑚𝑎𝑙) 𝑏 = 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑓𝑖𝑛𝑒𝑑 𝑎𝑞𝑢𝑖𝑓𝑒𝑟 𝐾𝑤 = 𝑏𝑢𝑙𝑘 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝐸𝑠 = 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 (𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦)𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙 𝑔𝑟𝑎𝑖𝑛𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑞𝑢𝑖𝑓𝑒𝑟. Since water is practically incompressible, expansibility of water as it comes out of the pores has a very little contribution to the value of the storage coefficient. The storage coefficient of an artesian aquifer ranges from 0.00005 to 0.005, while for a water table aquifer S = Sy = 0.05–0.30. The specific yield (unconfined aquifers) and storage coefficient (confined aquifers), values have to be determined for the aquifers in order to make estimates of the changes in the ground water storage due to fluctuation in the GWT or piezometric surface (ps) from the relation. 𝚫𝑮𝑾𝑺 = 𝑨𝒂𝒒 × ∆𝑮𝑾𝑻 𝒐𝒓 𝒑𝒔 × 𝑺 𝒐𝒓 𝑺𝒚 Where: Δ𝐺𝑊𝑆 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑔𝑟𝑜𝑢𝑛𝑑 𝑤𝑎𝑡𝑒𝑟 𝑠𝑡𝑜𝑟𝑎𝑔𝑒 𝐴𝑎𝑞 = 𝑖𝑛𝑣𝑜𝑙𝑣𝑒𝑑 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑞𝑢𝑖𝑓𝑒𝑟 ∆𝐺𝑊𝑇 𝑜𝑟 𝑝𝑠 = 𝑓𝑙𝑢𝑐𝑡𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝐺𝑊𝑇 𝑜𝑟 𝑝𝑠  17 3 | HYDROLOGY 𝑆 𝑜𝑟 𝑆𝑦 = 𝑠𝑡𝑜𝑟𝑎𝑔𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 (𝑐𝑜𝑛𝑓𝑖𝑛𝑒𝑑 𝑎𝑞𝑢𝑖𝑓𝑒𝑟)𝑜𝑟 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑦𝑖𝑒𝑙𝑑 (𝑢𝑛𝑐𝑜𝑛𝑓𝑖𝑛𝑒𝑑 𝑎𝑞𝑢𝑖𝑓𝑒𝑟) Example 1: In a certain alluvial basin of 100 km2, 90 Mm3 of ground water was pumped in a year and the ground water table dropped by about 5 m during the year. Assuming no replenishment, estimate the specific yield of the aquifer. If the specific retention is 12%, what is the porosity of the soil? Solution: Δ𝐺𝑊𝑆 = 𝐴𝑎𝑞 × ∆𝐺𝑊𝑇 𝑜𝑟 𝑝𝑠 × 𝑆 𝑜𝑟 𝑆𝑦 90 × 106 = (100 × 106 ) × 5 × 𝑆𝑦 𝑺𝒚 = 𝟎. 𝟏𝟖 𝑷𝒐𝒓𝒐𝒔𝒊𝒕𝒚(𝒏) = 𝑆𝑦 + 𝑆𝑟 = 0.18 + 0.12 = 𝟎. 𝟑𝟎 𝟎𝒓 𝟑𝟎% Example 2: An artesian aquifer, 30 m thick has a porosity of 25% and bulk modulus of compression 2000 kg/cm2. Estimate the storage coefficient of the aquifer. What fraction of this is attributable to the expansibility of water? Bulk modulus of elasticity of water = 2.4 × 104 kg/cm2. 1 1 𝑆 = 𝛾𝑤 𝑛𝑏 ( + ) 𝐾𝑤 𝑛𝐸𝑠 1 1 𝑆 = 1000 × 0.25 × 30 ( 8 + ) 2.11 × 10 0.25 × 2 × 107 𝑆 = 1.54 × 10−3 Storage coefficient due to the expansibility of water as a percentage of S above 7500 × 0.467 × 108 = × 100 = 2.28%, 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑛𝑒𝑔𝑙𝑖𝑔𝑖𝑏𝑙𝑒 7500 × 20.467 × 108 Note: In less compressible formations like limestones for which Es ≈ 2 × 10 5 kg/cm2, S = 5 × 10–5 and the fractions of this attributable to water and aquifer skeleton are 70% and 30%, respectively.  18 3 | HYDROLOGY 6.2.3 Transmissibility It can be seen from 𝑄 = 𝑇 𝑖𝑤, that T = Q, when i = 1 and w = 1; i.e., the transmissibility is the flow capacity of an aquifer per unit width under unit hydraulic gradient and is equal to the product of permeability times the saturated thickness of the aquifer. In a confined aquifer, T = Kb and is independent of the piezometric surface. In a water table aquifer, T = KH, where H is the saturated thickness. As the water table drops, H decreases and the transmissibility is reduced. Thus, the transmissibility of an unconfined aquifer depends upon the depth of GWT. 6.2.4 Radial Ground Water Flow in Confined and Unconfined Aquifers (Well Hydraulics) Steady radial flow into a well (Dupuit 1863, Thiem 1906) a. Water table conditions (unconfined aquifer) (Water table conditions (unconfined aquifer) Assuming that the well is pumped at a constant rate Q for a long time and the water levels in the observation wells have established, i.e., equilibrium conditions have been reached 𝜋𝐾(ℎ2 2 − ℎ2 2 ) 𝑄= 𝑟 2.303𝑙𝑜𝑔10 (𝑟2 ) 1  19 3 | HYDROLOGY Between the face of the well (r = rw, h = hw) and the point of zero drawdown (r = R, h = H): 𝜋𝐾(ℎ2 2 − ℎ2 2 ) 𝑄= 𝑅 2.303𝑙𝑜𝑔10 (𝑟 ) 𝑤 If the drawdown in the pumped well (Sw = H – hw) is small: 𝐻 2 − ℎ𝑤 2 = (𝐻 + ℎ𝑤 )(𝐻 − ℎ𝑤 ), 𝐻 + ℎ𝑤 = 2𝐻 = 2𝐻 (𝐻 − ℎ𝑤 ) Then, 2𝜋𝐾𝐻 (𝐻 − ℎ𝑤 ) 𝑄= , 𝐾𝐻 = 𝑇 𝑅 2.303𝑙𝑜𝑔10 ( ) 𝑟𝑤 2.72𝑇 (𝐻 − ℎ𝑤 ) 𝑄= 𝑅 𝑙𝑜𝑔10 ( ) 𝑟𝑤 (b) Artesian conditions (confined aquifer) If the well is pumped at constant pumping rate Q for a long time and the equilibrium conditions have reached 𝜋𝑘𝑏(ℎ2 − ℎ1 ) 𝑄= 𝑟 2.303𝑙𝑜𝑔10 (𝑟2 ) 1  20 3 | HYDROLOGY Between the face of the well (r = rw, h = hw) and the point of zero drawdown (r = R, h = H), simplifying and putting T = Kb 2.72𝑇 (𝐻 − ℎ𝑤 ) 𝑄= 𝑅 𝑙𝑜𝑔10 (𝑟 ) 𝑤 which is the same as equation for unconfined aquifer (for water table conditions under small drawdown). Note: The length of screen provided will be usually half to three-fourths of the thickness of the aquifer for obtaining a suitable entrance velocity (≈ 2.5 cm/sec) through the slots to avoid incrustation and corrosion at the openings; the percentage open area provided in the screen will be usually 15 to 18%. Dupuit’s Equations Assumptions The following assumptions are made in the derivation of the Dupuit Thiem equations: (i) Stabilized drawdown—i.e., the pumping has been continued for a sufficiently long time at a constant rate, so that the equilibrium stage of steady flow conditions has been reached. (ii) The aquifer is homogeneous, isotropic, of infinite areal extent and of constant thickness, i.e., constant permeability. (iii) Complete penetration of the well (with complete screening of the aquifer thickness) with 100% well efficiency. (iv) Flow lines are radial and horizontal and the flow is laminar, i.e., Darcy’s law is applicable. (v) The well is infinitely small with negligible storage and all the pumped water comes from the aquifer. The above assumptions may not be true under actual field conditions; for example, if there is no natural source of recharge nearby into the aquifer, all the pumped water comes from storage in the aquifer resulting in increased drawdowns in the well with prolonged pumping and thus the flow becomes unsteady (transient flow conditions). There may be even leakage through the overlying confining layer (say, from a water table aquifer above the confining layer) of an artesian aquifer (leaky artesian aquifer). The hydraulics of wells with steady and unsteady flow under such conditions as developed from time to time by various investigators like Theis, Jacob, Chow, De Glee, Hantush, Walton, Boulton etc. have been dealt in detail in the author’s companion volume ‘Ground Water’ published by M/s Wiley Eastern Limited New Delhi and the reader is advised to refer the book for a detailed practical study of Ground Water dealing with Hydrogeology, Ground Water Survey and Pumping Tests, Rural Water Supply and Irrigation Systems. For example, in the Theis equation for unsteady radial flow into a well it is assumed that the water pumped out is immediately released from storage of the aquifer (no recharge) as the piezometric surface or the water table drops. But in unconfined aquifers and leaky artesian aquifers (that receive water from upper confining layer with a free water table), the rate of fall of the water table may be faster  21 3 | HYDROLOGY than the rate at which pore water is released; this is called ‘delayed yield’ as suggested by Boulton. 6.2.5 Specific Capacity The specific capacity (Q/Sw) of a well is the discharge per unit drawdown in the well and is usually expressed as lpm/m. The specific capacity is a measure of the effectiveness of the well; it decreases with the increase in the pumping rate (Q) and prolonged pumping (time, t). Example 3 A 20-cm well penetrates 30 m below static water level (GWT). After a long period of pumping at a rate of 1800 lpm, the drawdowns in the observation wells at 12 m and 36 m from the pumped well are 1.2 m and 0.5 m, respectively. Determine: (i) the transmissibility of the aquifer. (ii) the drawdown in the pumped well assuming R = 300 m. (iii) the specific capacity of the well. Solution: 𝜋𝐾(ℎ22 −ℎ12 ) Dupuit’s Equation: 𝑄= 𝑟 2.303𝑙𝑜𝑔10(𝑟2 ) 1 ℎ2 = 𝐻 − 𝑠2 = 30 − 0.5 = 29.5 𝑚; ℎ1 = 𝐻 − 𝑠1 = 30 − 1.2 = 28.8 𝑚 1.800 𝜋𝐾(29.52 − 28.82 ) = 60 36 2.303𝑙𝑜𝑔10 ( ) 12 𝒎 𝑲 = 𝟐. 𝟔𝟐 × 𝟏𝟎−𝟒 𝒐𝒓 𝟐𝟐. 𝟕 𝒎/𝒅𝒂𝒚 𝒔𝒆𝒄 1. The transmissibility of the aquifer: 𝑇 = 𝐾𝐻 = (2.62 × 10−4 )30 = 78.6 × 10−4 𝑚2 /𝑠𝑒𝑐, = 22.7 × 30 = 𝟔𝟖𝟏𝒎𝟐 /𝒅𝒂𝒚 2. The drawdown in the pumped well assuming R = 300 m: 2.72𝑇 (𝐻 − ℎ𝑤 ) 𝑄= 𝑅 𝑙𝑜𝑔10 (𝑟 ) 𝑤  22 3 | HYDROLOGY 1.800 2.72(78.6 × 10−4 )𝑆𝑤 = 60 300 𝑙𝑜𝑔10 (0.10) ∴ 𝑑𝑟𝑎𝑤𝑑𝑜𝑤𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑤𝑒𝑙𝑙, 𝑺𝒘 = 𝟒. 𝟖𝟖 𝒎 3. The specific capacity of the well: 𝑄 1.800 = = = 0.0062 (𝑚3 𝑠𝑒𝑐 −1/𝑚) = 𝟑𝟕𝟐 𝑰𝒑𝒎/𝒎 𝑆𝑤 60 × 4.88 or 𝑄 𝑇 78.6 × 10−4 𝑚2 𝑠𝑒𝑐 −1 = = = 0.00655 = 𝟑𝟗𝟑 𝑰𝒑𝒎/𝒎 𝑆𝑤 1.2 1.2 𝑚 6.2.6 Travel Time of Confined Aquifer 𝐷 𝑡= 𝑉𝑠 Where: 𝑡 = 𝑡𝑖𝑚𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑑 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑉𝐷 𝑉𝑠 = 𝑠𝑒𝑒𝑝𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑛 Where: 𝑄 𝑉𝐷 = 𝑑𝑎𝑟𝑐𝑦 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝐴 𝑉𝑣 (𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑣𝑜𝑖𝑑𝑠) 𝑛 = 𝑝𝑜𝑟𝑜𝑠𝑖𝑡𝑦 = 𝑉𝑡 (𝑇𝑜𝑡𝑎𝑙 𝑉𝑜𝑙𝑢𝑚𝑒) Example 4: A confined Aquifer has a source of recharge, the piezometric head in the two wells 1000 m apart is 55 m and 50 m respectively, from a common datum. The average thickness of the aquifer is 5 km. The hydraulic conductivity for the aquifer is 50m/day, and the porosity is 0.20. a. Determine the rate of the flow through the aquifer. b. Time travelled from the head of the aquifer to a point 4 km downstream.  23 3 | HYDROLOGY Solution: n = 0.2, k=50m/day, width = 5000 m, thickness = 30m, h1 = 55m, h2=50m, L=1000m 55 − 50 1 𝑖= = 1000 200 𝐴 = 30(5000) = 150,000 1 𝑄 = 𝑘𝑖𝐴 = 50 ( ) (150,000) = 37,500 𝑚3 /𝑑𝑎𝑦 200 𝑞 37,500 𝑉𝐷 = = = 0.25 𝑚/𝑑𝑎𝑦 𝑎 150,000 𝑽𝑫 𝟎. 𝟐𝟓 𝑽𝒔 = = = 𝟏. 𝟐𝟓 𝒎/𝒅𝒂𝒚 (𝒓𝒂𝒕𝒆 𝒐𝒇 𝒇𝒍𝒐𝒘) 𝒏 𝟎. 𝟐 𝒅 𝟒𝟎𝟎 𝒕= = = 𝟑, 𝟐𝟎𝟎 𝒅𝒂𝒚𝒔 (𝒕𝒊𝒎𝒆 𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅) 𝑽𝒔 𝟏. 𝟐𝟓  24 3 | HYDROLOGY 6.3 Assessment Name: _______________________________________ Year & Section: ________ Instructor: ___________________________________ Date: ___ / ____ / _______ Test I. Identify the terms described in each item. Avoid Erasures. _________________ 1. It is a water bearing formation of the earth. It not only holds the water but yields it in sufficient quantity. Example are unconsolidated deposits of sand and gravel. _________________ 2. It is a formation through which only seepage is possible and thus the yield is insignificant compared to an aquifer. It is partly impermeable. _________________ 3. It is a geological formation which is essentially impermeable to the flow of water. It may contain large amounts of water through it. _________________ 4. It is the free water surface in an unconfined aquifer. _________________ 5. It is called the water table aquifer. _________________ 6. An aquifer between two impervious beds such as aquicludes and aquifuges. _________________ 7. It is the volume of water that can be extracted as percent of total volume of aquifer. _________________ 8. It is the volume of aquifer releases per unit surface area of aquifer per unit change in the component of head normal to that surface. _________________ 9. It is the discharge per unit draw down at the well. It is a measure of performance of the well. _________________ 10. It is the rate of flow of water through a vertical strip aquifer of unit width (1 meter) and extending to full saturation height under unit hydraulic gradient.  25 3 | HYDROLOGY Test II. Answer the following problems given below. Avoid erasures. Criteria for evaluation: Correct Answer w/solution - 80% Organization - 10% Cleanliness - 10% 100% 1. A 45 cm well penetrates an unconfined aquifer of saturated thickness 30 m completely. Under a steady pumping rate for a long time the drawdowns at two observation wells 15 and 30 m form the well are 5.0 and 4.2m respectively. If the permeability of the aquifer is 20 m/day, determine the discharge and the drawdown at the pumping well. (5 points) 2. Two channels are 2000 ft apart. The water level of two channel are 120 ft. and 110 ft., respectively. A pervious formation averaging 30 ft thick with hydraulic conductivity of 0.25 ft/hr. and porosity of n=0.25. Determine the flow rate of seepage from the river to the channel and the time travelled from the head of the aquifer to 5 ft downstream. (5 points)  26 3 | HYDROLOGY 6.4 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 Chapter 6: Runoff and Subsurface Flow. https://echo2.epfl.ch/VICAIRE/mod_1a/chapt_6/main.htm Mariane Encabo (Feb 7, 2020). Basic Subsurface Flow. Scribd. https://www.scribd.com/presentation/446060216/Basic-Subsurface-Flow-Steady- State-Condition 6.5 Acknowledgement The images, tables, figures and information contained in this module were taken from the references cited above.