CVE 523 Engineering Hydrology 1 Lecture Notes PDF

**ENGINEERING HYDROLOGY 1 -- CVE 523** **COURSE CONTENT** 1. **Quantity** -- Population forecasting, per capita consumption, Water requirements, domestic, public, commercial, industrial and agricultural purposes. 2. **Collection**: Rainwater from roofs, Determination of storage capacity for small water supplies, surface water from reservoirs, rivers, intake structures, Groundwater, Transmission conduits. 3. **Treatment:** Flow diagrams for the treatment of surface and ground water, preliminary treatment, screening, coagulation, flocculation and sedimentation, slow sand, rapid sand and pressure filters. Disinfection, water softening, Iron and Manganese removal, Chemicals for water treatment. 4. **Distribution:** Storage tanks and service reservoirs. Mains, pipelines and distribution network, Valves, meters and service pipes, pumps. Pumping stations 5. **Laboratory and course work.** **COURSE EVALUATION** a. **Continuous Assessment:** Classwork and take home assignments, term paper and test(s) = 30% b. **Examination:** 70% **REFERENCE TEXTS** 1. Water supply Engineering Design By M.Anis Al- Layla , S. Ahmad and Middlebrooks 2. Water Engineering Systems By J.C.Agunwamba 3. Water and Wastewater Technology 4. Environmental Engineering By Ruth F. Weiner and Robin Matthews 5. Water Supply Engineering By B.C.Punmia, Ashok Kumar Jain and Arun K. Jain **DEMAND FOR WATER** Whenever an engineer is required to design a water supply scheme for a community, or a section of a community, it is of important to determine the amount of water required (or the water demand) and then find the sources that will be used to meet the demand. **Water Requirement** The water requirement of a given community is dependent on the following factors: i. Population ii. Climatic conditions iii. Habit and mode of living iv. Plumbing facilities v. Sewerage vi. Industry vii. Water tax viii. Metering: In places where meters (water meters) are installed, less water is consumed in comparison to areas where flat rate is charged. The total water consumed in a community is estimated by adding domestic consumption. Water required for public use and water utilized by industry. The loss of water due to leaks and other sources are also considered when estimating the total water demand for a community. This may be taken as 10% of the total water demand. The average domestic water requirement of a community is computed based on water consumption per capita per day. In developing countries average water consumption may be about 135 *lpcd* and can be as high as 450- 500 *lpcd* in towns in developed countries with good plumbing facilities. **Municipal Water demand** Municipal water demand includes: (i) Domestic or residential water demand (ii) Commercial water demand (iii) Industrial water demand (iv) Public service water demand (v) Unaccounted for system losses and leakage. It is necessary to have an accurate estimate of the amount of water needed when water supply projects are being designed. **Components of Municipal Water demand:** The various components of municipal water demand previously itemized and which will be briefly discussed include: (i) Residential (ii) Industrial (iii) Commercial (iv) Public water uses and (v) Unaccounted for system losses. i. **Residential water demand:** This is the component of municipal water supply used in homes for such activities as toilet flushing, cooking, drinking, bathing, washing of plates and clothes, watering lawns and other uses. Average residential water demand may vary from 100 -- 450 *lpcd* ii. **Commercial water demand***:* This is the water used by commercial establishments such as hotels, office buildings, shopping centers, service stations, air ports. Commercial water demand. Commercial water demand depends on the type and number commercial establishments in the community. Commercial water demand is about 10 to 20 % of the total water demand. iii. **Industrial water demand:** Industrial water demand is the water used by industries. They are usually very large and hence many industries tend to develop their water supplies independent of public water supply systems. Only small industries depend on public utilities for their water supply thereby imposing demand on local municipal system. iv. **Public water demand:** This is water that is used in public buildings (such as city halls, jails, schools etc) as well as water used for such public services as fire protection, park irrigation and street washing. Public water use accounts for 5 to 10% of total municipal water demand. v. **Unaccounted system losses and leakage.** In a water supply system there is a certain amount of water that is lost or unaccounted for because of meter and slippage, leaks in mains, faulty meters and unauthorized water connections. In municipal supply systems, this may be 10 -- 20% of the water demand. **Fire demand** In thickly populated and industrial areas, fires usually break out and may lead to serious losses if not controlled effectively. Firefighting requires sufficient quantity of water for extinguishing fires in the event of an outbreak. The quantity of water required for extinguishing fires should be easily available and kept always stored in reservoirs. In the design of public water systems the rate of fire demand is sometimes treated as a function of population using empirical formulae. For example: i. Kuichling formula given by : \ [\$\$Q = \\ 3182\\sqrt{P}\$\$]{.math.display}\ Where Q = amount of water required in litres per minute (l/min) P = population in thousands. ii. National Board for Fire Underwriters (NBFU) formula is given by \ [\$\$Q = \\ 3860\\sqrt{P}\\left( 1 - 0.01\\sqrt{P} \\right)\$\$]{.math.display}\ Where Q = amount of water required in litres per minute (l/min) P = population in thousands. iii. **Insurance Service Office (ISO) Method** More recently, the fire flow requirement for a community have been established by Insurance office method (ISO). In rating cities and municipalities for insurance purposes, the following components are considered by Insurance Service Office (ISO) in the evaluation of adequacy and reliability: source of supply, pumping capacity, power supply, mains, spacing of valves, and location of fire hydrants. They are all the essential components of the firefighting facility or fire suppression infrastructure of a municipality (Sincero and Sincero, 1996) The required fire flow for a neighborhood depends upon the size of the area, type of construction, occupancy and exposure of the building within and surrounding the block or group complex. The equation adopted in the ISO method is: \ [*F* = 3.7*x*10^ − 3^ CA^0.5^]{.math.display}\ Where F = fire flow requirement in m^3^/s A= total floor area in m^2^ excluding the basement of the building. C= coefficient related to the type of construction. The C values for different types of construction are as follows: 1.5, for wood frame construction; 1.0 for ordinary construction; 0.8 for non-combustible construction and 0.6 for resistive construction. In addition it is recommended by ISO that the required fire flow be available for the expected duration of fire as given in Table 1 below. **Table 1:** Fire flow durations. **Fire flow (m^3^/s)** **Duration (hr)** ------------------------ ------------------- ≥ 0.61 10 0.54 -0.61 9 0.48-0.54 8 0.42-0.48 7 0.36-0.42 6 0.30-0.36 5 0.24-0.30 4 0.16-0.24 3 ≤ 0.16 2 **The average daily demand (Add)** represents the average daily demand over a period of one year and may be referred to as annual average daily demand. **The maximum day demand** represents the amount of water required during the day of maximum consumption in a year. This information is used in the analysis of peak capacity of production and treatment facilities. The peak hour demand represents the amount of water required during the maximum consumption hour in a given day. This information is used to analyze the peak capacity required of the distribution system, elevated reservoir and high service pumps to be able to deliver the peak water demand during the peak hour of the day. The annual average, maximum day and peak hour demands are important for planning and design of municipal water supply systems. These variations are conveniently expressed as a ratio of mean daily flow. **Example 1** The estimated hourly requirement for the maximum day in a small city is given in table below. The fire flow has been determined as 65l/s for a conflagration of 2 hours. If pumping is to start at hour 0800 and end at hour 1600. Calculate the storage requirement. **Solution.** Length of pumping 1600-0800 = 8 hours. +-----------------+-----------------+-----------------+-----------------+ | **Hour ending** | **Demand** | **Water | **Required | | | | pumped** | storage (m3)** | | | **(m3)** | | | | | | **(m3)** | | +=================+=================+=================+=================+ | 0100 | 160 | 0 | 160 | +-----------------+-----------------+-----------------+-----------------+ | 0200 | 129 | 0 | 129 | +-----------------+-----------------+-----------------+-----------------+ | 0300 | 146 | 0 | 146 | +-----------------+-----------------+-----------------+-----------------+ | 0400 | 138 | 0 | 138 | +-----------------+-----------------+-----------------+-----------------+ | 0500 | 159 | 0 | 159 | +-----------------+-----------------+-----------------+-----------------+ | 0600 | 196 | 0 | 196 | +-----------------+-----------------+-----------------+-----------------+ | 0700 | 274 | 0 | 274 | +-----------------+-----------------+-----------------+-----------------+ | 0800 | 390 | 0 | 390 | +-----------------+-----------------+-----------------+-----------------+ | 0900 | 480 | 961.25 | **\_** | +-----------------+-----------------+-----------------+-----------------+ | 1000 | 500 | 961.25 | **\_** | +-----------------+-----------------+-----------------+-----------------+ | 1100 | 473 | 961.25 | **\_** | +-----------------+-----------------+-----------------+-----------------+ | 1200 | 451 | 961.25 | **\_** | +-----------------+-----------------+-----------------+-----------------+ | 1300 | 444 | 961.25 | **\_** | +-----------------+-----------------+-----------------+-----------------+ | 1400 | 447 | 961.25 | **\_** | +-----------------+-----------------+-----------------+-----------------+ | 1500 | 434 | 961.25 | **\_** | +-----------------+-----------------+-----------------+-----------------+ | 1600 | 405 | 961.25 | **\_** | +-----------------+-----------------+-----------------+-----------------+ | 1700 | 417 | 0 | 417 | +-----------------+-----------------+-----------------+-----------------+ | 1800 | 421 | 0 | 421 | +-----------------+-----------------+-----------------+-----------------+ | 1900 | 406 | 0 | 406 | +-----------------+-----------------+-----------------+-----------------+ | 2000 | 343 | 0 | 343 | +-----------------+-----------------+-----------------+-----------------+ | 2100 | 270 | 0 | 270 | +-----------------+-----------------+-----------------+-----------------+ | 2200 | 223 | 0 | 223 | +-----------------+-----------------+-----------------+-----------------+ | 2300 | 199 | 0 | 199 | +-----------------+-----------------+-----------------+-----------------+ | 2400 | 185 | 0 | 185 | +-----------------+-----------------+-----------------+-----------------+ | **∑** | **7690** | **7690** | **4056** | +-----------------+-----------------+-----------------+-----------------+ Thus storage requirement = 4056 m^3^ + fire flow requirement = 4056 m^3^ + [\$\\frac{65}{1000}\\left( \\ 60 \\right)\\left( 60 \\right)\\ \\left( 2 \\right)\$]{.math.inline} = **4524 m^3^** **Example 2** The town as depicted in example 1 is populated by buildings of ordinary construction as general characteristics and with a total combined floor area of 2000m3 per story. Compute the storage requirement. Given that the average number of stories per building is 3. **Solution** [*F*= 3.7 (10^ − 3^) *CA*^0.5^]{.math.inline} = 3.7 [(10^ − 3^) (1.0)\[2000 (3)\]^0.5^]{.math.inline} = 0.29 m^3^/s From Table 1 the expected duration is 4 hours. From the previous example, the required storage for normal demand is 4056 m^3^. Therefore, The total storage requirement = 4056 + 0.29 x60x60x4 = **8232 m^3^** **Example 3** A water supply scheme is to be designed for a city with a population of 100,000 persons. Estimate the following: i. Average daily demand ii. Maximum daily demand if the maximum daily demand is 180% of average daily demand. iii. Fire demand using the NBFU formula if the average water consumption is 250 *lpcd.* i. Average daily demand = 100, 000 x 250 *lpcd* = 25 x 10^6^ = 25 Ml/d = 25,000 m^3^/d ii. Maximum daily demand if maximum daily demand = 1.8 x average daily demand. = 1.8x 25,000 m^3^/d = **45,000 m^3^/d.** iii. Fire flow using NBFU formula:[\$Q\\ = \\ 3860\\sqrt{P}\\left( 1 - 0.01\\sqrt{P} \\right)\$]{.math.inline} From the given parameters, Population = 100, 000, hence P= 100. Therefore: \ [\$\$Q\\ \\ = \\ 3860\\sqrt{P}\\left( 1 - 0.01\\sqrt{P} \\right)\$\$]{.math.display}\ **Q** (l/s) = [\$3860\\sqrt{100}\\left( 1 - 0.01\\sqrt{100} \\right)\$]{.math.inline} **Q** (l/s) = [\$3860\\sqrt{100}\\left( 1 - 0.01\\sqrt{100} \\right)\$]{.math.inline} **Q** (l/s) = {.math.inline}x10 (0.9) = 3860 x9 = **34,740 l/min** **ASSIGNMENT** A city of 60,000 residents has an average water demand of 350 *lpcd.* The institutional / commercial and the industrial average area in the city are 200 and 300 hectares respectively and water demand expected is 20and 23m^3^ per hectare per day. The public water use and water unaccounted for are 10 and 6% respectively. Calculate total municipal demand and each component as a percent of total municipal demand. **STEPS IN PLANNING A MAJOR OR PRINCIPAL WATER SYSTEM** The following steps are followed when a major water supply system is being planned: i. Estimate the future population of the community and the water consumption per capita per day in order to determine the quantity of water which is expected to be provided. ii. A reliable source of water of adequate quality is located. iii. Determine the physical, chemical and biological characteristics of the water. iv. Design any necessary water treatment facilities. v. Make provision for necessary storage of water and design the works required to abstract water from the source. vi. The transmission and distribution system as well as distribution reservoir, pumping station, elevated storage, layout and size of mains and location of fire hydrants. vii. Provision is made for the establishment of an organization for the operation and maintenance of the supply, distribution and treatment system. **Assignment 2** The water demand of each sector of an urban community is shown below: - Domestic demand \-\-\-\-- 150 *lpcd* - Irrigation (Annually) \-\-\-- 30,000 m^3^ - Recreation per month \-\-\-\-- 500 m^3^ - Requirement by each poultry per day = 0.02 m^3^ - Population = 100,000 persons Determine the total monthly consumption for the community assuming a 30 day month. **Solution** Assuming 1 month = 30 days i. Domestic consumption = [\$\\frac{150}{1000}x100,000x30days\$]{.math.inline} = **450,000 m^3^/month** ii. Irrigation = 30,000/ 12 = **250 m^3^/month** iii. Recreation = **500 m^3^/month** iv. Poultry = 0.02 x 10,000 x30 = **6000 m^3^/month** Total consumption **= 456. 750 m^3^/month** **Maximum day and Peak hour demand** The peak hour rate is the measure against which the adequacy of the distribution system, high service pumps and elevated storage tanks are established. The required fire flow must be available in addition to the coincident maximum daily flow rate. For the distribution system, the design flow rate is the larger of the peak hourly flow or maximum daily flow plus fire flow. The fire hydrant pressure during a fire should generally exceed 137.9 kN/m^2^ (20 psi) where motor pumps are used. Because a city will be penalized in its fire insurance rates if the needed flows cannot be met for specified durations, most cities provide emergency storage to meet fire demands. **Example** Estimate the water requirements for a community that will reach a population of 12,000 at the design year. The estimated municipal water demand for the community is 610 *lpcd.* Calculate the fire flow, design capacity of the water treatment plant and design capacity of the water distribution system. Use NBFU formula for the fire flow. **Solution** Average daily demand (Q) = 610 l/person/day x120, 000 persons = 610x120 m^3^/day Q = 73,200 m^3^/day **Q** (l/min) = 73,200/ 24x60 = **50,833.33 l/min** NBFU formula: [\$Q\\ \\ = \\ 3860\\sqrt{P}\\left( 1 - 0.01\\sqrt{P} \\right)\$]{.math.inline} Population = 120,000, therefore P= 120 \ [\$\$Q\\ \\ = \\ 3860\\sqrt{120}\\left( 1 - 0.01\\sqrt{120} \\right)\$\$]{.math.display}\ [Q = 3860*x*10.95(1−0.01*x*10.95)]{.math.inline}= **37,638.76 l/min** Maximum day demand (Mdd) = 1.8 x Average daily demand = 1.8x 50,833.33 l/min = **91,500 l/min** Peak hour demand = 1.55 x Mdd= 1.5 x 91,500 = **137,250 l/min = 137.250 m^3^/min** **Mdd + Fire flow =** 91,500 +37,652 = 129,152 l/min a. Design capacity of water treatment plant is Mdd, = **91, 500 l/min** **=** 91.5m^3^/min = 91.5x60x24 = **131,760m^3^/day** b. Design capacity of water distribution system is the larger of : i. Peak hour flow of 137250 l /min ii. FF + coincident maximum daily demand = 37,652 + 91,500 = 129,152 l/min = 129,152 m^3^/min. Fire Hydrants are placed throughout the service area to provide either direct hose connection for firefighting or connection to pumper trucks also known as fire engines. A single hose stream is generally taken as 1000l/min, called the standard fire stream and fire hydrants are typically located at street intersections or spaced 60 to 150 metres apart. However, in high value districts, additional hydrants may be needed in the middle of blocks to provide the required fire flows. The flow rates and pressures in the distribution system are analyzed under maximum daily plus fire demand and maximum hourly demand and the larger flow rate governs the design. Pumps are sized for a variety of conditions from maximum daily to maximum hourly demand depending on their function in the distribution system. **Table XX2: Required fire flow durations (AWWA, 1992)** **Required fire flow (l/min)** **Duration (hrs)** -------------------------------- -------------------- \ 1. **Graphical Method** **Year** 1880 1890 1900 1910 1920 1930 1940 ---------------- -------- -------- -------- -------- -------- -------- -------- **Population** 25,000 27,500 33,000 39,000 45,000 54,500 61,000 2. **Arithmetic Method** In this method, the rate of population change is assumed to be constant, that is; [\$\\frac{\\text{dP}}{\\text{dt}} = \\ K\_{a}\$]{.math.inline} [*dP*= *K*~*a*~ dt]{.math.inline}, where [*K*~*a*~]{.math.inline} is constant growth rate, P = population, t denotes time and [\$\\frac{\\text{dP}}{\\text{dt}}\$]{.math.inline} is the change of population with time. Integrating within limits of P (P~i~, P~f~) and t (t~i~, t~f~) we have: [∫~Pi~^Pf^dP]{.math.inline} = K~a~ [∫~ti~^tf^dt]{.math.inline} P~f~ -- P~i~ = K~a~ (t~f~ - t~i~) **P~f~** = **P~i~** + **K~a~** (t~f~ - t~i~). The method is commonly used for short to medium term estimates 1 to 10 years. **Worked Example** The census records of a city are as given below. Estimate the population of the city in 1970 assuming trend in growth. **Year** **Population (in, 1000)** ---------- --------------------------- 1930 62 1940 74 1950 85 1960 100 **Solution** [\$K\_{a} = \\mathbf{\\ }\\frac{74,000 - 62,000}{10}\\mathbf{\\ = \\ }\\frac{12,000}{10} = 1200\$]{.math.inline} For the 1940 and 1950 census records: [\$K\_{a} = \\mathbf{\\ }\\frac{85,000 - 74,000}{10}\\mathbf{\\ = \\ }\\frac{11,000}{10} = 1100\$]{.math.inline} For the 1950 and 1960 census records: [\$K\_{a} = \\mathbf{\\ }\\frac{100,000 - 85,000}{10}\\mathbf{\\ = \\ }\\frac{15,000}{10} = 1500\$]{.math.inline} Average [\$K\_{a} = \\mathbf{\\ }\\frac{1200 + 1100 + 1500}{3}\\ = \\frac{3800}{3} = 1266.66\$]{.math.inline} **P~f~** = **P~i~** + **K~a~** (t~f~ - t~i~). **P~1970~** = P~1960~+ 1266.66 (1970 - 1960). **P~1970~** = 100,000 + 1266.66x10 = 112,666.66 = **112,670.** **Assignment 2** The population of 5 decades from 1930 to 1970 for a city A is given below in table 2. Find the population in 1980, 1990 and 2000. Table x: Population of city A. **Year** 1930 1940 1950 1960 1970 ---------------- -------- -------- -------- -------- -------- **Population** 25,000 28,000 34,000 42,000 47,000 **Solution** [\$K\_{a1} = \\mathbf{\\ }\\frac{28,000 - 25,000}{10}\\mathbf{\\ = \\ }\\frac{3,000}{10} = 300\$]{.math.inline} [\$K\_{a2} = \\mathbf{\\ }\\frac{34,000 - 28,000}{10}\\mathbf{\\ = \\ }\\frac{6,000}{10} = 600\$]{.math.inline} For the 1960 and 1950 census records: [\$K\_{a3} = \\mathbf{\\ }\\frac{42,000 - 34,000}{10}\\mathbf{\\ = \\ }\\frac{8,000}{10} = 800\$]{.math.inline} For the 1970 and 1960 census records: [\$K\_{a4} = \\mathbf{\\ }\\frac{47,000 - 42,000}{10}\\mathbf{\\ = \\ }\\frac{5,000}{10} = 500\$]{.math.inline} Average [*K*~*a*~]{.math.inline} = [\$\\overline{K\_{a}}\$]{.math.inline}= [\$\\frac{300 + 600 + 800 + 500}{5}\$]{.math.inline} = 2200/4 = 550 **P~1980~** = **P~1970~** + [\$\\overline{K\_{a}}\$]{.math.inline} (t~f~ - t~i~). **P~1980~** = 47,000 + 550 x10 = 52,500 **P~1990~** = P~1970~ + [\$\\overline{K\_{a}}\$]{.math.inline} (t~f~ - t~i~). **P~1990~** = 47,000 +550 x20 = 47, 000 + 11,000 = 58,000 **P~2000~** = P~1970~ + [\$\\overline{K\_{a}}\$]{.math.inline} (t~f~ - t~i~). **P~2000~** = 47,000 + 550 (30) = 47,000 + 1650 = 63,500 3. **Geometric Method** In this method of projecting population, it is assumed that population will increase in proportion to the number present. That is the rate of population change is equivalent to the population at a given instant. That is: [\$\\frac{\\mathbf{\\text{dP}}}{\\mathbf{\\text{dt}}}\\mathbf{= \\ }\\mathbf{K}\_{\\mathbf{g}}\\mathbf{P}\$]{.math.inline} **,** in which [*K*~*g*~]{.math.inline} = geometric constant, integrating between the initial population Pi at the year t~i~ and population P~f~ at the forecast year t~f\.~ We have: [\$\\int\_{\\text{Pi}}\^{\\text{Pf}}\\frac{\\text{dP}}{P}\$]{.math.inline} = [*k*~*g*~∫~ti~^tf^dt]{.math.inline} [\[*lnP*~f~ \] ~Pf~^Pf^ ]{.math.inline}= K~g~ (t~f~ --t~i~) [ln Pf]{.math.inline} - [ln Pi]{.math.inline} = K~g~ (t~f~ --t~i~) K**~g~** = [\$\\frac{\\ln\\text{Pf}\\ - \\ \\ln\\text{Pi}\\text{\\ \\ }}{(tf\\ --ti)}\$]{.math.inline} [ln Pf]{.math.inline} = [ln Pi]{.math.inline} + K~g~ (t~f~ --t~i~) \-\-\-- (1) From equation (1) if population is plotted on logarithmic scale and time plotted in a linear scale a straight line will be obtained the slope of the graph gives K~g~. Before the choice of whether to use either the arithmetic or geometric method to forecast population, the past population values are plotted on ordinary graph paper. If the relationship between population and time is approximately linear, then arithmetic method should be used in forecasting the population. If the graph is concave upwards the geometric method may be employed. C:\\Users\\hp\\Desktop\\CVE 523- LECTURE NOTES\\Drawings\\Arith\_growth.jpg ![C:\\Users\\hp\\Desktop\\CVE 523- LECTURE NOTES\\Drawings\\Geom\_growth.jpg](media/image2.jpeg) **Worked Example** The census record of a city is given below. Estimate the population of the city in 1970 assuming geometric trend in growth. **Year** 1930 1940 1950 1960 -------------------------- ------ ------ ------ ------ **Population (**in 1000) 62 74 85 100 **Solution** For 1930 and 1940 census records K**~g1~** = [\$\\frac{\\ln\\left( \\frac{74}{62\\ } \\right)}{10}\\text{\\ \\ }\$]{.math.inline} = 0.01769 For 1940 and 1950 census records K**~g2~** = [\$\\frac{\\ln\\left( \\frac{85}{74\\ } \\right)}{10}\\text{\\ \\ }\$]{.math.inline} = 0.0138 For 1950 and 1960 census records K**~g3~** = [\$\\frac{\\ln\\left( \\frac{100}{85\\ } \\right)}{10}\\text{\\ \\ }\$]{.math.inline} = 0.0163 Average K~g~ = [\$\\overline{K\_{g}}\$]{.math.inline} = [\$\\frac{K\_{g1} + K\_{g2} + K\_{g3}}{3}\$]{.math.inline} = [\$\\frac{0.01769 + 0.0138 + \\ 0.0163\\ }{3}\$]{.math.inline} = 0.0159 [ln *P*1970]{.math.inline} = [ln *P*1960]{.math.inline} + [\$\\overline{K\_{g}}\$]{.math.inline} (10) [ln *P*1970]{.math.inline} = [ln 100, 000]{.math.inline} + 0.0159 (10) \ [ln *P*1970 = 5.069]{.math.display}\ **P~1970~** = [*e*^5.069^]{.math.inline} = 117,200 **Assignment** I. For the data given in the table below +---------+---------+---------+---------+---------+---------+---------+ | **Year* | 1930 | 1940 | 1950 | 1960 | 1970 | 1980 | | * | | | | | | | +=========+=========+=========+=========+=========+=========+=========+ | **Popul | 60 | 70 | 83 | 98 | 115 | 135 | | ation** | | | | | | | | | | | | | | | | (in | | | | | | | | 1000) | | | | | | | +---------+---------+---------+---------+---------+---------+---------+ a. Choose an appropriate method of population forecast and find the rate of population growth. b. Determine the population in the year 2050. II. The population statistics pertaining to Yenagoa capital territory is as given in the table below. Estimate the population expected in 1980 by Arithmetic and Geometric growth methods. **Year** 1930 1940 1950 1960 1970 ---------------- -------- --------- --------- --------- --------- **Population** 70,000 100,000 150,000 200,000 240,000 4. **Comparative Method** The future population can be predicted by plotting the population of several cities having similar pattern of growth. The population of the city under study is expected to grow in a similar manner to other older and larger cities. The forecast is made by extrapolating the population curve of the city under study into the future according to the trend of the population curve of other cities. C:\\Users\\hp\\Desktop\\CVE 523- LECTURE NOTES\\Drawings\\City\_growth.jpg 5. **Ratio and Component method** The growth of population of a smaller area is closely related to the growth of the population of the region in which the smaller area is situated. Thus the future population of the smaller area can be estimated by using the forecast of the future population of the region. Population department or commission usually forecast the future population of the region. Using the values, the future population of the area under study can be estimated. Thus [\$\\frac{P\_{\\text{f\\ }}}{P\_{f}\^{/}} = \\ \\frac{P\_{i}}{P\_{i}\^{/}}\$]{.math.inline} = k (constant) [*P*~f~ ]{.math.inline} = the population forecast for the area under study. [*P*~*i*~]{.math.inline} = the population of the area at the last census. [*P*~*f*~^/^ ]{.math.inline} = future population for the region. [*P*~*i*~^/^]{.math.inline} = population of the region at the last census. 6. **Employment Forecast or other utility connection forecast** In this method, future population is estimated using employment forecast. From the past data of population and employment, the ratio is plotted and population obtained from the projected employment forecast. This procedure is similar to that of ratio method. Similar procedure can be utilized from forecast of various utility service connections such as telephones, gas, water etc. **Worked Example:** Estimate the population of a city using employment forecast. The design year is 2010. Use the following data. Employment forecast for the year 2010 is 21,300. **Year** Population in 1000s Employment in 1000s Ratio of Population to employment ---------- --------------------- --------------------- ----------------------------------- 1960 20 6.8 2.94 1970 30 10.79 2.78 1980 39 14.77 2.64 1990 46 17.83 2.58 The procedure is illustrated in the figure below. ![C:\\Users\\hp\\Desktop\\CVE 523- LECTURE NOTES\\Drawings\\pop\_to\_employ.jpg](media/image4.jpeg) Estimated ratio of the design year = 2.54, Therefore the population in the design year is equal to Population = 21,300 x 2.54 **=** **54,109** **Example 1** A water supply system is to be designed for Gumel Municipality in Jigawa State, Nigeria for a planning period of up to 2040. The population census of Gumel Municipality from 1940 to 2010 is as recorded in Table 1. **Table 1: Population census of Gumel Municipality in Jigawa State, Nigeria** **Year** **Population** ---------- ---------------- 1940 100,000 1950 130,000 1960 160,000 1970 185,000 1980 220,000 1990 240,000 2000 280,000 2010 320,000 \(a) Estimate the population of the municipality in the year 2040 using (i) Graphical Method (ii) Arithmetic method and (iii) Geometric method \(b) Determine the design capacity of the water treatment plant that is adequate to meet the water requirements of the municipality in 2040 if the average per capita water consumption of the municipality in that year is 200 *lpcd.* **SOLUTION** a (i)**: Graphical method** [\[CHART\]]{.chart} **1 (aii)** Population **projection** By **Arithmetic method** [\$K\_{a1} = \\mathbf{\\ }\\frac{130,000 - 100,000}{10}\\mathbf{\\ = \\ }\\frac{30,000}{10} = 3000\$]{.math.inline} For the 1950 and 1960 census records: [\$K\_{a2} = \\mathbf{\\ }\\frac{160,000 - 130,000}{10}\\mathbf{\\ = \\ }\\frac{30,000}{10} = 3000\$]{.math.inline} For the 1960 and 1970 census records: [\$K\_{a3} = \\mathbf{\\ }\\frac{185,000 - 160,000}{10}\\mathbf{\\ = \\ }\\frac{25,000}{10} = 2500\$]{.math.inline} For the 1970 and 1980 census records: [\$K\_{a4} = \\mathbf{\\ }\\frac{220,000 - 185,000}{10}\\mathbf{\\ = \\ }\\frac{35,000}{10} = 3500\$]{.math.inline} For the 1980 and 1990 census records: [\$K\_{a5} = \\mathbf{\\ }\\frac{240,000 - 220,000}{10}\\mathbf{\\ = \\ }\\frac{20,000}{10} = 2000\$]{.math.inline} For the 1990 and 2000 census records: [\$K\_{a6} = \\mathbf{\\ }\\frac{280,000 - 240,000}{10}\\mathbf{\\ = \\ }\\frac{40,000}{10} = 4000\$]{.math.inline} For the 2000 and 2010 census records: [\$K\_{a7} = \\mathbf{\\ }\\frac{320,000 - 280,000}{10}\\mathbf{\\ = \\ }\\frac{40,000}{10} = 4000\$]{.math.inline} [\$K\_{\\text{average}} = \\mathbf{\\ }\\frac{3000 + 3000 + 2500 + 3500 + 2000 + 4000 + 4000}{7}\\ = \\frac{22,000}{7} = 3,142.9\$]{.math.inline} **P~f~** = **P~i~** + **K~average~** (t~f~ - t~i~). **P~2040~** = P~2010~ + 3142.9 (2040 - 2010). **P~2040~** = 320,000 + 3142.9 x30 = 320,000 + 94,287= 414, 287= **414, 300** **Design Period** A water supply scheme may include huge and costly structures (for example dams, reservoirs, treatment works etc) which cannot be replaced or expanded in their capacities easily or conveniently. Thus to avoid future complications of expansions , the various components of a water supply scheme are purposely made larger to satisfy the community needs for a reasonable number of years to come. *This future period or number of years for which a provision is made in designing the capacities of the various components of the water supply scheme is known as design period.* The design period should neither be too long or too short. The design period cannot exceed the useful life of the component structure. **Factors governing the selection of the design period.** The factors which guide the selection of design period are: i. The useful life of the component structures and equipment used. The design period should not exceed these respective values. ii. Initial capital, operation and maintenance cost. iii. The availability of capital. iv. Ease and difficulties to be faced in expansion if undertaken at a later date. v. The availability of the expected consumption at the end of the design period. vi. The change in the purchasing power of money during the design period. **Design period values** For water supply projects, the usually adopted design period is 20 to 30 years. This does not include the completion period for the project. The 20 to 30 year period may be modified in regard to certain components of the project depending upon (i) useful life of the component facility (ii) ease of carrying out extension when required and (iii) rate of interest -- so that expenditure far ahead of utility is avoided. The design period recommended for designing the various components of a water supply project are: +-----------------------------------+-----------------------------------+ | **Item** | **Design period in years** | +===================================+===================================+ | **Storage by dams** | 50 | +-----------------------------------+-----------------------------------+ | **Pumping** | 30 | | | | | i. Pump house | 20 | | | | | ii. Electric motor and pumps | 20- 50 | | | | | iii. Water treatment plant | 20-50 | | | | | iv. Distribution system | | +-----------------------------------+-----------------------------------+ **Table -xx-: Design periods and Capacity criteria for constituent structures (Gupta, 2008)** +-----------------------+-----------------------+-----------------------+ | **Structure** | **Design | **Required capacity** | | | period(Years)** | | +=======================+=======================+=======================+ | **1) Source of | Indefinite | Maximum daily | | supply** | | (requirements) | | | 10-25 | | | a\) River | | Maximum daily | | | 25- 50 | | | b\) Well field | | Average annual demand | | | | | | c\) Reservoir | | | +-----------------------+-----------------------+-----------------------+ | **2) Conveyance** | 25 -- 50 | Maximum daily | | | | | | a\) Intake conduit | 25 -50 | Maximum daily | | | | | | b\) Conduit to | | | | treatment plant | | | +-----------------------+-----------------------+-----------------------+ | **3) Pumps** | 10 | Maximum daily plus | | | | one reserve unit | | a\) Low lift | 10 | | | | | Maximum hourly plus | | b\) High lift | | one reserve unit | +-----------------------+-----------------------+-----------------------+ | 4\) **Treatment | 10 -- 15 | Maximum daily | | Plant** | | | +-----------------------+-----------------------+-----------------------+ | 5\) **Service | 20-25 | Working storage ( | | Reservoir** | | storage capacity | | | | component + fire | | | | demand for specified | | | | duration + Emergency | | | | storage) | +-----------------------+-----------------------+-----------------------+ | 6\) **Distribution** | 25 -- 50 | Greater of (1) | | | | Maximum daily plus | | a\) Supply pipe or | Full development | fire demand for a | | conduit | | specified duration or | | | | | | b\) Distribution | | 2\) Maximum hourly | | grid | | requirement. | +-----------------------+-----------------------+-----------------------+ **STORAGE RESERVOIRS** A water supply, irrigation or hydroelectric project drawing water from a stream may be unable to satisfy the demands of its consumers during extremely low flows. When the flow in the stream during dry weather is sufficient to the demand of a community, there is no need for a storage reservoir for water supply but when it is less than the demand, storage reservoirs need to be constructed. A storage or conservation reservoir retains such excess water from periods of rainy season for use during periods of drought. In addition to conserving water for later use, the storage of flood water may also reduce flood damage below the reservoir. Because of the fluctuations in both supply and demand of water most cities find it necessary to have reservoirs to help equilibrate /stabilize the flows. **Design characteristics of storage reservoirs** Since water storage is the main function of a reservoir, the determination of the storage capacity is very important and can be done either graphically or analytically. **Catchment yield and Reservoir yield** Long range runoff from a catchment is known as the yield of the catchment and generally, a period of one year is considered (i.e. Annual). The total yearly runoff, expressed as volume of water entering or passing the outlet point of the catchment is known as catchment yield. It is expressed in Mm^3^. The annual yield of a catchment up to the site of a reservoir located at the given point along a river indicates the quantum or amount of water the will annually enter the reservoir. This helps in designing the capacity of the reservoir and in fixing the outflows from the reservoirs. \ [*Inflows* = *f* ( *Inflows*, *reservoir* *losses* )]{.math.display}\ **Reservoir yield** The amount of water which can be drawn from a reservoir in any specified time interval is called *reservoir yield.* The reservoir yield is dependent on the inflow into the reservoir and reservoir losses which comprise of leakages from the reservoir and evaporation from the reservoir surface. \ [*Reservoir* *yield* = *f*( *inflow*, *reservoir* *losses*)]{.math.display}\ \ [*Reservoir* *losses* = *reservoir* *leakage* + *evaporation* *losses*]{.math.display}\ The construction of a reservoir increases the exposure areas of the water surface than that of the natural stream and thus increases the evaporation losses. In addition, seepage from the reservoir contributes to the total reservoir losses. The annual inflow to the reservoir called catchment yield is represented by the mass curve of inflow while the outflow from the reservoir called the reservoir yield is represented by the mass demand line. The two curves determine the reservoir capacity once the reservoir losses are separately accounted for. The inflows to the reservoir may vary or varies over the perspective life of the reservoir. The past available data of rainfall or runoff in the catchment is utilized in computing optimum value of catchment yield. For example if the past available record is for 35 years, the minimum yield from the catchment in the leanest rainfall year may be say 100 Mm^3^ and if the maximum yield in the in most rainfall year is as high as 200 Mm^3^. The question is should the reservoir capacity be provided to correspond to 100 Mm^3^ or 200 Mm^3^? If the reservoir capacity is provided to correspond to 100 Mm^3^ yield then eventually the reservoir will be fully filled every year with a dependability of 100% but if the capacity is provided to correspond to 200 Mm^3^ yield then the reservoir will be filled up fully only in the best rainfall year (that is once in 35 years with a dependability of about [\$\\frac{1}{35}x100 = 2.85\\%\\.\\ \$]{.math.inline} To achieve a balanced agreement for reservoir capacity a dependability percentage values of 50% to 75% may be used to compute the *dependable yield or design yield.* **Firm yield** The yields which correspond to the worst or most critical or leanest year on record is called *firm yield or safe yield.* Water available in excess of firm yield during years of higher inflows is called secondary yield. Hydropower may be developed from such secondary water. **Estimating yield of a Basin.** The runoff in a basin is the rainfall minus losses. The losses include: i. Evaporation losses including transpiration losses. ii. Percolation losses. **Khosla formula.** The formula may be applied to a catchment where rainfall is fairly uniformly distributed throughout the year. The formula is given as: \ [*Q* = *P* − 0.4813 *T*~*m*~]{.math.display}\ Where [*T*~*m*~ \ 4.5 ]{.math.inline}^o^C where Q = yearly or monthly runoff, P = yearly or monthly rainfall in cm. [*T*~*m*~]{.math.inline} = mean yearly or monthly temperature in degrees centigrade. **Computation of design or dependable yield value** The dependable yield corresponding to a given dependability percentage may be obtained from past available data of 20 years and above. The yearly rainfall data is of the reservoir catchment is generally used for this purpose. That is, the rainfall data of past years is used to compute the dependable rainfall value corresponding to the given dependability percentage (P). The rainfall is converted into runoff value by using the available empirical formula which connects yearly rainfall with yearly runoff. The procedure for computing the dependable rainfall value for a given dependability percentage (P) is as given below: 1. The available rainfall data for the past N years is arranged in descending order of magnitude. 2. The order number m is computed using : [\$m = \\ Nx\\frac{\\text{\\ P}}{100}\$]{.math.inline} 3. The rainfall value which corresponds to the order number in the tabulated data gives the dependable value. 4. If the computed value of m is a fraction, the arithmetic mean of rainfall values corresponding to whole numbers m values above and below this fraction value is taken as the dependable rainfall value. **Table 1: Annual rainfall for a catchment area (1956- 1990).** **Year** **Rainfall(cm)** **Year** **Rainfall(cm)** ---------- ------------------ ---------- ------------------ 1956 98 1974 88 1957 100 1975 94 1958 101 1976 107 1959 99 1977 110 1960 85 1978 208 1961 112 1979 114 1962 116 1980 104 1963 78 1981 120 1964 160 1982 108 1965 66 1983 102 1966 184 1984 80 1967 90 1985 109 1968 76 1986 122 1969 118 1987 115 1970 86 1988 140 1971 92 1989 138 1972 96 1990 60 1973 93 **SOLUTION:** The given data in Table 1 is arranged in descending order (order number, m) as shown in Table 2. **Rank** **Rainfall in descending order (cm)** **Rank** **Rainfall in descending order (cm)** **Rank** **Rainfall in descending order (cm)** ---------- --------------------------------------- ---------- --------------------------------------- ---------- --------------------------------------- 1 208 13 110 25 93 2 184 14 109 26 92 3 160 15 108 27 90 4 140 16 107 28 88 5 138 17 104 29 86 6 122 18 102 30 85 7 120 19 101 31 80 8 118 20 100 32 78 9 116 21 99 33 76 10 115 22 98 34 66 11 114 23 96 35 60 12 112 24 94 [\$m = \\ Nx\\frac{\\text{\\ P}}{100}\\text{\\ \\ }\$]{.math.inline} = [\$\\ 35x\\frac{\\ 60}{100}\$]{.math.inline} = 21 The rainfall value tabulated in table 2 at order 21 is 99cm. Thus the required dependable rainfall is 99 cm **Example 2.** The design annual rainfall for the catchment of a proposed reservoir has been computed to be 99 cm. The catchment area has been estimated to have mean annual temperature of 20 ^0^C. The catchment area contributing to the proposed reservoir is 1000 km^2.^ Calculate the annual design yield for the reservoir. Use the Khosla formula. **Solution.** Khosla formula connects design rainfall (P) with design yield (Q) as follow: [*Q* = 99 − 0.4813 *x*20]{.math.inline} = 99 -- 9.626 = 89.37cm. = 0.894 m. The total yield produced in m^3^ from the given catchment size 1000km^2^ is equal to 0.894 x 1000 km^2^ = 0.894 m x 1000x 10 ^6^ m^2^ = **8.94 x 10^9^ m^3^** Annual design yield = 894x 10 ^6^ m^3^ = 894Mm^3^ **Note:** unit for storage capacity is ha-m. 1ha-m = 10 ^4^ m^3^10 ^6^ m^3^ **Determination of storage capacity of a Reservoir** To determine the reservoir capacity, the average monthly inflow in the stream over a reasonable period and the probable demand for water during that period should be known. At the beginning of the dry period it is assumed that the reservoir is full. After that gradually the reservoir starts to empty. At the end of the dry period, there will be maximum deficiency. The storage capacity should be equal to the maximum deficiency which is expressed as follows: \ [*S* = *maximum* *deficiency* = *maximum* (*D* − *Q*)]{.math.display}\ Where[ *D* = *demand*]{.math.inline}, [*Q* = *flow* *in* *the* *stream*]{.math.inline}. The value of S can be obtained graphically or analytically. Storage capacity can be determined with the help of mass curve or diagram. Mass curve is a plot of cumulative flow against time. Inflow mass curve is the plot of cumulative inflow against time while outflow mass curve is the plot of cumulative outflow or demand against time. If a constant rate of withdrawal is required from the reservoir, the mass curve of demand will be a straight line with a slope equal to demand rate. Month **J** **F** **M** **A** **M** **J** **J** **A** **S** **O** **N** **D** -------------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- Mm3 of water 1.42 1.7 1.98 1.13 0.91 0.57 1.42 2.26 0.28 1.42 1.70 2.26 +-------------+-------------+-------------+-------------+-------------+ | Month | Stream flow | Cumulative | Water | Cumulative | | | (Mm^3^) | | demand | water | | | | stream flow | | demand | | | | (Mm^3^) | (Mm^3^) | (Mm^3^) | +=============+=============+=============+=============+=============+ | January | 1.42 | 1.42 | 1.1 | 1.1 | +-------------+-------------+-------------+-------------+-------------+ | February | 1.7 | 3.12 | 1.1 | 2.2 | +-------------+-------------+-------------+-------------+-------------+ | March | 1.98 | 5.1 | 1.1 | 3.3 | +-------------+-------------+-------------+-------------+-------------+ | April | 1.13 | 6.23 | 1.1 | 4.4 | +-------------+-------------+-------------+-------------+-------------+ | May | 0.91 | 7.14 | 1.1 | 5.5 | +-------------+-------------+-------------+-------------+-------------+ | June | 0.57 | 7.71 | 1.1 | 6.6 | +-------------+-------------+-------------+-------------+-------------+ | July | 1.42 | 9.13 | 1.1 | 7.7 | +-------------+-------------+-------------+-------------+-------------+ | August | 2.26 | 11.39 | 1.1 | 8.8 | +-------------+-------------+-------------+-------------+-------------+ | September | 0.28 | 11.67 | 1.1 | 9.9 | +-------------+-------------+-------------+-------------+-------------+ | October | 1.42 | 13.09 | 1.1 | 11 | +-------------+-------------+-------------+-------------+-------------+ | November | 1.70 | 14.79 | 1.1 | 12.1 | +-------------+-------------+-------------+-------------+-------------+ | December | 2.26 | 17.05 | 1.1 | 13.2 | +-------------+-------------+-------------+-------------+-------------+ Month **J** **F** **M** **A** **M** **J** **J** **A** **S** **O** **N** **D** -------------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- Mm3 of water 1.7 1.98 2.4 1.4 1.2 0.7 1.6 2.4 0.6 1.6 1.98 2.6 Losses in reservoir occur through evaporation and seepage. Because of these losses, some of the stored water is lost and hence not available for useful purposes. In the planning of reservoir for conservation, it is important that these losses are given consideration and various measures need to be adopted to minimize these losses as much as possible. **Evaporation losses** Evaporation depends upon a number of factors for example temperature, wind velocity and relative humidity. The greater the surface area of the reservoir, the greater will be the evaporation loss. Evaporation loss is expressed as depth of water lost in centimeters (or metres). Evaporation loss from a reservoir is generally estimated by measuring the evaporation in an evaporation pan and multiplying the same by a suitable pan coefficient. Thus Volume of water lost = Mean surface area x pan evaporation x pan evaporation coefficient. If evaporation pan data are not available, observed evaporation data of existing reservoir may be used for design of new reservoirs in the region. **Seepage losses:** seepage losses occur due to the continuous flow of water under pressure from the reservoir to the adjoining strata. These losses may be small but may be quite significant in porous strata and fissures in rocks. To prevent seepage losses, the proposed reservoir basin should be thoroughly investigated by the geologist and checked for water tightness. If necessary, suitable measures like grouting of the basin should be adopted to reduce seepage --from treatability studies. **Evaporation control in Reservoirs** The following measures are usually adopted to minimize evaporation losses from reservoirs. 1. Construction of deep reservoirs. As far as possible deep reservoirs should be constructed to reduce the surface area for a given storage volume. 2. Growing tall trees: Evaporation depends on wind velocity. If tall trees are grown on the windward side of the reservoir, they act as wind breakers and consequently reduce evaporation. 3. If huge trees are developed in the area adjoining the reservoir, the environment cools and evaporation is reduced. 4. Covering the reservoir. The evaporation from a reservoir can be reduced by covering it with a polythene sheet. This method is only applicable to small reservoirs. 5. Spraying chemicals. When certain chemicals or fatty acids are spread over the water surface a thin film is formed which reduces evaporation. Generally cetyl alcohol (hexadecanol) is used for reducing evaporation losses. This is invisible, nontoxic and permits free passage of rain, oxygen and sunlight. 6. By developing underground reservoirs, since the evaporation from a ground water table is very much less than the evaporation from a water surface. **RESERVOIR SEDIMENTATION AND CONTROL** The sediments are produced in the catchment of the river by erosion. Rivers carry a large amount of sediment load along with water. These sediments are deposited in the reservoir on the upstream of the dam because of the reduction of velocity. Deposition of sediments in the reservoir is known as reservoir sedimentation or siltation. Sedimentation reduces the available capacity of the reservoir. With continuous sedimentation, the useful life of the reservoir keeps on decreasing. As the deposition reduces the water storing capacity of the reservoir, if the process of deposition continues longer, a stage reaches when the entire reservoir becomes silted up and becomes useless. With passage of time, its capacity will go on reducing. In order to ensure that that the capacity does not fall short of requirements ever during the design period, sedimentation must be taken into account. The total volume of silts likely to be deposited during the design period of the dam is estimated ad that volume is left unused to allow for silting. This is known as dead storage while the remainder is known as effective storage or live storage. Dead storage is usually kept at about 20 -- 25% of the total storage. **Useful life of Reservoir** All reservoirs ultimately get filled with sediments. The river carries sediments to the reservoir which gets deposited in the reservoir. The deposition of sediments gradually decreases the available storage capacity of the reservoir. As more and more sediments are deposited in the reservoir, a stage reaches when the reservoir is not able to serve its intended purpose and its useful is over. If the annual sediment inflow is large compared with the reservoir capacity, the useful life of the reservoir would be very short. When planning a reservoir, it is essential to consider the rate of sedimentation to know whether the useful life of the proposed reservoir will be sufficiently long to justify the expenditure on the construction. The rate of sedimentation in the reservoir depends on trap efficiency [*η*~*t*~]{.math.inline}. Trap efficiency[ (*η*~*t*~]{.math.inline}) is defined as the percent of the total inflow sediment which is retained in the reservoir for example. \ [\$\$\\text{\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ }\\mathbf{\\text{Trap\\ efficiency\\ }}\\left( \\mathbf{\\eta}\_{\\mathbf{t}} \\right)\\mathbf{=}\\left( \\frac{\\mathbf{\\text{sediment\\ retained}}}{\\mathbf{\\text{Total\\ sediments\\ }}}\\mathbf{\\ } \\right)\\mathbf{x}\\mathbf{100}\$\$]{.math.display}\ From observation of the rate sedimentation of existing reservoirs, it has been found that trap efficiency of a reservoir depends on the Capacity -- Inflow rate. Thus [(*η*~*t*~) = *f* (*Capacity*/]{.math.inline}Inflow rate) ratio. Curves which relate trap efficiency and capacity -- inflow ratio based on sediment data obtained from existing reservoirs are given in Brune (1948). Generally, the greater capacity -- inflow ratio, the greater the trap efficiency hence sedimentation rate is high in relatively larger reservoirs. Trap efficiency decreases with the age of the reservoir because the available capacity gradually decreases due to sedimentation. The useful life of a reservoir is over when the capacity occupied by sediments is large enough to prevent the reservoir from serving its intended purpose(s). The useful life of a reservoir is considered to be over if the capacity reduces to about 20 % of the designed capacity. Most reservoirs are designed or planned to have a useful life of say 100 years. As guidance, approximate sedimentation may be assumed at a rate of 0.1 -- 0.2 hectare --metres per year per km^2^ (0.1 -- 0.2 x 10 ^4^ m^3^ per year per km^2^) of the drainage area of the river. **Measures to control Reservoir sedimentation** In order to increase the useful life of the reservoir, the rate of deposition of sediments in the reservoir must be decreased. Various measures are taken in achieving this aim. The different methods may be classified for convenience into two parts namely: 1. Pre -- construction measures and 2. Post -- construction measures. **Pre -- construction measures** They are measures which may be adopted before and during the execution of the project and this includes the following: i. Proper selection of suitable dam site- in this case the reservoir is suitable located such that sediment inflow is low. It should exclude runoff from easily erodible part of the catchment and the reservoir should not be built on the downstream of the confluence of a tributary which carries large quantity of sediment. ii. Proper design. The rate of sediment is low in reservoirs with small storage capacity. Hence reservoirs are to be designed such that the capacity increases in stages and raised subsequently as its capacity is silted up. iii. Provision of sluices- The dam is provided with sluices (openings) for the removal of silted water from the downstream side. Adequate number of sluices at should be provided at different levels in the dam to discharge sediment laden water to the downstream. iv. Creating large reservoirs: As far as possible large reservoirs should be created. The trap efficiency of large reservoirs is high it does not increase linearly with an increase in capacity. Therefore the useful life of large reservoirs is larger than that of small reservoirs if all other factors are constant. v. Control of sediment inflow: The inflow of sediments to a reservoir can be controlled by the following methods -- a. Check dam- A Check dam is a small dam constructed on a stream to trap the sediments carried by the stream. Check dam are constructed on the tributaries carrying large quantity of sediments. b. Vegetation screens -- vegetal cover on the catchment reduce the impact of rain drops and hence minimize erosion. Vegetal screen is developed by promoting the growth of vegetation. These measures are undertaken during the operation during the operation of the project. They are given below. i. **Removal of post floodwater** ii. **Mechanical stirring of sediments** The deposited sediment is scoured and disturbed by mechanical measures so as keep it in a moving state and thus help is pushing it towards the sluices. iii. **Erosion Control and soil conservation** This is includes all the general methods adopted to reduce erosion of soil in the catchment and make it more stable. When soil erosion is reduced, sedimentation problem is also reduced. Soil conservation methods commonly used are terraces, strip cropping and contour farming. These methods reduce velocity of flow of water and hence erosion. Other methods used are afforestation, control of deforestation, regressing, control of grazing, check of gully formation. **RAW WATER INTAKES** **Submerged Intake** Submerged intakes are used to withdraw water from streams or lakes that have relatively little change in water surface elevation throughout the year. These intakes are constructed as cribs or screened bell mouth. They may consist of simple concrete box or rock filled timber crib to support the influent end of the withdrawal pipe. The withdrawal pipe discharges either into a sump at the shore where pumping equipment is installed or into a gravity quality conveyance system. **Exposed or Tower Intake** Exposed intake or tower structures are generally used for large projects on rivers or reservoirs with large water level fluctuations. They may be built either sufficiently near the shore to be connected by a bridge walkway or at such distance that they can be only by boat. Gate controlled openings (also called ports) are generally provided at different levels in order to permit selection of the best quality water. There are two major types of exposed or tower intakes namely: *wet-intake towers* and *dry intake towers*. **Wet Intake towers** Wet intake consists of a concrete circular shell filled with water up to the reservoir level with an inner well that is connected to the withdrawal conduit. Ports leading to the inner well have gates that regulate the flow of water into the wells. The purpose of the water filled outer shell is to provide weight needed to balance the buoyant forces when the inner shell is dry. The details of the wet well intake are as illustrated in the following diagram. C:\\Users\\hp\\Desktop\\CVE 523- LECTURE NOTES\\Drawings\\Wet\_intake\_tower.jpg **Dry Intake tower** In a dry intake tower, water enters the well through gate controlled ports. In the absence of a water filled outer shell, the dry intake tower must be heavier in construction than a wet-intake tower to resist the buoyant forces when empty. However, these intakes are beneficial because water can be withdrawn from any selected level by using the proper port. The figure below is an illustration of a dry intake well. ![C:\\Users\\hp\\Desktop\\CVE 523- LECTURE NOTES\\Drawings\\Dry\_intake\_tower.jpg](media/image9.jpeg) **Selection of site for Intakes** Numerous factors are considered by engineers in the selection of site for a raw water intake. These factors include: (i) water quality (ii) water depth (iii) stream or current velocity (iv) foundation stability (v) access (vi) power availability (vii) proximity (viii) hazards to navigation and environmental impact. They as briefly discussed. i. **Water quality** ii. **Water depth** The intake is located in such a way that water can be withdrawn from the full range of water levels. The range of water levels being the lowest expected drought level and extreme flood level in the water source. iii. **Stream or current velocities.** Both the direction and magnitude of the stream or current velocities can have an impact in the operation of intake. Water current flowing past intake structure and ports at velocities greater than 0.6m/s can cause eddy currents that will affect the hydraulic function of the intake. Water velocity also affects the lateral stability and foundation stability of the structure. iv. **Foundation stability.** Typically, intake structures are tall narrow structures that often have extreme lateral loads caused by water velocity hence the foundation must be designed to resist overturning moments. v. **Access** vi. **Power Availability.** vii. **Proximity to water treatment plant** The cost of facilities to convey water from the source to the treatment plant is related directly to the distance between the source and the treatment plant. Therefore during selection of an intake site, the length of the conveyance system must be given due consideration. viii. **Environmental Impact** The entire water supply scheme consists of collecting water from the source or sources, carrying it or transporting it to the treatment plant in the city and finally distributing the treated water among the consumers. For a surface water source, the water is collected by intake structure and transported to the city either raw if the treatment plant is located within the city or transported after treatment if located away from the city near the source. When the source of supply is near the city limits the problem of transportation, conduit is less but if far off, cost of transport mains or conduit can be enormous. For the case of underground sources such as wells and boreholes where a separate intake is required but lift is required the problem of conveying water to the city does not arise or these sources are generally within the cities. If the city is located at an elevation higher than the elevation of the source, the water will have to be pumped and carried by pressure pipes. However, if the source is available at a higher elevation than the city, the water can be conveyed without lift through open or closed conduit either by simple gravity or under pressure. **Types of Conduits** i. **Gravity conduits** Gravity conduits are those in which the water flows under the action of gravity. In such a conduit, the hydraulic gradient line will coincide with the water surface and parallel to the bed of the conduit. This is because the water is at atmospheric pressure along the conduit, thus there is no pressure term in the Bernoulli equation: [\$\\frac{P\_{A}}{\\mathrm{\\Upsilon}} + Z\_{A}\\ + \\ \\frac{V\_{A}\^{2}}{2g} = \\ \\frac{P\_{B}}{\\mathrm{\\Upsilon}} + Z\_{B}\\ + \\ \\frac{V\_{B}\^{2}}{2g}\$]{.math.inline} [*P*~*A*~ = *P*~B~ = 0]{.math.inline} [\$Z\_{A}\\ + \\ \\frac{V\_{A}\^{2}}{2g} = \\ Z\_{B}\\ + \\ \\frac{V\_{B}\^{2}}{2g}\$]{.math.inline} Gravity conduits could be in the form of canals, aqueducts etc ii. **Pressure conduits** [\$Z\_{A}\\ + \\frac{P\_{A}}{\\mathrm{\\Upsilon}} + \\ \\ \\frac{V\_{A}\^{2}}{2g} + h\_{p}\\ = Z\_{B} + \\ \\ \\frac{P\_{B}}{\\mathrm{\\Upsilon}}\\ + \\ \\frac{V\_{B}\^{2}}{2g} + H\_{L}\\text{\\ \\ }\$]{.math.inline} Where Z is distance above arbitrary datum (horizontal), [\$\\frac{P}{\\mathrm{\\Upsilon}}\$]{.math.inline} is the pressure head, V = average velocity of flow, [*h*~*p*~]{.math.inline} = energy head imparted to the water by the pump, [*H*~*L*~]{.math.inline} is the total head loss between A and B. Flow velocities should be kept between 0.9 m/s and 1.5 m/s. The design procedure is done in such a way that the available pressure head between the source and the city is just lost in overcoming the frictional resistance offered to the flow by the pipe interior. In addition to the head lost due to pipe friction [(*H*~*L*~)]{.math.inline} there are minor losses caused by abrupt changes in the geometry as a result of changes in pipe sizes, bends, valves and fittings of all types. However in long pipes as typified by pipes used in conveying water from sources to the city these minor losses [(*H*~*m*~)]{.math.inline} may be neglected. **HEAD LOSS BY PIPE FRICTION** The head loss caused by pipe friction can be determined by use of many formulae but the most prominent of them are: (i) Darcy -- Weisbach formula (ii) Hazen- Williams. However, the most widely used for water flowing under turbulent conditions as typical in pumped water distribution systems is the Hazen Williams formula. **Darcy Weisbach formula.** The head loss by pipe friction may be found by using the Darcy --Weisbach equation given by: [\$h\_{f} = \\frac{fLv\^{2}}{2gD}\$]{.math.inline} in which [*L*]{.math.inline} and D are the length and diameter of pipe respectively and [*f*]{.math.inline} is the friction factor. This factor is a function of the relative roughness of the pipe and the Reynolds number [*R*~*E*~]{.math.inline} ([\$R\_{E} = \\ \\frac{\\text{V\\ D}}{\\mathrm{\\Upsilon}}\$]{.math.inline} ) of the flow. If [*R*~*E*~]{.math.inline} \< 2100 the flow is laminar and if over 4000 the flow is turbulent and between these values transitional flow exists. The relative roughness [\$\\frac{e}{D}\$]{.math.inline} of a pipe depends on the absolute roughness *e* of the pipe interior surface and the diameter D. Curves for determining the friction factor [*f*]{.math.inline} and values of absolute roughness *e* based on extensive tests of commercial pipes are available in the literature. Velocity of flow, V = [\$\\frac{\\text{Discharge}}{\\text{Cross\\ sectional\\ Area}}\$]{.math.inline} V = [\$\\frac{Q}{A}\$]{.math.inline} For circular pipes, V = [\$\\frac{Q}{\\frac{\\text{πD}\^{2}}{4}}\$]{.math.inline} [\$V = \\ \\frac{4Q}{\\pi D\^{2}}\$]{.math.inline} [\$V\^{2} = \\ \\frac{{16Q}\^{2}}{\\pi\^{2}D\^{4}}\$]{.math.inline} But [\$h\_{f} = \\frac{\\text{fL}v\^{2}}{2gD}\$]{.math.inline} Hence [\$\\mathbf{h}\_{\\mathbf{f}}\\mathbf{=}\\frac{\\mathbf{\\text{fL}}\\mathbf{Q}\^{\\mathbf{2}}}{\\mathbf{12.1\\ }\\mathbf{D}\^{\\mathbf{5}}}\\text{\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ }\$]{.math.inline} **Worked Example** Determine the size of supply conduit leading to a service reservoir serving a small town with a population of 25,000 persons if the per capita water consumption of the city is 120 *lpcd.* Assume that maximum daily demand is 180% of average daily demand. Also find the hydraulic gradient at which the pipelines are proposed to be laid. You may make any reasonable assumptions. (Take C= 110) **SOLUTION** Population = 25,000 Average daily consumption = 120 *lpcd* Average daily flow = 25000x120 = 3,000,000 l/d = 3,000 m^3^/d. Maximum daily flow (Q) = 1.8x 3000 = 5400 m^3^/d = [\$\\frac{5400}{24\\ x60x60}\$]{.math.inline} = 0.063 m^3^/s But Q = AV, Assume flow velocity = 1.2 m/s then 0.063 = A x 1.2, A = 0.063/1.2 = 0.0525 m^2^. Using a circular pipe, [\$\\ A = \\frac{\\Pi d\^{2}}{4}\$]{.math.inline} , [\$d\^{2} = \\ \\frac{4\\ x\\ 0.0525}{\\Pi}\$]{.math.inline} , d = [\$\\sqrt{\\frac{4x\\ 0.0525}{\\Pi}}\$]{.math.inline} = [\$\\sqrt{\\frac{0.\\ 21}{\\Pi}}\\ \$]{.math.inline} d = [\$\\sqrt{0.066836}\$]{.math.inline} = 0.258m = 0.26 m use 0.3 m (300mm) commercially available pipe size. **Hydraulic Gradient** Using the Hazen William formula. [*V* = 0.85*CR*^0.63^ *S*^0.54^]{.math.inline}. For circular pipe, [\$R = \\ \\frac{d}{4}\$]{.math.inline} [\$V = 0.85C\\left( \\frac{d}{4} \\right)\^{0.63}\\ S\^{0.54}\$]{.math.inline}. \ [\$\$V = 0.85x110x\\left( \\frac{0.3}{4} \\right)\^{0.63}\\ S\^{0.54}\\ \$\$]{.math.display}\ \ [\$\$1.2 = 0.85x110x\\left( \\frac{0.3}{4} \\right)\^{0.63}\\ S\^{0.54}\$\$]{.math.display}\ \ [*S* = 0.0078125]{.math.display}\ **Hazen Williams Formula** The Darcy- Weisbach equation for computing head loss is cumbersome and not widely used in waterworks design and evaluation. The most common pipe flow formula used in the design and evaluation of water distribution system is the Hazen Williams equation. The Hazen Williams formula is expressed as: [*Q* = 0.278*CD*^2.63^*S*^0.54^]{.math.inline}, where Q = quantity of flow in m^3^/s, C = Hazen Williams pipe carrying capacity factor whose values are as given in the table below. D = diameter of pipe in metres. S = Hydraulic gradient in m/m. The Hazen Williams formula is frequently used in analysis of pressure pipe system under turbulent flow conditions. Typical values of C are given in table -- below. **Table \-\--:** Typical values of C +-----------------------------------+-----------------------------------+ | **Pipe material** | **C** | +===================================+===================================+ | Asbestos cement | 140 | +-----------------------------------+-----------------------------------+ | **Ductile Iron** | **C** | | | | | Cement lined | | | | | | New unlined | | | | | | 5 year old , unlined | | | | | | 20 year old , unlined | | +-----------------------------------+-----------------------------------+ | | 130 - 150 | +-----------------------------------+-----------------------------------+ | | 130 | +-----------------------------------+-----------------------------------+ | | 120 | +-----------------------------------+-----------------------------------+ | | 100 | +-----------------------------------+-----------------------------------+ | Concrete | 130 | +-----------------------------------+-----------------------------------+ | PVC | 140- 150 | +-----------------------------------+-----------------------------------+ | Cast Iron | 130 | +-----------------------------------+-----------------------------------+ | New welded steel | 120 | +-----------------------------------+-----------------------------------+ | Galvanized Iron | 120 | +-----------------------------------+-----------------------------------+ The value of C depends on the pipe material, age and corrosiveness of liquid carried. The smoother the pipe, the higher the value of C. The Hazen Williams (HW) equation is also used to compute the head loss directly by substituting [\$S = \\ \\frac{h\_{f}}{L}\\ ,\\ \$]{.math.inline} we obtain; \ [\$\$h\_{f} = \\ \\frac{10.667LQ\^{1.852}}{C\^{1.852}D\^{4.87}}\\text{\\ \\ }\$\$]{.math.display}\ Where [*h*~*f*~=]{.math.inline} head loss due to friction (m), [*L*= ]{.math.inline}length of the pipe section (m). [*Q*=]{.math.inline} Discharge (m^3^/s), [ C ]{.math.inline}= HW pipe carrying capacity factor. [*D*]{.math.inline}= diameter of pipe (m) **Worked example 1** Ground water from a well field is pumped through a 350mm diameter transmission main of 3200m length to the treatment plant. Calculate the head loss for a flow rate of 105 l/s, if C = 140 and C= 100 **Solution** \ [\$\$h\_{f} = \\ \\frac{10.667LQ\^{1.852}}{C\^{1.852}D\^{4.87}}\\text{\\ \\ }\$\$]{.math.display}\ a. Substituting L = 3200 m, Q = 0.105m3/s, D = 0.350 m, C = 140 [\$h\_{f} = \\ \\frac{10.667x\\ 3200x{0.105}\^{1.852}}{140\^{1.852}{0.350}\^{4.87}}\\text{\\ \\ }\$]{.math.inline} = [\$\\frac{527.706}{56.224}\$]{.math.inline} = 9.385 m b. For the situation when L = 3200 m, Q = 0.105m3/s, D = 0.350 m, C = 100 [\$h\_{f} = \\ \\frac{10.667x\\ 3200x{0.105}\^{1.852}}{100\^{1.852}{0.350}\^{4.87}}\\text{\\ \\ }\$]{.math.inline} = [\$\\frac{527.68}{30.1714}\$]{.math.inline} =17.48 m **Worked Example 2** If a 200 mm water main (C=100) is carrying a flow of 30 l/s , what is the velocity of flow and head loss? **Solution** Q = 30 l/s = 0.030m^3^/s, d = 200 mm = 0.20 m. Cross=sectional area of flow = [\$\\ A = \\frac{\\Pi d\^{2}}{4}\$]{.math.inline} = 0.0314 m^2^ Q = AV, 0.030 = 0.0314 x v, v =[\$\\ \\frac{\\ \\ 0.030m3/s}{0.0314}\$]{.math.inline}, v = 0.955 m/s. Using Hazen -Williams formula [*Q* = 0.278*CD*^2.63^*S*^0.54^]{.math.inline}. 0.030 =[0.278*x*(100)(0.2)^2.63^*S*^0.54^]{.math.inline}. \ [\$\$S\^{0.54} = \\ \\frac{0.03}{0.278x100x0.01451}\$\$]{.math.display}\ [\$S\^{0.54} = \\ \\frac{0.03}{0.4034} = 0.07436\$]{.math.inline}, [\$S = \\ {(0.07436)}\^{\\frac{1}{0.54}}\$]{.math.inline} S = 8.127 x [10^ − 3^]{.math.inline} = 8.127m/ 1000m **MINOR LOSSES IN PIPELINES** Minor losses in pipelines are caused by abrupt changes in the flow geometry due to changes in pipe sizes, bends, valves and fittings of all types. In long pipelines these minor losses can often be neglected without serious error although they may be significant in short pipes. The magnitude of these losses is dependent primarily upon the shape of the fitting which directly affects the flow times in the pipe. The equation most commonly used for determining the loss in a fitting, valve, meter or other localized component is given by: [\$h\_{m} = \\ \\frac{kv\^{2}}{2g}\\ \$]{.math.inline} , where [*h*~*m*~= ]{.math.inline}loss due to the minor loss element (m), v = velocity (m/s), g = acceleration to gravity (m/s^2^), k = loss coefficient for the specific fitting. Table \-\--: Typical minor losses coefficient +-----------------------+-----------------------+-----------------------+ | **S/No** | **Item** | **Value of k** | +=======================+=======================+=======================+ | 1 | **Losses in | 0.04 | | | entrance** | | | | | 0.5 | | | i. Bell mouth entry | | | | | | | | ii. Square edge entry | | +-----------------------+-----------------------+-----------------------+ | 2 | Loss at exit | 1.0 | +-----------------------+-----------------------+-----------------------+ | 3 | Loss in sudden | 0 -- 0.5 | | | contraction | | +-----------------------+-----------------------+-----------------------+ | 4 | Loss in sudden | 0.17 to 1.0 | | | enlargement | | +-----------------------+-----------------------+-----------------------+ | 5 | Losses in Fitting and | 0.5 -- 1 (0.9) | | | Valve | | | | | 1.5 | | | i. Standard 90^o^ | | | | elbow | 5.6 | | | | | | | ii. Standard Tee | 0.2 | | | | | | | iii. Gate valve (half | | | | open) | | | | | | | | iv. Gate valve (wide | | | | open) | | +-----------------------+-----------------------+-----------------------+ | 6 | Venturimeter | 0.3 | +-----------------------+-----------------------+-----------------------+ | 7 | Orifice | 1 | +-----------------------+-----------------------+-----------------------+ **Flow meters** -- venturimeters and orifice meters are installed in pipelines to measure the discharge through them. **Valves:** valves are provided in pipelines to control the rate of flow or the pressure **Worked Example** A 5km long pipe with HW coefficient C= 100 transports water under a total head difference of 30 metres. Determine the diameter of pipe required to transport 300,000 litres of water in 1 hour. Take minor losses as 5 percent of the frictional head loss. Use C = 100. **Solution** Considering 5 percent minor losses and various other values, we obtain \ [\$\$h\_{f} = \\ \\frac{10.667LQ\^{1.852}}{C\^{1.852}D\^{4.87}}\\text{\\ \\ }\$\$]{.math.display}\ [*h*~*T*~ = *Total* *losses*= *h*~*f*~]{.math.inline} + [*h*~*m*~]{.math.inline} [\$h\_{T} = \\ \\ 1.05xh\_{f} = 1.05\\ x\\ \\frac{10.667LQ\^{1.852}}{C\^{1.852}D\^{4.87}}\\text{\\ \\ \\ }\$]{.math.inline} = [\$1.05\\ x\\ \\frac{10.667x5000x\\ Q\^{1.852}}{100\^{1.852}D\^{4.87}}\\text{\\ \\ \\ }\$]{.math.inline} [*Q*]{.math.inline} = 300,000 litres / hr \ [\$\$Q = \\ \\frac{300,000}{1000x60x60} = 0.08333m\^{3}/s\\ \$\$]{.math.display}\ \ [\$\$h\_{T} = \\ \\ \\ 1.05\\ x\\ \\frac{10.667x5000x\\ {0.0833}\^{1.852}}{100\^{1.852}D\^{4.87}}\\text{\\ \\ }\$\$]{.math.display}\ \ [*h*~*T*~ = 30 *m* (*given*)]{.math.display}\ \ [\$\$D\^{4.87} = \\ 1.05\\ x\\ \\frac{10.667x5000x\\ {0.0833}\^{1.852}}{30\\ x\\ 100\^{1.852}\\ }\\text{\\ \\ }\$\$]{.math.display}\ [\$D\^{4.87} = \\ \\frac{562.4121}{30\\ x\\ 100\^{1.852}}\$]{.math.inline} = [ 3.7062*x*10^ − 3^]{.math.inline} [\$D = \\ {{\\ (3.7062x10}\^{- 3})}\^{\\frac{1}{4.87}}\$]{.math.inline} = 0.3168m = 316.8 mm. In practice the next larger commercially available pipe size is selected. Thus use 350 mm pipe size. D = 350 mm **Worked Example** A storage reservoir is situated at a distance of 6km from a city with a population of 30,000 persons. The total loss of head from the source to city is not to exceed 20m. If the maximum daily demand is 200 *lpcd* and pumping is to be for 12 hours only. Determine the size of the supply main by (a) Darcy Weisbach formular taking friction factor (*f*) of 0.015. **Solution.** Maximum demand = 300,000 x 200 *l/d =* [60*x* 10^6^]{.math.inline}*l/d =* [ 60*x* 10^3^]{.math.inline} [*m*^3^/*day*]{.math.inline} \ [\$\$Q\_{\\max} = \\ \\frac{{\\ 60x\\ 10}\^{3}}{12x60x60\\ } = 1.389\\ m\^{3}/s\$\$]{.math.display}\ **Darcy Weisbach formula** [\$h\_{f} = \\frac{\\text{fL}v\^{2}}{2gD}\$]{.math.inline} , but [\$\\ V = \\frac{Q}{A}\$]{.math.inline} [\$h\_{f} = \\frac{\\text{fL}Q\^{2}}{12.1\\ D\^{5}}\\text{\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ }\$]{.math.inline} [*f*]{.math.inline} = 0.015 [\$20 = \\frac{0.015\\ x\\ 6000x{(1.389)}\^{2}}{12.1\\ D\^{5}}\\text{\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ }\$]{.math.inline} \ [\$\$D\^{5} = \\ \\frac{0.015\\ x\\ 6000x{(1.389)}\^{2}}{20x12.1\\ }\$\$]{.math.display}\ [\$D = \\ \\left( \\frac{0.015\\ x\\ 6000x{(1.389)}\^{2}}{20x12.1} \\right)\^{\\frac{1}{5}}\$]{.math.inline} = 0.936 m, use 1 m = 1000mm. **Worked Example:** A large service reservoir supplies water to two districts in Abuja which have the following population: (i) **Asokoro District** - Population = 12,000, (ii) **Bwari District-** Population 60,000. Determine the size of the supply conduits given the average daily water consumption is 200 *lpcd.* **Solution:** Assume maximum daily demand = 1.8 Add **For Asokoro District:** Quantity of water = 200 x 12,000 [*Q*~av~ = 2.4*x* 10^5^]{.math.inline} litres/ day. [*Q*~max~ = 1.8 *x* 2.4*x* 10^5^]{.math.inline} = [1.8 *x* 2.4*x* 10^2^ *m*^3^/*day*]{.math.inline} = [1.8 *x* 0.0278 *m*^3^/*s*]{.math.inline} Assume flow velocity = 1.2 m/s. Q = V A, V = Q/A, A = Q/V A =[\$\\frac{1.8\\ x\\ 0.0278}{1.2}\$]{.math.inline} = 0.0417, A = [\$\\frac{\\pi d\^{2}}{4}\$]{.math.inline} \ [\$\$\\frac{\\pi d\^{2}}{4} = \\ 0.0417\$\$]{.math.display}\ [*d*^2^]{.math.inline} = 0.05308 d= [\$\\sqrt{0.05308}\$]{.math.inline} = 0.2303 m use 0.25m. That is 250 mm diameter pipe required. **Bwari District-** Quantity of water = 200 x 60,000 = [ 12*x* 10^6^]{.math.inline} litres /day = [12*x* 10^3^]{.math.inline} [*m*^3^]{.math.inline}/day = [\$\\frac{12x\\ 10\^{3}}{24x60x60}\$]{.math.inline} [*m*^3^]{.math.inline}/s = 0.1389[*m*^3^]{.math.inline}/s [*Q*~av~=]{.math.inline}0.1389[*m*^3^]{.math.inline}/s [*Q*~max~ = 1.8*x*0.1389 = 0.25002*m*^3^]{.math.inline}/s Q = V A, Assume flow velocity = 1.2 m/s. A= Q/V = [\$\\frac{0.25002}{1.2}\$]{.math.inline} = 0.2084 A = [\$\\frac{\\pi d\^{2}}{4}\$]{.math.inline} = 0.2084 [*d*^2^]{.math.inline} = 0.26530 d= [\$\\sqrt{0.26530\\ }\$]{.math.inline} = 0.5150m. Use 600 mm diameter pipe. **Worked Example** A water supply system is to be designed for a city with a population of 400,000 persons. The storage reservoir is situated 8km away from the city and the loss of head from the source to the city is 16m. Calculate the size of the supply main using (a) Darcy-Weisbach (DW) formula (b) Hazen Williams (HW) formular assuming the maximum daily demand of 200 *lpcd* and half of the daily supply is to be pumped in 8 hours. Assume coefficient of friction for the pipe material is 0.012 in the DW formular and [*C*~*H*~ = 130 ]{.math.inline}in the HW formular. **Solution** Population = 400, 000 Maximum per capita demand = 200 *lpcd* Maximum daily water demand = 400,000 x 200 = [ 80 *x* 10^6 ^ *l* /*day* ]{.math.inline} = [ 80 *x* 10^3^ *m*^3^/*day* ]{.math.inline} Maximum demand for which the supply main is to be designed is half of daily supply [\$Q\_{\\text{design}} = \\ \\frac{{\\ 80x10}\^{3}\\ m\^{3}/d}{2}\$]{.math.inline} = [ 40*x*10^3^ *m*^3^/*d*]{.math.inline} This daily supply is to be pumped in 8 hours [\$Q\_{\\text{design}} = \\ \\frac{{\\ 40x10}\^{3}\\ }{8\\ x\\ 60x60\\ }\\ m\^{3}/s\$]{.math.inline} = 1.388 [*m*^3^/*s*]{.math.inline} L = 8 km = 8000 m **Using Darcy --Weisbach Equation** [\$h\_{f} = \\frac{\\text{fL}Q\^{2}}{12.1\\ D\^{5}}\\text{\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ }\$]{.math.inline} = 16 m *f* = 0.012, L = 8,000 [\$h\_{f} = \\frac{\\text{fL}Q\^{2}}{12.1\\ D\^{5}}\\text{\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ }\$]{.math.inline} = 16 m [\$\\frac{\\text{fL}Q\^{2}}{12.1\\ D\^{5}}\\text{\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ }\$]{.math.inline} = 16 m [\$\\frac{0.012x\\ 8000x(1.39\\ )\^{2}}{12.1\\ D\^{5}}\\ \\ \\ \\ = \\ \\ \\ 16\\ \\text{\\ \\ \\ \\ \\ }\$]{.math.inline} [*D*^5^]{.math.inline} = [\$\\frac{0.012x\\ 8000x(1.39\\ )\^{2}}{12.1\\ x16}\$]{.math.inline} = 0.958066 [\$D = \\ {(0.958066)}\^{\\frac{1}{5\\ }}\$]{.math.inline} = 0.9914m use 1m, that is D = 1000mm **Using Hazen Williams Equation** \ [\$\$h\_{f} = \\ \\frac{10.667LQ\^{1.852}}{C\^{1.852}D\^{4.87}}\\text{\\ \\ }\$\$]{.math.display}\ \ [\$\$16\\ = \\ \\frac{10.667x8000x\\ {1.39}\^{1.852}}{130\^{1.852}xD\^{4.87}}\\text{\\ \\ }\$\$]{.math.display}\ [\$D\^{4.87} = \\ \\frac{10.667x8000x\\ {1.39}\^{1.852}}{130\^{1.852}x\\ 16\\ }\\ \\ = \\ \$]{.math.inline}1.195 [\$D = \\ {(1.195)}\^{\\frac{1}{4.87\\ }}\$]{.math.inline} = 1.037, use 1.2 m. That is, the diameter of the pipe is 1200mm **WATER TREATMENT SYSTEMS** **Introduction** The objective of municipal water treatment is to provide a potable supply -- one that is chemically and microbiologically safe for human consumption. For domestic uses treated water must be aesthetically acceptable, - free from apparent turbidity, colour, odour and objectionable taste. Quality requirements for industrial uses are frequently more stringent than for domestic supplies. The purpose of water treatment systems is to bring raw water up to drinking water quality. The particular type of treatment equipment required to meet these standards will depend to some extent on the source of water. Most large cities rely more heavily on surface water while most small towns or communities depend more on ground water. Surface water tends to have more turbidity and a much greater chance of microbial contamination. So filtration is almost always a necessity. Groundwater, on the other hand is uncontaminated and has little relatively little suspended solids hence filtration is often less important. Groundwater however may have objectionable dissolved gases that will have to be removed and hardness (ions of calcium and magnesium) removal may also be needed. A typical treatment plant for surface water might include the following sequence of steps. The primary process in surface water treatment is chemical clarification by coagulation, sedimentation and filtration. **Figure \-\--:** Flow diagram of a typical water treatment plant for surface water. i. **Screening** : is used to remove relatively large floating and suspended debris ii. **Mixing:** mixing the water with chemicals that encourage suspended to coagulate into larger particles which will settle more easily. iii. **Flocculation:** is the process of gently mixing the water and coagulant allowing the formation of large particles of flocs. iv. **Sedimentation:** sedimentation or settling tank is a tank with short detention time used for the removal of suspended flocs. In sedimentation process, flocs are slowed down so to enable them settle by gravity. v. **Filtration:** used for the removal of finely divided particles, suspended flocs and microorganism. vi. **Sludge processing**: it is the unit operation in which the mixture of solids and liquids collected from the settling is dewatered and disposed of. vii. **Disinfection** of the liquid effluent to ensure that the water is free of harmful pathogens. Hardness removal may be added to this generalized flow diagram if it is needed. The following methods are used to remove certain chemicals or taste or odour from water: 1. ***Adsorption :*** used for the removal of taste and odour 2. ***Chemical precipitation:*** used for the removal of dissolved substances like iron, manganese and hardness 3. ***Ion exchange:*** used for the exchange of certain salts present in water for ions present in an exchange medium. **COAGULATION AND FLOCCULATION** Raw water may contain suspended particles of colour, turbidity, and bacteria that are too small to settle in a reasonable time period and cannot be removed by simple filtration. The object of coagulation is to alter these particles in such a way as to allow them to adhere to each other. Thus they can grow to a size that will allow the removal by sedimentation and filtration. Coagulation is considered to be a chemical treatment process that destabilizes colloidal particles (particles in the size of about 0.001 to 1ϻm as opposed to the physical treatment operations of flocculation, sedimentation and filtration that follows. Most colloids of interest in water treatment remain suspended in solution because they have a net negative surface charge that cause the particles to repel each other. The intended action of the coagulant is to neutralize that charge, allowing the particles to come together to form larger particles that can be more easily removed from raw water. The usual coagulant is alum (Al~2~ (SO4)~3~. 18H2O, although FeCl3, FeSO4 and other coagulants such as polyelectrolytes can be used. **Coagulation** Suspended particles cannot be completely removed from water by plain settling, even when they are subjected to very long detention times and low over flow rates. Some of the very small turbidity causing particles called colloids will not settle without some help. If certain chemicals called coagulants are rapidly mixed in the water and the mixture slowly stirred before allowing sedimentation to occur, the particles will settle. An important property of colloidal particles that keeps them perpetually in suspension is the fact that they carry small electrostatic charges which make them repel each other. The coagulant chemical neutralizes the effect of colloidal charges. Once the colloidal charges are neutralized, the colloidal particles can collide and agglomerate (stick together) forming larger and heavier particles called flocs. After the initial flash mix of the coagulant with the water, a gentle agitation achieved by slow stirring helps to enhance the growth of the flocs by increasing the number of particle collisions. The slow mixing or stirring process is called *flocculation.* The combined rapid mix- slow mix process is usually referred to *coagulation.* Most of the flocs formed during coagulation are settleable and can be removed from water in a sedimentation tank. Coagulation generally takes place before the sedimentation process in a water treatment plant. There are many chemicals that can be used for coagulation but the most common is aluminum sulphate, Al~2~ (SO4)~3~ also called alum. Sometimes certain synthetic organic chemicals called polymers (polyelectrolytes) are added along with the alum to act as *coagulant aids* improve its effectiveness especially for the treatment of turbid waters. These long chain high molecular weight compounds help with the formation of larger, heavier floc particles*.* The coagulation process depends on several factors such as chemical dosage, water temperature, pH and alkalinity. As the quality of surface water often vary with time, it is usually necessary to adjust the dosage as appropriate. The optimum coagulant dose is determined in the laboratory by a procedure called Jar test. **SEDIMENTATION AND FILTRATION** **Introduction.** Sedimentation is a physical treatment process that utilizes gravity to separate suspended solids from water. The process is widely used as the first step in surface water treatment to remove turbidity --causing particles after coagulation and flocculation. Pre-sedimentation is also carried out in some cases to remove settle able solids such as gravel, grit from river water before it is pumped to the treatment plant. After flocculation, the water flows through a sedimentation basin or clarifier. A sedimentation basin is a large circular or rectangular concrete tank designed to hold the water for long enough time to allow most of the suspended solids to settle out. Typical detention time range from 1 to 10 hours. The longer the detention time, the bigger and more expensive the tank must be but correspondingly the better will be the tank performance. Solids that collect on the bottom of the tank must be removed manually by periodically shutting down the tank and washing out the collected sludge or the tank may be continuously and mechanically cleaned using a bottom scrapper. The effluent from the tank is the filtered. Therefore sedimentation or clarification is the removal of particulate matter, chemical floc and precipitates from suspension through gravity settling. The common criteria for sizing settling basins are detention times, overflow rate, weir loading and with rectangular tanks, horizontal velocity. Settling tanks for water or waste water treatment are designed to operate as the flow slowly moves from the inlet to outlet. The movement of water is slow enough to allow quiescent settling for a large percentage of the suspended particles. Thus in general terms, the water remains in the tank for only a few hours before it reaches the tank outlet. *The theoretical amount of tank remains in a settling tank is called detention time.* It is computed using the relationship: \ [\$\$T\_{\\text{D\\ }} = \\ \\frac{V}{Q}\$\$]{.math.display}\ Where [*T*~D~ is ]{.math.inline}the detention time and [*Q*]{.math.inline} is the average flow rate. Detention time is expressed in hours Detention time expressed in hours is calculated by dividing the basin volume by average daily flow. [\$T\_{\\text{D\\ }} = \\ \\frac{Vx\\ 24}{Q}\$]{.math.inline} Where [*T*~D~ ]{.math.inline} -- detention time (hours) V = basin volume (m^3^), Q = average daily flow (m^3^/d), 24 = number of hours per day. The overflow rate (surface loading rate -SLR) is equal to the average daily flow divided by total surface area of the settling basin expressed as. \ [\$\$\\ SLR\\ (V\_{o}) = \\ \\frac{Q}{A}\$\$]{.math.display}\ **Settling tank design** Settling tanks can be either circular or rectangular. *The actual depth of water in the tank is called side water depth (SWD)*. The height of tank wall is designed to have a free board of 0.45 to 0.5 m above the side water depth. Free board is provided to prevent the splashing of water over the wall sides. **Worked Example:** A circular sedimentation tank is designed to have a minimum detention time of 4 hour

CVE 523 Engineering Hydrology 1 Lecture Notes PDF

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