MINISTRY OF SCIENCE AND TECHNOLOGY DEPARTMENT OF TECHNICAL AND VOCATIONAL EDUCATION Sample Questions & Worked Out Examples for CE-04026 ENGINEERING HYDROLOGY B.Tech. (Second Year) Civil Engineering MINISTRY OF SCIENCE AND TECHNOLOGY DEPARTMENT OF TECHNICAL AND VOCATIONAL EDUCATION CE-04026 ENGINEERING HYDROLOGY Sample Questions B.Tech. (Second Year) Civil Engineering 1 Problems for CE 04026 Engineering Hydrology Chapter 1 * * * * * * * * What is the hydrologic cycle? Sketch the hydrologic cycle. Explain the hydrologic cycle. Describe the liquid transport phases of the hydrologic cycle. Name the vapour-transport phases of the hydrologic cycle. What is a catchment? Give a brief description of different components of the hydrologic cycle. How can you get the catchment area? Chapter 2 * * * * What are the different forms of precipitation and rainfall? Distinguish between the precipitation and rainfall. What are the different methods for the measurement of precipitation. Describe various types of recording type rain gauges. What are the advantages and disadvantages of these gauges? * Explain the method for estimation of missing rainfall data. * When is the normal ratio method used to fill in missing precipitation records? What is a double mass analysis? * What are different methods for the estimation of average rainfall depth over an area? * Describe the methods for plotting the mass rainfall curve and the hyetograph. * What is a double-mass curve? What is its use? * Differentiate between the infiltration capacity and infiltration index. * * Differentiate between φ - index and w- index. * * Explain the method for the determination of φ - index. * What are the various losses which occur in the precipitation to become runoff. * Draw the intensity duration curve from the following data. Duration (mts) Precipitation (cm) 5 0.8 10 1.2 15 1.4 30 1.7 60 2.1 90 2.4 120 2.8 ** Storm precipitation occurred from 6 AM to 10 AM on a particular day over a basin of 1500 ha area. The precipitation was measured by 3 rain gauges suitably located on the basin. The rain gauge readings and the areas of the Thiessen polygons are as follows: 2 Rain gauge no. Area of Thiessen polygon (ha) Rain gauge reading 6 to 7 AM 7 to 8 AM 8 to 9 AM 9 to 10 AM 1 450 0.8 1.4 6.2 4.4 2 750 1.2 3.6 5.8 4.6 3 300 2.0 3.2 5.6 2.8 Compute and draw the storm hyetograph and mass rainfall curve of the basin. * * The computation of an isohyet map of a 2000 ha basin following a 6 hr storm gave the following data. Determine the average precipitation for the basin. Isohyet Area (ha) 35-40 cm 40 30-35 80 25-30 170 20-25 310 15-20 480 * * Compute the φ–index from the following data: Total runoff Estimated ground water contribution Area of basin 10-15 670 below 10 cm 250 77 x 106 m3 2 x 106 m3 250 km2 = = = The rainfall distribution is as follows: Hour Rainfall(cm/hr) ** 0-2 2.5 2-4 5.0 4-6 5.0 6-8 3.5 8-10 2.0 10-12 2.0 12-14 1.5 14-16 1.5 The following rainfall distribution was measured during a 6-hour storm Time (hr) Rainfall intensity (cm/hr) 0 1 0.5 2 1.5 3 1.2 4 0.3 5 1.0 6 0.5 The runoff depth has been estimated at 2 cm. Calculate the φ- index. * The precipitation gage for station X was inoperative during part of the month of January. During that same period, the precipitation depths measured at three index stations A, B, and C were 25, 28, and 27 mm respectively. Estimate the missing precipitation data at X, given the following average annual precipitation at X, A, B and C: 285, 250, 225 and 275 mm, respectively. * The annual precipitation at station z and the average annual precipitation at 10 neighbouring stations are as follows: 3 Year 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 Precipitation at z (mm) 35 37 39 35 30 25 20 24 30 31 35 38 40 28 25 21 10 station Average (mm) 28 29 31 27 25 21 17 21 26 31 36 39 44 32 30 23 Use double-mass analysis to correct for any data inconsistencies at station z. ** The following rainfall distribution was measured during a 12-h storm: Time (hr) Rainfall intensity (cm/hr) 0-2 2-4 4-6 6-8 8-10 10-12 1.0 2.0 4.0 3.0 0.5 1.5 Runoff depth was 16 cm. Calculate the φ-index for this storm. *** Using the data of above problem, calculate the w-index, assuming the sum of the interception loss and depth of surface storage is 1 cm. * The isohyets for annual rainfall over a catchment were drawn and the area enclosed by the isohyets are given below. Determine the average depth of annual rainfall over the catchment. Isohyet (cm) Area enclosed (km2) * 40 - 35 20 30 70 25 150 20 320 15 450 10 600 Precipitation station X was inoperative for part of a month during which a storm occurred. The respective storm totals at three surrounding stations A, B and C were 98, 80 and 110 mm. The normal annual precipitation amounts at station X, A, B and C are, respectively, 800, 1008, 842 and 1080 mm. 4 *** The annual precipitation at station X and the average annual precipitation at 15 surrounding stations are shown in the following table. (a) Determine the consistency of the record at station X. (b) In what year is a change in regime indicated? (c) Compute the mean annual precipitation for station X for the entire 30 year period without adjustment. (d) Repeat (c) for station X at its 1979 site with the data adjusted for the change in regime. Yean 1950 1951 1952 1953 1954 1955 1956 1957 1958 1959 1960 1961 1962 1963 1964 Sta. X 47 24 42 27 25 35 29 36 37 35 58 41 34 20 26 15 Sta. Avg 29 21 36 26 23 30 26 26 26 28 40 26 24 22 25 Year 1965 1966 1967 1968 1969 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 Sta. X 36 35 28 29 32 39 25 30 23 37 34 30 28 27 34 15 Sta. Avg 34 28 23 33 33 35 26 29 28 34 33 35 26 25 35 * * A rain gage recorded the following accumulated rainfall during the storm. Draw the mass rainfall curve and the hyetograph. Time (AM) Accumulated rainfall (mm) ** 8:00 8:05 8:10 8:15 8:20 8:25 8:30 0 1 2 6 13 18 19 An isolated storm in a catchment produced a runoff of 3.5 cm. The mass curve of the average rainfall depth over the catchment was as below: Time (hr) Accumulated Rainfall (cm) 0 1 2 3 4 5 6 0 0.05 1.65 3.55 5.65 6.80 7.75 Calculate the φ index ( constant loss rate ) for the storm. 5 ** A 10 hour storm occurred over 18 sq.km basin . The hourly values of rainfall were as follows 1.8,4.2,10.4,5.8,16.4,7.7,15.2,9.6,5.4, 1.2 cm. If the surface runoff was observed to be 705 6 ha-m, determine the infiltration index φ Time from start (hour) Incremental rainfall (cm) 1 2 3 4 5 6 7 8 9 10 1.8 4.2 10.4 5.8 16.4 7.7 15.2 9.6 5.4 1.2 Chapter 3 * A class A pan set up adjacent to a a late. the depth of water in the pan at beginning of a certain week was 195 mm . In that week , there was a rainfall of 45 mm and 15 mm of water removed from the pan to keep the water level within the specified depth range. If the depth of the depth of the water in the pan at the end of the week was 190 mm, estimate the lake evaporation in that week. * A reservoir had an average surface area of 20 km2 during June 1992 . In that month , the mean rate of inflow is 10m3/s , out flow is 15 m3/s, monthly rainfall is 10 cm and change in storage is 16 Mm3 Assuming the seepage losses to be 1.8 cm, estimate the evaporation in that month . * * The average annual discharge at the outlet of a catchment is 0.5 m3/s . The catchment is situated in desert area (no vegetation) and the size is 800 Mm2 .The average annual precipitation is 200 mm / year (a) Compute the average annual evaporation from the catchment in mm / year. In the catchment area an irrigation project covering 10 Mm2 is developed. After some years the average discharge at the outlet of the catchment appears to be 0.175 m3/s. (b) Compute the evapotranspiration from the irrigated area in mm/ year, assuming no change in the evaporation from the rest of the catchment. * Determine the monthly evaporation (mm) from a free water surface using the Penman’s method for a given weather station locality Month Temperature C° Relative humidity (%) Mean wind speed Mean daily sunshine hours Mean daily possible sunshine hours Yangon N16° 07' March 36.2(max) and 19.8 (min) 64 (max)and 48 (min) 200 Km /day 9.6 12 6 Reflection coefficient (albedo) Psychrometer constant * 0.05 0.49 At a reservoir in a certain location, the following climatic month of June by Penman’s method, assuming that the lake evaporation is the same as P.E.T. Latitude Elevation Mean monthly temperature Mean relative humidity Mean observed sunshine hour wind speed at 2m height * 28°N 230m above MSL 33.5° C 52° % 9 hr 10 Km /hr For an area (latitude 12° N) the mean monthly temperature are given Month Temp(C°) June 31.5 July 31.0 Aug 30.0 Sept 29.0 Oct 28.0 Calculate the seasonal consumptive use of water for the rice crop in the season (June to October) by using the Blaney- Criddle method . Monthly daytime hour percentage, P N lat 12° June 8.68 July 8.94 Aug 8.76 Sept 8.26 Oct 8.31 Chapter 4 * The following data were collected during a stream gaging operation in a river. Compute the discharge. Distance from bank (m) 0.0 1.5 3.0 4.5 6.0 7.5 9.0 Velocity (m/s) Depth 0.0 1.3 2.5 1.7 1.0 0.4 0.0 at 0.2d 0.0 0.6 0.9 0.7 0.6 0.4 0 at 0.8 d 0.0 0.4 0.6 0.5 0.4 0.3 0 7 *** Given below are data for a station rating curve Extend the relations and estimate the flow at a stage of 14.5 ft by A D method and logarithmic method. Stage (ft) 1.72 2.50 3.47 4.02 4.26 5.08 5.61 5.98 6.70 6.83 7.80 8.75 9.21 9.90 14.50 Area (ft2) 263 674 1200 1570 1790 2150 2380 2910 3280 3420 3960 4820 5000 5250 8200 Discharge (ft2/s) 1070 2700 4900 6600 7700 9450 10700 13100 15100 16100 19000 24100 25000 27300 Depth (ft) 1.5 1.8 2.1 2.8 3.2 3.9 4.6 4.9 5.2 5.4 5.7 6.0 6.1 6.5 9.0 ** The following data were collected at a gauging station on a stream. Compute the discharge by (a) the mid-section method (b) the mean-section method. Distance from one 0 bank (m) Water depth (m) 0 Mean velocity (m/s) 0 3 6 9 12 15 18 21 24 27 1.5 0.12 3.2 0.24 5.0 0.25 9.0 0.26 5.5 0.24 4.0 0.23 1.6 0.16 1.4 0.14 0 0 * Calculate the discharge of river from the following measurements made with a flow meter. Distance from one 0.0 15 30 45 60 75 90 105 bank (m) depth of water (m) 0.0 0.8 1.2 1.5 1.8 1.5 0.9 0.0 average velocity (m/s) 0.0 0.15 0.24 0.30 0.36 0.33 0.24 0.0 Chapter 5 * * The data given below are the annual rainfall, X and annual runoff, Y for a certain river catchment for 16 years. It has been decided to develop a linear relation between these two variables so as to estimate runoff for those years where rainfall data only are available. 8 Year 1 2 3 4 5 6 7 8 X, cm 150 141 184 205 131 222 181 133 Y, cm 124 123 134 178 127 158 147 106 Year 9 10 11 12 13 14 15 16 X, cm 135 184 119 150 192 179 156 182 Y,cm 116 151 104 113 164 133 140 162 Find the equation of regression line. Is the linear relationship appropriate for the above data? ** The following table gives the mean monthly flows in a river during a year. Calculate the minimum storage required to maintain a demand rate of 90m3/s. Month Jan 80 Flow 3 (m /s) ** Feb 60 Mar 40 Apr 30 May June July 25 60 200 Aug Sept Oct 300 200 150 Nov Dec 100 90 The average annual discharge of a river for 11 years is as follows: Year 1960 1961 1962 1963 1964 1965 1966 1967 1968 1969 1970 Discharge 1750 (cumecs) 2650 3010 2240 2630 3200 1000 950 1200 4150 3500 Determine the storage capacity required to meet a demand of 2000 cumecs throughout the year. * * * The runoff from a catchment area during successive months in a year is given below. Determine the maximum capacity of the reservoir required if the entire volume of water is to be drawn off at a uniform rate, without any loss of water over the spillway. Month Jan Feb March Apri May June July Aug Sept Oct Nov Dec 3 Runoff(Mm ) 1.3 2.0 2.7 8.5 12.0 12.0 19.0 22.0 2.5 2.2 1.9 1.7 * * * The average monthly runoff that flowed down a river during a critical year is given below 9 Month Jan Feb March Apri May June July Aug Sept Oct Nov Runoff 500 350 650 600 300 650 7500 6000 3500 2500 600 (hect-m) Dec 700 (a) If the monthly demands are as under, determine the required storage capacity. Assume that the reservoir is full on Jan 1. Month Jan Demand 5 (Cumecs) Feb 6 March Apri May June July 7 5 8 10 5 Aug 6 Sept 4 Oct 8 Nov 10 Dec 12 (b) If there is a uniform demand of 6 m3/s, determine the required storage. Chapter 6 * What is a hydrograph? What are its different segments? * Explain various methods for the separation of base flow a hydrograph. Why the separation of flow is required? * Explain the procedure for the derivation of a unit hydrograph from an isolated storm hydrograph. * What is S-hydrograph? How would you derive a S-hydrograph? Discuss the procedure of derivation of the unit hydrograph from a S-hydrograph. * How would you obtain a storm hydrograph from a unit hydrograph? * The ordinates of 3 hour unit hydrograph of a basin at 6 hour interval are below 0,3,5,9,11,7,5,4,2,1,0 Cumecs. Derive the storm hydrograph due to a 3 hour storm with a total rainfall of 15 cm. Assume an initial loss of 0.5 cm and φ - index of 1 cm/hr. Take base flow = 4 cumecs. * A 3 hour duration unit hydrograph has the following ordinates: Time(hour) 0 3 6 9 12 15 Q(cumec) 0 3.08 4.94 8.64 9.88 7.41 Develop a unit hydrograph of 6 hour duration. 18 4.94 21 3.70 24 2.47 27 1.23 30 0 * Find out the ordinates of storm hydrograph resulting from a 9 hour storm with rainfall of 2.0, 5.75 and 2.75 cm during subsequent 3 hour intervals. The ordinates of 3 hour U.H at 3 hour intervals are as follows: 0,100,355,510,380,300,200,225,165,120,85,55,30,22,10,0 (cumecs) Assume an initial loss of 0.5 cm, an infiltration index of 0.25 cm/hr and a base flow of 10 cumecs. 10 * Given below are observed flows from a storm of 6 hour duration on a stream with a catchment area of 500 km2. Time(hr) 0 3 Flow(m /s) 20 6 12 18 24 30 120 270 220 170 120 36 90 42 70 48 55 54 45 60 35 66 25 72 20 Assuming a constant base flow of 20 m3/s, derive the ordinates of a 6 hour unit hydrograph. * Given below is the 4 hour UH for a basin. Compute the S-curve ordinate and find the 6 hour UH. Hour 0 4 hr 0 UH 2 150 4 500 6 610 8 450 10 320 12 220 14 140 16 80 18 40 20 10 22 0 * Calculate the streamflow hydrograph for a storm of 6 inches excess rainfall, with 2 inches in the first half-hour, 3 inches in the second half-hour and 1 inch I the third half-hour. Use the half-hour unit hydrograph and assume the base is constant at 500 cfs throughout the flood. Check that the depth of direct runoff is equal to the total excess precipitation (watershed area = 7.03 sq.mile). The ordinates of half hour unit hydrograph are given below. Time(hour) 0 Q(cumec) 0 ½ 404 1 1½ 2 2½ 3 1979 2343 2506 1460 453 3½ 381 4 274 4½ 173 5 0 * * * A catchment of 5 km2 has rainfall of 5.0, 7.5 and 5.0 cm in three consecutive days. The average φ - index is 2.5 cm/day. The surface runoff extends over 7 days for each rainfall of 1 day duration. Distribution graph percentage for each day are 5,15,35,25,10,6,4. Determine the ordinates of the storm hydrograph. Neglect base flow. Chapter 7 * The annual rainfall for 10 years are as follows: 40, 35, 55, 65, 70, 25, 45, 30, 50 and 42 cm. Determine the rainfall which has a recurrence interval of 12 years. * The maximum values of 24 hr rainfall at a place from 1960 to 1980 are as follows: 11 12.7 11.7 19.7 13.2 13.3 19.7 12.8 13.6 18.9 11.6 13.9 17.4 16.9 16.4 15.8 17.2 14.7 14.9 14.0 8.4 18.3 14.2 12.5 17.7 17.8 11.2 18.6 18.8 20.7 19.2 Estimate the maximum rainfall having a recurrence interval of 10 years and 50 years. ** The maximum annual observed floods for 20 years from 1950 to 1969 for a catchment are given below. Determine the maximum flood with a recurrence interval of 30 years by the following methods: (a) Probability plotting, using a log-log paper (b) Gumble's method Year Discharge (Lakh cumecs) Year Discharge (Lakh cumecs) Year Discharge (Lakh cumecs) ** 1950 1951 1952 1953 1954 1955 1956 1957 1.38 1958 1.25 1959 1.92 1960 1.45 1961 1.65 1962 1.43 1963 1.84 1964 1.74 1965 1.32 1966 1.86 1967 1.20 1968 1.82 1969 1.70 1.95 1.60 1.32 1.41 1.78 1.80 1.50 The maximum annual floods for 23 years are given below, arranged in the descending order 1960 Year Flood discharge(m3/s) 720 1952 1970 1954 1972 1971 1968 1964 710 705 665 570 490 450 440 1966 Year Flood discharge(m3/s) 425 1973 1956 1957 1951 1965 1951 1961 410 405 400 395 390 385 375 1955 Year Flood discharge(m3/s) 360 1963 1958 1962 1969 1959 1967 345 340 330 320 310 300 Find the magnitude of 100-Year flood, using Gumbel's method. 12 * A well of diameter 30 cm fully penetrates a confined aquifer of thickness 15 m. When pumped at a steady rate of 30 lps, the draw downs observed in wells at radial distances of 10 m and 40 m, are 1.5 and 1.0 m respectively. Compute the radius of influence, the permeability, the transmissibility and the draw down at the well. * A 0.4 m diameter well fully penetrates an unconfined aquifer whose bottom is 80 m below the undisturbed ground water table. When pumped at a steady rate of 1.5 m3/min, the draw downs observed in two observation wells at radial distances of 5 m and 15 m are, respectively, 4 m and 2 m. Determine the drawn down in the well. * A well penetrates in the centre of an unconfined aquifer bounded externally by a circle of radius 600m along which the height of water table is 8m. If at a distance of 10m from the centre of well, the height of the water table is 7.5m when steady conditions are established, determine the discharge of the well. Take k = 10-4 m/s. * * A well fully penetrating a confined aquifer was pumped at a constant rate of 0.03 cusecs. During the pumping period, the draw down S in an observation well measured at different instants of time are given below. If the distance of the observation well from the pump well was 50m, determine the formation constants S and T by (a) Theis' method (b) Cooper Jacob's method. Time(t)minutes Drawdown 0 0.0 14 067 100 1.02 1 0.4 18 0.71 140 1.07 2 0.32 22 0.75 180 1.12 3 0.38 26 0.78 240 1.15 4 0.43 30 0.81 5 0.49 40 0.85 6 0.52 50 0.89 8 0.57 60 0.94 10 0.61 80 0.97 12 0.64 Chapter 9 *** Tabulated below are the elevation storage and elevation discharge data for a small reservoir. Elevation(ft) Storage(sfd) Discharge(cfs) 0 0 0 5 30 5 10 50 10 15 80 20 20 110 25 25 138 30 30 160 80 35 190 130 From the inflow hydrograph shown below, Compute the maximum outflow discharge and pool level to be expected. Assume initial outflow = 20 cfs 13 * The following inflow and outflow hydrographs were observed in a river reach. Estimate the values of k and x applicable to this reach for use in the Muskingum equation. Time (hr) Inflow (m3/s) Outflow (m3/s) * 0 5 5 6 20 6 12 50 12 18 50 29 24 32 38 30 22 35 36 15 29 42 10 23 48 7 17 54 5 13 60 5 9 66 5 7 The inflow hydrograph for a stream channel reach is tabulated below. Compute the outflow hydrograph using Muskingum method of routing with k = 36 hr and x = 0.25. Assume initial outflow as 30 cumecs. Date hr Inflow (cumec) 1 6 AM 30 Noon 50 6 PM 86 MN 124 2 6 AM 155 Noon 140 6 PM 127 MN 103 3 6 AM 95 Noon 76 6 PM 65 MN 54 4 6 AM 40 Noon 30 6 PM 24 MN 20 MINISTRY OF SCIENCE AND TECHNOLOGY DEPARTMENT OF TECHNICAL AND VOCATIONAL EDUCATION CE-04026 ENGINEERING HYDROLOGY Worked Out Examples B.Tech. (Second Year) Civil Engineering 1 CE04026 Engineering Hydrology 1.(2-7)* The isohyets for annual rainfall over a catchment were drawn and the area enclosed by the isohyets are given below. Determine the average depth of annual rainfall over the catchment. Isohyet (cm) Area enclosed (km2) 40 - 35 20 30 70 25 150 20 320 15 450 10 600 Solution Isohyet (cm) 40 35 30 25 20 15 10 Area enclosed (km2) 20 70 150 320 450 600 Net Area (km2) 20 50 80 170 130 150 ∴ Average depth of annual rainfall over the catchment Average depth (cm) 37.5 32.5 27.5 22.5 17.5 12.5 = ppt. volume 750 1625 2200 3825 2275 1875 12550 12550 600 = 20.92 cm 2.(2-7)* Precipitation station X was inoperative for part of a month during which a storm occured. The respective storm totals at three surrounding stations A, B and C were 98, 80 and 110 mm. The normal annual precipitation amounts at station X, A, B and C are, respectively, 800, 1008, 842 and 1080 mm. Estimate the storm precipitation for station X. Solution PA = 98 mm, NA = 1008 mm, PB = 80 mm, NB = 842 mm, PX = NX M PA PB PC + + NA NB NC PX = 800 3 98 80 110 + + 1008 842 1080 PX = 78.42 mm PC = 110 mm, NC = 1080 mm, PX = ? NX = 800 mm 2 3.(2-7)** An isolated storm in a catchment produced a runoff of 3.5 cm. The mass curve of the average rainfall depth over the catchment was as below. Time (hr) Accumulated Rainfall (cm) 0 1 2 3 4 5 6 0 0.05 1.65 3.55 5.65 6.80 7.75 4 2.1 5 1.15 Calculate the φ index (constant loss rate) for the storm. Solution Time from start (hr) Incremental rainfall (cm) 1 0.05 Total rainfall direct runoff ∴ Total infiltration = 7.75 cm = 3.5 cm = 7.75 – 3.5 = 4.25 cm 1st trial Assume te = 6 hr φ1 = infiltration loss = te 2 1.6 3 1.95 4.25 6 = 0.708 cm/hr 4.2 5 0.84 cm/hr 6 0.95 φ1 is not effective the 1st hr 2nd trial Total infiltration = 4.25 – 0.05 = 4.20 cm Assume te = 5 hr φ2 = infiltration loss = te ∴ infiltration index φ = 0.84 cm/hr 4.(3-7)* A class A pan set up adjacent to a lake. The depth of water in the pan at beginning of a certain week was 195 mm. In that week, there was a rainfall of 45mm and 15mm of water removed from the pan to keep the water level within the specified depth range. If the depth of the water in the pan at the end of the week was 190 mm, estimate the lake evaporation in that week. Solution pan evaporation = 195 + 45 – 15 – 190 = 35 mm pan coeff. for class A pan = 0.7 ∴ lake evaporation = pan evaporation x pan coefficient = 35 x 0.7 = 25.5 mm 3 5.(3-7)* A reservoir had an average surface area of 20 km2 during June 1992. In that month, the mean rate of inflow is 10m3/s, outflow is 15m3/s, monthly rainfall is 10cm and change in storage is 16 Mm3. Assuming the seepage losses to be 1.8 cm, estimate the evaporation in that month. Solution Surface area = 20 km2 = 20 x (103) = 20 x 106 m2 Inflow for June 3 = 10 m /s x 30 x 24 x 3600 x 100 = 129.6 cm 20 x 106m2 outflow for June = storage change = 16 Mm3 = seepage losses = 1.8 cm ( assume ) rainfall (monthly) = 10 cm 15 m3/s x 30 x 24 x 3600 x 100 = 194.4 cm 20 x 106m2 16 x 106 m3 20 x 106 = 0.8 m x 100 = 80 cm Evaporation for June = ∆S + I + P – O – Os ( Water balance method ) = 80 + 129.6 + 10 – 194.4 – 1.8 = 23.4 cm 6(4-7)* The following data were collected during a stream gaging operation in a river. Compute the discharge. Distance from bank (m) 0.0 1.5 3.0 4.5 6.0 7.5 9.0 Depth Velocity (m/s) at 0.2 d at 0.8 d 0.0 0.0 0.4 0.6 0.6 0.9 0.5 0.7 0.4 0.6 0.3 0.4 0.0 0.0 0.0 1.3 2.5 1.7 1.0 0.4 0.0 Solution Distance Width Depth Meter depth (m) (m) (m) (m) 0 1.5 0 1.5 3.0 1.5 Velocity (m/s) at point 0 1.3 0 0.3 1.0 0 0.6 0.4 2.5 0.5 2.0 0.9 0.6 Area Discharge mean velo. 0 (m2) (m3/s) 0 0 0.5 1.95 0.975 0.75 3.75 2.8125 4 4.5 1.5 1.7 0.34 1.36 0.7 0.5 0.60 2.55 1.53 6.0 1.5 1.0 0.2 0.8 0.6 0.4 0.50 1.5 0.75 7.5 1.5 0.4 0.1 0.3 0.4 0.3 0.35 0.6 0.21 9.0 0.75 0 0 0 0 0 6.2775 7.(5-7)** The data given below are the annual rainfall, X and annual runoff, Y for a certain river catchment for 16 years. It has been decided to develop a linear relation between these two variables so as to estimate runoff for those years where rainfall data only are available. Year 1 2 3 4 X,cm 150 141 184 205 Y,cm 124 123 134 178 Year 5 6 7 8 X,cm 131 222 181 133 Y,cm 127 158 147 106 Year 9 10 11 12 X,cm 135 184 119 150 Y,cm 116 151 104 113 Year 13 14 15 16 X,cm 192 179 156 182 Y,cm 164 133 140 162 Find the equation of regression line. Is the linear relationship appropriate the above data? Solution Year 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Σ X, cm 150 141 184 205 131 222 181 133 135 184 119 150 192 179 156 182 2644 Y, cm 124 123 134 178 127 158 147 106 116 151 104 113 164 133 140 162 2180 X2 22500 19881 33856 42025 17161 49284 32761 17689 18225 33856 14161 22500 36864 32041 24336 33124 450264 Y = a + bX b= N (ΣXY) – (ΣX) (ΣY) N (ΣX2) – (ΣX)2 = 16 (368896) – 2644 x 2180 16 (450264) – (2644)2 = 0.648 Y2 15376 15129 17956 31864 16129 24964 21609 11236 13456 22801 10816 12769 26896 17689 19600 26244 304354 XY 18600 17343 24656 36490 16637 35076 26607 14098 15660 27784 12376 16950 31488 23807 21840 29484 368896 5 ΣY - bΣX N a = = ∴ Y = 29.10 + 0.648 X Correlation coefficient r = 2180 – 0.648 x 2644 16 = 29.10 ← Regression line N (ΣXY) - (ΣX) (ΣY) [NΣX [NΣX22––(ΣX) (ΣX)22]][NΣY [ΣNY22––(ΣY) (Y)22]] 16 (368896) - 2644 x 2180 r = r = 0.875 > 0.6 [16 x 450264 – (2644)2] [16 x 304354 – (2180)2 ] ∴ good correlation Linear relationship is appropriate for above data. 8.(6-8)* The ordinates of a 3hr unit hydrograph of a basin at 6 hr interval are given below. 0, 3 , 5 , 9 , 11 , 7 , 5 , 4 , 2 , 1 , 0 cumecs. Derive the storm hydrograph due to a 3 hr storm with a total rainfall of 15 cm. Assume an initial loss of 0.5 cm and a φ index of 1 cm/ hr. Take base flow = 4 cumecs. Solution Effective rainfall depth R = 15 – 0.5 – 1 x 3 = 11.5 cm Time (hours) 0 6 12 18 24 30 36 42 48 54 60 Unit hydrograph ordinates (cumecs) 0 3 5 9 11 7 5 4 2 1 0 Direct runoff ordinates (cumecs) 0 34.5 57.5 103.5 115.0 80.5 57.5 46.0 23.0 11.5 0 Base flow 4 4 4 4 4 4 4 4 4 4 4 Ordinate of storm hydrograph (cumecs) 4 38.5 61.5 107.5 119.0 84.5 61.5 50.0 27.0 15.5 4.0 6 9.(6-8)* A 3 hr duration unit hydrograph has the following ordinates: Time (hr) Q (cumec) 0 3 0 3.08 6 4.94 9 8.64 12 9.88 15 7.41 18 4.94 21 3.70 24 2.47 27 1.23 30 0 Develop a unit hydrograph of 6 hour duration Solution Time (hr) Ordinate of 3 hr unit hydrograph 0 3.08 4.94 8.64 9.88 7.41 4.94 3.70 2.47 1.23 0 0 3 6 9 12 15 18 21 24 27 30 33 3 hr U.H lagged 3hr Combined hydrograph 6 hr UH 0 3.08 4.94 8.64 9.88 7.41 4.94 3.70 2.47 1.23 0 0 3.08 8.02 13.58 18.52 17.29 12.35 8.64 6.17 3.70 1.23 0 0 1.54 4.01 6.79 9.26 8.65 6.18 4.32 3.09 1.85 0.62 0 10.(8-8)* A well of diameter 30 cm fully penetrates a confined aquifer of thickness 15m. When pumped at a steady rate of 30 lps, the drawdowns observed in wells at radial distances of 10m and 40m, are 1.5 and 1.0 m respectively. Compute the radius of influence, the permeability, the transmissibility and the drawdown at the well. Solution Q= 30 x 10-3 = 2π bk (Z2 – Z1) Log (r1 / r2) 2π x 15 x k (1.5 – 1.0) Log (40/10) k = 8.8 x 10-4 m/s T = bk = 15 x 8.8 x 10-4 = 1.32 x 10-2 m2 /s Let Zw be the drawdown at the well face Q= 30 x 10-3 = 2π bk (Zw – 1.0) Log (40/0.15) 2π x 15 x 8.8 x 10-4 (Zw – 1.0) 5.586 7 Zw = 3.02 m Let R be the radius of influence ( ie drawdown Z = 0 ) Q= 2π bk Zw Log (R/rw) 2π x 15 x 8.8 x 10-4 x 3.02 30 x 10 = Log (R/0.15) -3 R = 634.0 m 11.(8-8)* A 0.4 m diameter well fully penetrates an unconfined aquifer whose bottom is 80 m below the undisturbed ground water table. When pumped at a steady rate of 1.5 m3/min, the drawdowns observed in two observation wells at radial distance of 5m and 15m are, respectively, 4m and 2m. Determine the drawdown in the well. Solution Q = 1.5 60 = 2 πk (h 12 - h 2 ) Log (r1 /r2) πk [ (80 – 2)2 – (80-4)2] Log 15/5 = 2.84 x 10-5 m/s k Let hw = the depth of water in the well Q = πk (h 22 – h 2w ) Loge (r2 /rw) 1.5 60 = π x 2.84 x 10-5 (782 – h 2w ) Loge (15/0.2) hw = 69.82 m Zw = 80 – 69.82 = 10.18 m 8 12(9-8)*** Tabulated below are the elevation-storage and elevation-discharge data for a small reservoir. Elevation (ft) 0 5 10 15 20 25 30 35 Storage (sfd) 0 30 50 80 110 138 160 190 Discharge (cfs) 0 5 10 20 25 30 80 130 From the inflow hydrograph shown below, compute the maximum outflow discharge and pool level to be expected. Assume initial outflow = 20 cfs. Data Hour Inflow, cfs 1 MN 20 2 NOON 50 3 MN 100 NOON 120 4 MN 80 NOON 40 MN 20 Solution t = 0.5 day Elevation 0 5 10 15 20 25 30 35 Date Hour 1 2 MN NOON MN NOON MN NOON MN NOON 3 4 5 Discharge (cfs) 0 5 10 20 25 30 80 130 Inflow (cfs) 20 50 100 120 80 40 20 10 Storage (sfd) 0 30 50 80 110 138 160 190 2s t (2s/t) - Q (2s/t) + Q 300 328 426 544 572 560 534 340 370 478 646 744 692 620 564 ∴ Maximum outflow discharge = 86 cfs at pool level = 30.5 ft + Q , cfs 0 125 210 340 465 582 720 890 outflow Q , cfs 20 21 26 51 86 66 43 29 5 NOON 10